electrical engineering - forgotten books

o PR INC IPL ES

O F

ELECTRICA L

ENGINEERING

BY

HAROLD EENDER , PH .D .

PROFESSOR OF THEORETICAL AND APPLIED ELECTRI CITYMASSACHUSETTS INSTITUTE OF TECHNOLOGY

MEM . A . I . E . E .

FIRST EDITION

M CGRAW -H I LL BOOK COMPANY

239 WEST 39TH STREET, NEW YORK

6 BOUVERIE STREET, LONDON, E . C.

1 9 1 1

COPYRIGHT, 1910, 191 1BY THE

M CGRAW-HILD BOOK

The P l impton P ress , Norwood,M ass . , U.S .A .

PRE FA C E

TH IS book is an exposition of the physical principles uponwhich the art of electrical engineering is based , together with adiscussion of the application of these principles in some of thesimpler forms of electric apparatus and machineryf The firstfour chapters of the book may be looked upon as an introductionto the detailed study of continuous current machinery, the lastfive to the detailed study of alternating current machinery andthe transm ission and distribution of energy by alternating cur

rents . The text is the development of a course of lecturesgiven by the author to the junior c lass in electrical engineering at the M assachusetts Institute of Technology. The timeallotted to this course , exclusive of home preparation , is fiftyfive hours of lectures, twenty- three hours of recitations

,and

twenty- three hours for the solution of problems under thesupervision of an instructor .

A preliminary edition of the book was published in 1910,

primarily for the use of the students at the Institute,and this

preliminary edition was preceded by m imeographed notes covering practically the same ground . The present edition differsfrom the prelim inary edition chiefly in the addition

,at the end

of each chapter,of a summary of the important defin itions and

principles developed in that chapter and of a list of problems,

with answers,illustrative of these principles . Certain sections of

the text have also been rewritten,pa rticularly the sections on

energy, inductance , and capacity ; the conception of linkagesbetween the electric and magnetic circuits has also been morefully developed . The typographical errors of the preliminaryedition have been corrected and additional steps have beeninserted in some of the mathematical deductions .

There will undoubtedly be those who,after an examination

of the text , will pronounce it more of a treatise on physics thanon electrical engineering . But electrical engineering is primarilythe application of physics, and of necessity the principles of

electrical engineering are the principles of physics . This fact isV

2 3 5471

vi PREFACE

too frequently overlooked , and the student is rushed into thestudy of electric machinery and other advanced subj ects withbut the vaguest conception of the physical principles upon whichthe Operation of electric apparatus is based . In the author ’sOpinion

,a clear conception of the principles of physics and the

ability to apply these principles in co-ordinating the experimentalfacts of physics, both qualitatively and quantitatively

,is abso

lutely essential before one can get a clear understanding of themore complicated reactions that take place in electric machineryand transm ission c ircuits . It is with the hope that others may

find this text useful in filling the gap between elementary physicsand applied electricity that the author offers it to the public .

The method of treatment adopted throughout is to describefirst certain Simple and typ ical experiments wh ich illustrate agiven principle , second , to state the principle in an exact mannerin its general form ,

and then explain the application of the principle in one or more practical cases . The problems given at theend of each chapter serve as a further illustration of the principlesdeveloped . The attempt has also been made to make each sec

tion lead naturally into the next and to Show how the variousphenomena of electricity and magnetism are interrelated . Theanalogy between the flow of electric ity and hydraulics is broughtout repeatedly

,and emphasis is laid upon the sim ilarity of

magnetic and electrostatic phenomena . In the discussion of

alternating currents sine functions are used until the meaning/of the various terms

,such as effective value

,phase differencey

etc .

,has been made clear ; the vector method is then introduced

,

and finally the symbolic—

method is developed .

The calculus is employed from the very beginn ing of the book .

The calculus is taught the students of electrical engineering inpractically every technical school in the country . The authorhas adopted the common-sense V iew that Since the student isprovided with so useful a tool he should learn how to use it

,

particularly as this tool is one of the greatest labor-saving devicesever invented . In every case

,however

,the. physical meaning

of the formulas developed has been clearly stated .

The . problems at the end of each chapter are of two kinds,

which for convenience may be designated as practical andtheoretical . To fix principles in the student ’s m ind problems

may be devised which , though seldom met with in practice,are

PREFACE vii

much more effective than purely practical ones . Practical problems

,however

,should not be neglected , for the student should

also gain facility in making the simple calculations of ordinarypractice . In addition , many of the problems of the latter classhave been selected to bring out the principles

'

of certain specialphenomena and methods of practice which are not treated indetail in the text .

To those who may use the book in the c lass- room the following suggestions are offered . When the time available . IS

lim ited,the artic les printed with c lose spacing, for example ,

Article 42,may be om itted. The student should be made to

understand that the summaries at the end of each chapter arenot to be memorized

,but are given merely as a bird ’s-eye view

of the chapter . The solution of as many problems as possiblein conjunction w ith the study of the text will also enable thestudent to test his understanding of the latter ; it is with thisobj ect in view that the answers to the prob lems have been given .

The student should be required to follow c losely in all writtenwork the recommendations of the American Institute of ElectricalEngineers , given in Appendix A .

The author takes this opportunity of expressing his indebtedness to M essrs . Cary T . Hutchinson

,W . A . Del M ar

, and H . S.

Osborne and to the various members of the E lectrical Engineering Department of the M assachusetts Institute of Technologyfor many valuable suggestions and criticisms . The maj ority of

the problems in this edition were prepared by M r . R . G. Hudson,

to whom the author is also indebted for his assistance in thereading of proof .

HAROLD PENDER

EAST BLUE HILL,M AINE

,

August 14, 19 1 1

C ON TE N T S

CHAPTER I

FUNDAMENTAL IDEAS UNITS

IntroductionLengthSurfaceVolume

AngleTime

DisplacementVectorsComposition ofVectorsVelocity .

AccelerationM assC . G . S . or absolute System ofUnitsDensity and Specific GravityCenter ofM assLinear M omentum and M oment ofM omentumConservation of M ass, Conservation of Linear M omentum

, Con

servation ofM oment ofM omentumForceMoment ofForce or TorqueM oment ofa System ofParticles Actedupon by Several ForcesWork and EnergyPowerHarmoni c M otionTemperatureHeat EnergyEfficiency and LossesNewton’s Law ofGravitationProblems

28 . M agnets . M agnetic Poles29 . Paramagnetic and Diamagnetic Substances30 . Attraction and Repulsion ofM agnetic Poles31 . M agnetic Charge . A M agnetic Pole as a Force-Producmg Agent

ix

CONTENTS

Induced MagnetisationPoint-PolesProperties OfM agnetic PolesPole Strength per Unit AreaM agnetic Field ofForce . Field Intensity or M agnetising ForceExample Illustrating the Application ofAbove DefinitionsForce Required to Separate Two Equal and Opposite Poles inContact

The Earth’s M agnetic FieldM agnetic M oment . Equivalent Length ofa M agnetEquality Of the Poles ofa M agnet . Equilibrium Position ofa M ag

net in a Uni form M agnetic FieldM easurement of the Horizontal Component of the Intensity of a

M agnetic Field .

Flux Of M agnetic Force Due to a Single Pole . Lines ofM agneticForce Due to a Single Pole

Gauss ’s TheoremLines Of Force Representing the Resultant Field Due to any Number

ofM agnetsIntensity Of M agnetisationLines ofM agnetic Induction. Flux ofInductionFlux DensityThe Normal ComponentsOf the Flux Density on the Two Sides Of

any Surface are EqualThe Tangential Components of the Field Intensity on the Two Sides

Ofany Surface are EqualConditions which must be Satisfied at Every Surfac’

e in a M agneticField

Induced M agnetisationM agnetic PermeabilityRefraction of the Lines of Induction at the Surface ofSeparation of

Two Bodies ofDiflerent PermeabilitiesValue of the Pole Strength per Unit Area Inducedon the Surface ofSeparation ofTwo Bodies ofDifferent Permeabilities

Field Intensity at any Point in a M agnetic M edium ofConstant Perm eabil ity Completely Surrounding a Point-Pole and Filling al l

SpaceM agnetic Hysteresis .

Normal B H Curves . M agnetic SaturationM agnetic Potential .

Difference ofM agnetic PotentialEquipotentia l SurfacesSummary ofImportant Definitions and PrinciplesProblems

CONTENTS

CHAPTER III

CONTINUOUS ELECTRIC CURRENTS

The Electric CurrentConductors and Insulators or D ielectricsElectricity Analogous to an Incompressible Fluid Filling a l l SpaceAWire as a Geom etrical LineDefinition of the Strength Of an Electric Current . Definition Of aContinuous Current

Definition Of the Direction ofan Electric Current, Left-HandRuleConductors In Series and In ParallelTotal Force Produced by a M agnetic Field on a Wire Carrying an

Electric CurrentForce Produced by a Uniform M agnetic Field on a StraightWireCarrying an Electric Current

M agnetic Field Produced by an Electric Current in aWireD irection ofthe Lines ofForce Produced by an Electric CurrentM agnetic Field Due to a Current in a Long StraightWireM agnetic Field Due to a Current in a C ircular Coil ofWireAbsolute M easurement ofan Electric CurrentComparison Of the Strengths ofElectric Currents . Galvonometers

and Amm etersElec trolysis and ElectrolytesElectrochem ical Equivalent ofa SubstanceThe International AmpereQuantity OfElectricityElectric Resistance . Joul e ’S LawAbsolute M easurement ofElectric ResistanceSpecific Resistance or ResistivityElectr ic Conductance and ConductivityM atthiessen

’

s Standard ofConductivityTemperatur e Coefficient ofElectric ResistanceDifference ofElectric Potential . Electric EnergyM easurement ofDrop ofElectric PotentialOhm

’s Law

Electric Power andElectric EnergyThe WattmeterElectromotive ForceGeneralized Ohm ’

s Law

Contact Electromotive ForceChem i cal BatteriesDefinition ofthe InternationalVoltThermal Electromotive ForcesKirchhoff s LawsTheWheatstone BridgeThe PotentiometerStream Lines ofElectric Current

xi

1 CONTENTS

Resistance Drop and Electric Intensity . Electric EquipotentialSurfaces

Insulation Resistance ofa Single Conductor CableField Intensity at any Point due to a Current In a Wire Of C ircularCross Section

Summary ofImportant Definitions and PrinciplesProblems

CHAPTER IV

ELECTROMAGNETISMElectromotive Force Due to Change in the Number Of Lines OfInductionLinking a C ircuit. Linkages between Flux and Cur

rent106. Work Required toChangetheNumber OfLinesOf InductionLink

ing an Electric C ircuitElectromotive Force Induced by the Cutting Of Lines OfInduction .

Right-Hand RuleIntensity ofthe Magnetic Field Inside a Long SolenoidDetermination of the Number Of Lines Of Induction Linking an

Electric Circuit . M easurement ofQuantity ofElectricityDeterm ination ofthe B-H Curve and Hysteresis LoopThe Continuous Current DynamoCalculation of the Electromotive Force Induced in the Armature

ofa Continuous Current Dynamo

M agnetomotive ForceM agnetic ReluctanceCalculation OfAmpere-TurnsRequ1red to Establish a G iven FluxSelf and Mutual InductionProof ofthe Relation between Inductance and LinkagesM agnetic Energy ofan Electric CurrentAnalogy between the M agnetic Energy Of an Electric Current and

the Kinetic Energy ofa M oving Column ofWaterMagnetic Energy ofTwo or M ore Electric CurrentsCalculation ofInductance . Skin EffectSelf Inductance ofTwo Long ParallelWiresSelf Inductance ofa ConcentratedWindingSelf Inductance ofa Long SolenoidTotal Energy ofa M agnetic Field in Terms of the Field Intensity

and Flux DensityHeat Energy Due to HysteresisTractive Force ofan ElectromagnetSummary ofImportant Definitions and PrinciplesProblems

. CONTENTS

CHAPTER V

ELECTROSTATICS

Electric ChargesPositive and Negative Charges . Attraction and Repulsion of

Charged BodiesCharging by Contact and by InductionPoint-ChargesProperties ofElectric ChargesElectrostatic Field ofForce . Electrostatic IntensityLines ofElectrostatic Force . Flux ofElectrostatic ForceLines OfElectrisationLines of Electrostatic Induction. Flux of Electrostatic InductionDielectric ConstantA Closed Hollow Conductor an Electrostatic ScreenThe Electrostatic Intensity Within the Substance of a Conductor

inwhich there Is no Electric Current 1s always Z eroThe Total Charge Within any Region Completely Enclosed by a

Conductor is always Z eroThe Electrostatic IntensityJust Outside a Charged Conductor isNormal to the Surface ofthe Conductor

The Electrostatic Flux Density Just Outside a Charged Conductoris Independent ofthe Nature ofthe SurroundingDielectric

Conditions which must be Satisfied at any Surface in an Electrostatic Field

Electrostatic PotentialDifference ofElectrostatic PotentialElectrostatic Equipotential SurfacesParallel Plate ElectrometerRelation between Electrostatic Charge and Quantity Of Electric ityRelation between Electrostatic Potential and Electric PotentialDisplacement CurrentDielectric Strength. Discharge from Points . CoronaElectric Capac ity . Electric CondenserSimple Forms ofCondensersSpecific Inductive CapacityCondensers in Series and in ParallelAbsorption in a Condenser . Residual ChargeEnergy ofan Electrostatic FieldDiele ctric HysteresisSummary ofImportant Definitions and Princ iplesProblems

Iriii

x iv CONTENTS .

CHAPTER V I

V ARIABLE CURRENTS

Total Energy Assoc iated with an Electric C ircuitGeneral Equations ofthe Simple Electric C ircuitEstablishment ofa Steady Current In a C ircuit ContamingResistance and Inductance

Decay of Current in a C ircuit Containing Resistance andInductance

Charging a Condenser through a ResistanceDischarge ofa Condenser through a ResistanceDischarge ofa Condenser through an InductanceSummary ofImportant RelationsProblems

CHAPTER VII

ALTERNATING CURRENTSIntroduction .

The Simple AlternatorDefini tion of Al ternating Current and Alternating Electromotive

ForcePeriod, Frequency, Alternations, Periodicity and PhaseDifference in PhaseInstantaneous, M aximum and Average ValuesEffective ValuesThe Use ofAlternating CurrentsAlternating Current TransformerAverage Power Corresponding to a Harmonic P. D . and a Har

monic Current . Power FactorPower Corresponding to a Non-Harmomc P . D . and a Non Har

monic CurrentEfiective Value ofaNon-Harmonic Current or P . D .

Equivalent Sine-Wave Current and P . D .

Determ ination ofthe M aximum Value and Phase ofthe Harmonicsin aWave ofany Shape

The Fisher-HinnenM ethod for Analysing a Non-Harmonic Wave .

Form FactorAmplitude FactorPower and Reactive Components Of P . D . and Current .

Apparent PowerReactive PowerAddition ofAlternating Currents and Of Alternating P . D .

’s

Representation ofa Harmonic Function by a Rotating VectorThe Vectors Representing Any Number Of Harmonic Currents andP . D .

’s Of the same Frequency are Stationary with Respect to

One Another

CONTENTS XV

Page

The Lengths ofthe Vectors Representing Harmonic Functions TakenEqual to their Effective Values

P . D . due to Harmonic Current in a C ircuit of Constant Resistanceand Inductance

Effective Resistance , Reactance and Impedance of an AlternatingCurrent Circuit

Reactance and Impedance of a Coil Of Constant Resistance and

Inductance to a Harmonic CurrentReactance and Impedance of a Coil of Constant Resistance andInductance to a Non-Harmonic Current

Current through a Condenserwhena Harmonic P . D . is Establishedacross it . Capacity Reactance

Current through a Condenser when a Non-Harmonic P . D .

Establ ished across itImpedance Ofa Resistance, Inductance andCapacity In Series to aHarmonic Current

ResonanceImpedances in SeriesImpedances In ParallelConductance, Susceptance and Adm ittanceAdm ittance of an Inductance and Capacity In Parallel to a Har

monic Current . ResonanceTransient Effects Producedwhen a Harmonic E . M . F . is Impressed

on a C ircuitD ischarge of a Condenser having Negligible Leakance through aResistance and Inductance

Summary ofImportant Definitions and PrinciplesProblems

CHAPTER V III

SYMBOLIC M ETHOD OF TREATING ALTERNATING CURRENTSSymbolic Representation Of a VectorAddition ofVectorsSubstraction Of VectorsThe Symbol “ j ” as aMultiplier Signifying RotationSymbolic Expression for a Vector Referred to Any Other Vector as

the Line ofReferenceDifference in Phase between two Vectors Expressed in SymbolicNotation

Symbolic Representation ofa Harmonic FunctionSymbolic Expression for the Derivative Ofa Harmonic FunctionSymboli c Notation for Impedance . Impedance as a ComplexNumber .

Symbolic Notationfor AdmittanceDivision ofa Rotating Vector by a Complex NumberKirchhoff’s Laws 1n Symbolic Notation

xvi CONTENTS

Expression for Average Power in Symbolic NotationExpression for Rea ctive Power in Symbolic NotationExpression for Power Factor in Symbolic NotationExamples OfUse Of Symbolic M ethodSummary ofImportant Definitions and PrinciplesProblems

CHAPTER IX

THREE-PHASE ALTERNATING CURRENTSPolyphase Alternating CurrentsVector Sum of the Induced E . M . F .

’s in the Thr ee Windings Of a

Three-Phase Generator Equals Z eroRelation between Coil E . M . F . and E . M . F . between Term inals

in a Y-Connected Three-Phase GeneratorCurrents from a Three—Phase Generator—Generator and Load

both Y ConnectedCurrents from a Three-Phase Generator—Generator A Con

nected and Load Y ConnectedCoil Current in a A-Connected Generator for Balanced LoadCoil E . M . F . and Impedance ofthe Equivalent Y—Connected Gen

erator and of the EquivalentY-Connected LoadReduction ofall Balanced Three-Phase C ircuits to EquivalentY ’

s

Power In a Balanced Three-Phase SystemExample of Three-Phase C ircuit CalculationRating of Three-Phase ApparatusM easurement of Power in a Three-Phase C ircuit . Two-Watt

meter M ethodTwo—Wattmeter M ethod Applied to a Balanced Three-PhaseSystem

Summary ofImportant Definitions and RelationsProblems

Appendix A—Institute Style and AbbreviationsAppendix B—B . S . Gauge

Elec tric al Engineering

FUNDAMENTAL IDEAS AND UNITS

1 . Introduction .

—The fundamental conceptions with which“we have to start are the ideas Of space , time and matter . Wenotice all about us that matter is changing its position in Space,or moving, and that each motion requires a certain interval oftime : By measuring the amount ofmatter involved, the amountof motion that takes place and the tim e required , scientists havefound that these quantities are invariably related in definitequantitative ways ; in other words, that e very change in themotion of matter takes place in accordance with definite laws .

Many Of the fundamental conceptions used in sc ientific workare based on assumptions which are incapable ofproof , but which,on account of their simplic ity and plausibil ity

,we accept as true .

For example,we accept as true that the interval Of time required

for a given change in the position of a given portion ofmatter willalways be the sam e , provided the conditions under which thechange takes place are exactly duplicated ; again , we accept as

true that whenever the veloc ity Of a particle ofmatter changes ,this change is due to the influence Of some other particle or par~ticles of matter or to som e agent assoc iated with the particleor particles . Such assumptions have been called Articles of

Sc ientific Faith .

”

In order to express the laws Of nature in a quantitative man

ner,it is necessary to define c learly 1 . What shall be taken as a

measure of each quantity,2 . What is m eant by equal amounts

of this quantity,and 3 . What shall be taken as the unit Of this

quantity .

2 . Length .

— Two straight lines which can be superimposedone upon the other

,so that the ends of the two lines exactly coin

cide,are said to be equal m length . The dis tances between any

two pairs of points AB and A ’B ’ respectively are said to be equalwhen the straight lines drawn between A and B and between A '

and B ’ are equal . Ifwe choose the distance between any twoarbitrary points as a unit or standard of length, then any other

o'

s’

2 ELECTRICAL ENGINEERING

length may be expressed as the number of these equal unit lengthsinto which the line between the two given points may be divided .

The unit or standard of length employed in all sc ientific workis the centimeter

,abbreviated cm . ; it is the rtt th portion Of the

length of a certain platinum - iridium bar known as the International Meter and preserved at the International Bureau

“

ofWeightsand Measures near Paris ; the length of the bar is m easuredwhen it is at the temperature ofmelting ice, t. e.

,at 0° centigrade .

Some of the other common units of length employed in scientificand engineering work are related to one another as follows :

1 m eter = 100 centimeters1 m illimeter centimeter1 kilometer —1000 m eters1 inch centimeters1 m il inch1 foot centim eters1 yard centimeters1 m ile 5280 feet1 m ile kilometers1 m ile m eters3 . Surface . The unit of surface is the area of a square each

side Of which is one unit in length ; som e of the common units Ofsurface are related to one another as follows

1 square inch square cent imeters1 c ircular mil square inch1 c ircular m il square m illimeter1 square foot square centimeters1 square yard square centimeters1 acre square feet1 acre Square meters1 square m ile square feet1 square m ile 640 acres1 square m ile square kilometers4 . V olume . The unit ofvolum e is the volum e of a cube each

edge ofwhich is one unit in length . Some of the common unitsofvolum e are related to one another as follows

1 liter - 1000 cubic centim eters1 cubic inch cubic centimeters1 cubic foot cubic centim eters1 cubic foot —1728 cubic inches

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under which the displacement occurs are exactly duplicated . Thesimplest motion Of this kind is that of a pendulum suspended ata given point with reference to the earth and oscillating so thatany given point Of the pendulum passes over the sam e distanceduring each oscillation . We m ay then take as the numericalm easure of any interval of time the number of vibrations m ade inthis interval by such a pendulum . The unit or standard of tim eordinarily adopted in scientific work is the tim e required for one

osc illation of a pendulum,which

,when kept under absolutely

constant conditions,would m ake oscillations in a m ean

solar day ; this unit is called the second . The solar day is theinterval of time between two successive transits of the sun acrossthe meridian of the earth at the point of observation ; this interval varies in length at different tim es during the year butthe average length Of the interval for one year is constant as faras we know . Some of the common units of time are related toone another as follows :

1 hour —3600 seconds1 day seconds1 c ivil or calendar year - 8760 hours1 c ivil or calendar year seconds1 solar year days1 solar year seconds1 leap year —8784 hours7 . Displacement. Let a point which had a position P at

any instant have at som e later instant a position Q . Thestraight line drawn frOm P to Q is called the l inear di sp lacement

of the point . Let AB be any other straight line in space andimagine a plane drawn through P and the line AB and anotherplane through Q and the line AB ; the angle between the planesP AB and QAB is called the angu lar di sp lacement of the pointabout the axi s AB .

8 . V ectors .

—The line P Q representing the linear displacement of a point has both length and direction ; the length Of thel ine may be represented by a number and its direction by theangles m ade by P Q with any three arbitrarily chosen axes Ofco—ordinates . In the maj ority of problems that arise in engineering work the various points

’

of a body move in parallel planes ;in this case the direction Of the linear displacement P Q Of anypoint can be

.

expressed numerically by the angle made by the

FUNDAMENTAL IDEAS AND UNITS 5

line P Q with an arbitrary l ine of reference fixed in any one Of

these planes . For example,in Fig . 2 let the plane Of the paper

be the plane in which the point moves , let 0X be a line fixedin this plane and let P Q be the l inear displacement of the point .Draw a line through P parallel to OX and let 0 be the anglebetween this line and P Q then both the amount and dIrec

300

QD I

I

Fig . 2 .

tion of the linear displacement of the point are completely determ ined when we know the m agnitude Of the length P Q and itsdirection . A quantity Of this kind which requires for its com

plete representation a magnitude and a direction is called avector quanti ty and the line representing such a quantity is calleda vector . A quantity which has m agnitude only , but not direction

,such as time

,m ass

,etc .

,is called a sca lar quanti ty.

When a point moves from P to Q and then back from Q to P ,

the final displacement of the point is zero ; this m ay be expressedmathem atically by the formula P Q+ QP =0 , or P Q = QP

That is,the vector PQ

is equal but Oppositeto the vector QP ,

’ the 1

vector QP is said to bein the opposite sense

to the vector P Q . It

is therefore necessaryin dealing with vectorsto spec ify definitelythe sense of the vector ; this m ay be donebv writing the lettersrepresenting the endsOf the vector in the order such that motion from the first to

Fig . 3 .


the second is in the positive sense of the vector , or by placingon the line representing the vector an arrow pointing in thepositive sense of the vector . In dealing with vectors which liein the same or in parallel planes ~ co-

p lanar vectors) , i t isusual to select an arbitrary line drawn from an arbitrary pointas the ax is of reference and to take as the positive sense of eachvecto r its sense away from this arbitrary point Of origin . Thedirection of each vector m ay then be expressed by the anglemeasured around to the vector in the counterclockwise directionfrom the line Of reference . For example

,in Fig . 2

,the vector P Q

makes an angle of 30° with the line of reference 0X . In Fig . 3 the

vector P Q makes an angle of 330° with OX

,which is equivalent

to an angle of 30° with 0 X .

9 . Composition of V ectors. Since by definition the lineardisplacement of a point when it moves from a pos ition P to aposition Q is the straight line P Q, this displacement is independent Of the actua l path over which the p oint m oves fromP to Q . For example , in Fig . 4

,thepoint maymove in astraight

line from P to any other point B , then in a straight line to a pointand finally to the point Q . The lines PB

,BC and CQ are

called the components of

the vector P Q,and P Q

is called the resu ltant Ofthe vectors PB

,BC

,

C Q . P Q m ay thereforebe considered to be madeup of any number of

component vectors, pro

vided that when all of

these components areplaced end to end theyform a continuous pathfrom P to Q such that

a point moving from P to Q over this path moves in the pos itivesense Of each component successively . Sim ilarly the resultantof any number Of vectors PB

,BC

,CQ,

etc .

,is found by

plac ing the lines representing these vectors end to end in sucha m anner that a point m oving along the path form ed by theselines always moves in the positive sense Of the vectors ; the linedrawn from the beginning of the first vector to the end of the

Fig . 4 .

l


last of the series is then the resultant“.

The above facts may be

represented"

by a formula thus : P Q= PB + BC+ CQ where theline over the second term indicates that the lines PB ,

BC, and

CQ must be“ added

in the manner j ustdescribed . Additionof this sort is usual lycalled geometric or

vector addition ; thel ine over the secondterm then indicatesthat PB

,BC and CQ

must be added geometrically or vectorially .

Sim ilarly,we may subtract a vector PB from any other vector

P Q by adding toP Q a vector QB' equal to PB and in the opposite

or negative sense . This m ay be represented by a formula thus :PB

’ =PQ,

PB . (See Fig.

The addition or more co-planar vectors may also bee x p r e s s e d analytically .

For example,let it be re

quired to find the resultantR of the three vectors A

,

B,and C

,Fig . 6 . Choose

any arbitrary line OX as

a line of reference and let9a ,9b and 90 be the

angles made by R,A

,B

,

and C respectively withthe line OX

,and call these

angles posi tive when measx ured in the counter- c lock

Fig ' 6° wise direction around fromOX and negative when measured in the clockwise direction . Thenthe component Of R parallel to OX is

R cos 9,= A cos 0

a + B cos 9b+ C cos 9,

and the component of R which makes an angle of 90° with OX is

R sin dr=A sin H

a

—l—B sin 9b+ C sin 96

Hence the length of R is

B Fig . 5.


and the angle which R makes with OX is Hr , where

A sin 0a + B sin fib—l- C sin 9

0t 0an

r

A cos Owl—B oos 9b+ C cos (9

6

For example,let

A = 3

B = 2

C=5

Then

a sset 2 x 07 07+ 5 x (3 x 05+ 2 x x

072 44

1 1 124m"

e,

We shall see later on that many of the quantities met with Inengineering problems

,such as velocity

,force

,mom ents

,electric

current,etc .

,can be represented by vectors . In certain cases

it will be seen that it will be unnecessary to Specify the locationOf the vector

,but that any two vectors which are equal in length

and are parallel may be considered equivalent ; in other caseswe shall find that we m ay consider two equal and parallel vectorsas equivalent only when they lie in the same plane

,or in the

sam e line, or it m ay be that the vector cannot be consideredequivalent to any other vector . When the location of a vector isthus lim ited it is said to be locali zed in a plane , on a line or at apoint , as the case may be . The above laws for the com position

and resolution ofvectors applyto localized vecto rs only in casethe vectors , or vectors equivalent to them ,

m eet in a point .The angular displacement

Of a point about any axis is aquantity which requires for itsrepresentation a vector locali zed in a line . For example

,let

P and Q, lying in the plane OfFig . 7.

the paper , be the initial andfinal positions of the

'

point ; letthe axis of rotation be a line drawn perpendicular to this plane atA . By definition , the angular displacem ent of the point about this


axis is then the angle PAQ . A line drawn in the axis throughA having a length equal num eri cally to the angle PAQ will serve to represent both the axis Ofrotation and the numeri cal value of the angulardisplacement : TO represent the sense of the rotation , i . e. , whetherfrom P to Q or from Q to P ,

it is custom ary to choose arbitrarily thepositive sense of the axis, and then call the rotation positive whenthe motion of the point is in a c lockwise or right-handed directionwhen viewed by a person looking in the sense of this line . In

the case i l lustrated , i f we choose the positive sense of the axisto be toward the reader the line representing the angular displacem ent from P to Q wi ll be drawn upward perpendicular to theplane Of the paper ; the line representing the displacement from Qto P will be drawn downward . The angular displacement Of apoint about any ax is is therefore completely defined by a line having(1 ) a definite length , (2) a definite direction , and (3) locali zed in theaxis of rotation . We m ay also represent the displacement of thepoint from the position P to the position Q by an angular displacem ent about any other axis

,such as an axis through B parallel to the

axis through A , but the vector representing this angular displacement about the axis through B will not be equal to the vector representing the angular d isplacement about the axis through A , sincethe angles PAQ and PBQ are not equal . Again , two equal andparallel vectors do not represent equivalent angular displacements unless these two vectors lie in the sam e line .

10 . Velocity . Let,in any small interval of time di

,a point

P be displaced a distance d l with respect to any other point 0 ;

then the lim iting value at any instant of the ratio ill-

l, when dt is

dt

taken extremely small,is called the linear veloci ty of the point P

at that instant relative to the point 0 . Representing linearveloc ity by v, we have

d l

dt

When the point moves in a straight line over equal distances inequal small intervals of time its linear veloc ity is said to be uni

form ; in this case the linear veloc ity Of the point may be definedas the linear displacement Of the point in unit time . Since lineardisplacement is a vector quantity, i .e .

,has both magnitude and

direction ,and time is a scalar quantity, linear velocity is also a


vector quantity ; for a vector quantity divided by a Scalar is avector quantity . Therefore a change either in the magni tude or

in the direction of the linear veloc ity constitutes a change in thelinear veloc ity . The m agnitude Of the linear velocity Of a point isfrequently called the linear speed of the point ; that is , the l inearspeed Of a point is simply a number expressing the distance overwhich the point m oves in unit time ; linear Speed is therefore ascalar quantity . For example

,a point m oving in a c irc le in such

a m anner that in equal small intervals of time it passes over equaldistances measured along the c ircum ference Of the Circle, is said tobe movingw ith a constant speed ; its veloc ity , however, is changingat every instant

,S ince the direction Of motion is continually

changing .

Linear speeds may be expressed in various units , such as thenumber of centimeters per second

,feet per second

,m iles per hour

,

etc . The more common units are related as follows :1 kilometer per hour foot per second1 kilom eter per hour feet per m inute1 m ile per hour meter per second1 m ile per hour feet per second1 m ile per hour 88 feet per m inute

Angu lar ve locity is defined in exactly the same way as linearvelocity

,i .e.

,the angular velocity of a point about any axis

ISd 0

dt (4)

where dB is the angular displacement Of the point about thataxis in the small interval of tim e dt. Angular velocity is represented by a localized vector in the same way that angulardisplacement is represented by a localized vector . The termangular speed is used to express the m agnitude of the angularveloc ity in the sam e way that linear speed is used to expressthe m agnitude Of linear veloc ity . Angular speeds m ay be expressed in degrees per unit tim e, radians per unit tim e or revolutions per unit time . The more common units are related as

follows1 radian per second degrees per second1 radian per second revolution per second

1 1 . Acceleration . The rate of increase of linear veloc ity withrespect to time is called the linear acceleration; i .e.

, when the linear


1 kilometer per hour per secondcentimeters per second per secondm ile per hour per second

gravity1 m ile per hour per second

centim eters per second per secondfeet per second per second

gravity—3600 m iles per hour per hour

Gravity centim eters per second per secondGravitv 2 feet per second per secondThe variation Of gravity with altitude and locat l on is very

slight , and the approximate values 98 1 centim eters per second persecond and feet per second per second are as a rule suffi cientlyaccurate for engineering work, independent of altitude or location .

The angu lar accelera ti on of a point is s im ilarly defined as therate of change of its angular velocity . In case the point is rotatingin a c irc le about a fixed axis , its angular accelerat ion m ay bedefined as the rate of change of its angular speed about this axis .

Ifw is the angular speed then the angular acceleration isdw

a

dt

d dSince ( 1) —

E,where d d IS the angular d isplacement In tIm e dt

,

we also have,under the same conditions

,that

dt2

1 2 . Mass . The quantity of matter in a body or its mass

can be defined only in terms Ofsom e effect produced on the body bysom e other body or bodies exterior to it . It has been found byexperim ent that two bodies

,which appear to our senses to be

identical in every respect,wil l exactly counter- balance each other

when suspended one from each end of an equal- armed balance in avacuum . We may, then , go a step further and define the m ass ofany two bodies as equal irrespective Of their volume, shape or

chem ical composition,if,when they are suspended simultaneously

in a vacuum,one from each end Of an equal- armed balance, there

is no tipping of the beam of the balance from its original position .

This criterion for the equality Of two masses holds only in case the


bodies and the balance a re neither electrically charged nor mag

netised, and both bodies are supported at the same distancefrom the earth

,and the equil ibrium Of the balance is not affected

by the presence of any other bodies (except the earth) in thevicini ty .

This is an entirely arbitrary definition, but it has been foundthat mass as thus defined is a fundamental property of matter .Any arbitrary portion Ofmatter m ay be taken as the unit ofm ass ;the mass of any given portion of m atter may then be expressed asthe number of such equal units, which taken together, and sus

pended from one arm Of an equal- armed balance in a vacuum, willj ust cOunter-balance the given body suspended from the other arm .

Note that m ass is a scalar quantity .

The unit or standard of mass adopted in all scientific work isthe gram,

abbreviated g; it is the wlw th portion of a certain

platinum - iridium cylinder,known as the International Kilogram

and preserved at'

the International Bureau Of Weights and Measures near Paris . Some Of the common units of mass are relatedto one another as follows :

1 metric ton —1000 kilograms

1 centigram gram1 m illigram gram1 pound (avoirdupois) grams

1 Short ton —2000 lbs .

1 short ton kilograms1 Short ton metric ton1 long or gross ton 2240 lbs .

1 3 . C. G. S. or Absolute System Of Units . We have seen thatthe standard units adopted in sc ientific work for the m easurementOf the fundamental quantities Of length , mass and time are thecentimeter

, gram and second . Units for the measurement of allother quantities such as surface

,volume

,veloc ity

,acceleration

,

force,etc .

,can

‘be expressed in terms of these units ; such units arecalled derived units in contradistinction to the three fundamenta l

or absolute units of length , mass and tim e . The system Of unitsderived from the units of centimeter

, gram and second is known asthe absolute system,

or the c . g. 8 . system,from the initials Of the

three fundamental units .

14 . Density and Spec ific Gravity . The density of a uniformsubstance is defined as the mass Of the subs tance per unit volume .


In the c . g. 8 . system density is the weight in grams of one cubiccentimeter Of the substance . When the substance is not uniform

,

its density at any point is defined as the m ass of an infinitesim ally small volume taken about the point divided by this volume ;i .e.

,calling dv the volume and dm the m ass of this volume

,the

density is

( 10)

The specific gravi ty Of a substance is defined as the ratio of the

weight Of a given volum e of the substance to an equal volum e ofwater at standard temperature . Sixty- two degrees Fahrenheit isusually taken as the standard temperature

,although there is no

general agreement on this point . Dens ity on the c . g. 8 . systemand spec ific gravity are practically numerically equal .

1 5 . Center o f Mass .

—A body Of mass M which has any siz e orshape may be considered as made up Of a number Of small particles of masses m,,

m, ,m, ,

etc .

,such that M .

These particles may be cons idered as small as we w ish,that is

we m ay consider each particle so small that it occupies but a pointin space . If we consider three mutually perpendicular planesX, Y,

Z,fix ed in space

,and represent by at] , y,, and 2 , the

perpendicular distances Of the particle m 1 from these planesrespectively

,and by 332, y, , and 2 2 the perpendicular distances of

the partic le m2 from these three planes respectively, and so on

for the other partic les,then the point whose distances from these

three planes are respectively

77225132 7723153

M

m1y1 mzyz msys

M

77m. m zz z

M

is defined as the center ofmass of the body . The center ofmassOf a body is therefore the point the distance of which from eachof three mutually perpendicular planes is the average distance ofthe matter in the body from each of these planes . It can be shownthat the position of the center of mass Of a body relative to any


point in the body is independent Of the pos ition Of the p lanes of

reference .

The center of mass of a system of any number of bodies isdefined in exactly the sam e m anner, except that M in this caseis taken as the tota l mass of a l l the bodies .

1 6. Linear Momentum and M oment of Momentum .

—Whena body as a whole is in m otion

,or when there is any relative

motion of parts Of the body,the various points Of the body will

in general move with different veloc ities . It is therefore con

venient,in analysing the motion Of a system of bodies , to con

sider each body as m ade up Of a number of individual partic lesand to take these particles so small that the m ass of each m ay

be considered as occupying but a point in space . The product ofthe mass (m) of each particle times its linear veloc ity

'

(v) is definedas the linear momentum ofthe particle, i .e.

,linear m omentum =mv.

Linear momentum is a vector quantity,since v is a vector quan

tity and m is a scalar quantity . The vector sum of the linearmom enta Of all the particles Of a rigid body is called the totallinear m omentum of the body . From the above definition ofcenter ofmass it can be shown that the total linear m omentumof a body is equal to its total mass times the linear veloc ity of

its center Of m ass .

Consider a fixed axis , and a particle ofmass m at a distancer from the axis, and let the particle be moving with a velocity v.

.Then the product Of m,r and that component u of the velocity

v perpendicular to the plane passing through the p article andthe axis

,is defined as the moment ofmomentum of the particle

about the fixed axis ; i . e.

,m oment Of mom entum =mru . The

moment of momentum Of a particle is to be taken positive ifthe particle moves in a c lockwise direction as seen by an Observerlooking along the axis in the positive sense

,negative if in the

oppos ite direction . The component u of the linear veloc ity of

a partic le at right angles to a plane passing through the particleand any fixed axis

,is equal to the product

‘

of the distance r of

the particle from the axis times the angular veloc ity w of theparticle about this axis ; therefore the moment of momentumof the partic le may also be written mr

’w. In the case of a rigid

body each partic le ofwhich has the same angular veloc ity to abouta given axis , the m om ent Of m om entum about this axis is

equal to wEmrz, the summation including all the particles"

of the


body . The quantity Emr2is called the moment of inertia Of the

body about the given axis of rotation, and is usually written I ;then I =2 mr2 ( 12)

The m oment Ofmom entum of a solid body rotating in this man

ner is called the angu lar momentum Of the body ; angular momentum then equals Ico . The moment Of inertia of a body of givendimensions and given distribution of m aterial about a givenaxis may be written M k

” where M is the total mass of the bodyand k is a length such that 2 m71

M

Thelength It is called the radius ofgyrati on of the body about the

given axis .

1 7. Conservation ofMass,Conservation of Linear Momentum

,

Conservation ofMoment ofM omentum . It has been found thatevery phenom enon of nature

,which has so far been tested by

experiment,takes place in such a way that the three following

conditions are invariably satisfied :1 . Matter cannot be created or destroyed . As a consequence

of this condition,if any number of bodies are kept entirely sepa

rate from the rest Of the universe,for example

,in a c losed vessel

through the walls ofwhich no m atter can pass, then the total m assof these bodies must likewise remain constant

,irrespective of any

changes that mav take place in these bodies . This condition isknown as the law or principle of the conservation of mass .

2 . Whenthe linear momentum Of one or more bodies changesrelative to any fixed point

,then there must be an equal and

opposite change in the linear momentum Of some other body orbodies relative to this point . This condition is known as thelaw or princ iple of the conservation of linear momentum .

3 . When the moment of momentum Of any body or bodiesabout any fixed axis changes , then there must be an e qual andopposite change in the moment of momentum of som e other bodyor bodies about this same axis . This condition is known as thelaw or principle of the conservation of moment of momentum .

18 . Force . In general terms,a force is that which produces

or tends to produce a change in the motion Of a body . Since,by the principle of the cons ervation Of linear mom entum , anychange in the motion of a particle A is accompanied by a changein the mot ion of some other particle or partic les B,

it is convenient

( 12a)


to consider the force acting on A as due to the presence of theparticle or particles B . Or

,we m ay say that the change in the

m otion of A is due to a force produced on A by the partic le orparticles B . In other words

,we may consider the changes in

motion ofmaterial particles as being due to a property possessedby these partic les themselves .

* We take arbitrarily as themeasure Of the force with which a particle A is acted upon bya particle B the tim e rate Of change of the linear momentum OfA relative to any fixed point

,when the relative m otion Of the two

particles with respect to each other is unaffected by the presenceof any other particle or partic les ; the direction Of the force on

A due to the partic le B is defined as the direction of the rate ofchange of the linear m omentum of A . From the principle Ofthe conservation Of linear m omentum we then conclude that theforce with whi ch B is acted upon by A is equal and oppos iteto the force with which A is acted upon by B ,

that is,

“ actionand reaction are equal and oppos ite .

” When the masses of

the two partic les A and B rem ain unchanged , the time rate Of

change of l inear momentum of each particle is equal to the productof its mass and its acceleration

,i . e.

,the force with which A acts

on B is ma and the force with which B acts on A is m l a l , wherem and m l are the masses OfA and B respectively and a and a l arethe accelerations of A and B respectively . The direction of theforce acting on A is then the direction of the linear accelerationof A

,and the direction of the force acting on B is the direction

Of the linear acceleration Of B . The acceleration Of A will then*This conception of the something that causes changes in the m otion of

matter as localized in material particles m erely gives us a convenient wayof describing ex perimentally Observed fa cts . It is also conceivable that thechanges in the motion Of matter are due, not to a property inherent in mat

ter itself, but to a property possessed by the medium or the ether” in

whi ch the partic les ofmatter are imm ersed. However,a particle ofmatter

is a tangible thing , whereas the ex istence Of the ether is purely hypothetical.For thi s reason we shall adhere to the Older conception that every force,whether it be gravitational , elec tric , or magnetic , has its origin in a materialparticle and that it can make its influence felt on other particles at a distancefrom it . As far as engineering is concerned, we need not concern ourselveswith the mechanism by whi ch this action takes place , about which, at best,we can only theorize . Ana logies , where theyhelp one to form a mentalp icture of how certain effects might be produced, are ex tremely useful andwill be frequently employed ; but it must be rem embered that a’nalogies donot explain anything .


be in the opposite sense to the acceleration of B,and the ratio

of the acceleration of A to that of B wil l be equal to the inverseratio of the respective masses . Since acceleration is a vectorquantity

,force is likewise a vector quantity ; therefore , when

there are several forces acting on a particle,the resultant force

on the particle is the vector sum of all the forces acting ; thisresultant force is also equal to the product of the mass of theparticle by its acceleration in the direction of the force . Theline through a particle in the direction of the force acting on itis called the line of action of the force .

The unit of force in the c . g . 3 . system of units is that forcewhich will give unit linear acceleration (one centim eter per secondper second) to a mass ofone gram ; this unit force is called the dyne .

A s im ilar unit in the foot-pound- second system,called the pounda l ,

is sometimes used ; it is defined as the force which will accelerateone pound one foot per Second per second .

It is a matter of' experience that each particle Of matter whichhas neither electric nor magnetic properties (to be described later) ,when allowed to fall in a vacuum from any given height at anypoint near the earth ’s surface

,falls to the earth in a vertical line

with a constant acceleration independent of the size,shape

,or

m aterial of the body . We therefore say that the earth exertsa force ‘

on each partic le Of the body proportional to its mass,

and that therefore the total force with which the earth attractsa given body is equal to the total m ass of the body multiplied byits acceleration when falling freely to the earth in a vacuum .

We have seen (Article 1 1) that the acceleration due to gravityis constant for any given point with reference to the earth

’s

surface,but varies s lightly with the location and also w ith

the elevation above the earth . For most practical purposes theacceleration due to gravity may be taken as 98 1 . The attractionof the earth on a body offers a ready means for m easuring a force,since it is only necessary to balance the force to be m easuredagainst the force exerted by the earth on a known m ass

,or by

measuring the change produced by the force to be m easured inthe shape of som e body

,e .g.

,a Spiral spring, which has been pre

viously“calibrated by suspending known masses from it . These

are the usual ways of m easuring forces , S ince the determ inationof the acceleration due to any other force than gravity

” is

extremely difficult .

ELECTRICAL ENGINEERING

The moment of a force about any axis is frequently called thetorque about this axis ; we then have that the algebraic sum of allthe torques (T,,

T T, ,etc .) about any axis is equal to the product

of the moment Of inertia (I) of the body about this axis and theangular acceleration (a ) about this axis , that is

Ia = T1 + T2+ T3+ (13b)When the line of action of any _

force passes through the axis , thetorque corresponding to this force is of course zero . When twoforces are acting on a body , the condition for constant angularveloc ity is that the corresponding torque be equal and opposite .

The unit torque in the c . g. s . system of units is the torque produced on a particle by a force of one dyne acting perpendicular tothe plane determined by the position of the partic le and theaxis of rotation and at a distance of one centimeter from thelatter —this unit is called the centimeter-dyne . Other commonunits of torque are the poundfoot and the centimeter-

gram. Therelations of these units one to another are the same as for theunits of energy (see A rtic le 2 1) having corresponding names .

20 . Motion of a System of Partic les Acted upon by Several

Forces .

—When only one of a system ofpartic les is acted upon by aforce external to the svstem (for example, a stretched string at

tached to a point in a solid body) the partic le on which the forceis acting will in general exert a force 0 11 the other partic les of thebody and those partic les in turn will each exert a force on the particle to which the force is applied . It can readily be shown that

,

as a consequence of the princ iple Of the conservation of linearmomentum

,the resultant acceleration of the system ofparticles in

this case will be such that the center ofmass Of the system w ill begiven an acceleration equal to the external force acting on theparticle divided by the total mass Of the body ; that is, a S ingleexternal force acting on any partic le of a system will produce thesame acceleration Of the center ofmass of the system as would beproduced by the same force ac ting on a s ingle partic le located atthe center of m ass of system and having a mass equal to thetotal mass Of the system . In the case of several external forcesacting on the system

,the acceleration of the center Of mass of

the system will be equal to the resultant force divided by thetotal m ass

,or calling M -the mass of the system

,F the resultant

of all the external forces acting on it , and A the linear accelerationof the center of mass

,then

FUNDAMENTAL IDEAS AND UNITS

F =M A . ( 14)In general , each partic le of the system will also have its angular

veloc ity about any axis changed due to the action of a force on anyone partic le . In the case of a rigid body, it can be shown that inaddition to the change in the veloc ity of the center Of mass produced by this force, each partic le of the body will be given anangular acceleration about an axis through the center Of mass per

pendicu lar to the plane determ ined by the center Of mass and theline Of application of the force

,and that this angular acceleration

will be equal to the mom ent of the force about this axis dividedby the m om ent of inertia of the body about this axis . The condition for no angu lar acceleration is then that the line of actionOf the force (or of the resultant force when there is m ore thanone force acting) pass through the center Of mass ; vi ce versa , whenthere is no angular acceleration of the body

,the line of action

of the force (or of the resultant force when there is more than oneforce acting) must pass through the center ofmass .

When there are two equal and oppos ite external forces actingon a rigid body, there can be no linear acceleration Of the body,i .e.

,no acceleration of its center of mass . However

,when the

lines of action of these two forces do not coinc ide,the m oment of

the two forces about any axis will not be equal,hence there will be

an angular acceleration of the body about its center of mass . Theaxis of this acceleration will be through the center Of m ass per

pendicular to the plane determ ined by the line Of action of thetwo forces

,and the resultant moment Of the two forces about this

axis will be equal to the product of either force by the perpendicular distance between their lines Of action . Two such equal andoppos ite forces are called a coup le, and the value of the resultantmoment is called the strength of the couple . Two couples balanceeach other

,i .e.

,there is no angular acceleration , when their

strengths are equal .21 . Work and Energy. Whenever one portion of matter

effects a change in some other portion of m atter,the form er is

said to do work on the latter . The attribute or condition of

matter in virtue of which one portion ofmatter can effect changesin other portions of m atter is called energy; that is , energy isthe capac ity for doing work . One means by which a body Amay produce a change in another body B is by exerting a force onthe latter and producing, as a result of this force, a displacement


of this body as a whole or a displacement Of its individual parts ;in this case the body A is said to do mechani ca l work on thebody B . As the measure of the amount of mechanical workdone on a particle of matter by the force which causes its displacement is taken the product of the displacement of the particleby the component Of this force in the direction of the displacem ent

,provided this component of the force

.

remains constantduring the displacement .

In general, the direction of the displacement as well as theamount and the direction Of the force will change as the

.

positionof the particle is changed ; in such a case , the above definitionapplies only to an infinites im al displacement of the particle

,i .e.

,to

a displacement so small that while the particle is being displacedthis small am ount

,the force may be considered constant both in

amount and direction . The mechan ical work W done on aparticle displaced any finite distance l is then the sum of theproducts Of the force for each infinitesimal displacement timesthe component of this displacement in the direction of this force ;t .e.

,

W (fcos d ) dl

wheref represents the force during any infinitesimal displacement ,d l the displacement and 9 the angle between the direction Of thedisplacement and the direction of the force .

As the measure Of the amount of work done on a body whena change is produced in it by other m eans than by a displacementof the body as a whole or by a displacement of its individualparts under the action of a force

,is taken the amount of me

chanical work which would be required to produce exactly thissame change were this change effected solely by a displacem entOf the body as a whole or by a displacement of its individual partsby means of a force exerted on it . For example

,when the tem

perature of a body is increased by any m eans whatever, thebody is said to have an amount Of work done on it equal to theamount of m echanical work which would have to be done on itt o produce exactly this sam e rise of temperature .

The results of all known experim ents j ustify the assumptionthat whenever work is done on a body this body is in turn giventhe capacity for doing an exactly equal amount of work on otherbodies

,and

.

that the capacity Of some other body or system of


bodies for doing work is dim inished by an exactly equal amount .

That is,whenever work is done

,som e body or system Of bodies

loses an amount of energy equal to the amount Of work doneand some body or system of bodi es gains an exactly equal amountof energy . This assumption, which is j ustified by all knownexperim ents

,is known as the principle of the conservation of

energy. This princ iple,together with the principles of the Con

servation of Mass,the Conservation of Linear Momentum and

the Cons ervation of Mom ent Of Mom entum,are the four cardinal

principles of all natural science and engineering.

It is found convenient in discussing the vari ous propertiesOf matter to look upon each property of a portion of m atter asrepresenting a definite am ount of energy and to give a specialname to the energy associ ated with each property . For example ,the energy associated with a body in virtue of its speed is calledits kineti c energy ; the energy associated with a body in virtueof its position with respect to other bodies which exert forceson it is called its potentia l energy ; the energy possessed by abody in virtue of its temperature is called its heat energy or its

therma l energy ; the energy assoc iated with a body in virtue ofits chem ical nature is called its chemi ca l energy, etc . In thisterm inology the princ iple of the conservation of energy m ay

be stated thus : The only possible changes which can take placein the energy associated with a system of bodies which neitherinfluences

,nor is influenced by

,any other bodies

,are changes

in form; the tota l amount ofenergy in the system rem ains unaltered .

It can readily be shown that the kinetic energy Of a particleof mass m moving with a linear speed s is é m s

’. A rigid

body which is rotating with an angular speed or,about an axis

fixed in the body and passing through its center of m ass,which

axis at the same time has a linear speed s (for example , the arm ature of a railway motor moving relatively to the earth) has atotal amount of kinetic energy equal to

t M sz+ t l w

2

( 16)

r elative to the earth,where it! is the total mass Of the body and

I is the moment of inertia Of the body about the axis Of rotationfixed in the body . The express ions for other forms of energywill be given when the corresponding properties of m atter arediscussed .


The unit of work or energy on the c . g. 3 . system is the workdone when a force of one dyne displaces a particle a distance ofone centim eter ; this unit is called the erg. Other common unitsof work and energy are related to one another and to the erg as

follows :1 gram - centim eter ergs (at sea level

and 45° lat .)1 j oule —1 watt- second1 j oule —107 ergs1 kilogram -m eter 10

5

gram - centim eters1 foot—pound j oules1 foot-pound kilogram -m eter1 sm all calorie large calorie1 small calorie j oules1 small calorie foot- pounds1 British thermal unit small calories1 British therm al unit j oules1 British thermal unit foot-pounds1 kilowatt-hour j oules1 kilowatt- hour foot—pounds1 horsepower—hour j oules1 horsepower- hour foot-pounds22 . Power . Power is defined as the time rate of doing work ,

or as the tim e rate of change of energy ; the two definitions areequivalent .When an am ount of energy dW is transferred in an infinitesimal

interval of time dt,then the corresponding power is

dt ( 17)

When this rate of transfer Of energy is constant,then the power

may be defined as the amount of energy transferred, or thework done

,in unit time . It should also be noted that since energy

or work is the product of force and displacem ent,power may also

be defined as the product of force and veloc ity .

In the c . g. 3 . system the unit Of power is the power requiredto do work at the rate of one erg per second . This unit is seldomused as it is an extrem ely sm all quantity ; instead is employeda unit which is ergs per second, called the watt; one

watt is therefore equal to one j oule per second . Other common units


of power are related to one another and to the watt as follows :1 kilowatt — 1000 watts1 megawatt 1000 kilowatts1 m etric horsepower 75 kilogram -meters per second1 metric horsepower 35448 kilowatt1 horsepower foot- pounds per m inute1 horsepower kilowattIt is to be noted that torque multiplied by angular veloc ity

(in radians per unit time) gives the power of a rotating body ex

pressed in the corresponding units . In stating the performanceof electric machines use

- is frequently made of the expressiontorque at onefoot radius this is equivalent to expressing thetorque in pound- feet . (In the expression

“ torque at one- footradius

,

” the word torque is incorrectly used,since torque is in

dependent of the radius — what is m eant is the force at onefoot radius . ) The following are useful relations

Power in kilowatts - l .4197X 10"Xtorque in pound

feetx revolutions per m inute .Power in horsepower X X torque in pound

feet revolutions p er m inute .23 . Harmonic Motion .

- A particular typeof motion which is not only of considerableimportance itself

,but also serves as a useful

analogue in the discussion Of electric c ircuits ,is tha t known as harmoni c motion . Harmonicm otion is defined as motion such that theacceleration of the moving body is proportional to

,and in the Opposite direction to ,

the displacement of the body from its positionOf equilibrium . (The equilibrium position of abody is that position in which the resultantforce acting on the body is zero .) Harm onicmotion is illustrated by the motion of a pendulum

,the motion of a weight attached to a

spring , the motion Of each particle Of a tuningfork

,the motion of the balance wheel Of a F ig - 8

watch,etc . The first three are examples of harmonic motion Of

trans lation,the latter of harm onic m otion of rotation .

Cons ider the case of a s imple pendulum . Let B be the bobof the pendulum and let it be so small that it may be cons idered

o - c


as a single particle of mass m,and let this bob be suspended by

a weightless thread of length l from a fixed point P . Let thebob be displaced from its

,position of equilibrium and at any

tim e t let its displacem ent , measured along the are in which itmoves

,be a: arc OB . The linear velocity at which the bob

dxis moving at any instant is then

dtand its kinetic energy is

therefore dx

dt

The potential energy of the bob at this instant ismg 0 Q ==mgl ( 1 —cos B)

x

where B=7and g is the acceleration due to gravity .

energy of the bob is then

W= % m +mgl ( 1—cos B)0

and if there is no fri ction this energy must be cons tant , and therefore the rate of change of this total energy with respect to tim emust be zero . Hence

dW dx d’xl B

dB0m 0

dt dt dizmg 8m

dt

dB 1 dxh 1w ence

,srnce

dt l dl

d2

diffi

g sin BFor B small

,sin B B when B is expressed in radians ; whence

for g; small , sin B and therefore

2d x gx .

diz l ( 18)

Equation tells us that the linear acceleration of the bobalong its path of motion is proportional to , and in the oppos itedirection to

,its displacement ; hence this is a case of harmonic

motion .

*This equation may also be deduced directly from a consideration Of theforces acting on the bob at any instant .


marked v is for 0 7T. Note that the angle 9 in any caseis also equal to the distance to the left of the origin at whichthe curve first crosses the base line in the posi tive direction .

A function of the form X0 sin (w t -F9) , that is , a function whichcan be represented single sine wave

,is som etimes called

Fig . 9 .

a simp le harmoni c function , or briefly, a harmonic function . Otherperiodic functions are called non-harmoni c functions .

In the case of the simple pendulum let the bob be held outa distance A from its equilibrium position

,and at a given instant

let it be released . Let time be counted from this instant ; thenfor t=0 we have

s

and the velocity at this instant is

dsr

dt

Substitution Of these values in ( 19) gives

A =X0 sin 9

0 —X0 cos (9

whence 07

2

—7: and XO

= A . Therefore the displacement

bob from its equilibrium position at any instant is

x = A sin =A cos w t

and its velocity at this instant is

dxz —wA sin w t

di

FUNDAMENTAL IDEAS AND UN ITS

where 01 Note that the displacement has its first positive

maximum after the start for m i = 2 n‘ and the veloc ity its first

3 7Tpositive m aximum for wt

3hence the velocrty reaches a pos

itive max imum a quarter of a period ahead of the displacement, orthe velocity leads the displacem ent by 90° (see Fig . Two S inewaves of the same frequency which reach their positive max imaat different times are said to differ in phase by the angle corresponding to the time interval between successive positive maximaof the two waves . When both waves are expressed as sinefunctions

,the difference in phase is the difference between the

phase angles of‘ the two functions . In the above example

x =A sin

v=wA sin (w t -P77)

whence the phase angle of as 1s gand the phase angle of v is 77 ;

therefore the difference in phase between the two is

7

5 or 90 degrees .

The function w ith the larger (algebraically) phase angle alwaysleads . Note that the l eading curve is to the l eft.

24 . Temperature .

—The physical properties of any . piece of

matter depend,among other things , upon its temperature , i .e.

,

upon its relative hotness or coolness referred to some standard substance under standard conditions . The idea of temperature isfam iliar to every one

,and one ’s So~ called temperature sense en

ables one to form a rough j udgment of the relative hotness or coolness of two or more bodies . For sc ientific purposes

,however

,a

more reliable and more delicate means of measuring tem

perature is desirable . Any device which serves this purpose iscalled aThe standard temperature—measuring device is the constant

volume hydrogen therm ometer, which consists essentially of asuitable receptacle containing a constant mass of hydrogen gaskept at constant volume

,with means provided for measuring any

*A thermometer des igned to measure very high temperatures is called apyrometer. ”


variation that may be caused to take place in the pressure of the

gas . The numerical value of the temperature of any substanceis then defined in terms Of the relative pressure of this gas when thereceptacle is immersed in the substance, referred to the pressureof this gas when the receptacle is immersed in m elting ice at a pressure of 760 mm . ofmercury, the pressure Of the gas in each casebeing measured after it has reached a constant value . The tem

perature of melting ice at a pressure Of 760 mm . of m ercury isarbitrarily taken as zero deg rees, and the temperature ofsaturatedsteam at a pressure of 760 mm . Of mercury is taken as 100 degrees .

Call ing p 1 the pressure of the hydrogen gas when the receptac le isimmersed in the melting ice and p2 its pressure when immersedin the saturated steam

,and p its pressure when imm ersed in any

given substance S (the pressure in each case being m easured afterthe lapse of a suffic ient time for it to reach a constant value) , thenumerical value Of the temperature of the given substance is

p PI

275—271

The Fahrenheit scale of temperature is derived in the samemanner

,except that the temperature of the melting ice is taken as

32 degrees and that of the saturated steam as 212 degrees . A

temperature of t, degrees Fahrenheit is then equal to

defined as t degrees centigrade .

to (t, 32) degrees centigrade .

For practical purposes,the volume expans ion of mercury in

glass , or of alcohol in glass , is used as a m easure Of temperature .

For strictly accurate m easurements,such m ercury- in-

glass oralcohol- in-

glass therm ometers should be calibrated by comparisonwith a standard hydrogen thermometer, but for ordinary work itis suffic ient to determ ine the two points on the thermometer scalecorresponding to the temperature of m elting ice and saturatedsteam under standard conditions

,and to assume that the volume

expansion Of the thermometer fluid is proportional to the temperature .

25 . Heat Energy . Al l known experiments lead us to believethat whenever the temperature of a body increases

,energy is

transferred to it,and that whenever the temperature of a body

decreases energy is transferred from it ; further , that there isa fixed numerical relation between the quantity of energy transferred to a given body and its change in temperature . This nu


merical relation can be determ ined directly only in case thechange in temperature is produced by mechanical work , s ince

,

by definition , the measure of energy is the product of a forceby a distance . Careful experiments have shown that the workrequired to raise the temperature of one gram of water from0° to 100° centigrade is ergs ; We may then take as

the unit of heat energy the one- hundredth part of the workrequired to raise one gram of water from 0 to 100 degrees centi

grade . This unit is known as the mean sma l l ca lori e,and in

terms of it,the amount of energy involved in various heat effects

may be measured with comparative ease . It should be notedthat the work required to raise one gram of water one degreeis different at different temperatures ; the variation is negl igible,however

,except in the most refined work .

The above numerical relation between the m ean small calorieand the erg is called the mechani ca l equiva lent ofheat on the c . g. 3 .

system . Another way Of expressing the mechanical equivalentof heat is

,the number of foot-pounds Of work required to raise

the temperature of one pound of water one degree Fahrenheitor at near its maximum dens ity Fahrenheit) ; this may

likewise be used as a unit of heat energy, and is known as theBritish Thermal Unit . ” See Article 2 1 for the relations betweenthe various units of heat energy .

Practically every phenom enon in nature is ac companied by achange in temperature of one or m ore bodies . In certain casesthe entire amount of energy transferred in the process m ay becaused to appear as heat energy, and when this can be done thetotal amount Of energy transferred can be m easured with a fairdegree of accuracy . Usually

,the heat energy developed in any

process is not in a useful form ; in such cases the heat energyis said to be dissipated .

”

26 . Efficiency and Losses .

- In any m achine or apparatuswhich is employed for transform ing energy from one form intoanother or for transferring energy from one place to another

,

a certain amount of energy is always converted into forms whichcannot be readily utilized . In general , this useless energy appearsas heat energy . The rate at which energy is put into a m achineis called the power input into the m achine and the correspondingrate at which the m achine gives out usefu l energy is called the

power output, or the load on the machine . The difference between


the power input and the power output is called the power loss .

The ratio of the output P0to the input P, for any given output

of a m achine is defined as the efii ci ency Of the machine at .thisoutput ; this ratio is usually expressed as a percentage . Theper cent efficiency e is then

e= 400 (29

The ratio of the difference between the input P, and the outputP0to the input P

,expressed as a percentage, is called the per cent

power loss p; that is,

p= 4m}-

75—e.

1

Sometimes it is more convenient to express the power loss as apercentage of the output P

0. Let p

’ be the percentage lossexpressed in this manner ; then

l 00p

100—pThe efficiency of a machine varies as a rule with the lo ad

or output . For no load,i .e.

,no output

,the efficiency is zero

,

since in general energy must be supplied to the m achine to operateit

,even though the machine does no useful work ; as the load

comes on the effic iency increases up to a certain output,depend

ing on the design of the m achine,and then decreases .

The rated load or rated power output of a m achine whichis designed for continuous service is the maximum rate at whichuseful energy m ay be transferred through the m achine con

tinuously w ithout inj ury to any Of its parts . In most electricmachinery this is determ ined by the rise of temperature produced by the energy dissipated or lost ” in the windings andiron cores . As a rule

,the insulation of the windings will deterio

rate if the temperature of the machine exceeds 75° centigrade .

27 . Newton ’s Law ofGravitation . From the observations by

various astronomers of the revolution of the planets about the sun,

and of the revolution Of the m oon about the earth,Newton was

led to the belief that every particle of m atter in the universeattracts every other particle with a force proportional to the product of the masses Of the two partic les

,and inversely proportional

to the square of the distance. between them ,namely with a force

mm’

r2


where h is a constant depending upon the units in which m,m

’

, r,

and f are measured . Further Observations and careful laboratoryexperiments by Cavendish and others have confirm ed this beliefwhich is now accepted as one of the fundam ental laws ii of

nature . The constant It has been determ ined by experiment to bewhen m and m ’ are expressed in grams

,r in centi

meters,and f in dynes . For small m asses

,therefore

,such as one

ordinarily deals with in the laboratory,this force is extremely

small,and can be detected only with the most delicate instrum ents .

As we shal l see presently,the mutua l attraction or repu lsion

between magnetic poles and a lso the mutua l attraction or repu l

sion between electri c charges obey a law of exactly the same form.

Hence the study of forces which obey this law becomes of primeimportance . It Should be borne in m ind that any conclusionsderived from this law of force action will apply to any one of thethree agents , gravitational m asses , m agnetic poles , and electriccharges . In order to express such deductions in terms of phys icalquantities

,we shall cons ider first the forces produced by magnetic

poles .

PROBLEMS

1 . Find the speed in meters per second and in feet per secondof an electric locomotive travelling at a uniform speed Of 60 m ilesper hour . If the locom otive is travelling at this Speed arounda curve of 1000 feet radius

,what is the direction and amount

of the acceleration in feet per second per second and in m ilesper hour per second?

Ans . : m eters per second ; feet per second ; feetper second per second ; m iles per hour per second . Acceleration toward center Of curvature .

2 ; In Problem 1,if the distance between rails is 4 feet

,

inches,how many inches would the outer rail have to be elevated

above the inner rail in order that the flanges of the locomotivewheels exert no side thrust on the rails?

Ans . : inches . (In practice the elevation would beabout half this ; that is , the actual force on the outer rail wouldbe about half what it would be were both rails in the same horizontal plane .)

3 . If in Problem 1 the outer rail of the curve is elevated 6inches and the locomotive weighs 100 tons

,what will be the


average shearing force exerted on each of the spikes holdingthe rails to the ties? Assume the total force to be exerted against15 spikes

,and neglect the friction of the rail against the ties .

Gauge of track 4 feet, inches .

Ans : 1794 pounds .

4 . A man stands on the floor of a trolley car which is travellingat the rate of 20 m iles an hour around a curve of 100 feet radius .

How m any degrees from the vertical must the m an lean to preventhis falling over? Assum e the man may be represented by arod standing on its end . Draw a complete vector diagram of

the forces acting .

Ans . : 150°

5 . The tangential force exerted by a brake on a pulley whichis rotating at a speed of 500 revolutions per m inute is 200 pounds .

If the diam eter of the pulley is one foot , calculate the torquein pound- feet developed by the pulley , and the amount of workin horsepower-hours done by the pulley in 10 m inutes .

Ans . : 100 pound- feet ; horsepower-hours .

6. The radius of gyration of a cylinder about its own axisD

is equal to4where D is the diameter of the cylinder . What is

the mom ent of inertia of an iron cylinder 5 feet in diam eter and4 feet long? Specific gravity of iron give answer in poundfoot units .

Ans : poundfoot .7. If the cylinder in Problem 6 is rotating at a speed of 1000

revolutions per m inute what is its kinetic energy in foot—poundsand in horsepower- hours?Ans ; foot-pounds ; horsepower- hours .

8 . If the force driving the cylinder in Problem 7 is removed,

how many hours will it take the cylinder to com e to rest if theopposing torque due to friction is cons tant and equal to 300

pound- feet?Ans hours .

Note : The data given in Problems 6 to 8 are fairly representative of a large turbo-generator, except that the friction torqueis not cons tant . Unless a brake of som e kind is used the rotatingpart of such a m achine will continue in motion for severalhours after the steam is shut off.

9 . A weight of 500 pounds falls from rest a distance of 100


vertical t seconds later, neglecting the effect of friction ; what willbe its maximum velocity in inches per second and what will beits displacement from the vertical at this ins tant ; what willbe the frequency of the pendulum ?

Ans . : x =3 cos (3 .27t) inches ; inches per second everytim e the displacement from the vertical is zero ; cycles persecond .

13 . Plot to scale the values of the displacem ent,veloc ity

,

and acceleration of the vibrating bob in Problem 12 . Whatare the phase relations between the veloc ity and the displacement between the velocity and the acceleration?

Ans ; The veloc ity leads the displacement by 90° and lags90

°behind the acceleration .

14 . What will be the kinetic energy of the ball in Problem 12

when it is half way between its equilibrium position and its pointof m aximum deflection? What will be its potential energy atthis instant? Give answers in j oules and in foot-pounds .

Ans . : Kinetic. energy foot-pounds or j oules ;potential energy foot-pounds or j oules .

1 5 . Two lead spheres each 1 foot in diameter are placed neareach other with their nearest points 1 inch apart . The densityof lead is Calculate the force (gravitational) in dynes withwhich the two spheres attract each other

,assum ing the tota l

mass of each sphere concentrated at its center .Ans . : dynes .

MAGNETISM

28 . Magnets . Magnetic Poles . It has long been knownthat a certain m ineral, cal led the loadstone, has the property of

attracting pieces of iron or steel with a readily perceptible force,

even though the loadstone and the piece of iron or steel be com

paratively small . In other words,between loadstone and iron or

‘

steel there is a force of attraction many times greater than theforce of attraction between like m asses of ordinary matter . It

has also been known for many centuries that a steel bar can begiven this same property by stroking it lengthwise with a piece ofloadstone

,and that when the bar thus treated is freely suspended

,

it takes up a definite position with respect to the earth,one end

of the bar pointing approximately toward the north geographicalpole and the other end toward the south geographical pole . A

steel bar having this property is called a magnet, the north pointingend is called its north or posi tivepole and the south pointing end itssouth or negative pole. In general, am agnet may be defined as abody which possesses the property of attracting with a readily perceptible force pieces of iron or steel

,and which

,when freely sus

pended,takes up a definite position with respect to the geographical

meridian . As we shall see later on,it is possible t o magnetise

a steel bar in other ways than by stroking it with a piece of loadstone ; in particular, when an insulated wire is wrapped aroundsuch a bar and an electric current is established in the wire, thebar becomes magnetised ; this is the m odern way of making amagnet and the only way ofmaking a powerful one .

29 . Paramagnetic and Diamagnetic Substances. It is foundby experiment that a magnet attracts not only iron and steel, butto a less extent nickel and cobalt . B ismuth

,on the other hand

,is

repelled by a magnet . Substances which of them selves are not mag

nets but which are attracted by a magnet are called paramagneti csubstances, or simply magnetic substances ; substances which of

themselves are not magnets but which are repelled by a magnet arecalled diamagneti c substances . Except in the case of iron

,steel

,

37


nickel , and cobalt , the force of attraction for paramagnetic bodiesis extremely small ; of the diamagnetic substan ces bismuth is theonly one which is strongly repelled by a magnet, and t he repu l

s ion even ofb ismuth is weak compared to the attrac t i on of iron .

Whether or not a substance is attracted or repelled by amagnet depends upon the nature of the m edium surrounding themagnet and the substance . For example

,it is found by experi

ment that a solution of perchloride of iron is attracted by a magnetwhen the solution and the magnet are surrounded by air . On theother hand

,if the magnet is immersed in the solution and a bubble

of air formed in the latter,the air will be repelled by the magnet .

Therefore perchloride of iron is paramagnetic with respect to air , .and air is diamagnetic with respect to perchloride of iron . In

other words,paramagnetic and diamagnetic are purely relative

terms . It is customary to take air as the standard of reference,

although in certain cases it would be preferable to take free spaceor the ether ”

as the standard . However,the difference be

tween air and free space in respect to their m agnetic qualities ispractically inapprec iable , so that a body which is m agnetic withrespect to air is likewise m agnetic w ith respect to free Space . In

the following discussion , when the terms paramagnetic and diamagnetic are used, air is to be understood as the standard of

reference ; i .e .

,air is assumed to be non-m agnetic .

30 . Attraction and Repul sion of Magnetic Poles . A piece ofiron or other magnetic substance which of itself is not a m agnet ,will be attracted by either pole of a m agnet when placed near thispole . However

,when two m agnets are placed near each other,

the mutual force between the two may be either an attraction or arepulsion . When the two like poles of the two m agnets arenearer together than their unlike poles , the forceis a repulsion ;when the unlike poles are nearer together, the force is an attraction .

This is readily tested by suspending one of the magnets by a threadattached to its m iddle point and bringing the two ends of thesecond magnet successively near one end of the suspended m agnet .We therefore conc lude that like m agnetic poles repel each otherand unlike poles attract each other . Experiment also shows thatthe force of attraction or repulsion of two magnets falls off rapidlyas the distance between them is increased .

3 1 . Magnetic Charge . A Magnetic Pole as a Force-ProducingAgent . I

‘

t is found,

by experiment that the forces produced by

MAGNETISM 39

magnets on one another and upon m agnetic bodies placed in theirvic inity m ay be accounted for in terms of a something confinedsolely to the external surfaces of such bodies . In the case of along , slim magnet this force -produc ing agent is confined almostentirely to the ends of the m agnet . Since the two ends of a mag

net possess opposite properties , it is necessary to look upon thisforce-produc ing som ething as different at the two ends of themagne t . This something at the north pointing end of a magnetmay be called a pos itive m agnetic charge ,

” and the somethi ngat the south pointing end a negative magnetic charge .

” How

ever,it is customary to use the expression “ north (or pos itive)

magnetic pole to signify this something at the north pointingend of a magnet and the expression south (or negative) magneticpole to signify this something at the south pointing end . Whenused in this sense

,the word pole signifies the forceproduc ing

agent assoc iated with the surface of a magnet rather than the

particular part of the magnet at which this force -produc ing agentis located .

It is to be noted that although the external forces produced bya magnet may be accounted for in terms of a som ething confinedsolely to its external surface

,there is als o a change produced

throughout the substance of a body when it is magnetised]

. For,

when such a body is broken in two,the two broken ends are

found to be the seat of equal and opposite magnetic poles .

There are also certain very special cases in which the poles of amagnetised body must be considered as distributed throughoutits interior

,but the theory developed below for surface poles may

be readily extended to such cases .

32 . Induced Magnetisa tion . In order to express in an exactquantitative m anner the value of the mutual forces produced onone another by magnetic poles it is always necessary to take intoaccount the effect of any m agnetic bodies which may be in theirvic inity . As we have seen

,when a m agnetic body which itself is

not a magnet (a small rod of soft iron,for example) is placed in

the vic inity of a m agnetic pole, this substance isattracted whetherthe pole is a north or a south pole . In fact , the body acts exactlylike a m agnet, except that the strength and position of its polesdepend on its position with respect to the m agnet attracting it .The m agnetic body is therefore said to be m agnetised by induction .

”Since the magnetised body , is always attracted

,the loca


tion of the poles induced on it is always such that two of the un likepoles on the attracting magnet and on the magnetised body re

spectively are nearer together than the like poles on these twobodies . A diamagnetic body placed in the vic inity of a m agnetis s im ilarly magnetised by induction , except that in this case

,

since there is a repuls ion between the inducing m agnet and thediamagnetic body, the li ke poles are nearer together . For

example,a small rod AB of soft iron placed in the vic inity of a

magnet in the various positions shown in Fig . 10,is magnetised as

Fig . 10 .

indicated . When the rod is removed from the vicinity of themagnet the induced poles disappear almost entirely . In case ahard steel rod

,or1g1nally unmagnetised, is placed near the mag

net,it becom es likewise magnetised by induction, though to a

less extent than the soft iron . If, however , the steel rod is rem ovedfrom the vicinity of the m agnet it is found to retain to a consid

erable degree its magnetisation, that is , the rod becom es a permanent magnet .

The phenomenon of induced m agnetisation is a most importantone both in practice and in theory

,and we shall return to this sub

j ect for a more detailed study . The chief fact to be borne in m indfor the present is that any magneti c or diamagneti c body p laced in

the vi cini ty of a magnet has magneti c po les. induced on i ts surface,

and that these induced poles produce forces on every magneti c pole,

permanent or induced,whi ch may be in the vi cini ty. For example

,

when two m agnets are separated from each other by a given distance and there are no m agnetic or diam agnetic bodies in thevicinity

,they will exert a certain force upon each other . When a

magnetic body , such as a piece ofunmagnetised soft iron, is placed

MAGNETISM 41

near them (see Fig . the force acting on each magnet wil levidently be changed, since the poles s and n indu ced on the

Fig . 1 1 .

soft iron also exert forces upon each magnet . In the case illustrated , the resultant force acting on each magnet will be increasedand also changed in direction . Again , when the entire regionaround the two magnets is filled with a magnetic liquid (perchloride of i ron , for example) , magnetic poles are induced on

the surface of the liquid where it comes in contact with the polesofthemagnets, Fig . 12

,and these

induced poles are ofthe oppositesign to those of the magnet atthis surface . The effect of theseinduced poles in this case is todecrease the resultant force on

each magnet , Art . 56. The as

sumption that a ir i s non-magneti c

n b 2

i s equiva lentto assuming thatthere

are no poles , ei ther permanentMagnetic L'qu'd

or induced, produced on the a ir

in contact wi th a magnet. Fig . 12 .

33 . Point-Poles . Experim ent shows that it is phys ically impossible to have a magnetic charge or pole offinite amount concentrated in a point . However, for the purposes ofmathematicalanalysis of the forces produced by magnets , it is frequently con

venient to cons ider a magnetic pole as occupying but a point inspace . We may call such a pole a point

-

pole. A physicalapproxim ation to such a point-pole is the pole on a very small area .

34 . Properties ofMagnetic Poles . We are now in a pos ition.to state the properties which must be attributed to magneticpoles in order to account for the experimentally observed factsconcerning their mutual action . When the induced magneti c po lesas wel l as the permanent poles are taken into account

,and a ir i s

assumed to be non-magneti c these properties are the following :


1 . Like poles repel each other ; unlike poles attract each other .

2 . Whenever a m agnetic pole ofone sign exists on a body thereis always an equal pole of the opposite sign on some other part ofthe same body . Neither kind ofpole can be produced by itself .

3 . Two point-poles of “strengths” m andm’ respectively located

a distance r apart repel each other with a force proportional to theproduct ofthe strengths m and m

’ and inversely proportional to thesquare of the distance between them ; but independent of the

medium between them; that is , with a force

lm m

,

where k is a constant depending upon the units in which m,m r

and f are m easured .

Since we have not as yet spec ified what we shall take as ameasure of the strength of a magnetic pole ,we may select the unitofpole strength so that this constant k is unity when r is m easuredin centim eters and f in dynes . Equation (1) then becomes

im m

’

In this expression for the force between two magnetic polesthe strengths o f the poles m and m

’ are to be expressed as positivenumbers when they are north or positive poles and as negativenumbers when they are south or negative poles . Hence when m isa north pole and m’ a south pole, or vi ce versa

,the force of repu l

sion f is likewise negative , that is, - the force f is an attracti on ; thisis in accord with the first property attributed to these poles . Theexpression (1a) as thus understood may be looked upon as adefinition of the m easure of the strength of a m agnetic pole . For ,

in accordance with this law of the mutual action of two point—poles,

we may define the strengths of two such poles m and m ’as equal

when each repels with the same force f, a third point- pole m”

placed at the sam e distance from each . A uni t poi nt-

po le i s then

a point-

pole which repe ls wi th a force ofone dyne an equa l point-

pole

p laced one centimeter away. This unit is cal l ed the c . g. s . electromagnetic unit of pole strength .

The above law ofmutual action of two magnetic poles appliesdirectly only to point- poles

,for when the poles are extended over

a surface the distance r between the poles has no definite m eaning,since the distance between every -pair of points taken respectively


electric current (see Chapter III ), will have a force exerted upon it ;the region of space in which a magnetic pole is acted upon by aforce is therefore called a magneti c field offorce . In general,whena magnetic pole is placed in the vic ini ty of a m agnet or of a mag

netic body this pole will induce new poles on the magnet or mag

netic body . Again, in order to place a magnetic pole ins ide amagnet it is necessary to cut a hole in the magnet , and, as pointedout in Artic le 46

,whenever a hole or gap is cut in a magnet,

magnetic poles in general appear on the wal ls of this gap . Hence,

when a m agnetic pole is placed near or ins ide a magnet or magnetic“

body,new poles will in general be form ed . However

,the force

which would be exerted by the poles origina l ly in the field on a poleplaced at any point in the field

,whether ins ide or outside the mag

net,may be calculated from the fundamental law of the mutual

action of two magnetic poles (equation 1a) . Sim ilarly,the force

exerted by an electric current on a magnetic pole may be cal cu

lated from the fundamental law of the mutual action of a currentand a pole (see Chapter III) . The force in dynes which wou ld beexerted on a uni t north point

-

pole p laced at any point in a magneti c.

fie ld,due solely to the poles and currents origina l ly in the fie ld

,i s

defined as the intensi ty ofthe magneti c field at that point due to these

agents . The field intens ity at any point is also called the mag

netising force at that point .It should be noted that the unit of field intensity is not the

dyne,but is a dyne per unit po le; compare with power, the unit of

which is not the erg , but is an erg per second. No spec ific name hasbeen given to the c . g. 3 . unit

_

of field intensity but it is usuallyexpressed as so many gilberts per centimeter, the m eaning of

which expression is explained inArticle 60 . Magnetic field intensityor magnetising force is also expressed in terms of ampere- turnsper centim eter or ampere- turns per inch

,

” the m eaning ofwhich expressions is explained in Chapter IV . The relations between these various units are

1 c . g. s . electromagnetic unit 1 gilbert per centimeter1 c . g. s . electrom agnetic unit 9578 ampere- turns per

centimeter1 c . g. s . electromagnetic unit ampere—turns per

inchSince the force in dynes produced by a magnetic point-pole of

strengthmat a distance r centimeters away on apoint-pole ofstrength

MAGNETISM

mX 1unity is the intensity of the magnetic field at a point P a

distance of r centimeters from a point—pole of strength m,due solely

to thi s pole, 1sm (3)

whether this point be outside or inside a m agnet . The field intensity H will be in the direction of the line drawn from m to Pwhere m is positive, and in the direct ion of the line drawn from P

to m when m is negative . Field intensity is therefore a vectorquantity .

The field intensity at any point due to any number of pointpoles is then the vector sum

'

HJET

where the symbol 2 wi th a line over i t is used to represent a vector

sum . E ach pole then contributes a component ln- to the result

ant intensity,where m is the strength of this pole and r is its

distance from the given point .The force exerted on any point-pole ofstrength m in a magnetic

field isF =mH (4)

where H is the field intensity at the point occupied by the polem due to all the poles in the field except the pole m. The field dueto any given pole can ofcourse produce no mechanical force on thispole itself ; a man cannot lift himself by his boot- straps .

”

When the field intensity has the same value at every pointthroughout a given region , the field is said to be uniform throughout

'

that region ; we shall find several examples of practically uniform magnetic fields .

37 . Example Il lustrating the Appl ication ofAbove Defini tions .

The following example will serve to illustrate the ideas j ust discussed . The problem chosen is to find the field intensityat any point on the axis of a magnet having the shape of

a long right cylinder and having its poles confined entirely tothe end surfaces of the cylinder and of the same strength per unitarea at each point of these end surfaces . Let 0 be the strengthper unit area of the north pole of this magnet and—o the polestrength per unit area of its south pole . Consider first the field


intensity due to its north pole . Let the point P,Fig . 1 4

,at which

the intens ity is to be determ ined , be at a distance a from this pole,and let the radius of the magnet be r . Consider a ring drawn in theend surface with its center at N on the axis of the magnet, andlet the radius of this ring be x and its width be dx . The elem entarypole at any small area ds of this ring will produce a field intensityat P

,and thi s force will have a component in the direction NP and

a component perpendicular to NP . The elementary pole on anequal sm all area ds ' diametrically opposite ds will likewise producea field intensity at P numerically equal to that produced by the

Fig . 14 .

pole at ds,but in the direction shown . Hence the components of

the field intensities,due to the poles at ds and ds ’

,perpendicular to

NP are exactly equal and opposite,and therefore exactly neutral

ise each other ; sim ilarly for any other two diametrically oppositepoints in this ring . Hence the resultant intensity due to theentire ring is in the directionNP and is equa l to

a

02

-14 82

x/a 2 —x 2

For 2 7Txdx is the area of the ring, and therefore 0 X2 7Txdx is thetotal pole strength of the ring ; a2+ x

2 is the square of the distance

of the point P from each element of the ring , and —i 18 theV a

2+ x

2

cosine of the angle between the field intensity at P due to eachpoint in the ring and the line NP . The total field intensity Hn ,

at P,due to the entire north pole of the m agnet is then

the sum of the field intensities due to all the contiguous elementary rings into which the pole face N may be divided . Hence

ZE = T

= 7r 0‘

a (dz

-H z”

)

MAGNETISM 47

l

— 77 0'

a 2 (a2+ x

2)2

1

—2 7r 0'

(l —cos 0) (5)

where d is the angle between the axis of the magnet and the linedrawn from P to the edge of its north pole . The field intensitydue to the south pole of the m agnet is sim ilarly

H8=—2 7T —cos

where 0’is the angle between the axis of the magnet and the line

drawn from P to the edge ofits south pole .

Consequently the total field intensity is9'

- cos (5a)

Each end of this magnet is a circular disc over which is uniform ly distributed a magnetic pole . It is interesting to notewhat are the lim iting values of the field intensity due to a singlemagnetically charged disc * when the point P is ( 1) very c lose to thedisc

,and (2) when the point P is at a considerable distance from

the disc . In the first case,cos (9becomes negligibly small in compari

son with unity,since 9 becomes practically Hence at the

point on the ax is of a uniformly magnetically charged disc at aninfini tesimal distance from its surface

,the field intensity due to this

disc a lone is

H 2 7T0 (5b)and is independent of the size of the disc

,but depends only upon

its pole strength per unit area .

It can also be shown that a m agnetic pole distributed in anymanner over any plane surface

,produces a field intensity at a

point j ust outside this surface which has a component norma l tothis surface at this point equal to

Hn

2 7r 0'

(50)

independent of the shape or s ize of the surface,where (r is the pole

strength per unit area of the surface directly opposite this point .In general , however, there will also be a tangential component tothe field intensity

,that is

,a component parallel to the surface ; also

the other pole of the magnet will be sufficiently c lose to produce afield intens ity at this . point

,whi ch intensity may have both a

normal and a tangential component .

*e . g], the end ofa very long cylindrical magnet when the other end is soremote that it produces no appreciable effect.


When the point P is at a great distance from the pole face, sothat a is large compared with r

,the angle (9 becomes very small

and therefore this angle in radians is equal to its tangent, i .e.,

a

Whence,expanding cos 9 into a series and neglecting all terms

of a higher order than the second , we have

whence

1 cos 9

and therefore equation (5) becomes

Hwa r

?a m

02

a2

where m = 7r r20

' is the total pole strength .

Hence the fiel d intensity at a point on the axis ofa magnet a considerable distance from the magnet may be calculated approximately by assum ing the poles concentrated in points at the ends of themagnet . In most practical problems a sufficiently accurate valueof the field intens ity at any point due to a bar magnet m ay be

obtained by considering the poles of the magnet to be concentratedin points at its ends

,except when the point under consideration

is very close to a pole,but the l im itations ofthis assumption should

always be borne in m ind .

38 . Force Required to Separate Two Equa l and Opposite Polesin Contact. A useful application of equation (5b) is the cal culation of the force required to separate two equal and oppositepoles distributed uniform ly over plane surfaces

,such as the two

poles of two equal bar magnets placed end to end . Call the twosurfaces A and B and let them be separated by an infinitesimal ly

narrow gap . Then,if 0 ' is the pole strength per unit area of the

north pole on the wall of the gap ,the field intensity inside the gap

produced by either surface is H =2 7T and therefore this sur

face attracts each unit pole on the other ‘

surface B with a force of2 7T 0 " dynes . Hence , calling S the area of each surface

,the total

pole strength of the surface B is —0'

S and consequently the surface A attracts the surface B with a force of

F =2 7Ta'2 8

MAGNETISM 49

dynes . This formula is deduced on the assumption that the onlyforces acting are those due the two poles which are separated ; ifthere are any other forces present (for example, the forces due tothe other ends of the magnets and to the direct action of theelectric current in the coil surrounding the m agnets if the latter areelectromagnets) their effect also must be taken into account . Theformula is also based upon the assumption that the poles are confined solely to the end surfaces and are uniform ly distributed overthese end surfaces

,which is practically never realized . The

formula does give a rough means , however, for determ ining theaverage value of the pole strength per unit area , since both theforce F and the area S are readily measured .

39 . The Earth ’s Magnetic Field . Since a magnet suspended

at any point near the earth ’s surface tends to point in a definitedirection

,indicating the existence of a force acting on the magnet,

the region of space in the vic inity of the earth is a magnetic field offorce . For points at a considerable distance from any large massesof iron or wires carrying electric currents

,the intensity of the

earth ’s field is practically uniform over a large area , though t hereis a considerable variation in both the magnitude and the directionof the earth ’s field with longitude and latitude, and also a smallvariation from day to day and from year to year . The earth ’sfield at any point in general has both a horizontal and a verticalcomponent (downwards in the northern hem isphere, approx i

mately) . The horizontal intensity of the earth ’s field in the longitude of Washington and at latitude 45

° north is roughly c . g. s .

electromagnetic units (defined above) and the vertical componentc . g. s . electromagnetic units . In any laboratory, however,

particularly in the vic inity of electric trolley lines,the value of

the m agnetic field may differ considerably from this and variescontinually .

40 . Magnetic Moment . Equ ivalent Length of a Magnet .To hold a magnet in a uniform magnetic field in any other directionthan that which it would take if acted upon only by the forcesdue to this field

,requires a certain torque or moment

,and the

maximum value of this moment will correspond to a’ pos ition of

the magnet at right angles to its equilibrium position when actedupon only by the forces due to the field . The ratio of the value ofthe maximum mom ent which can be exerted by a uniform mag

netic field on a given magnet to the value of the field intensity, is


called the magnetic moment of the magnet . In the ideal case of amagnet which has its poles concentrated in two points a distance lapart

,the magnetic mom ent is equal to m l where m is the strength

of the north pole of the magnet . For,the total force acting on

each pole of such a m agnet when placed in a uniform field of intensity H is mH ,

and when the magnet is at right angles to this fieldthe moment acting on it is mH l , whence the ratio of this maximummom ent to the field intensity H is m l . The equiva lent length ofany m agnet is defined as the ratio of its magnetic moment to thetotal strength of its north pole .

41 . Equal ity of the Poles of a Magnet . Equil ibrium Positionofa Magnet in a Uniform Magnetic Field . When a magnet of anyshape is supported in a uniform m agnetic field

,for example

,that

due to the earth , in such a manner that it is free to movefloated on a piece of cork) . the magnet takes up a definite positionand remains at rest . Hence the two poles of the magnet must beof equal and opposite strength ; for, calling H the intens ity of thefield and m and m ’ the total strengths of the two poles respectively ,the force acting on one pole of the magnet is mH and the forceacting on the other pole is m

’ H,and therefore the .total force

on the magnet is (m+m’

) H . But s ince there is no motion of themagnet relative to the earth , which produces the field

,this total

force must be zero ; hence m = —m ’. Further

,since there is no

rotation of the magnet, its position of equilibrium must be suchthat the line of action of the two forces mH and m ’H coincides .

Hence,when the magnet is in the shape of a long rod or needle , a

line drawn from its south to its north pole will give the directionof the field . This fact is usually expressed by saying that theneedle points in the direction of the m agnetic field . When themotion of the m agnet is lim ited to a definite plane, for example ,when it is floated on a piece of cork or m ounted on a pivot of thekind used in a m agnetic compass

,it will point in a direction cor

responding to the component of the field intens ity in this plane .

42 . Measurement of the Horizontal Component of the IntensityofaMagnetic Field . When a barm agnet ormagnetic needle is supported in a m agnetic field in such a m anner that it is free to vibrateabout a vertical axis , and is set vibrating about this axis , it will oscillate with a definite period depending upon the horizontal component ofthe intensity of them agnetic field ; due to the fri ction of theair and of its support it will ultim ately com e to rest in line withthe hori zontal component of the field intensity . Let this hori


inertia I and the magnetic moment M of the magnet . Them om ent of inertia I can be readily calculated when the magnet is inthe form of a bar of rectangular cross section or in the form of aright cylinder, or it may be determ ined experimentally . A secondrelation between the moment M and the field intensity H can beobtained from the following experiment . Remove the bar magnetfrom the point P and suspend in its place a small magnetic needle(Fig . Place the bar magnet with its center at a distance afrom P with its axis perpendicular to the direction of H and inthe sam e hori zontal plane as the point P . The magnetic needlewill now be acted upon by two magnetic fields, the original fieldHand the field due to the magnet . The intensity of this latterfield will be

Fig . 16 .

1When a is chosen large in comparison with _

2then will be

practically negligible in comparison with unity ; hence to a closeapproximation

2m l 2M

a3

_

5

where M is the moment of the bar magnet . When the m agneticneedle is taken sufficiently small the value of the field intensity atthe needle due to the bar magnet will be practically uniform overthe space occupied by the needle , and therefore the resultant fieldintensity at the needle will also be uniform and will m ake an angle

h9= tanf

1

Hwith the direction of the original field H . This angle

may be readily m easured by noting the deflection of the needlewhen the bar magnet is removed to a great distance , for when

MAGNETISM 3

the bar magnet is present , the needle points in the directionof this resultant field intensity, h+ H , and when the bar magnetis rem oved the needle points in the direction of the original intensity H . Hence, the value of h in terms of 0 and H is

h= H tan 0whence 2M

which substituted in the above equation for H gives

H27rf

and all the quantities in the right- hand member can be m easuredor calculated . Sim ilarly, the substitution .of thi s value for H inthe equation just deduced for M gives

M = 7Ta f\/2a 1 tan 0 (7b)from which the magnetic mom ent m ay be calculated .

When the hori zontal component of the field intensity has thusbeen determ ined , the direction and num eri cal value of the resultant intensity can be found by placing at the point P a m agneticneedle which can turn only about a hori zontal axis (a so- called“ dip circle and setting thi s horizontal axis perpendicular tothe hori zontal component to the field . This needle will then pointin the direction of the resultant field intensi ty

,and if the angle it

makes with the hori zontal be a, the value of the resultant inten

sity is

H

cos a

It should be noted that there are other and more convenientmethods for the m easurement of the intensity of a m agnetic field

,

based on the m easurem ent of an electric current . But since as the m easureof an electric current is taken the forceproduc ed by a m agnetic field on a wirein which the current is established , it isnecessary to start with an independentmethod for the m easurement of magnetic field intensity .

43 . Flux of Magnetic Force Dueto a Single Pole . Lines of MagneticForce Due to a Single Pole .

—Con

sider a point- pole m at any point P(Fig . The field intensity at any Fig . 17

point Q at a distance r from m is ftand is in the direction P Q


when the pole is positive or in the direction QP when the poleis negative . Consequently at every point in the surface of asphere of radius r drawn about m as a center the field intensity

will have the same valuemand will be normal to the surface of

the sphere . Hence the product of the area of this Sphere andthe field intensity at its surface is

-

n—4 77 m

which is independent of the radius of'

the Sphere but depends onlvupon the strength of the pole m. The product of the area of anysphere drawn about a point-pole as its center and the field intens ityat the surface of this sphere is called the flux offorce due to thispole

,and this flux of force is said to be outward from the pole when

the pole is positive and inward towards the pole when the pole isnegative . A magnetic pole of strength m then produces a flux offOI'CG

1P=4 Wm

outward . (When m is negative the flux outward is also negative ,which is equivalent to a posi tive flux inward .)The flux of force from a magnetic point-pole can be represented

graphically in a S imple m anner . Imagine the surface of anySphere surrounding the pole divided into 4 n m equal areas andcones drawn with these areas as their bases and their commonvertex at the pole m

,and let the lateral walls of these cones extend

out indefinitely . The number of these cones then represents theflux of force from the pole m . Each of these cones may be repre

sented by a line coinc iding with its axis , and the number of thesel ines will also represent the flux offorce from the pole m

,and these

lines wil l coincide in direction withthe fi eld intensity at every point intheir path . Such lines are calledmagneti c l ines of force.

From the definition of theselines of force

,it follows that the

number of these lines per squarecentim eter of area normal to theirdirection at any point will be equal

fig , 18 .to the field intensity at that point ;

MAGNETISM 55

for the unit area normal to their direction at any point Q at a distancer from the pole will be unit area in the surface of a sphere of radiusr,nand since a total of4 7Tm equally spaced lines are considered asradiating out from the pole m,

the number crossing this unit areais equal to the total number of these lines divided by the total

4surface of the Sphere, that 1s

W m 7_

n. Hence the l1nes of force

4 n F—fi

due to a point-pole m coincide in direction at every point withthe direction of the field intensity at that point due to this poleand the number of these lines crossing unit area at right angles to

thei r di rection is equal to the field intensity at this area due tothe polem . The lines due to a S ingle north pole radiate out toinfinity ; those due to a south pole radiate in from infinity .

It should be borne in'

m ind that these lines really representcones

,and therefore it is entirely logical to Speak of a fraction of

a line of force . For example,from a pole of strength 1 there

4 71“

would -be but a single cone,which would fill all space ; that is ,

the cone would be a sphere with m at its center . A unit areanormal to the direction of the field intens ity at a distance of 10

centimeters from this pole would cut out1 1

th4 7T>< 10

21256

of this cone,or this unit area would be cut by only

1th of

1256

this cone of force,which

,from the manner in which these cones

are drawn,is equal to the field intens ity at this point .

A s imple relation also exists between the number of these linesof force crossing

'

any area and the field intensity norma l to thatarea . Let ds ( Fig . 1 9) be any plane area at any point P taken

Fig . 19 .

so small that the field intensity at every point in this area may be

considered as having the sam e value and the same direction,and


let the normal to this area make the angle a with the direction of

the field intensity,that is

,with the direction of the lines of

force through ds . Let ds’be the proj ection of ds on a plane

through P perpendicular to the direction of the field intens ity .

Then the number,d ll}, of lines of force crossing ds

’ must be the sameas the number crossing ds , s ince the perpendicu lars dropped fromthe periphery of ds on the plane of ds ’ are parallel to thelines of force . But the field intensity H at the point P is equalto the number of lines of force per unit area crossing ds ’ , that is

H i tds

But ds’ =ds . cos a

,hence

Hd ‘i'

ds . cos a

0141

ds

Therefore the number of lines of force per unit area at any point ina m agnetic field

,due to a point- pole of strength m,

1s equal to thecomponent of the field intensity at thi s point norma l to this area .

An important point to be noted in regard to a line of force isthat a line cross ing a surface in the direction from the side A to thes ide B is to be considered as equivalent to a negative line c rossing

H oos a

Fig . 20 .

this surface in the direction from the s ide B to the s ide A . For,

calling a the angle between the direction of the field intensity Hand the normal drawn outward from B

,then the angle between

the direction of the field intens ity and the normal drawn outwardfrom A is 180°—a . Hence the number of lines of force crossingthe surface ds is (H cos a ) ds in the direction from A to B

,or is

[H cos (180°

a ) ] ds (H cos ds in the direction from B to A .

44 . Gauss’s Theorem .

- An extrem ely useful relation in the

MAGNETISM 57

theory of magnetism is that the a lgebraic sum of the l ines of forceoutward across any c losed surface of any shape whatever is equa l

to 4 17' times the a lgebra ic sum of the strengths of a l l the poles

inside this surface . This follows imm ediately from the factthat from any north pole of strength m there radiate out 4 77 m

lines of force and therefore all these 4 77 m lines of force froma north pole ins ide such a surface will cut this surface in the direction from the inside outward, while to a south pole of strength m

’

there radiate in 4 7Tm ’ lines of force,and therefore all these 4 7Tm ’

lines of force radiating into a south pole ins ide such a surface cutthis surface in the direction from the outs ide inward . A lso

,the

lines of force due to any pole outs ide this surface , whether the polebe a north or a south pole

,will either not cut the surface at all or

will cut it an even number of times, as m any t imes in the directionfrom outs ide inward as in the direction from ins ide outward .

Since a line of force inward is equivalent to a negative l ine of forceoutward

,in calculating the algebraic sum of the lines of force out

ward the lines of force due to the south poles inside the surface areto be substracted from the lines of force due to the north poles inside the surface ; the lines of force due to any pole outside the sur~

face contribute nothing to the algebraic sum of the lines of forceoutward

,since each line leaves the surface as many times as it

enters it . Hence the algebraic sum of the lines of force which cut

a closed surface in the direction from the inside outward is equalto 4 77 tim es the a lgebra i c sum of the strengths of all the poles insidethis surface .

45 . Lines of Force Representing the Resu ltant Field Due to

any Number of Magnets. So far we have been cons idering thelines of force due to each point- pole as a separate set of lines

,the re

being as m any sets of these lines as

there are individual point- poles . In

any ac tual case,however

,whenever

there is one magnetic pole theremust be som ewhere else in the fieldan opposite pole of equal strength .

That is,in any actual c ase

,the sum

of the strengths of all the northpoles in the field is equal to the sumof the strengths of all the south ds

poles in the field . A field due to Fig .

‘


any number of such equal and opposite poles m ay also be repre

sented by a single set of lines,drawn out from the north poles in

the field,4 7T of these f‘resu ltant ” lines being drawn from each unit

north pole , and each line coinc iding in direction at each pointw ith the resultant field intens ity at that point . Al l of these lineswill also end on south poles

,4 7T of them on every unit south pole .

Also,the number of these resultant lines of force per unit area

c rossing any elementary area ds a t any point in the field is equalto the component H

nof the resultant field intens ity normal to

this area,i .e.

H0144

n

ds

when the area ds is taken norm al to the direction of the field intensity the number of lines of force per unit area is equal to theresu ltant intensity H

,since in this case the norm al component is

equal to the resultant .Consider first the number of lines of force

,considered as sepa

rate sets oi radial lines from each pole,which cross any elemen

tary area ds in a m agnetic field due to any number of poles . In

Fig . 2 1 let H 1 H 2 etc .

,be the field intensities at ds due to the in

dividual poles produc ing the field , and let a , ,a , ,

etc .

,be the angles

between the normal to ds and the directions of the respective field intensities at ds . Then the number of lines of force cross ing ds in thedirection of this norm al to ds due to the first pole is (H 1 cos a ,)ds; the number of lines of force c ross ing ds in this sam e directiondue to the second pole is (H 2

cos a ,) ds ; etc . Therefore the totalnumber of these lines of force crossing ds in direction of the normal1S

d ip= (H , cos a 1 + H 2 cos a 2 + ds

But = H cos a

where H is the resu ltant field intensity at ds and a is the anglebetween the normal to ds and the direction of this resu ltant fieldintens ity . Hence the net number of lines of force c ross ing ds inthis direction may also be written

d ip= (H cos ds (10)

and therefore the net number of lines of force crossing anysurface S is

H cos a ) ds (10a)

where H is.the resu ltant field intens ity at any element ds of th is


that point . Let a second surface B be drawn in the field 1n sucha manner that it encloses all the south poles but none of the northpoles . Let S be the area cut out of the surface A by any one ofthese tubes

,and S ’ the area cut out of B by this tube .

Cons ider the c losed surface formed by the lateral walls of thistube and the two areas S and S ’

. Since there are no poles betweenS and S ’

,we have from Gauss ’s Theo1 em that the net number of

lines of force leaving this closed section of the tube is zero . But

the net number of lines of force c rossing the lateral walls of this tubeis zero

,S ince for each area ds 1n these lateral walls ( H cos a ) ds

( H cos ds =0 . But this tube 1S drawn m such a m anner thatentering i t at S the net number of lines of force 1s unity

,s i nce the

integral of ( H cos ds over this surface S 1s unity,where a is the

angle between the direction of the resultant field intensity at dsand the norm al drawn from ds into the tube . Hence the netnumber of lines of force leaving the tube at S

’ must also be unity.

Consequently the number of these tubes entering the closed sur

face B surrounding all the south poles is equal to the net numberof lines of force (considered as individual sets of lines) enteringthis closed surface . But since the total strength of all the southpoles inside this surface B is equal to the total strength of all thenorth poles ins ide the surface A

,the total number of these tubes

entering the closed surface B is equal to 4 W M . Hence all thetubes leaving the c losed surface A surrounding the north polesenter the closed surface B s urrounding the south poles . Thissame relation holds when the two surfaces A and B are drawninfinitely close to the north poles and south poles respectively ;cons equently 4 n

'

of these tubes must originate at every unitnorth pole in the field and 4 71

' of them end on every unit southpole in the field . A lso

,the net number of lines of force cross

ing any surface in the field other than that on which there is apole

,is equal to the number of these tubes crossing this surface .

Hence these tubes are mathematically identical with the linesof force cons idered as separate sets radiating from or to eachindividual point-pole in the field

,and therefore if we represent each

of these tubes by a line coinciding with its axis, we m ay cons iderthese latter lines as resultant lines of force . From equation( 10) the number d iff of these resultant lines of force per unit areacrossing any elementary area ds at any point in the field is equalto the component of the resultant field intens ity normal to thisarea

,that is

H cos a

ds

where H is the resultant field intensity and o. is the angle betweenthe norm al to this area and the direction of the resultant fieldintensity at this point . When the elementary area is taken normalto the di rection of these lines the resultant field intensity is equal

MAGNETISM 61

to the number of these lines per unit area crossing this elementaryarea

,that is

H Ell i!ds

,,

where ds,, represents an elementary area norma l to the directionof these resultant lines of force .

It Should be noted that the idea of lines of force is based uponthe idea of field intens ity

,and consequently the statement that

the number of resultant lines of force per unit area perpendicularto their direction is equal to. thefield intens ity i s not a defini tion .

A definition presupposes a knowledge of a l l the terms in it,and

consequently the use of this statement as a definition is a spec iesof arguing in a c irc le . The definition of field intensity at anypoint is the force in dynes due to the agents producing the fieldthat would be exerted upon a unit north point- pole placed at thatpoint (see Artic le These resultant lines of force are s implya convenient means of representing the resultant magnetic field

,

these lines being drawn in such a manner that their direction ateach point coinc ides w ith the direction of the resultant field intens ity at that point and their number per unit area at that pointcross ing a surface perpendicular to their direction is equal to theresultant field intens ity at that point .

A rough picture of the direction of the lines of force representmg the horizontal component of the resultant field outside any

Fig . 23 .

number ofmagnets can readily be obtained by plac ing the magnetson a table and sprinkling fine iron filings over the table in their


vic inity . When the table is tapped lightly, the filings arrangethemselves in the direction of the horizontal component of thefield intens ity . This is due to the fact that the fil ings are magnetised by induction and each then acts like a sm all m agneticneedle , setting itself in the direction of the component of the fieldintens ity parallel to the surface of the table . It should be noted

,

however,that in general the field intensity has a value inside the

magnets als o and that therefore there are lines of force inside themagnets as well as outs ide .

In Fig . 23 are shown the lines of force due to a north pointpole and an equal south point-pole cons idered as two separate setsof lines

,one set radiating out from the north pole and one set

radiating into the south pole , and in Fig . 24 are shown the resultant lines of force due to these same two poles . In each case thelines are drawn so that their number per unit area at any pointperpendicular to their direction is proportional to the field intensityat this point .

From the definition of the lines of force representing the re

sultant field,it follows that where

the field intensity is great these linesof force wil l be c lose together, andwhere the field intensity is weakthese lines of force will be far apart .Again , S ince 4 71 of the lines originatefrom each unit north pole and endon each unit south pole

,it follows

that these lines will be c lose togetherwhere they enter or leave a surface

on which the pole strength per unit area is large , and will be farapart where they enter or leave a surface onwhich the pole strengthper unit area is small . Again , s ince a surface has two S ides

,these

lines will in general leave both s ides of a surface on which thereis a north pole and enter both Sides of a surface on which thereis a south pole .

46 . Intensity ofMagnetisation . When a narrow gap is cut ina magnet , either permanent or induced , it is found that in generalpoles appear upon the walls of this gap and that these poles areequal and Opposite

,a north pole appearing on the wall of the gap

nearer the south pole of the original magnet, and a south pole onthe wall of the gap nearer the north pole of the original m agnet .

Fig . 24.

MAGNETISM 63

The strength per unit area of the poles which appear on the wallof such a gap is found to depend upon the direc tion in which the

gap is cut . There is one,

direction at each point in the magnetfor which this pole strength per unit area is a maximum ,

andwhen the gap is cut perpendicular to this direction no poles appearon its walls . (In the case of a long bar m agnet, a gap cut a t

its center perpendicular to its axis will have maximum polestrength per unit area , while on a gap cut at this point parallelto its axis there will be no poles form ed .)A magnetised body may then be cons idered as m ade up of

magnetic filam ents such that were the lateral walls of any one ofthese filam ents separated from the rest of the m agnet by a narrowair gap ,

no poles would be formed on these lateral walls . Thetwo ends of any such filament where i t term inates in the surfaceof the magnetised body must then have equal and opposite mag

netic poles ; if the filament is cut transversely by a narrow air gapat any point

,equal and opposite poles will be formed at the two

walls of this air gap ,and these poles will in turn be numerically

equal to the poles at the ends of the filam ent in the original surface of the m agnetised body . (For, S ince each filament is a mag

net,it must have equal and opposite poles

,and by definition each

filament has no poles on its lateral walls .) Such a filam ent is considered as existing only in a magnetised body ; the filament isbroken by the transverse air gap j ust as a string is broken when itis cut in two .

Cons ider such a filament cut at any point by an air gap per

pendicular to its axis , and let ds ,, be its cross section at this point ,and let dm be the pole strength of the north pole formed where itends in the gap . Then the pole strength per unit area of this pole1s

dm(T

ds,,

or vi ce versa,the strength of - dm

the pole dm formed where thefilament is cut by the gap is

If the filament is broken atany other point

,poles of exactly the sam e strength will be form ed

on the broken ends , a north pole on the end nearer to the originalsouth pole of the magnetised body and a south pole on the

F ig . 2 5 .


opposite end . Also,the strength of the poles where this filament

ends in the original surface of the magnetised body will benum erically equal to dm . Consequently where the cross

'

sectionds

,,of the filament is great the strength per unit area of the pole

which would be formed were it broken in two is small,and where

the cross section is small the pole strength per unit area is large .

The strength per unit area of the pole which would be form ed on

the walls of a gap‘

cut at any point in a magnetised body perpendicular to the direction of the m agnetic fil aments of which thebody may be considered as m ade up is defined as the intensity ofmagneti sation

*of the body at this point

,and may be represented

by the symbol J . That is,

where ds,,is the cross section of the filament at the point under

consideration and dm is the strength of the north pole which wouldbe form ed on one wall of a narrow gap coinciding with the crosssection ds

n. The direction of the intensity of magnetisation J is

chosen arbitrarily as the direction of the magnetic filament atthis point and the posi tive sense of J is chosen as the sense of theline drawn into the gap from the wall of the gap on which the northpole is form ed .

When a magnetic filament is cut by a gap ds which is not atright angles to the axis of the filament

,the num erical value of the

strength of the pole form ed on either wall of the gap must be equalto the strength of the pole which would be formed on a gap ds,,cut normal to the filament

,since the pole formed on either gap

must be equal to the strength of the pole at either end o f the filam ent in the original surface of the m agnetised body . Let 0

' be the

pole strength per unit area of the north-pole end of the filam entin the surface of the gap , let a be the angle between the directionof this filam ent and the normal drawn outward into the gap from

*Intensity ofmagnetisation at any point may also be defined as the m ag

netic m om ent per unit volum e of an infinitesimal length of the m agneticfilam ent passing through that point . For

,calling dsn the cross section of

the filam ent,(Tn the num erica l value of the pole strength per uni t area at

each end of the infinitesimal length dl of the filam ent ; the m agnetic mom entofthis elem ent of the filam ent is (O'

n dsn ) dl and the volum e is dsnd l , whencethe magnetic mom ent per ,

un it volume is O'

n ; and therefore th is definition isequivalent to‘that given above .

MAGNETISM 65

this north pole , and let ds,, be the proj ection of ds on a planenormal to the direction of the filament . Then from ( 1 1)

dm= 0' ds =s

,,

ds,,=ds cos a

whence0

' =Jcos (1 ( l l a)

That is,the pole strength per unit area which would be formed on

the wall of a narrow gap cut in any direction at any point in amagnet is equal to the component of the intens ity ofmagnetisa

tion in the direction of the normal drawn outward from this wallinto the gap .

Intens ity of magnetisation may be represented by lines j ustas magnetic field intensity may be represented by lines . This isdone by choosing arbitrarily the size of a unit magnetic filamentand representing each uni t filament by a line coinciding with itsaxis . As the unit magnetic filament is taken a filament such thatwere it cut by a narrow air gap at any point , the strength of the

pole formed on either wall of the gap would be equal to i . The4 7r

line representing such a filament is called a line ofmagneti sation;the direction of this line coincides with the direction of the inten

Fig . 26. Lines of Magne tisat ion in a Bar Magnet .

sity ofmagnetisation . Hence from every unit south pole in thesurface of a magnetised body 4 77 lines of magnetisation originate,these lines run through the magnetised body to the surface overwhi ch the north pole is distributed

,4 77 of them ending in every

unit north pole in the surface of the magnet .

The reason for introduc ing the factor —1is to have number of

4 7

these fi laments leaving a south pole equal to the lines of forceentering that pole and the number of these filaments entering a northpole equal to the lines of force leaving that pole . It should be


noted that these filaments are confined entirely to magnets or

magnetised bodies , while lines of force in general exist in all thespace surrounding a magnetic pole , whether this space be occupiedby a non-magnetic or by a magnetic bodyThe relation between the number of lines ofmagnetisation dN

crossing any elementary surface ds and the intens ity ofmagnetisa

tion at ds is given by the formuladN =4 77 (J cos a ) ds (12)

where a is the angle between the direction of these lines at ds andthe direction ofthe normal to ds . J cos a is the component of theintens ity of magnetisation normal to the surface ds . Comparewith the mathematical expression ( 10) for the number of lines of

force crossing an area .

Since in the case of a long bar magnet the poles are confinedalmost entirely

'

to its ends,the lines : of magnetisation inside the

magnet near its center must be parallel to the S ides of the magnet .When such a magnet is cut in two by a p lane surface perpendicularto its axis

,the magnetic poles which appear on the walls of the

gap thus formed must then, from equation have a strength perunit area equal to the intens ity of magnetisation at this surface .

(This is strictly true only in case the gap between the two parts ofthe magnet is of infinites imal width .)This fact suggests a method for the experimental determ ination

ofthe intensity ofmagnetisation ofa bar which can be separated intwo parts . For

,by equation the force required to separate the

two equal and oppos ite poles which appear on the walls of the gapformed by separating the two parts of the bar is

F =2 7r <72 S

where S is the cross sect ion of the bar and 0 ' the strength of thesepoles per unit area

,which may be assumed constant over each wall

of the gap . Hence in this case the intensity of magnetisation isF (13)

2 7TS

and both F and S are readily measured . In employing thismethod certain corrections have to be made for the effect of theaction of other forces on the poles which are separated . A

description of the method and apparatus will be found in Foster’sPocket Book

,p . 94 .

(In the above discussion the matter form ing a magnet is con


body or a non-magnetic body or in free space,is defined as the

number of lines of magneti c induction crossing that surface , or the

flux ofmagneti c induction across that surface . The unit of flux ofmagnetic induction is called the maxwel l , that is, 1 maxwell1 c . g. 3 . line ofm agnetic induction .

From equations ( 10) and ( 12) the m athem atical expression forthe number of lines ofmagnetic induction crossing any elementaryarea ds is

d Jcos a ,+ H cos a .) ds ( 14)where J and H are the intensity of m agnetisation and the fieldintens ity respectively at ds and a 1 and a 2 are the angles between thenorm al drawn to ds and the directions of the intens ity of mag

netisation and field intensity respectively . The direction of thelines of induction through the elementary area ds is taken as thedirection of the vector which is equal to the vector sum of 4 77JandH . As a rule

,in all practical applications when the field inten

sity and intensity ;of m agnetisation are due to the same cause

(e. g .

,when a piece of s

'

oft'

iron is m agnetised by the action of anelectric current) , these two quantities are 1n the same direction

,

and this expression m ay then be writtend (13 77J+ H) cos a . ds . ( l 4a)

where a is the common angle made by J and H with the norm alto ds; in this case the lines of force

,lines of magnetisation, and

lines of induction all coincide in d irection .

From the definition of lines of magnetisation , there can b e nolines of m agnetisation in air or in any other non-magnetic sub

stance . Hence in air or in any othernon-m agnetic substance the lines of

force and the lines of induction are identical

,but this i s never the case in a mag

neti c or diamagneti c substance,for when

A such a substance is placed in a magneticFig . 27 . field it becomes m agnetised by induction

(see Art icle 32) and consequently lines of m agnetisation , as wellas lines of force , are produced in the substance .

Cons ider first the lines of m agnetisation in a single m agnet andthe lines of force due to its poles . Let N be the total number oflines of magneti sation in this m agnet , the number of l ines of

force outsi de the m agnet , and if}, the number of lines of force insidethe m agnet . Outside the m agnet the total number of lines of in

MAGNETISM 69

duction is then simply (to. Ins ide the magnet , across the section

A taken perpendicular to its axis at its m iddle point , pass all thelines of magnetisation N and the if}, lines of force

,the former in

the di rection S to N and the latter in the direction from N to S.

Hence the total number of lines of induction through the magnet

across this area A is

¢ = N—tBut Since the tota l number of lines of m agnetisation is equal tothe tota l number of lines of force , we also have

N = 1Il o 441whence xii , N—1p, .

But it, is also equal to the total number of lines of induction out

side the m agnet . Hence the total number of lines of inductionpass ing through a perm anent m agnet from its sduth to its north

pole is equal to the total number of lines of induction pass ing backoutside the magnet from its north to its south pole . Therefore eachline of induction must be a c losed loop , part of which lies ins idethe magnet and part outs ide . The magnet may then be looked

Fig . 28 . Lines of Magnetic Induction due to a Bar Ma gnet ,

upon as a sheath which binds these lines of induct ion closelytogether (Fig .

these lines Spread out from one end of the Sheath ,bend around and re- enter the Sheath at the other end . Fig . 28

Should be compared with Fig . .29 which Shows the lines of force dueto a s ingle magnet . It should be noted that s ince some of the lines

of magnetisation end in the lateral walls of the magnet , part ofthese lines of induction pass through its lateral walls . In the caseof a long sl im magnet , however, practically all the lines of induc


tion pass through its ends . In general , wherever there is a mag

netic pole on the surface of a magnetic substance there must also

Fig . 29 . Magnetic Lines of Force due to a Bar Magnet .

be a line of induction passing through this surface ; or,vice

versa,wherever a line of i nduction passes through the surface of

a magnetic substance there mus t be a pole on its surface .

Since each line of induction due to a S ingle magnet forms ac losed loop it follows that such a line of induction will always cut

a c losed surface an even number of times,as many times in the

direction from the outside to the ins ide as in the direction fromthe ins ide to the outside . Hence

,adopting the same convention as

Line 0! Induction

Cl osed SurfaceFig . 30 .

in the case of lines offorce,namely

,that a line entering a surface is

equivalent to a negative l ine leaving that surface, it follows that thealgebraic sum of the lines of induction outward ac ross any c losedsurface is always zero

,even though this surface encloses the pole of

MAGNETISM 71

a magnet . The difference in this respect between lines offorce andlines of induction should be carefully noted ; 1 , lines of forceend on magnetic poles , while lines of induction pa ss through mag

netic poles ; 2,the lines of force due to a magnet therefore have

ends,while the lines of induction are c losed curves; 3

,the alge

braic sum of the lines of force outward across a closed surface isequal to 4 77 times the algebraic sum of the magnetic poles enclosedby this surface

,whil e the algebraic sum of the lines of induction

across any closed surface is always zero .

48 . Flux Density. We have seen that the number of lines of

force per unit area cross ing any elementary surface ds is ‘equal tothe component of the intens ity of the magnetic field normal to thisarea (equation We shall also find that the number of linesof induction per unit area crossing any elementary surface playsa very important role in the theory of magnetism and electromagnetism . When the surface is taken normal to the l ines ofinduction through it , the number of these lines of induction perunit area c rossing this surface is called the densi ty of the lines

of induction at the point occupied by this elementary surface ors imply the flux density at that point , and the direc tion of thisflux dens ity is defined as the direction of the lines of induction atthis surface . Flux dens ity is therefore a vector quantity

,s ince

it has both m agnitude and direction . The symbol usually em

ployed for flux dens ity is the capital letter B .

The mathematical expression for the flux density in terms ofthe intens ity of magnetisation J and the field intensity H whenboth J and H are in the sam e direction is therefore

B =4 77J+ H

See equation ( 14a) .

When J and H are not in the same direction 4 77Jand H mustbe added vectorially see equation This can be done by re

solving J and H along two mutually perpendicular axes , addingthe components of 4 7rJ and H along these respective axes , andtaking the square root of the sum of the squares ; see Article 9 .

The unit of flux density on the c . g. 8 . system is one line of induction per square centimeter

,or one maxwell per square centi

meter ; this unit is called the gauss . Flux dens ity may also beexpressed as so many 0 . g. 3 . lines per square inch

,or so m any

thousands of‘

c . g. 3 . lines per square inch . A thousand lines is


called a kilo- line . Hence the following relations between thevarious units of flux density

1 gauss - 1 maxwell per square centimeter1 gauss 1 c . g. 3 . line per square centimeter1 line per square inch gauss1 kilo—line per square inch gaussesFrom the definition of flux density

,the number d (f) of lines of

induction c ross ing any elementary surface ds may be expressed interms of the flux dens ity by the formula

d cos ds (1 6)

where B is the flux dens ity at ds and a the angle between thedirection of the lines of induction through ds and the normal tods; this follows immediately from equation Compare thiswith the expression for the number of lines of force across anyelementary surface ds (equationThe above definition of flux dens ity and equations ( 14) and

( 15) are applicable to any magnetic field,no matter how this field

may be produced , whether by a S ingle permanent magnet , by anynumber ofmagnets , or by an electric current . Moreover

,j ust as

the separate sets of lines of force due to any number of magnetsmay be represented by a single set of resultant lines of force

,

so the separate sets of lines ofinducti on due to any number of mag

nets may be represented by a single set of resultant lines ofinduction

,and these resultant lines of induction always form

closed loops,j ust as the lines of induction due to a Single magnet

are closed loops .

For,let ds be any elementary surface in the field

,B, B2 etc .

,

the flux densi ties at ds due to the respective magnets , and a ,, (1 2 ,

etc .

,the angles between the normal to ds and the directions of the

lines of induction through ds due to the respective m agnets . Thenthe number of lines of induction across ds due to the first magnetis (B, cos a ,) ds; the number of l ines of induction across ds in thesame direction due to the second magnet is (B, cos (12) ds; etc .

Hence the net number of lines of induction across ds in the sam esense due to a l l the magnets is

d ch (B , cos a ,+ B 2 cos a ,+ ds = (B cos a ) ds

and the total number of lines of induction across any finite surfaceis then

cos a ,) ds —i- fS

(B 2cos a ,) ds —i =f (B cos a ) ds

where B is the resultant or vector sum of B 1 B 2 etc .,and a is the

angle between the norm al to ds and the direction of the resultant

MAGNETISM 73

flux density B . Hence we may‘look upon the independent sets of

lines of inductionfs <B l 00 8 a i ) ds ,f (B 2 cos a ,) ds, etc .

,as combin

3

ing and form ing one set of lines equal in number tof (B cos a ) ds,s

the direction of each of these lines at each point in its pathcoinciding with the direction of the resu ltant flux density at thatpoint .Since the net number of lines of induction due to a single m ag

net outward across any closed surface is zero (Article 47, last paragraph) , the algebraic sum of the lines of induction due to anynumber of m agnets, outward across any closed surface

,must also

be zero , that is

f (B cos a.) ds z f (B , cos a ,) ds +f (B , cos a ,) ds+ISI IS I ISI

since each term of the m iddle member of this equation is zero .

Hence the lines of induction due to any number of magnets ,whether these magnets be perm anent or induced , must form closedloops , and therefore the number of these lines com ing up to anysurface on one side must always be equal to the number of thesel ines which leave the other side of this surface . When we cometo the study of electric currents we shall also see that the lines ofinduction produced by an electric current are also c losed loops .

Hence a l ine ofinduction a lways forms a c losed loop, no matter how

i t may be produced .

49 . The Normal Components of the Flux Density on the TwoSides ofany Surface are Equal . The fact that the number of linesof induction com ing up to side surface must equal the

Normal

number of these lines which leave the other Side,is a very impor

tant one in the theory of magnetism ,and m ay be expressed m athe

matical ly as follows . Let B , and B, be the flux densities on the


two sides of any surface directly opposite any point P m this surface

,and let a , and a, be the angles between the direction of the

normal drawn through the surface at this point and the directionsof these flux dens ities . Let ds be any elementary area of the surface at P . Then the number of lines of induction com ing up tods on one side is (B, cos a ,) ds, and the number of lines of inductionleaving ds on the other S ide is (B, cos a ,) ds . Therefore

,since the

number of lines of induction com ing up to ds must equal the number leaving ds,

(B , cos a ,) ds = (B , cos a ,) ds

B, cos a ,=B , cos a , (17)

That is,the norma l components ofthe flux densi ties on the two sides

ofthe surface are equa l , and this is true even though the surface isthe seat of a m agnetic pole . When there is no pole at the surface

,

the tangential components of the flux densities (B , sin a , and B 2

sin (1 2) are also equal , but when there 1S a pole at the surface , thetangential components will not in general be equal . In the lattercase the lines of induction make an abrupt change in direction .

See Article 54 .

50 . The Tangential Components of the Field Intensity on the

N o Sides ofany Surface are Equal . While lines of induction arealways continuous lines form ing closed loops

,and therefore always

pass through any surface in their path , lines of force originate at orend on magnetic poles . Hence the normal components of the fieldintensiti es on the two S ides of a surface on which there are m agnetic poles are not equal , the tangentia l components of these fieldintensi ties are however a lways equal , even though the surface bem agnetised . This may be proved as follows . Let AB be thenormal through the surface at any point Q and let P , be a point on


52 . Induced Magnetisation . We have seen that when a mag

netic body which itself is not a m agnet is placed in a magneticfield

,this body becomes a magnet ; this phenomenon is de

scribed by saying that the body becom es magnetised by induction . To fix ideas

,let this field be that in the vic inity of the

north pole ofa barmagnet , and let the magnetic body be a soft iron

Fig . 33.

rod placed as shown . As we have already seen,the end A of

this rod becomes a south pole and the end B a north pole . Con

sequently inside the rod AB there will be (1) a field of force dueto the original field of the bar magnet in the direction from A

to B and (2) a field of force due to the magnetic poles in

duced on the rod,which field will be in the direction from B to

A . Let H be the num erical value of field intens ity at any pointP inside the rod due to the original magnetic field

,_

H ’ thenumerical value of the field intens ity at this point due to the polesinduced on the rod

,and J the numerical value of the intensity of

m agnetisation of the rod at this point . The induced intens ity ofm agnetisation J m ay be cons idered as produced by the originalfield H . The intensity H ’ which is due to the induced poles andwhich therefore is approximately in the opposite direction to theoriginal field H ,

would of itself tend to m agnetise a body in theoppos ite direction

,and is therefore called the demagnetis

ing force due to the ends of the rod . Experiment Shows thatthis demagnetising force H

’ is always less than the m agnetis

ing force H but is not negl igi ble unless the field in which othe rod

is placed is unifo rm and the rod itself is very long . The resultantfield intensity at P will then be the vector difference

a = H—H '

When H and H’ are in exactly opposite directions the resultantfield intensity H,

is the arithm etical difference between H andI

F orIL= H —H

’

This condit1on 1s approximately realized in the case of a long,

MAGNETISM 77

slim bar placed in a uniform m agnetic field parallel to the direction of the field . Experiment also shows that when a bodywhich of itself is not a magnet the soft iron rod we arecons idering) is placed in a magnetic field

,the intens ity of mag

netisation produced at any point in the body is in general in thedirection of the resultant field intensity . (Certain exceptions willbe noted later, Article Consequently

,across any elementary

surface ds drawn normal to the direction of the resultant field intens ity there will be H

,ds lines offorce and 4 77 J ds lines ofmag

netisation,both as a rule m the same direction

,namely from A to B

(see equations 10 and Hence the total number of lines of

induction across this area will be (see equation 1 4a)-i- 4 77J) ds

and therefore the flux dens ity at the point P,i .e .

,the number of

lines of induction per unit area normal to the direction of thesellines , will be

d (f)s

7TJ (1 9)

53 . Magnetic Permeability . The resultant intens ity of mag

netisation J produced in a piece of iron Or steel when it is placedin a m agnet ic field is in general inany times greater than theoriginal intens ity of the field

,and ‘ therefore the number of lines

of induction through the space occupied by this piece of iron or

steel is greatly increased by its presence ,* although the numberof lines of force through this space is in general decreased (thelatter is a lways true ex cept

’

when there are no poles produced on

the surface of the body which is m agnetised by induction — we

shal l see later when we com e to the s tudy ofelectric currents how abodymay bem agnetised without produc ing any poles on its surface) .Hence iron or steel

,or in fact any magnetic substance , is said

to be more permeable to lines of induction than a non-mag

netic body (diamagnetic bodies are less perm eab le) , and the ratio*The fact that when a magnetic body 1s placed in a magnetic field the

number of lines of induction through the space occupied by the body is increased , is frequently described by saying that the lines of induction tendto crowd into a magnetic body when placed in a magnetic field . In general ,however, the presence of such a substance in a magnetic field not onlycauses the

.

original lines of induction to crowd into the substance , but alsogives rise to an additiona l number of these lines . This is particularly tru ein the case of an iron rod placed in the magnetic field produc ed by an electric current flowing in a coi l ofwire .

78 ELECTRICAL E NGINEERING

of the resultant flux density at any point of the body to the resultant field intens ity is called the permeabi li ty of the body at thatpoint, and is usual ly represented by the symbol M That is

B

where H,is the resu ltant field intens ity

,which in case j ust con

sidered is equal to H—H ’. Since there can be no lines of m ag

netisation in a non-m agnetic body , the flux density and field intensity in such body are numerically equal

,and therefore for all

non-magnetic bodies u= 1 . For magnetic bodies ft is always greaterthan 1 and for diamagnetic bodies less than 1 . Strictly speaking,the perm eab ility is unity only for “air, s ince all bodies are slightlymagnetic or diamagnetic with respect to air . However

, IA is

practically unity for all substances other than iron,steel

,nickel

,

cobalt and b ismuth ; for the last M 1s less than unity and for therest greater than unity .

The permeab ility of any of these substances is not a constant,

but depends upon the intens ity of the resultant magnetic field,

and also upon its previous history,whether it is already a magnet

before being placed in the field of force and upon how thefield of force i nduc ing the magnetisation is established . In

fact,the permeab ility may be negative, that is , the resultant flux

dens ity may be in the oppos ite direction to the resultant fieldintens ity . This is always true of a S ingle permanent magnet byitself . In practice

,however

,the permeab ility of a body is taken

to m ean the ratio of the flux dens ity to the resultant field intensitywhen the body i s origina l ly unmagneti sed and then p laced in a field

whi ch i s continuous ly increased from zero to i ts fina l va lue.

54 . Refraction of the Lines of Induction at the Surface of

Separation of Two Bodies of Different Permeabil ities . It can

readily be shown , by making use of the surface conditions givenin A rticle 51 , that wherever a line of induction crosses the surfaceof separation between two bodies of different perm eabilities, thisli ne is refracted toward the norm al at thi s surface in the bodyof lesser perm eability . Let the perm eabilities of the two bodiesdirectly opposite any point Q in the surface separating them be

p , and [L2 respectively let H, and H2 be the field intensities in thetwo bodies respectively at points infinitely close to Q on the twosides of the surface ; let B, and B, be the flux densities at these twopoints and a , and a, the angles between the norm al drawn throughthe surface at Q and the directions on the two Sides of the surfaceof the l ine of induction through Q . The line of force through

MAGNETISM 79

Q will coincide in direction with the lineof induction (provided the bodies are notcrystals and are m agnetised solely byinducti on) , hence a , and a, will also bethe angles between the normal at Q andthe direc tions of the lines of force on thetwo sides of the surface . Hence from thesurface conditions given in Article 5 1 ,

B, cos a ,=B2 cos a ,

H, sin a ,= H2 sin a ,

HenceH1 H2

B l

tan tan (1 2 .

13116 B l fl ] H1 and 8 2: H2 H2 . Whence

” 1tan a , tan org) . (21)

Fe

Hence if u, is less than HQ , then a , is less than (1 2; that is, the lineof induction is bent toward the normal in the body of lesser permeability . For example , the perm eability of soft iron is about3000 under ordinary conditions . Hence a line of induction com ingup to the surface inside the iron at an angle of 45

°say, com es out

into the air a t an angle of a = tan’ 1 with the norm al , that

is, comes out into the air practically at right angles to the surface .

Therefore,when a piece of soft iron is placed in a magnetic field in

air practically all the lines of induction which pass through it enterand leave its surface approximately at ri ght angles to the surface .

55 . V al ue of th e Pole Strength per Unit Area Induced on theSurface of Separation ofTwo Bodies ofDifferent Permeabi lities .

From the surface condition that the number of lines of inductioncom ing up to the surface is equal to the number of lines leavingthe sufface , can also be deduced an expression for the net polestrength per unit area induced on the surface of separation between the two bodies . Let ds (Fig . 35) be an elem entary area inthis surface at Q and draw about ds a closed cylinder having thecross section ds and its ends infinitely close to ds . ApplyingGauss ’s Theorem (Art . 44) to this c losed cylinder, we have

4 7r 0'

ds = H, cos a , ds H2 cos a 2 ds

whence or

4 77“

where 0 '

is the pole strength per unit area on ds . But since thenumber of lines of induction com ing up to ds is equal to the number of lines of induction leaving ds , we also have

B, cos a ,=B, cos a ,

or, Since B,=u,H, and B2

=711 2H2


H—2 COS (1 281 00 8 a ,

H 1 [12

whi “h substituted in the above equation , givesH, cos a ,

4 77

Fig . 35

Hence if the line of induction at Q passes from a m edium of low

to one of high perm eability the pole induced on thesurface is negative , whi le if the line of induction passes from a m ed~

ium ofhigh to one oflow permeability u, < p.2) the pole inducedon the surface is positive . This explains the fact , noted in A rticle29

,that the direction of the force produced on an originally unmag

netised body placed near a m agnet depends upon the nature of them edium between the body and the magnet , and that whether thisforce is an attraction or a repulsion depends upon the relative permeabil ities of the body and the surrounding medium . When u,and p, are both d ifferent from uni ty there will be a pole inducedon each of the substances in contact ; equati on (22) gives the a lge

braic sum of these pole st rengths per unit area .

56. Field Intensity at anyPoint inMagnetic Medium ofConstant

Permeability Compl etely Surrounding a Point-Pole and Fill ing al l

Space— An important relation in the theory ofmagnetism is that

the resu ltant field intensity at any point in a magnetic m ediumof constant permeability u completely surrounding a point-pole

of strength m isHofi where r is the distance of the pointW

”

from the pole . This is equivalent to stating that the polem induces on the surface of the medium in contact w ith ita pole of opposi te S ign , the num erical strength of whi ch is equalto m

,for then the resu ltant field intensity will be

m m

r2

ja r"

MAGNETISM 8 1

To prove this, consider a m agnetic filament of infinite lengthand infinitesimal cross section , and let the poles of this filamenthave the constant strengths m and —m. We m ay imagine the pole—m at infinity and therefore producing no field intensi ty at anypoint in the vicinity ofm . Again , Since the fil ament 1S assumedto have infinitesimal cross section , we may neglect the non

uniform ity produced by it in the medium surrounding m . Thelines of force due to the pole m will therefore be equally spacedlines radiating out from m in all directions . When the pole issurrounded by a non-magnetic m edium ,

i .e .

,one having unit per

meabil ity, the lines of induction in the surrounding m edium wil lcoincide with these lines of force ; the lines of induction com e up tothe pole through the magnetic filament

,as lines of magneti sation ,

and then radiate out from,

the pole , as lines of force, into’

the surrounding m edium . Since the pole m is constant by hypothesis

,

the intensity ofmagnetisation of the filam ent must also be constant ; whence the number of lines ofm agnetisation , 4 77 m

,in the

filament is constant , and since the filam ent has an infinitesimalcross section , the lines of force inside the filament will be negligiblein comparison with the number of lines ofmagnetisation . Hencethe number of lines of induction coming up to the pole m throughthe filament must also be constant and equal to 4 7r m ; con

sequently the number of lines of induction radiating out into thesurrounding m edium is 4 7rm independent of the nature of them edium .

When the pole is surrounded by a non-magnetic medium ,the

field intensity at any point isH7

11 2 , and Since the m edium 1S non

magnetic the flux density is also B =g Since the lines of

induction are the same whether the surrounding m edium is

magnetic or non-magnetic , the flux density at any point in anymedium whatever completely surrounding a point - pole of strength

mm is B

F' Consequently, when the m edium has a perm eability p ,

the field intensity at any point in it due to the pole m must be

H E m. The number of lines of force outward across the

.u

.a r

’

4 7r msurface of any Sphere surroundmg m 1s therefore

,but by

p.

Gauss ’8 Theorem this must be equal to 4 71 times the algebraicsum of the poles inside this sphere ; hence calling m, the valueof the strength of the pole induced by m on the surface of the

medium in contact with it , we have , 4 77 (m mi )4 W m

and there

From the fact that the resultant field intensity at any point in a

82 ELECTRICAL E NGINEERING

medium of constant perm eability completely surrounding a pointm

M”

sion on a second point p—ole of strength m’ placed in the m edium at

mm'

Wis the perm eability of the m edium . It Should be noted that thisis the resul tant force acting on m

’ due to both the pole m and the

pole ofstrength m is i t follows that the resultant force ofrepul

any point is f where r is the distance between the poles and ,u

m on the surface ofthe m edium in con

tact with it . The force due to m alone ism

r

mand 1S independent

ofthe nature ofthe surrounding m edium .

The above deductions hold only when the poles are completelysurrounded

'

by the m edium of constant perm eabili ty p . In anyactual case this can never be true , Since a magnetic pole can existonly where there is a surface of separation between two substancesofdifferent perm eabi lities .

57 . Magnetic Hysteresis .

— As noted in Article 53,the flux

density produced m a given piece of iron or other magnetic sub

stance by a given field intensity depends upon the previous history of the sample . To make this fact c learer

,consider the Spec ial

case of an originally unmagnetised rod of soft iron placed in a uniform magnetic field with its axis parallel to the direction ofthe field

,

and let this field be gradually increased from zero up to some maximum value Hm and then decreased to zero , then increased in thisreversed direction to an equal negative maximum value Hm,

decreased to zero again and then again increased to the sam e maximum value Hm in the original direction . It is found by experi

ment that the relation between the flux density and the fieldintens ity during the various steps in this process may be represented by a curve of the form shown in Fig . 36. (The experi

mental m ethod of determ ining such a curve is described in ChapterIV .) From this curve it is seen that at first

,when the field

intensity is sm all,the flux density increases relatively slowly as the

field intensity inc reases . When the field intensity is increasedfurther

,the flux density increases very rapidly ; when the field in

tens ity becom es stil l greater, the flux dens ity increases more andmore S lowly

,and finally any further increase in the field intens ity

produces only a comparatively slight change in the flux density .

When the field intens ity is now reduced the flux dens ity insteadof returning to the sam e values it had for the increasing values ofthe field intens ity

,decreases less rapidly than i t increased, that - is

,


value in the positive direction,the flux density likewise reaches a

maximum in the negative direction equal to the maximum valueit had in the pos itive direction . When the field intensity is nowreduced to zero and then increased again in the positive directionto its original m aximum value, the flux density passes through theseries of values represented by the right-hand branch of the curve,which is perfectly symmetrical with the left-hand branch . Thec losed curve form ed by these two branches is called the hysteresisloop .

It Should be noted that when the sample is originally un

magnetised the curve giving the relation between the flux density and the field intens ity when the field intensity is first increasedfrom zero up to the m aximum value

,does not coinc ide with either

branch of the hysteresis loop , but is a curve which in general l iesbetween the two branches of the loop . After the completion of acycle of changes in the field intens ity from a positive maximum toan equal negative maximum and then back again to the positivemaximum

,the field intensity may then be reversed back and forth

any number of tim es between these equal positive and negativemaximum values

,and the relation between the flux dens ity and

the field intens ity will be the sam e for each cyc le of changes in thefield intens ity as for the first cyc le . If the iron is not originallyunm agnetised, the first hysteresis loop will be shifted ‘above orbelow the axis of field intens ities

,but after a number of reversals

of the field intensity between given pos itive and negative values ,the loop will become practically symm etrical with this axis

,par

ticularly if the iron is continually j arred . In the armatures ofelectrical machines and the cores of transform ers

,in which the field

intens ity reverses a large number of times every second and theiron is continually j arred

,the relation between flux density and

field intensity,after a Short interval of time

,is represented by

a symmetrical loop of the form shown in Fig . 36.

The area enc losed by the hysteresis loop depends upon themaximum value of the flux dens ity reached during the cyc le, butthe general Shape remains about the same .

’

Fig . 37 shows a seriesof loops corresponding to various values of the maximum fluxdensity . The area of the loop is also different for various kinds ofironor steel . AS we Shall see when we come to the study ofelectriccurrents (Chapter IV ) , the area of this loop represents a certainamount ofenergy dissipated as heat energy in the iron ; in fact, the

MAGNETISM

ergs of heat energy dissipated per cyc le is equal to -

1- tim es the

4 71“

area of this loop , when both the flux density and the field intensity are plotted to the same scale .

Fig . 37.

On account of this energy loss due to hysteresis, it is desirableto have all parts of the magnetic c ircuit of an electric machine inwhich there is a varying magnetic flux made of iron or steel in '

which this hysteresis loss is a m inimum . It has been found that iniron which contains about three per cent silicon thi s hysteresisloss is about half what it is in the best grade of low carbon steel .This so—called silicon steel is now being extens ively used in thecons truction of electric machines , particularly transformers .

An exam ination of this hysteres is loop also makes clear how abar of steel may be permanently magnetised by plac ing it in amagnetic field . For

,when the bar is removed from the field it

retains an intensity ofmagnetisation approximately equal to theflux density where the hysteresis loop cuts the axis offlux dens ities ,that is

,an intens ity of magnetisation approximately equal to the

remanent magnetism . Experiment Shows that a hard steel bar


its poles , but in the case of a soft iron bar even the slightest j arwill cause it to lose its magnetism almost entirely .

The propertypossessed by a hard steel bar of retaining its m agnetisation iscalled its retentiveness .

”

58 . Norma l B-H Curves .

—Magnetic Saturation .The curve

giving the relation between the flux dens ity and the resultant fieldintensity when the latter is increased from zero up to success ively

H in Electromagnetic Uni tsFig .

greater values is called the normal B-H curve . In Fig . 38 aregiven the norm al B-H curves for cas t iron

,cast steel and annealed

Sheet steel ( low carbon) such as is ordinarily used in the construotion of electric machines . In Fig . 39 are given the correspondingcurves Showing the relation between the intensity ofmagnetisationand the field intensity (calculated from the B—H curves by equation

and in Fig . 40 the corresponding perm eability curves (caleulated from the B-H curves by equation It will be notedthat the intensity of m agnetisation corresponding- to values of theflux density above the Sharp bend or knee in the B -H curves

,

*Standard curves used by General Electric Co.


increases very Slowly and soon becomes practically constantindependent of the value of the field intens ity . In fact

,experi

m ent shows that it is imposs ible to produce an intens ity of

magnetisation in a given substance greater than a certain definite value

,which is different for different substances . When

a magnetic substance is thus magnetised to its maximum in

tensity of magnetisation it is said to be saturated .

”Such a

substance is practically saturated for any value of the field intensity well above the knee of the B-H curve .

It Should be noted that the B-H curves for iron and steel depend to a very great extent upon the physical structure and chem ical constitution of the sample

,and the heat treatment to which

it has been subj ected . It has also been recently discovered thatwhen iron is annealed in an alternating magnetic field

,the per

meabil ity is increased in certain cases as much as 50 per cent . TheB-H curves of two samples taken from the same lot of materialmay even differ considerably . The curves also depend upon thetemperature of the sample at the time the observations are taken

,

though the variation due to ordinary changes of temperature isSlight . For very h igh temperatures, however, all m agnetic sub

stances become practically non-magnetic . This temperature cor

responds to the maj or recalescence point,which is about 750

degrees for steel of the quality used in armature and transformer

punchings . When iron is kept continuously at a m oderately hightemperature the hysteresis loop also gradually increasesin size

,and therefore the energy loss in the magnetic c ircuits of

electric machines due to hysteresis increases with time . Thiseffect is called aging .

” There is practically no agrng of siliconsteel .

59 . M agnetic Potentia l . —Whenever the position of a mag

netic pole with respect to another pole' is‘

changed, work must in

general be done, either by the poles themselves or by the externalagent which produces this change in the relative position of the twopoles . For , every m agnetic pole produces a force on every otherpole

,and consequently when one pole moves with respect to the

other there 13 a force and a displacement , and , by definition , workis equal to the product ofthe displacem ent by the component of theforce in the direction of the displacem ent (Article The totalwork which two permanent poles of like S ign at a given distanceapart can do on each other

,is the work done when the two poles

MAGNETISM 89

move from their original positions to an infinite distance apart .

This amount ofwork may be looked upon as the relative potentialenergy of the two poles, since it is the work they are capable of, orhave the potentiality ”

of,doing on each other . Similarly, the

work done by any number of magnetic poles in m oving a unitnorth point- pole from a given point to an infinite distance may be

looked upon as the relative potential energy with respect to thesepoles of a unit north pole located at this point . This quantityis called simply the potentia l at the point ; i .e.

, the potentialat any point in a magnet field is the work done by the polesproduc ing the field in moving a unit north point-pole

‘

from that pointto infinity

,provided these poles remain constant in strength and

their relative pos itions remain unaltered . Potential is then workdivided by pole strength, and since both work and pole strengthare scalar quantities

,potential is also a scalar quantity . Hence

potentials may be added algebraically .

The potential at any point P due to a point- pole of strengthm at a distance r away can be readily calculated . Consider any

p>é) l

l N

Fig . 41 .

point Q on the line drawn through the pole m and the point P ;let this point be at a distance x fromm . Then the force which would

act on a unit north pole at Q is 3 in the direction from m to Q .

When the unit pole moves a distance dx along this line, the work

done by m on it is Consequently,when the unit polex2

moves along this line from P to infinity the work done is


Hence the potential at any point due to a point-pole m at a distance r away is

m

When m is a south pole,122 is negative, which means that ther

unit pole does work against the force due to m,instead of this

force doing work on the unit pole . It can readily be Shown thatthe work done by the pole m on the unit north pole , when the lattermoves from the point P to infinity , is independent of the pathover which the unit pole moves . For , the work done on the unitpole when it moves a distance al l in any direction making an angle

6’with the l ine from m to Q isgut cos But dl cos 0 is equal

to the proj ection of dl on the line from m to Q . Calling thisproj ection dx

,we have that the work corresponding to the dis

Placement al l is fix

,which integrated between the limits x =r

x2

and x = 00, gives the same value of the potential as .before .

Hence the potential at any point due“ to a point-pole of strengthm depends only on the position of this point with respect to mand the value ofm .

The potential at any point due to any number of point-polesm, m, etc .

,at the distances r , r 2 etc .

, from this point is then thealgebraic sum

V _

’i ’

and the potential at any point due to any magnetised surface is

(T ds

r (24b)

where ds is any elementary area in this surface, 0 ' the pole strengthper unit area at ds

,and r the distance of ds from the point con

sidered,and fs indicates the sum of the expressions 0

'

dsfor 3 11

t he elements into which the surfaces are divided .


That is,the component in any direction of the field intens ity at

any point 1S equal to the negative ofthe space rate ” ofchange of thepotential in that direction . When the elementary length is takenin the direction of the field intensity equation (25a) becomes

dV (256)

dz,

where dl, means an elementary length taken tangent to the lineof force through this point . Hence a large field intensity cor

responds to a rapid fall of potential in the direction of the fieldintensity

,and a small field intensity a gradual fall of potential in

the direction of the field intens ity . Consequently,the express ion

potential gradient is frequently used for field intensity,where

by potential gradient is meant the drop of potential per unitlength in the direction of the lines of force . The unit ofmagneticpotential difference in the c . g. 8 . system is called the gi lbert.

Hence magnetic field intensity may be expressed as so many

gi lberts per centimeter .

When the magnetic field is produced by an electric current itis usual to express the drop of magnetic potential as SO m anyampere- turns (see Chap ter The relation between gilbertsand ampere- turns is

1 gilbert ampere- turns .

61 . Equipotentia l Surfaces . A m agnetic equipotential surface is a surface drawn in a m agnetic field in such a manner thatthe drop of m agnetic potential along any path in the surface iszero

. Such a surface is perpendicular at each point to the line offorce through that point, otherwise the field intens ity at thepoint in question would have a component along the surface andconsequently there would be a difference of potential betweenthis point and the neighboring point in the direction of this com

ponent . The lines of force representing a magnetic field aretherefore always normal to any equipotential surface which may

be drawn in the field . Calling dn an elementary length measuredoutward along the norma l to such a surface at any point , thefield intensity at this point is

dV

dn

which is the same as equation (25b) , only expressed in a different form .

MAGNETISM

SUMMARY OF IMPORTANT DEFINITIONSAND PRINCIPLES

1 . A unit point-pole is a pole which repels with a force of onedyne an equal point- pole placed one centim eter away

2 . Two point-poles of strengths m and m’ at a distance r centi

m eters apart repel each other with a force ofm m ’

ir,

3 . The field intensity at any point of a m agnetic field is definedas the force in dynes which would act on a uni t north point- poleplaced at that point due solely to the agents (magnetic poles orelectric currents) producing the original field . The uni t of fieldintensity in the c . g. 3 . system is the gilbert per centimeter .

4 . The field intensity at a distance r from a point-pole of

strength m is

dynes .

Z7}gilberts per cm .

T2

5 . The m echanical force exerted on a point-pole of strength m

F =mH dyneswhere H is the field intensity in gilberts per cm . at the pointoccupied by m due to all the poles (and electric currents) in thefield except the pole m .

6. The normal component of the field intens ity at any pointP j ust outside a magnetically charged surface due solely to thepole on this surface is

H = 2 77 0 '

gilberts per cm .

where 0' is the

'“pole strength per unit area at the point on the

surface directly opposit e P .

7; The magnetic moment of a magnet is defined as the ratioof the maximum mom ent exerted on a m agnet , when placed in auniform m agnetic field

,to the intensity of this field . The magnetic

moment of “ a bar magnet of length l with poles of strength mand—m concentrated in points at its two ends is m l .

8 . The frequency of vibration of a magnetic needle suspendedin

v

a magnetic field is proportional to the square root of the fieldintens ity .

9 . Lines of magnetic force are lines drawn in a magnetic fieldin such a manner that they coinc ide in direction at each point Pwith the field intensity at P and their number per unit area at each


point P across a surface at right angles to their direction at P isequal to the field intensity at P .

10. The number of lines of force,or flux of force

,crossing any

elementary area ds isd lIJ= ( H cos a ) ds

where H is the field intensity at ds and a is the angle between thedirection of H and the normal to ds; or H cos a is the componentof H normal to ds .

1 1 . Gauss’s Theorem The algebraic sum of the number oflines of force outward from any closed surface is equal to 4 77 timesthe algebraic sum of the poles inside this surface , i . e.

,

cos a ) ds =4 77 2 m

where represents the surface integral over the closed surface S.

12 . A magnetised body is considered to be made up of mag

netic fi laments such that were the lateral walls of any one of thesefilaments separated from the rest of the body by a narrow air gap ,no poles would be formed on these lateral walls .

13 . The intensity of magnetisation J at any point of a mag

netised body is the value of the strength per unit area of the polewhi ch would be formed on the walls of a gap cut in the body at thispoint perpendicular to the direction of the

.

magnetic filamentthrough this point . The direction of the intensity of magnetisa

tion is the direction of the filam ent,and the positive sense of the

filam ent is the Sense of the line drawn into the gap from the wal l onwhi ch the north pole is formed .

14 . A line of magnetisation, or unit magnetic filam ent,is

filament of such a S ize that were it broken by a narrow gap at anypoint

,the strength of the pole formed on either wall of the gap

would be -

1

4 77

15 . The number of lines of magnetisation crossing any elementary area ds is

dN = 4 77 (Jcos a ) ds

where J is the intensity of m agnetisation at ds and a is the anglebetween the direction of J and the normal to ds .

16 .The number of lines of magnetic induction, or flux of in

duction,crossing any surface is defined as the algebraic sum of


22 . The magnetic potential V at any. point in a magnetic fieldis the work which would be done by the agents producing the fieldin moving a unit north point-pole from this point to infinity . Theunit of m agnetic potential in the c . g . 3 . system is the gilbert . Thepotential at any point due to a point-pole of strength m at a distance r centimeters away is

The resultant potential due to any number ofpoles is the algebraicsum of the potentials due to all the individual poles .

23 . The drop of potential from any point 1 to any point 2 is

V , V 2 H cos 9) dl gi lberts

where dl is the length in centimeters of any element of the pathfrom 1 to 2

,H the field intens ity in gilberts per cm . at dl

,and

9 the angle between the direction of H and d l . When the field isdue solely to m agnetic poles the drop ofpotential is independent ofthe path from 1 to 2 . This is not true when the path links anelectric current .

24 . A magnetic equipotential surface is a surface drawn in amagnetic field in such manner that the drop of magnetic potentialalong any path in the surface is zero . Such a surface is per

pendicular at each point to the line of force through that point .

PROBLEMS

(Note : In calculating forces and field intensities when a Slenderbar magnet is Specified , the poles are to be cons idered concentratedin points at its ends .)

1 . Two slender bar magnets each with poles of 100 c . g. 3 . unitsand 10 inches in length lie upon the same straight line with theircenters 15 inches apart . What is the force in dynes exerted by onemagnet on the other if the nearest poles on the respective m agnetsare of Opposite signs?

Ans . : dynes .

2 . Two Slender bar magnets A and B are placed parall el toeach other and 10 inches apart with their centers on a line per

pendicular to their axes and with opposite poles on the respectivemagnets adj acent . The poles of A are 50 c . g. 3 . units and thepoles ofB are 30 c. g. s . units . A is 20 inches in length and B is 10

MAGNETISM 7

1nches in length . What is the amount ( in dynes) and the directionof the total force exerted by one magnet on the other?

Ans . : dynes in the direction perpendicular to the axis ofeach magnet .

3 . Find the amount and direction of the field intens ity due toa slender bar magnet at a point 10 cm . from the magnet on a linenormal to the axis of the magnet through its north pole . Themagnet is 10 cm . in length and the strength of each pole is 50 c . g. 3 .

units .

Ans . : gilberts per cm . at an angle of with the axis ofthe magnet .

4 , A slender bar magnet with poles'

of 25 c . g. 3 . units is placed ina uniform field the intens ity of whi ch is 100 gilberts per cm . Whatis the force in dynes acting upon each pole? What is the totalforce acting on the magnet?

Ans . : 2500 dynes on each pole ; total force zero .

5. Two srm ilar bar magnets A and B are placed end on withtheir two nearest pole- faces inch apart and of Opposite S ign .

Each m agnet is 20 inches long and the pole strength of each pole is100 c . g. s . units uniform ly distributed over the respective end surfaces of the m agnet . If the c ross section of each magnet is 1Square inch

,find the force in dynes exerted on each other by the

two adj acent poles . What is the force produced by one magneton the other due to the action of the other poles?

Ans . : An attraction of 9736 dynes . A repulsion of dynes .

(Note that this latter force is less than of the force due to theadj acent poles .)

6 . A Slender bar magnet 30 cm . in length and with poles of

80 c . g. 3 . units is pivoted at its center and placed in a uniform mag

netic field,the intensity ofwhich is 300 gilberts per cm . What is

the torque acting upon the magnet about its center, when the axisof the magnet is perpendi cular to the field intensity? What isthe magnetic moment of the magnet ?

Ans . : cm- dynes . 2400 c . g. 3 . units.

7. A Slender bar magnet l centimeters long produces a fieldintensity of H gilberts per cm . at a point on a line through theaxis of the magnet at a distance of 10 l centimeters from its center .,What is the field intensity produced by this magnet at a point on aline through the center of the m agnet perpendicular to its axis andat the same distance l from its center?


2

In solving this problem note that is negligible compared

Hg1lberts per . cm . Note : The ratl o of these two

intens ities depends solely upon the exponent of r in the fundamental formula

,equation for the mutual action of two poles .

Gauss determ ined this ratio experimentally to be which wasthe first accurate experimental proof of the inverse squarelaw .

8 . A slender bar magnet 5 square centimeters in cross sectionand 30 centimeters in length has a pole strength of 1500 c . g. 8 .

units . Find the fiel d intensity and the flux density at the centerof the magnet ; state the direction of each .

Ans . : gilberts per cm . in the direction from north to southpole of the m agnet . 3757 gausses in the direction from south tonorth pole of the magnet .

99 . A S lender iron bar 3 sq . cm . in cross section and 40 cm .

in length is placed in a uniform magnetic field,the intensity of

which before the introduction of the bar is 160 ampere—turns perinch . After the bar of iron is placed in this field with its axisparallel to the direction of the field intensity

,a uniform intensity

of m agnetisation of 1000 c . g. 3 . units is established in the barthe lines of m agnetisation are to be assumed straight lines paral lelto the axis of the bar) . Calculate the field intensity in gilberts percm . and the total flux of induction at the center of the. bar ; statethe direction of each .

Ans . : gilberts per cm . in the direction from the south tothe north pole of the bar . m axwells in the same directionas field intensity .

10 . The field intensity at the center of a slender bar magnet100 cm . long and 4 sq . cm . in cross section is 500 gilberts per cm .

If the flux density at the center of the magnet is 5000 gausses andis parallel to the direction of the field intensity

,what is the strength

of each pole,assum ing the lines of m agnetisation to be straight

lines? What would be the field intensity in the region occupied bythe magnet if the magnet were removed?

A ns . : 1432 c . g. 3 . units . gilberts per cm .

A slender magnet has a m agnetic moment of 5000 c . g . 8 .

units . The magnet is 50 cm . in length and 2 sq . cm . in cross


center of the north pole of the magnet . In making this calculationfind first the potential at this point due to the north pole con

sidered as a uniform ly magnetically charged disc and then add

(algebraically) to this the potential due to the south pole con

Sidered as a point-pole . Why is this approxim ation j ustified?What is the drop ofm agnetic potential from one pole to the other?Is the drop through the magnet the same as the drop between thesetwo points through the air around the magnet?

Ans . : 1249 gilberts. Since the angle subtended by the southpole is small , see Article 37. 2498 gilberts . Yes .


62 . The E lectric Current. —When a strip of z inc is dippedinto a dilute solution of sulphuric ac id in water, hydrogen gasis given off from the strip

,but if the zinc is pure, this action

ceases almost immediately . A copper st rip dipped into thesame solution is not appreciably affected , provided the two strips

Z inc StripCoppe r Str ip

D i luteSu l ph ur ic A cid

Fig . 43.

do not touch each other . Now connec t the ends of the two stripsby a copper wire (Fig . the following phenomena are thenobserved :

1 . The z inc gradually wastes away and hydrogen gas is l iberated at the copper strip .

2 . The wire,the strips

,and the solution all become heated .

3 . A m agnetic field is produced around the wire,the strips and

the solution .

4 . A magnet placed in the vic inity of the apparatus will exerta mechanical force on the wire, the strips , and the solution .

5 . When the copper wire connecting the two strips is broken intwo and the two ends a and b (Fig . 44) are both dipped into asolution of copper sulphate, it is found that the wire a was tesaway where it is immersed in the copper sulphate solution andcopper is deposited on the wire b where it is immersed in the solu

1 0 1


tion . In addition , this solution becomes heated,has a magnetic

field produced around it,and a m agnet p laced near it exerts a

mechanical force upon it .6. When the ends A and B of the copper w ire are reversed

,so

W i re

Z inc Strip

Copper SulphateSol ution

Fig . 44.

that A is connected to the zinc strip and B is connected to thecopper strip

,all the phenomena remain the same as before except

thata . The direction of the magnetic field around thewire and the

copper sulphate solution is reversed .

b. The direction of the m echanical force produced on the wireand the copper sulphate solution by a magnet placed in theirvic inity is also reversed .

c . The end of the wire a in the copper sulphate solution now

has copper deposited upon it , while the end of the wire b in thecopper sulphate solution wastes away .

These various phenomena are said to be due to a current

of briefly,to an electri c current

” in the wire,the

strips and the solution . The various parts of the apparatustaken together are said to form a c losed e lectri c circui t

,and any

one of the parts is spoken of as part of an electric c ircuit . Thecombination of the copper and zinc strips and the dilute sulphuricac id solution is called an electric battery, and the copper and zincstrips are called the poles of the battery . Experiment Shows thatwhen the phenomena j ust described occur in and around the wire ,the battery loses chem ical energy, consequently the battery isto be looked upon as the cause of these phenomena

,and since the

phenomena are attributed to a flow of electric ity through thevarious parts of the apparatus

,the battery is said to produce

an electricity moving force or,more briefly ,

an electromotive

force . There are other devices which are capable of causingthese phenpm ena to occur in and around a wire connected to

1O4 ELECTRICAL ENGINEERING

rubber- covered wire, for example, with its ends connected to thepoles of a battery

,is spoken of as an “ insulated wire connected

to the battery .

”

64 . E lectricity Ana logous to an Incompressible Fluid Fil lingal l Space .

— The following analogies will be found helpful inunderstanding the significance of the various properties whichexperiment leads us to assign to the something called electricity .

In the first place, experiment shows that this som ething musthave many properties analogous to those which wouldbe possessed by an incompressible fluid filling all Space,including the Space occupiedby matter . In this analogy ,

the force of gravitation act

ing on the fluid is to be neglected . The properties pos

Fig ~ 45 sessed by free space or anyspace occupied by a dielectric are found to be analogous to theproperties which would be possessed by this space were theincompressible fluid in th is space enc losed in m inute cell s withelastic walls

,form ing a cellular or honey—comb structure with

cont inuous walls completely filling this space ; whil e the properties possessed by a conductor are analogous to the propertiespossessed by a space in which the walls of the cells are completely

'

destroyed . Any particle of this fluid in the space correspondingto a dielectric can then move only as the result of a strain produced in the elastic structure which enmeshes it

,whil e any partic le

of the fluid in the space corresponding to a conductor can movefreely throughout this space . A conductor may then be lookedupon as analogous to an elastic sack (or tube , in the case of a wire)completely filled with this incompress ible fluid

,the elastic walls

of the sack being formed by the walls of the cellular structurerepresenting the dielectric surrounding the conductor .The property possessed by a battery of being capable of pro

duc ing a flow of electric ity is analogous to the pressure produced by a pump . The analogy to a simple battery of the formdesc ribed above is a pump which maintains a constant pressure

(approximately) in the same direction whether or not there is aflow of the fluid through the pump . Since a battery is always

CONTINUOUS ELECTRIC CURRENTS 105

made of conductors , the walls of this pump must also be con

sidered as elastic .

A conductor connected to the two poles of a battery is thenanalogous to the rubber hose completely filled with water (whichmay be taken as an approxim ate ly incompressible fluid) withits two ends connected respectively to the outl et and intake of thepump

,which is also completely filled with water . The pump will

then force a current of water through the hose, and the strength

re ssure PumpOutl et u lar Structure

InEl as tic Wal ls

Rubber HoseFig . 46.

of this water current the quantity of water per unit tim e flowing through any cross section of the hose) will depend upon thepressure developed bv the pump and the res istance due to thefriction of the water against the walls of the hose . The effectsproduced in and around the wire connected to the battery aresim ilarly found to depend upon a property of the battery (calledits electromotive force) analogous to the pressure developed bythe pump

,and upon a property of the wire (called its res istance)

analogous in certain respects to the resis tance of the rubber hoseto a water current .When the current ofwater is established in the hose

,a fall of

pressure is produced in the hose in the direction of the flow of thewater

,and consequently the pressure acting on the walls of

the hose will vary from point to point . Hence, near the outlet ofthe pump the walls of the hose will expand and near the intakethe walls wil l contract . Consequently , there will be more water inthe portion of

‘the hose near the outlet of the pump and less nearthe intake

,than there was in these portions of the hose before it

was connected to the pump . Experiment shows that an analogous


phenomenon occurs when a wire is connected to an electric battery . The portions of the wire near the two poles of the batterymanifest properties which they did not possess before ; these properties are such that they m ay be attributed to an excess of

electric ity,or a positive charge of electric ity on the portion

of the wire nearer the pole of the battery from which the electriccurrent is said to flow

,and a defic it of electric ity

,or a negative

charge of electric ity,on the portion of the conductor nearer the

pole of the battery toward which the current is said to flow.

On account of the inertia of a mass of water , tim e is required toset the water in the hose in motion when the latter is connected tothe pump . It is also found that time is required for the effectswhich we ascribe to the flow ofan electric current to reach a steadystate . As we shall see later

,the property of electric ity which is

analogous to inertia may be expressed in terms of the magneticfield produced by electric ity in motion . There is no experim entalevidence to lead us to assign to electricity a property analogous toweight .

The expansion of the walls of the rubber hose by the pressureproduced by the pump likewise -produces a pressure on the cellularstructure representing the dielectric surrounding the W i re

,and

produces a displacement of the cells and the fluid ( e . g .

,water)

which they contain . When the current of water in the hose becomes steady , the motion of these cells ceases

,but while the cur

rent in the hose is being established , there is also a motion of thesecells and the water they contain . As we shall see later

,an effect

is produced in the dielectric while the current in the wire is beingestablished or changes in any way, which may be attributed toa flow or displacement of electric ity through the dielectric , analogous to the disp lacement of the water in the cellular structuresurrounding the rubber hose , which displacement ceases when theelectric current reaches a constant value .

The displacement of the cells of the structure surrounding therubber hose produces a change in the shape of these cells

,and

therefore a strain in their walls . Analogous effects are observedin the dielectric surrounding a wire connected to a battery or

other device which is capable of produc ing the phenom ena whichwe attribute to the flow of an electric current . This effect is particu larly noticeable when the electromotive force of the cur

rent-producing device is high , and may even cause a rupture of

l 08 ELECTRICAL ENGINEERING

or sack which we have considered as representing a conductor,

changes in shape when connected to the pump , but there is no evidence that a w ire or other conductor changes in shape when itmanifests the properties which are a ttributed to a charge of

electric ity on it .65 . A Wire as a Geometrical Line . In the discussion of the

phenomena which are attributed to the flow of an electric currenta wire will usually be considered as equivalent to a geometricalline . This

,of course

,is not strictly accurate

,since a wire always

has a finite cross section . However,in many cases the error

involved is practically inappreciable ; when this is not so,atten

tion will be called to the fact . The exact expression for a wireof finite cross section may always be derived when we havededuced the relation which holds for a geometrical line, for wemay express this same relation for the W ire of fini te cross sectionby considering the wire made up of an infinite number of filaments of infinitesimal cross section

,each of which filaments is

equivalent to a line,and then determ ine the resultant effect due

to all these filaments . In general , such an expression is extrem elydiffi cult to evaluate ; only in one or two simple cases will it benecessary to do this .

66. Definition of the Strength of an E lectric Current—Defini

tion of a Continuous Current. —The quantity of electric ity thatflows through any section of a wire in unit time may be calledthe strength of the current of electric ity in this section of thewire

,j ust as the quantity of water flowing through any section

of a pipe in unit time may be called the strength of the watercurrent in this section of the pipe . By

“ quantity of waterflowing through any section of a pipe is meant the volume of

water flowing through this section ; therefore quantity of water “

has a perfectly definite meaning and can be readily determ ined,

either directly or by measuring its mass . There is no experi

mental evidence,however

,to lead us to attribute to electricity

either m ass or volume,in the ordinary sense of these terms . To

define the strength of an electric current in this m anner,there

fore,it would be necessary first to define what is to be m eant by

“ quantity of electricity . It is more convenient,however

,to

define the strength of an electric current in some other way,

and then to define quantity of electric ity in terms of theelectric current, particularly as this method of procedure is in


accord with the usual experimental methods employed in engineering work in the determ ination of the quantity of electric ity .

We m ight take any one of the effects described above as them easure of the strength of the current flowing in the wire . Seientists have agreed, however , to take as the measure of the strengthof an electric current flowing in aW ire , the mechanical force whichis exerted on the wire when it isplaced in a m agnetic field . (Thiseffect was illustrated above by d l

the mechanical force produced on

the wire by a m agnet placed nearit .) When a constant m agnetic Fig , 47 ,

field is established at the wire by some external agent (for example

,by means of a permanent magnet) it is found that in gen

eral the force exerted on the wire when it is connected to a battery of the kind described in A rticl e 62 remains constant (atleast apprec iably so for several seconds or more

,though , unless

spec ial precautions are taken this force will gradually change) .

The current in the wire is sai d to be continuousf< as long as thisforce remains constant . This force is found to depend uponthe flux density at the wire of the m agnetic field produced by thisexternal agent and also upon the direction of this flux densitywith respect to the direction of the wire

,and upon the length of

the wire . For a continuous current in the wire,the force dF

produced on any elementary length d l of the wire (see Fig . 47)by a magnetic field when the flux density at d l is B and makesan angle 9 with the direction of d l , is found to be proportionalto the product of the flux density B

,the s ine of the angle (9, and

the length d l ; that is , the mechanical force dF is proportionalto (B sin (9 ) dl . The di rection of this force is found by experim entto be perpendi cu lar to the p lane determined by the direction of the

flux densi ty B and the directi on ofthe length d l .

V arious modifications may be made in the rest of the c ircuitwhich will cause the force acting on a given length of wire tochange, even though the flux density B and the angle 9 at eachpoint is kept unaltered . For example

,another piece of wi re may

be inserted between one end of the original wire and the pole of*The following definitions are those adopted by the Am erican Institute

of Electrical Engineers .

1 10 ELECTRICAL ENGINEERING

the battery to which it was originally connected,that is

,placed

“ in series with the original wire ; or a second battery m ay be

connected in the c ircuit . Any modification which causes a changein the force acting on a given length ofwire when the flux densityB and the angle 0 at each point is kept constant , is attributedto a change in the strength of the electric current in the wire . In

Short,the strength of the current in any section of the wire is

arbitrarily defined to be proportional to this force when the fluxdensity B and the angle 9 at each element of thi s section is keptconstant . We then have that the force on any elementary lengthdl

,besides being proportional to (B sin 0) dl is also proportional

to the strength of the current ; that is

dF =kI (B sin 0) al l

where I is’

the strength of the current in the elementary length dlof the wire

,B is the flux density of the m agnetic field at d l pro

duced by any external agent , 0 the angle between the directionof this flux density and the direction of d l

,and k is a factor

of prop ortionality . Experiment shows that this quantity k isindependent of the nature of the substances which form the wireand the medium surrounding the wire, but depends only uponthe units in which the force

,the flux density

,and the length are

m easured . The unit of current strength may therefore be chosensuch that when the force dF is measured in dynes , the flux density B ,

in gausses , and the length d l in centimeters,this factor

of proportionality k is equal to unity . The above equation thenbecomes *

dF =I (B sin 9) d l

from whichdF

(B sin 0) al l

Note thatd

d

}

; is the m echanical force per unit length of wire

at d l,and that B sin 0 is the component of the flux dens ity at

d l perpendicular to the wire at this point . Hence as the measureof the strength of the electric current in a W i re i s taken the ratio

of the force per uni t length of the wi re whi ch wou ld be produced

by a magneti c field to the component of the flux densi ty of this field

perpendi cu lar to the wire . This definition and its mathematical'

*Thi s relation is known as Biot and Savart’ s Law.


by noting the corresponding letters in m iddle and I, thumb andm ove

,forefinger and flux .

68 . Conductors in Series and in Para l lel . Experim ent Showsthat when a number of conductors are connected end to end

(Fig . 48) and are completely surrounded by insulators , then thestrength of the current as above defined is the sam e in all theseconductors

,provided the current strength does not vary with

Battery

Fig . 48 .

time . Two or more conductors thus connected end to end aresaid to be connected in seri es .

Experiment also shows that when any portion of an electricc ircuit between any two points A and B is formed by two ormore insulated conductors (Fig . the strength of the currentcom ing up to the junction point A or leaving the junction pointB is equal to the Sum of the strengths of the currents in the con

Fig . 49 .

ductors j oining A and B,provided the current strength does

not vary with time . For example,in Fig . 49

I = I l + 1 2+ l 3

and the conductors j oining the two points A and B are saidto be connected in para l le l .When any portion of a c ircuit is made up of one or more con

ductors connected in series with one or more groups of conductorsin parallel

.this portion of the '

c ircuit is said to be connected inseri es—paral lel .

CONTINUOUS ELECTRIC CURRENTS 1 13

69 . Total Force Produced by a Magnetic Field on a Wire Carrying an E lectric Current. From the above discuss ion and equation it follows that the total mechanical force F produced byany external magnetic field on an insulated wire carrying anelectric current I , when the wire is of any length l and bent intoany shape whatever, is equal to the vector sum

P =If0

l

(B sin 9) d l (2)

where d l is a ny elem entary length of the wire measured in thedirection of the current , B the

'

flux dens ity of the field at d l,and

0 the angle between d l and the direction of B and f1indicates

the vector sum of the expressions (B sin 0) d l for all the elementarylengths into which the wire is divided . Al l quantities in thisequation are in c . g. s . units . In general , the flux dens ity andthe angle (9 will be different for each point of the wire . Thedirection of the m echanical force dF acting on each element willalso be different for each elem ent

,s ince the plane determ ined by

the direction of the flux density and the direction of the elementwill in general be different ; hence the necessity for taking thevector sum .

70. Force Produced by a Uniform Magnetic Field on a Stra ightWire Carrying an E lectri c Current. One of the s implest casesis that of a straight w ire in a uniform

_magnetic field . Cons ider

such a wire carrying a current I,placed at an angle 9 with the

direction of the lines of induct-ion ; let the flux dens ity of thismagnetic field be B . In this case B and 0 are constant for allpoints along the wire, and the mechanical force on all elementsof the wire is in the same direction ; the integration is then asirnple algebraic one and therefore the total force on the wire is

l

F =IB sin 9 f0d l =IB l sin 9. (2a)

When the wire is perpendicular to the direction of the field theforce acting on the wire is

F I B l (2b)S ince (9 = 90° and sin 90° = 1 . In case the wire is in a non-m agneticmedium

,equation (2b) becom es

F =I HZ

All quantities in these equations are in c . g . 3 . units .

71 . Magnetic Field Produced by an E lec tric Current in a

Wire . We have seen that when a w ire carrying an electric cur


rent is placed in a magnetic field,this field exerts a mechanical

force on the wire . It is also found by experiment that an equaland opposite mechanical force is exerted on the magnet or otheragent produc ing this field ; this, of course

,is in accord with the

general princ iple of nature that action and reaction are equal “

and opposite . The region around a wire carrying an electric .

current is therefore a m agnetic field of force,for by definition

a m agnetic field of force is any region in which a m agnetic polewill be acted upon by a mechanical force . The intens ity of thefield of force due to a wire carrying an electric current m ay bereadily determ ined from equation

Consider an elementary length d l of the wire in which the current is I abamperes and let this length be m easured in the direrition

of the current .

Let the m agnetic fi e l d w h i chp r o d u c e s t he

mechanical forcedF on th is elem entary lengthbe that due to aunit north pointpole at any poin tP at a distance

of r centimeters away . The flux density of the magnetic field at dl

due to this unit pole is then 12independent ofthe permeabi l i ty of the

r

medium surro

i

unding the pole and the wi re (see Article Theangle (9 between the direction of this flux density and the lengthd l is the angle between the direction of dl and the direction of

the line drawn from the pole to d l . The mechanical force exertedby the unit point-pole on the length al l is then

Fig . 50.

dF ='

I

and is perpendicular to the plane determ ined by the point P andthe length dl . The direction of this force is determ ined by theleft-hand rule

,see A rticle 67 ; in the diagram this force is down

into the plane of the paper . The current I in the elementarylength d l exerts an equal and opposite force on the unit point


coincide in direction with the field intensity at this poin t ; see

the preceding artic le . From the dedu ctions of the precedingarticle it al so follows that the positive sense of these lines of fo rceis the same as that in which a right-handed screw placed alongthe w ire at dl must be turned to advance it in the positivesense of the cu rrent . When the positive senses of two quantitiesare thus related , the quantities are said to be in the righthanded screw direction with respect to each other .The current in every other elementary length of the c ircuit

w ill likewise produce a m agnetic field, the lines of force of whichare c ircles which have a like relation to the current in the elementary length which produces them . The lines of force representing the resultant field due to a current in any fini te length of wirewill not in general be c ircles , but it can be shown that each of theresultant lines of force is a c losed loop which links the wire inthe right

-handed screw directi on with respect to the current.

Note an important difference between the lines of force due toan electric current and those due to magnetic poles : the l ines offorce due to magnetic poles are not c losed but end on the pol es , whi l e

the l ines of force due to an electri c

current are a lways c losed loops , l ink

ing the circui t whi ch produces them .

When there is any magnetic body inthe vic inity of an electric current thisbody will in general have poles induced on it by the magnetic fielddue to the current

,and part of the

lines of force representing the resultant field will end on these poles andthe rest will b e c losed loops linkingthe electric ci rcuit . The lines of

induction are always closed loopswhether or not there are magneticpoles in the field .

73 . Magnetic Field Due to a Cur

rent in a Long Straight Wire .

-A

useful application of equation (3) isthe calculation of the intensity of

the magnetic field at any point due to a current in a longstraight wire .

Fig . 52 .

CONTINUOUS ELECTRIC . CURRENTS

In Fig . 52 let BA be a straight wire and let P be any point ata distance r f rom it (m easured perpendicular to the wire) , I thecurrent in the wire from B to A in abamperes

,d l any elementary

length in the direction of I at a distance l from the point 0 Wherethe perpendicular from P cuts the wire

,0 the angle between d l

and the line from P to dl , and a: the distance from P to d l . E achelement of the wire will produce a field intensity at P perpendicular to the plane of the paper, downward , and therefore the fieldintensities at P due to all the elements of the wire may be addedalgebraically . The field intens ity at P due to the current in thelength d l is, from equation

(I sin 0) dl

and therefore the total field intensity at P due to all the elementsof the wire is

11 (1 sin 9) al l

where l 1 = 0 A and l z= 0 B . The simplest way to evaluate thisintegral is to express the different variable quantities in terms of

the variable angle a (See Figure From the figure we have

=cos a x :

Tand l =r tan a . Differentiating the latter

00 3 a

r d“Let a 1 be the num erical value ofex ression we et d lp g

cos2a

the angle GPA and a 2 the numerical value of the angle OPE .

Substituting these values in the above equation we get

(1 00 3 a ) d ot I

ra ,+ sin a

Al l quantities in this equation are in c . g. 3 . units .

When the wire is of infinite length (or practically ,when the distance of the point P from the ends of the wire is great compared tothe perpendicular distance r) a 1 and a 2 become equal to andtherefore sin a ,

= 1 and sin a ,= l . We then have that the field

intensity at a distance r from a straight wire of infinite lengthcarrying a current I is

H :


All quantities in this equation are m c . g. 3 . units . The directionof the field intensity is perpendicular to the plane drawn throughthe wire and the point ; the lines of force are therefore c ircleswith their centers along the wire and their planes perpendicularto the wire . The relation between the direction of these linesof force and the direction of the current is conveniently remembered by the rule that the lines of force are in the direction inwhich a right—handed screw must be turned to advance it in thedirection of the current . Equations (4) apply to a variable as

well as to a continuous current .The above formula has been deduced on the assumption that

the wire may be considered as a geometrical line . It can also beshown that the field intensity at any point P at a distance from r

from a long circu lar wire of finite cross section is also given by theabove formula provided the point P lies outside the wire

,and the

current density is uniform over the section of the wire ; see Article104. At any point inside such a wire the field intensity is also per

pendicu lar to the plane through the point and the axis of thewire but is equal to

2 I r

2H .

fa

where r is the distance of the point from the center of the wire,a

is the radius oi the wire and I is the total current in the wire ,and all quantities are in c . g. 3 . units . Experiment justifiesassumption that the current dens ity in a wire is cons tant

,pro

vided the wire is of uniform structure and the current is a con

tinuous current . This is not true when the current varies rapidlywith time

,but is approximately true for ordinary variable or

alternating currents used in practical work , provided the wireis non-magnetic and has a diameter less than one inch . Theabove formula for the field intensity outside a wire is also ap

proximately true for a wire whi ch has a cross section of any shape,

provided the distance of the point from the wire is great compared to the greatest diam eter of the wire .

74 . Magnetic Field Due to an E lectri c Current in a Circu larCoil

’

ofWire .

—Another useful application of equation (3) is thecalculation of the intensity of the magnetic field at any point dueto a current in a ”

c ircular coil ofwire . The solution of this problemexcept for points on the axis of the coil is quite difficult ; thesolution for a point on the ax is of the coil is obtained as follows .


S ities at P due to all the elementary lengths in the c i rcumferenceof the c ircle . The component parallel to OP of the field intens itydH due to any element is

dH . sin a

I d l.

7"

l/ a 2+ r2

which , integrated along the c ircumference of the c irc le , gives as

the value of the resultant field intensity at Pl= 2 7r r

'

Ir dlg

(aZ

(a2+ r

2

)f

A l l quantities in this equation are in c . g. 3 . units .

The d irection of the field intensity at any point along the ax 1sis the direction in which a right- handed screw p laced at thispoint advances when it is turned in the direction of the current .The field intensity at the center of the coil is found by putting

a equal to zero ih ' the above equation,which gives

2 7r IH

or

The field intensity at any point on the axis of a ci rcular coilwhich has any number of concentri c turns m ay also be calculatedfrom equation (5) by calculating the intensity due to each turnseparately and adding these separate intens ities . In the case ofa c ircular coil with a concentrated winding of N turns , i .e.

,when

the N turns are so c lose together that they m ay all be consideredas occupying but a single geometrical line, the field intensity atthe center of the coil is

,from (5a) ,

_

2 7r N IC

Hr

The winding of a coil may be considered as concentrated whenthe radius of the coil is large compared with the linear dimens ionsof the cross section of the winding .

In Article 108 is deduced by a different method the field intensity for a point inside a long coil wound in the form of a longcylindrical helix or solenoid .

”

75 . Absolute Measurement of an E lectric Current . —When a

current is established in an insulated wire wound into a c ircularcoil of N turns

,the strength of this current can be determ ined

experim entally in terms of quantities which can be m easured or


calculated . Let,such a coil (Fig . 54) be setup with the plane ofits

windings parallel to the direction of the earth ’s magnetic fieldand let a smal l

l

magnetic needle be suspended at the center of thecoil in such a manner that it is free to turn about a vertical axis .

When there is no current in the coil,this needl e will then point in

a direction parallel to the plane of the .coil . When a current isin the coil

,this current will set up a magnetic field at

the plane of the coil,and therefore the direction

field at the center of the coil will be changed -and.therefore be deflected a certain angle Q: Thisual to the angle between the direct ion of the resultthe direction of the horizontal component of thet the center of the coil . This latter field intens ity

,

may call He ,can be determ ined by the method described

Article 42 . The intensity of the field at the center of the coilCis given by equation (5b) .

then have thatH

c2 7TNI

Her H

.

r H8tan (9

2 7TN

where all the quantities are in c . g . 3 . units .

Since all the quantities in the right-hand -m ember of equa


tion can be measured , the strength of the current I can be cal

culated . Such a device is called a tangent ga lvanometer .The

accuracy of the instrument depends upon the accuracy with whichthe horizontal component of the earth ’s field may be measured , but

an accurate measurement of the latter is difficult . Besides,in any

ordinary laboratory or testing room the electric c ircuits in thebuilding and the surrounding streets also produce magneticfields which act on the needle

,and these fields are continually

changing . Such an instrument is therefore seldom used nowadays .

Amuchmore accuratem ethod of determ ining the absolute valueof an electric current is to cause the same current to flow throughtwo parallel coils

,one of which is suspended from one arm of a

del icate balance . The stationary coil then produces a magneticfield which produces a pull on the m ovable coil and the amount ofthis pul l can be readily measured . From equations (2) and (3)the amount of this pull in terms of the current and the dimensionsof the coils can also be deduced

,and s ince both the pull and the

dimens ions of the coil can be measured,the strength of the

current can be calculated . Su ch an instrum ent is call ed an ab

solute current balance . This is also the princ iple of the Kelvincurrent balances ; in the latter , however , the current strength is riotdeterm ined directly from the pull and the dimensions of the coils

,

but is calculated from the pos ition of a rider on an arbitraryScale ; the latter is calibrated by comparison either with anabsolute balance or with a s ilver voltam eter (see Article

76. Comparison of the Strengths of E lectric Currents . Gal

vanometers and Ammeters . We have j ust seen how the strengthof an electric current flowing through the coils of a current balancemay be measured in terms of the dimensions of the coils and thepull of the fixed coil on the movable coil . The accurate m easurement of an electric current by this method requires a balance con

structed with great care and numerous precautions have to betaken in us ing it . Much simpler and cheaper instruments can

be constructed which will indicate by“

the deflection of a needleor spot of light the relative m agnitude of the electric currents whichmay be established in them .

The s implest form of such an instrument is a device whichconsists essentially of a m agnet suspended ins ide a coil of insulatedw ire which may be connected in series with the conductor in which


by comparing it with a standard instrument in the sameway that a moving needle

,or Thomson

, galvanometer . is

calibrated .

An instrument extens ively used for the measurement,of

electric currents , and known as the Weston ammeter,is essen

tial ly a D’Arsonval galvanometer . In this instrument the

movable coil,instead of being suspended by a metal strip , is pro

vided with a fine pivot which rests in a j ewel bearing . Thecurrent is led to and from ” the instrument by means Of twosmall flat spiral Springs connected to the two ends of the coilrespectively and also to the term inals on the outside of the caseOf the instrument . These springs also serve to hold the coilnormally in a definite pos ition . A light metal pointer is attachedto the coil

,and the position of the pointer is read off on a scale

which is m arked to read directly in amperes . The scale of suchan instrument is seldom exactly correct

,and for accurate work

the instrum ent ni'

ust be calibrated in the manner described above .

There are other makes Of amm eters based upon the princ ipleof the D’

Arsonval galvanometer, and still others in which thecurrent in the coil of the instrument sets up a magnetic field whichexerts a force on a piece of soft iron and thereby causes the deflection of a pointer over a graduated scale . For a detaileddescription of amm eters

,the reader is referred to any text -book

on electric m easuring instruments .

77 . E lectrolysis and E lectrolytes . Having now seen howthe strength of an electric current may be measured, we are readyto investigate som e of the other phenomena which are attributedto the flow ofan electric current . An important group ofphenome na are those which take place in a conducting solution when acurrent is established through the solution . In the first place , inany solution which is a conductor

,chem ical decomposition always

takes place . The change that takes place in the copper sulphatesolution described in Artic le 62 is an illustration . Any substancethe constituents ofwhich are separated when an electric currentis established in it , is called an e lectrolyte, and the process of separation is cal led e lectrolysi s . Al l conducting liquids other thanmolten metals are electrolytes ; gases , when they becom e conducting , are also electrolytes . Some solids are al so electrolytes

,an exam

ple Of which is silver iodide . The decompos ition that takes place inthe elec trolyte is found to be confined entirely to the portions of


the electrolyte in contact with the metal plates or wires (called theelectrodes) which connect the electrolyte to the rest of the c ircuit .The results of

"the decomposition are therefore usually said to bedeposi ted at the electrodes ; though , of course , if the substanceliberated is a gas , it will immediately escape through the surfaceof the liquid , or if the substance liberated is soluble , it will go intosolution

,or if it is a solid which is not soluble , it may fall to the

bottom Of the vessel containing the electrolyte . In c ertain caseswhen the result of the decompos ition is a m etal

,the metal becom es

firm ly attached to the electrode where it is liberated . The copperdepos ited on the copper wire in the copper Sulphate solution inArtic le 62 is an example . The important industry of electroplat

ing is based upon this fact .The electrode through which the current enters the solution

is called the anode,and the electrode through which the current

leaves the solution is called the cathode . With very few exceptions,

an element,or such a group of elements as is called a radical

,

”

is always .deposited at the same electrode; no matter from whatcompound it m ay be liberated . Hydrogen and metals are alwaysdepos ited at the cathode .

As the result of a series Of careful experiments Fa r aday foundthat a Simple relation exists between the rate at which a substance is deposited from an electrolyte ,

‘

when an electric currentis established in it , and the strength of the current . This relation is

1 . The rate (mass per unit time) at which a substance is de

posited from an electrolyte , when an electric current is establishedin it

,is directly proportional to the strength of the current estab

l ished .

Faraday also found that a s imple relation exists between themasses Of the various substances deposited

,when the same current

is established in several electrolytes , and the chem ical «

equiva

lents"< Of these substances . This relation is2 . When the same current is established through different*The chem ical equivalent of an element or radical is the atom ic weight

of the element or radical divided by its valence . By valence is m eant thenumber of atoms of hydrogen with which one atom of the elem ent or radical will form a stable combination . For example

,the valence of oxygen is

2,since 2 atoms Of hydrogen combine with 1 atom of oxygen to form

water. The valence of copper in cuprous compounds is 1 and in cupriccompounds 2 .


electrolytes the rates (mass per unit time) at which the varioussubstances are deposited are directly proportional to the chem icalequivalents of these substances .

These two statements of experimental facts are known as

Faraday’s Laws of E lectrolysis .

78 . Electrochem ical Equiva lent Of a Substance . Faraday ’sfirst law may be expressed mathematically by the equation

m

t

where I is the current established in the electrolyte , m themass Of a given substance deposited in time t

,and h a constant

Of proportionality,the value of which depends only upon the

nature of the substance and the units in which m,t and I are

measured . This constant is called the e lectrochemi ca l equiva lent

of the substance . Its value for any substance may be readilydeterm ined experimentally by measuring the current I by meansof an absolute current balance

,and measuring the m ass of the

substance deposited when this current flows through the electrolytefor a given length of time . The values of the electrochem icalequivalent for silver

,copper

,h ydrogen and oxygen , when m is

measured in grams,t in seconds and I in amperes

,are

SilverCopper (from cupric solutions)HydrogenOxygenIn equation (7) the current is assumed to be continuous ; when

the current varies with time the mathem atical expression ofFaraday ’s first law is

or the total mass of the substance deposited in time t is

m = k dt (7a)

79 . The International Ampere . Knowing the value of theelectrochem ical equivalent of a given substance

,for example

,

S ilver , one can readily determ ine the strength of the current flowing through an electrolyte from which this substance is deposited ,by measuring the number of grams of this substance deposited


is the integral with respect to time of the variable current atthat section

,that is

Q (80 )

where i represents the value of the current at any instant,dt an

infinitesimal interval Of tim e measured from this Instant,and t

the total time during which the current flows . From the relationexpressed by equation (7a) the quantity of electric ity whichflows through an electrolyte in any interval of time may be

determ ined by measuring the m ass of the substance depositedat either electrode in this interval . In Article 109 is describedthe m ethod usually employed for m easuring the quantity ofelectric ity corresponding to a variable current .Since in the case of a continuous current the total current

strength is the sam e at all cross sections of a given conductor,it

follows that the quantity of electric ity flowing across each crosssection of the conductor in any given interval of time must alsobe the sam e at each cross section . Note the analogy with anincompress ible fluid . Hence the flow Of a current Of strength Ithrough the conductor for t seconds may be looked upon as

equivalent to the transfer of I t units of electric ity from one endOf the conductor to the other

,j ust as a current of ten cubic feet

of water per second in a pipe for five seconds is equivalent to atransfer of 10 x 5 =50 cub ic feet Of water from one end Of the pipeto the other . In the case Of water in a pipe the 50 cubic feet Ofwater which enter the pipe m ay not be the same as the 50 cubicfeet which leave the pipe at the other end . In the same way

,it

is not necessary that we l ook upon the electricity which entersone end Of a conductor as being the sam e elect ricity as leaves theconductor at its other end .

The unit Of quantity Of electric ity in the c . g. s . electromag

netic system of units is the quantity Of electric ity transferred bya current Of one abampere for one second . This unit is calledthe absolute unit of quantity or the abcou lomb. The practicalunit of quantity Of electric ity is the cou lomb

,which was defined

by the International Congress Of E lectric ians as foll owsAs the unit of quantity (shall be taken) , the International

Coulomb,which is the quantity of electric ity transferred by a

current of one international ampere in one second .

”

The ampere-hour, i .e .

,the quantity Of electric ity corresponding


to a current ofone ampere for one hour is also employed in practice .

In accordance with these definitions we then have1 abcoulomb = 10 coulombs1 ampere—hour = 3600 coulombs

81 . E lectric Resistance .

—Jou1e ’s Law.

— One Of the phenomena always assoc iated with an electric current in a conductoris the dissipation of heat energy in the conductor . In the caseof a wire of uniform structure kept at a constant uniform temperature experim ent shows that the rate at which heat energy is dissipated in a given length of the wire

,between any two points 1

and 2 say, when a continuous current is established in the wire,

is directly proportional to the square of the strength of the currentin this wire . That is

,calling Ph the rate at which heat energy

is developed in the wire between the points 1 and 2,and I the

strength Of the current in this portion of the wire,then

Ph=R 1 2 (9)

where R is a constant depending upon the dimensions and tem

perature of the wire, the nature of the substance form ing the wire,and the units in which Ph and I are measured , but is independentOf the current strength . This factor R is called the resistance of

the wire,and the statement of the experimental fact represented

by this equation is called Joule ’s Law,from the name of the

sc ientist who first clearly enunc iated the fact . The resistanceof a given length of wire may then be defined as the rati o of the

rate at whi ch a continuous current produces heat energy in the wire

to the square of the strength ofthi s current. When the rate Ph atwhich the heat energy is dissipated is expressed in ergs per secondand the current I in abamperes , a resistance of one unit is equal tothe ratio Of one erg per second to one abampere- squared . Thisunit Of res istance is called the c . gi s . electromagnetic unit of resistance or the abohm. When the rate Ph at which heat energy is dissipated is expressed in j oules per second (one j oule by definition isequal to 107 ergs) and the current I in amperes , a res istance of oneunit is equal to the ratio of one j oule per second to one amperesquared . This unit is called the practical unit of resistance or theohm. The relation between the Ohm and the abohm is therefore

I ohm = 109 abohms .

To express small resistances a unit one-m illionth of the sizeof an Ohm is ordinarily used ; this unit is called the mi crohm . To

-express large res istances a unit one million times the S i ze of an


Ohm is frequently used ; this unit is called the megohm. Hence1 ohm = 10

“m icrohms

1 n1egohm = 106 Ohms

The resi stance of a wire to a variable current is the same as

its resistance to a continuous current provided the wire is smalland the current does not vary rapidly with tim e . In the case ofa large wire the resistance of any current filam ent (see A rticle 101)is the same to the variable as to a continuous current

, but the resistance of the wire as a whole is greater . The rapid variationof the current with time causes a different distribution of thestream lines of the current and thereby produces a greater heatingthan that which takes place due to a continuous current . See

Article 1 21 .

82 . Absolute Measurement of E lectric Resistance . Thedeterm ination Of the electric resistance in terms of the quantitiesspecified in the above definition would require the measurementOf the heat energy dissipated in the wire in a g iven interval Oftime when a continuous electric current of known strength isestablished in the wire and the wire kept at constant temperature .

The interval of tim e can be readily m easured,and the current

strength m ay be determ ined directly by an absolute current balance or by means of any kind of galvanometer or ammeter whichhas been calibrated . The heat energy dissipated in the wirecould be determ ined by som e calorimetric measurement . Calorimetric measurements

,however

,are difficult and at best are not

susceptible of a high degree of accuracy . A much m ore accuratemethod of measuring the electric resistance Of a wire in terms of

quantities which may be measured or readily calculated, is thefollowing , which is based upon the princ iple of electromag

netic induction (see Chapter IV ) , but the only quantities to bemeasured are those which have already been defined . A metaldisc D

,mounted on a m etal axis coinc ident with the axis of a coil

Of wire C,is arranged so that it can be rotated with a constant

angular veloc ity or. The coil C’ and the res istance R to be m easured are connected in series with a battery B . A galvanom eterG has one of its term inals connected by a wire to the axis of thedisc and the other term inal to one end of the res istance R

,the other

end of R is connected by a wire to a metal“ brush m aking

contact with the rim of the disc . We have then two c losed c ircuitsin each ofwhich the resistance R forms a part ; one c ircuit is the


The results of a large number ofmeasurements by these variousmethods Show that a column Of m ercury centimeters longwhich has a uniform cross section of 1 square m illim eter has a resistance ofone ohm at zero degrees centigrade . A column Of m ercuryof these dimens ions has a m ass Of grams . Hence theadoption ofthe fol lowing definition ofthe Ohm by the InternationalCongress Of E lectric ians :ifAs a unit of resistance (shall be taken) the International

Ohm,which is based upon the Ohm equal to 10 units of res istance

of the C. G. S. system Of electromagnetic units , and is representedby the res istance Offered to an unvarying electric current by acolumn of m ercury at a temperature of melt ing ice

,

grammes in m ass , of a constant cross sectional area,and of the

length centimeters .

”

This definition,l ike that Of the ampere

,has been legalized by

most c ivilized countries .

A lthough the absolute'

measurement Of a res istance is com

paratively difli cult , the compari son of the values of two or moreresistances is a very S imple m atter and can be carried out with ahigh degree Of accuracy . A l l these m ethods Of comparison arebased upon Kirchhoff’s Laws . (See Article

83 . Specific Resistance or Resistivity. As already noted,the

res istance Of a conductor depends upon its dimensions . It is

found by experiment that the resistance of a conductor of uniformc ross section throughout its length , when the conductor is keptat a uniform temperature throughout and the current density

(see Article 101) is uniform over its cross section,varies directly

as the length Of the conductor and inversely as the area Of its

cross section (perpendicular to its length) , but does not dependupon the shape of its cross section . That is

,calling l the length

of the conductor,A the area of its cross section , and R the resist

ance of this conductor,then

l

A

where p is a factor of proportionality which depends upon themateria l of which the conductor is m ade and the temperature atwhich the conductor is kept

,and also upon the units in which

R,l,and A ‘are expressed . The value Of this factor p for any

conductor,at any temperature

,is called the specific resistance

or the resi stivi ty of the conductor at that temperature . When R

R= P


is expressed in m icrohms,l in centimeters and A in Square centi

meters,this factor p is equal to the resistance in m icrohms of

a cube Of the conductor 1 cm . on each edge , provided the streamlines of the current are parallel to four parallel edges Of the cubeand the current density over the section at right angles to thesestream lines is uniform (see Article The specific resistanceof the conductor may then be expressed as so m any m icrohmsper centimeter- cube . Sim ilarly

,when R is expressed in m icrohms

,

l in inches and A in Square inches , this factor p is equal to theres istance in m icrohms of an inch - cube of the conductor ; thespec ific res istance of the conductor may then be expressed as SOmany -m icrohms per inch- cube . Again , when R is expressed inohms

,l in feet and A in c ircular m ils * this factor p is equal to

the resistance in ohms of a portion of the conductor one foot longand one c ircular m il in cross section ; the spec ific res istance of aconductor may then be expressed as so many ohms per m il - foot .

’

There is still another way of express ing the spec ific res istanceof a conductor

,which was formerly much used and is stil l em

ployed sometimes in wire spec ifications . The cross section of abar or wire of uniform cross section is equal to the volume of thebar divided by its length, and the volume Of the bar is In turnequal to its mass divided by the density of the material of whichit is made . Hence

,us ing the same symbols as above

,and in

addition calling m the mass of the conductor and 8 its densityor spec ific gravity, we have

m l8and therefore R :

For a given material at a given temperature p 8 is also constant,

henceR=h

*A mi l i s defined as one-thousandth of an inch and a circu lar mi l i s defined as the area of a c ircle one m il in diameter . Since the area Of a c irclevaries a s the square of its diameter , the area of a c ircle in circular m ils isequal to the square Of its diameter in m ils . This unit of area is thereforeverv convenient for expressing the area of the cross section of a circularwire ,Si nce the factor 7r is eliminated . The area of a c ircle one—thousandth of an

inch in diameter is equal to41

square inches or to

square centimeters ; hencel c ircular m il= 0 .78540 inches

x 10‘ °square centimeters


where k is a constant (equal to p 8) for a given material at a giventemperature . The Spec ific resistance of a conductor may then alsobe expressed indirectly in terms of this factor k . When R is expressed in Ohms

,l in meters and m in grams

,this factor k is equal

to the resistance of one gram of the conductor m ade into a wire ofuniform cross section and one meter in length ; this factor k is thensaid to be the specific resistance of the conductor in ohms per meter

gram . It Should be noted that this method of expressing the specific resistance ofa conductor involves the density of the conductor,while the other methods do not . The determ ination of the spec ificresistance of a conductor in m eter-grams

,however

,does not

require a determ ination of the density,but merely the measure

ment of the resistance , length and mass of a given wire of the

conductor, since k —T—I-

g. However

,to calculate from the value

Z2

Of k the Spec ific res istance in m icrohms per centimeter- cube,or

per inch- cube,or ohm s per m il - foot

,does require a knowledge of

the density of the conductor .

The various units of spec ific resistance are related as follows

m icrohm per inch- cube m icrohms per centimeter- cube

1 m i c rohm per centimeter- cube ohms per m il - foot1 Ohm per meter-gram L0

8

2

m icrohms per centimeter- cube

1 Ohm per m eter-gram = 6 0 1

8

5 3 Ohms per m il—foot

TO calculate the resistance of a wire of uniform cross section whenits specific resistance is given , the formulas

lor

A

are directly applicable, but care must be employed to express allthe quantities entering into the formula in the same system ofunits . For example

,the resistance in ohms of a wire which has

a spec ific resistance of m icrohms per centimeter cube,a

length of 1000 feet and a cross section of square inch,is

R X 10 “1000X 12X

ohms .

x

Experim ent shows that the resistance Of a given length ofw ire of un iform cross section is independent of the shape into

R = p R = k


practical conductor . Steel, however, in the form of track rails orthird rails is largely used in railway work, but is seldom used

as a conductor in the form of wires , except for short telephonelines

,and even here it is being largely supplanted by copper .

Any impurity in a metal increases its res istance . All alloyshave a greater resistance than that Of the best conductor in them .

For certain purposes, particularly in the construction Of resistance boxes and rheostats ,

” a high spec ific resistance is des irable . V arious alloys are m ade for this purpose, the spec ificresistances Of which can be found in any electrical engineer ’shandbook . For the resistance of copper and alum inum wires ofvarious sizes see Appendix B .

84 . E lectric Conductance and Conductivity . The rec iprocalof electric resistance is defined as electric conductance . Thatis

,when R is the res istance Of a conductor

,the conductance is

1 I 2

R Ph

where I is the continuous current establ ished in the conductor andPh the rate at which heat energy is developed in it . The unit ofconductance on the c . g. s . electromagnetic system

‘

of units isone abampere- Squared per erg per second ; this unit is called theabsolute unit of conductance or the abmho. The practical unitof conductance is one ampere—squared per j oule per second

,or one

amperesquared per watt ; this unit is called the mho. (Notethat the word mho

” is s imply the word Ohm written backwards . )The reciprocal of spec ific resistance or res is tivity is called

specific conductance or conductivi ty. For example,a spec ific

resistance of 10 ohms per m il - foot is the sam e as a conductivityof mhos per m i l - foot .

85 . Matthiessen’s Standard of Conductivity .

— M atthiessen'

,

about 50 years ago, found that the spec ific res istance Of thepurest copper at that time obtainable was ohms permeter-gram at 0

°cent . M atthiessen

,however

,failed to state the

dens ity of his copper,so that the exact value of this spec ific re

tance in m icrohms per centimeter- cube or in Ohms per m il - footis not known . However

,assum ing as the probable value

Of this density,M atthiessen ’

s value of ohms per m eter

gram is equivalent to Ohms per m il - foot . The Corresponding value of the conductivity

,namely mhos per

m il- foot,has been adopted by electrical engineers in this coun


try and in England as the standard conductivity thisstandard is

'

~

usual ly referred to as JI atthi essen’s standard . The

re lative conductivi ty of any conductor is then equal to the percentage ratio Of its conductivity per mil - foot at 0° centigrade to theconductivity ofMatthiess

'

en’s standard ; that is , the rela

tive conductivity of a conductor which has a resistance of

Ohms per m il - foot at 0° centigrade mhos per m il - foot)is Note that the lower the relative conductivity the higheris the spec ific res istance . It is of interest to note that copper canbe made at present of a greater purity than that used by Matthiessen and consequently having a relative conductivity greater than100% ofMatthiessen

’s standard . Comm erc ial copper wire usually

has a conductivity of from 96% to the harder the wirethe lower its conductivity . Commerc ial alum inum has a con

ductivity of about86 . Temperature Coeffi cient of E lectric Resistance . It is

found by experiment that the variation of. the res istance of aconductor with temperature may be expressed by t he formula

R=RO (1 +Bt) (13)

where R0is the resistance of the conductor at any given stand

ard temperature,R the res istance Of the conductor at a tem

perature t degrees above this temperature, and B a coeffic ientwhich is approxim ately constant independent of the temperaturerise t

,but does .depend upon the temperature corresponding to R0

.

Z ero degrees centigrade is usually taken as the standard tem

perature, and the value Of this coeffic ient B when the rise of

temperature is referred to 0° cent . is called the resi stivi ty tem

perature coefii ci ent of the conductor,or briefly , the temperature

coefii ci ent of the conductor . The value of B for comm erc iallypure copper depends som ewhat upon the purity Of the m etaland also upon the rise of temperature t. The Am erican Instituteof E lectrical Engineers , however, has adopted the constantvalue as suffic iently accurate for all grades of commerc iallypure copper and for any value of the temperature rise ordinarilymet with in electric m achinery

,that is

,for any rise of temperature

not greatly in excess of 100°cent . The temperature coeffic ient

of any other commerc ially pure metal,except the magnetic

m etals such as iron,nickel

,cobalt and b ismuth

,has practically

the same value . The temperature coefficient for iron isper degree cent .


The temperature coefficient of carbon is negative and is not aconstant . For example

,the resistance of a carbon- filament lamp

is less when the lamp is burning than when it is cold . The tem

perature coefli cient of m ost insulators is also negative and verylarge ; moreover , it is far from being constant .The temperature coeffic ient Of alloys depends largely upon

their constituents . It is poss ible to make alloys for which thetemperature coeffic ient is zero over a considerable range of tem

perature . Such alloys are extrem ely useful in the constructionOf standard resistance coils

,since the res istance of such a standard

rem ains constant for any ordinary vari ation of temperature andconsequently its res istance does not have to be corrected for

temperature .

Equation (13) enables one to calculate the resistance of a conductor at any temperature when its resistance at any othertemperature and its temperature coeffic ient are known . For

,

calling R, the res istance of the conductor at t° cent . and R,. itsres istance at t’ degrees centigrade , we have

(1 +Bt)Rt’

=Ro

Hence,taking the ratio of these two equations, we have

R, 1 +Bt

1 +,Bz

’

Substituting for B the numerical value of and dividingthe numerator and denom inator of the fraction in the right- handmember by gives

238+ t’

238 -H.

The operation expressed by this equation is readily performed byOne setting of a S lide rule

,when a temperature scale is marked

Off on the lower scale of the slide,marking 238 as 248 as

258 as etc . Then,if the resistance at 25° cent . is 3 Ohms

,

say,the res istance at any other temperature

,say is found by

setting 25 on this new scale oppos ite 3 on the lower sc ale of therul e and reading Off on the lower scale the number correspondingto 60 on the new scale

,i .e .

,Ohms .

R .


That is,the drop of pressure between any two points along a pipe

through which a water current is flowing is equal to ratio of thepower developed by the current to the strength of the current .This relation holds for a varying current as well

,as for a steady

current ; i .e .

,the instantaneous pressure drop is equal to the instan

taneous power divided by the instantaneous value of the currentstrength .

When there is no pump or motor connected in this givenlength of pipe

,all the work done by the water current appears as

heat energy, that is , the potential energy lost by the water in thislength Of pipe is converted into heat energy . When the water in

going from 1 to 2 passes through a water-motor, part of the potential energy lost by the water is converted into mechanical energyby the m otor . Hence , when the same quantity of water persecond as before is forced through the pipe, the water between thepoints 1 and 2 will lose potential energy at a greater rate , andconsequently the drop of pressure between these points w ill begreater than before . Again , the water in going through a pumphas work done on it

,and consequently the water in passing

through the pump has its potential energy increased . Hencethe drop in pressure through the pump in the direction of theflow ofwater is negative, that is , there is a ri se of pressure throughthe pump in the direction of the current . In general , wheneverthe water does work

,or loses potential energy , there is a drop of

pressure in the direction Of the current,and whenever work is

done on the water,or the water gains potential energy, there is a

rise of pressure in the direction of the current .In any given system of pipes connecting any number of pumps

and motors,in which the pipes

,pumps and motors are all com

pletely filled with water,the rate at which work is done by the

water in driving the m otors, in producing heat energy in the pipesand the water passages of the pumps and m otors, and in accelerating itself , is exactly equal to the rate at which work is done on thewater by the pumps . That is

,the total am ount of potential

energy of the water rem ains unaltered ; the gain in potential energyby any particle ofwater as it is forced through a pump is exactlyequal to the simultaneous loss of potential energy by som e otherparticle of water somewhere else in the system . The water currentthen acts simply as a m eans for the transfer of energy ; the work

CONTINUOUS ELECTRIC CURRENTS I

done onthe water at each ins tant is exactly equal to the workdone by it at that instant .Since the pressure in a current of water has but a s ingle value

at each point in this current , it follows that the total drop ofpressure around any closed path formed by any number Of suchpipes

,pumps and motors must be zero . A lso

,s ince the water is

incompress ible (at least , practically so) the quantity or volumeof water flowing up to any j unction of two or more pipes in anyinterval Of time must be exactly equal to the quantity of waterwhich flows away from this j unction in this sam e interval of time .

Or, cal ling the quantity Of water which flows away from any givenpoint equivalent to an equal negative quantity flowing to thatpoint

,an equivalent statement of this same fact is that the a lge

braie sum of the strengths of the water currents flowing up to

any junction of the pipes is always zero .

Similarly, it is found by experim ent that whenever there occurany of the phenomena which are attributed to the flow of anelectric current

,there is always a loss Of energy by one or more

parts of the c ircuit or by some body in the vicinity of the c ircuit,

and a gain of energy by other parts of the c ircuit or by bodies inthe vic inity of the c ircuit . For example

,in the simple c ircuit

formed by a wire connecting the poles of a battery, the batteryloses chem ical energy, and the wire and the conductors form ingthe battery become heated

,or gain heat energy . The energy lost

by any part of the c ircuit or by any body in its vic inity is said toresult in a gain of an equal amount Of electric energy by the electriccurrent

,and the energy gained by any part Of the c ircuit or by

any body in its vicinity is said to be due to the loss of an equalamount of electric energy by the electric current . That is

,the

electri c energy gained by the current in any portion of the c ircuit isdefined as the amount Of energy lost by this part of the c ircuit orby any other bodies as the result of the existence Of the current inthis part of the c ircuit ; and the e lectri c energy lost by the current

in any portion of the circuit is defined as the amount of energy

gained by this portion of the Ci rcuit or by any other bodies as theresult Of the ex istence of the current in this portion of the c ircuit . *

*Electric energy as thus defined is to be distinguished from electrostaticenergy ; electric energy is the work done on or by electric currents

,While

electrostatic energy is energy possessed by electric charges . Electrostaticenergy is discussed in detail in Chapter V .


Experiment shows that the net electric energy , as thus defined ,gained by the electric current in any c losed electric c ircuit in anyinterval of tim e Is zero. (This is also true for any stream line ofelectric current

,whether the current be continuous or variable . )

That is,any electric energy gained by the current in any part of its

path,e .g.

,the energy lost by a battery or other source of electromo

tive force,is lost by the current in some other part of its

_path ,

e .g .

,as heat energy , m echanical energy ,

chem ical energy , or in thecase of a variable current , as m agnetic and electrostatic energy .

The electric energy of an electric current is entirely analogous tothe potential energy of a current Of water in a pipe or system ofpipes completely filled with water . The potential energy gainedby the water in any part of its path , e .g .

,from a pump

,is all given

out by the water in some other form of energy , e .g.

,the heat energy

developed in the pipe and the work done on any hydraulic motorthat may be connected in the pipe . An electric current may thenbe cons idered as a . m eans for the transfer of energy, j ust as astream of water completely filling a c losed pipe or system of pipesis a m eans for the transfer Of energy . The net amount of potentialenergy in the water current remains constant ; sim ilarly the netamount of electric energy in the electric current rem ains constant .

We have seen how the rate at which potential energy is lost or

gained by a current of water flowing through a given length ofpipe the power developed by this current) m ay be expressedin terms of the strength of the current and the drop of pressurealong the given length of pipe . We are therefore led to theconception of e lectri c pressure, or , as it is also called

,electri c

potentia l , as a property of an electric current analogous to hydraulicpressure . AS in the case Of the flow ofwater

,we are concerned only

with the difference of pressure and not with the absolute pressure Ofthe water

,so in discussing the flow of electric currents we need

concern ourselves only with the difference of electric pressureor difference of electric potential . We may then define thedrop ofe lectri c pressure, or the drop of e lectri c potentia l , from anypoint 1 to any other point 2 along a wire carrying an electric cur

rent as the ratio of the rate at which electric energy is lost by thecurrent between these two points to the value of the current from1 to 2 . That is

,calling P the rate at which electric energy is lost

by the current between the points 1 and 2 the power developedby this current) and I the value of the current from 1 to 2

,the


88 . Measurement ofDrop of E lectric Potential . ExperimentShows that the drop Of electric potential as defined by equation

(14) is strictly analogous to the hydraulic pressure in a systemof pipes

,pumps and motors completely filled with water . In

particular,it is found that the drop of electric potential around

any c losed c ircuit is always zero , whether this c ircuit be a s implec ircuit like that formed by a battery with its poles connected bya wire

,or whether this c ircuit be part of any network of c ircuits

,

no matter how complicated the network may be . Hence,when

a wire is connected between any two points 1 and 2 Of anyc ircuit whatever, the drop of potential from the poin t 1 to thepoin t 2 through the wire will be exactly equal to the drop Of

potential from the point 1 to the point 2 through the conductorsform ing the original c ircuit . In general, when a wire is con

nected to an electric c ircuit,the current established in the wire

will cause a change in the strength of the current in the originalc ircuit

,which will in turn cause a change in the drop of potential

between the two pbints . Hence,in general , the drop of potential

in a wire when it is connected to any two points of a c ircuitis not the drop which origina l ly ex isted between these two points .

However,by making the resistance Of the wire suffic iently great ,

the current which is established in it can be made negligiblysmall , and the change produced in the current in the originalc ircuit by the presence of the wire may be neglected . Again ,(see A rticle 94) when any two dissim ilar substances are placedin contact a small difference Of potential is produced betweenthem . Hence if the wire is m ade Of a different material fromthat of the conductors form ing the c ircuit, the difference of potential between the two ends of the wire will not necessarilybe equal to the difference of potential between the two points of

the conductor in contact with which the two ends of the wireare placed . This difference

,however

,is inapprec iable in ordinary

practical measurements .

The fact that the drop of electric potential between any twopoints is the sam e over any path connecting these points, leads toa S imple method of actually measuring the potential drop betweenany two points of an electric c i rcuit in which is established acontinuous current . For

,experim ent shows that when a continu

ous current is established in a wire which is of uniform structurekept at uniform temperature in an unvarying magnetic field

,then


the only form of energy produced in or around the wire is heatenergy . Hence from the definition of electric energy given above ,the total loss of electric energy by the current in this wire is equalto the heat energy which is produced in the wire . But by Joule ’sLaw (Article the rate Ph at which heat energy is developedbetween any two points 1 and 2 of such a wire under these conditions

,is equal to RP

,where R is the res istance of the wire be

tween the points 1 and 2,and I is the current in the wire . Hence

the drop of potential from the point 1 to the point 2 is,from the

definition given above (equation1 2

VI

f, RI . (15)

That is,the drop of potential between any two points 1 and 2

in the direction of the current in a wire of uniform structure andof uniform temperature throughout , when this wire is kept in anunvarying m agnetic field

,is equal to the product of the resistance

of the wire between these two points by the strength of the current in the wire ; this product RI is frequently called the resistance drop between the two points . We have already seen howthe current strength and the resistance may be measured ; consequently the above relation gives a method for actually measuringthe drop of electric potential between two points of such a wire .

To measure the drop of potential between any two points1 and 2 of any c ircuit

,it is then only necessary to connect

to these points a wire Of known res istance and measure thestrength of the current established in it . Then

,

'

if the res istance ofthe wire is suffic iently high to make the current in the wire negl igibly -small

,the difference of potential between the two points

is equal to the product of the res istance Of the wire by the currentestablished in it . This is the princ iple Of the ordinary continuouscurrent voltmeter

,which is simply an ammeter with a high res ist

ance coil in series with it . The scale of the instrument is calibrated to read directly in volts instead of in amperes ; that is ,the scale reads directly the product Of the strength of the currentthrough the wire form ing the coils of the instrument and the highresistance coil in series with it and the value of the total resistanceOf this wire . For accurate measurements with such an instrument the effect of changes of temperature in changing the resistance of the coils has to be taken into account . Also , care must betaken not to place the instrument in a strong magnetic field, since


such a field may alter the magnetic field in which the coil of theinstrument is designed to m ove . The effect of using wires of

different m aterials in the various parts of the instrument, or for

the connections to the instrument,is usually inapprec iable in

ordinary practical measurements .

89 . Ohm ’s Law.

—Equation (15) may also be expressed inthe form

V

I

and in this form is known as Ohm’s Law

, from the name Of thescientist who discovered the relation expressed by this formula .

In words,Ohm ’

s Law is that,when a continuous current is estab

l ished in a wire of uniform structure throughout , kept at a con

stant temperature in an unvarying m agnetic field,then the rat io

of the potential drop between any two points along this w ire tothe strength of the current established in the w ire is constant

,

i ndependent of the strength of the current .

90 . E lectric Power and E lectr ic Energy. From the definitionof drop of electric potential given in Article 87, it follows that therate P at which electric energy is - lost by an electric current ofStrength I between any two points ll and 2 of an electric circuiti s equal to the drop In potential V from the point 1 to the point 2 '

multiplied by the value of the current from 1 to 2,that 1s

P = VI ( 17)

When an electric current loses electric energy it is said to devel opan amount of power equal to the rate at which it loses energy .

Hence the power developed by an electric current in any portionof .an electric circuit is equal to the product Of the strength of thecurrent in this portion of the c ircuit by the drop of potential inthis portion of the c ircuit .When the cu rrent I is measured in abamperes and the potential

drop V in abvolts,the product V I gives the power in ergs per

second . When I is measured in amperes and V involts the produot VI gives the power in watts . Large am ounts Of electric powerare usually expressed in kilowatts

,i .e .

,thousands of watts . See

Article 22 for the relation between the various units of power .Since both the current and the potential drop may be readily

measured,the amount Of power developed by a current can be

readily determ ined ; in fact , electric power can be measured with


m easure the power, and the other coil C,, called the currentcoil ,

” is connected in series with the c ircuit . The current coilis usually stationary and the voltage coil is movable . A highresistance R is connected in series with the voltage coil

,so that

only a very sm all current is established through this coil ; thiscurrent then depends only upon the difference of potential between the points 1 and 2 . The current in the current coil , whichis the same as the current in the line, sets up a magnetic fieldthe strength of which is proportional to the strength Of this cur

rent (See Article this magnetic field produces a moment or

Fig . 56 .

torque on the voltage coil,which torque is proportional to the

strength of this magnetic field and to the strength of the currentestablished in the voltage coil . Hence the torque produced onthe movable coil is proportional to the product Of the differenceof potential from 1 to 2 and the current in the c ircuit betweenthese two points . If the opposmg torque produced by the sus

pension of the movable c oil,or by a suitable Spring attached to it,

is proportional to the angular twist given the coil,its deflection

will then be proportional to the power transferred to the c ircuitbetween these two points . A suitable pointer attached to themovable coil and arranged to move over a suitably graduateds cale will then indicate directly the value of the power .

92 . E lectromotive Force . The heat energy developed in anelectric circuit due to the resistance of the conductors form ing thecircuit

,that is

,the heat energy expressed quantitatively by the

formula Wh RPt, is only one of a'

number Of formsofenergy intowhich electric energy is converted . For example

,electric energy

may m anifest itself by producing chem ical changes, that is, bybeing converted into chem ical energy ; or by producing a mag


netic field , which in turn can produce m echanical motion,that is,

electric energy m ay be converted into magnetic energy which inturn is converted into mechanical energy, etc . Again, whenever anelectric current is established som e body or bodies lose some formof energy, whi ch , in accordance with the definition of electricenergy, is converted into electric energy . For example

,when a

wi re is connected to the two poles of a battery , chem ical changestake place in the battery which result in a loss Of chem i cal energyby the battery ; that is, the chem ical energy Of the battery is con

verted into electric energy . Again, it is found by experimentthat work is required to change the number Of lines of magneticinduction linking any part Of an electric ci rcuit ; the work which .

is thus done is , by the definition of electric energy, converted intoelectric energy .

In any portion Of a c ircuit in which work is done on an electriccurrent or in which the current does work, other than that . doneas a consequence Of the resistance of this portion of the c i rcuit

,

there is said to exist an electromotive force. An electromotive forcemay then be looked upon as that which produces or opposes theflow of electric ity, other than the opposition due to the resistanceOf the conductor in which it flows . As the measure of the electromotive force in any portion Of a c ircuit is taken the ri se Of potential which it would produce in the direction Of the current in thisportion Of the c ircuit were there no resistance drop in this portionof the c ircuit . As noted in Article 88

,the resistance drop is

always in the direction of the current ; hence in any port ion of ac ircuit in which there is a resistance drop the electromotive forc eis equal to the resu ltant ri se of potential in the direction Of thecurrent in this portion ofthe c ircuit plus a lgebrai ca l ly the resistancedrop in this portion of the

'

circuit . An electromotive force is thencons idered posi tive with respect to the current when it produces arise of potential in the same direction as that of the current ;negative when it produces a ri se of potential in the opposi te direc

tion to that of the current . An electromotive force in the oppositedirection to that of the current is frequently called a back or

counter electromotive force . From the definition of potential drop

(Article 87) it follows that wherever the current and the electromotive force are in the same direction there is a gain of electricenergy by the current , or the current has work done on it ; wherever there is a back electromotive force the current loses electric

l 50 ELECTRICAL ENGINEERING

energy , or does work . E lectromotive force is frequently abbre

viated e. m . f.An electromotive force is analogous to the pressure developed

by a pump , while a back electromotive force is analogous tothe back pressure due to hydrostatic head or to the backpressure due to an hydraulic motor . The resistance drop in aconductor is analogous to the drop of pressure , or

“ loss of head,

”

due to the friction of a pipe . When the term inals of any devicewhich develops an electromotive force are connected by a con:

ductor,the electromotive force may be looked upon as the cause

of the flow Of the electric current,j ust as the pressure developed

by a pump may be looked upon as the cause of the flow Of thewater current through a pipe connecting its outlet and intake .

Just as the pressure developed by the pump is equal to the dropin pressure in the pipe and the water passages of

"

the pump due totheir frictional resistance (provided there is no acceleration of thewater or other source ofback pressure in the pipe) , so is the electromotive force developed by any device equal to the drop of electricpressure

,or potential drop

,due to the electric resistance of the

conductors form ing the electric c ircuit (provided the electric cur

rent is continuous,i .e.

,does not vary with time

,and there is no

back electromotive force in the c ircuit) . Again , when the outletand intake of a pump are connected respectively to the intakeand outlet of an hydraulic motor

,the pressure developed by the

pump is no longer equal to the drop in pressure due to the frictional resistance of the pipe and water passages of the pump andmotor

, but is equal to this drop p lus the back pressure due to themotor ; S imilarly , when any device which develops an electromotiveforce is connected by conductors to another device which developsa back electromotive an electric motor) the electromotiveforce of the first device is not equal to the resistance drop in thiscircuit , but is equal to this drop p lus the back electromotive forcedeveloped by the second device .

Since an electromotive force is measured by the difference ofpotential it produces

,the units of electromotive force are the same

as those of potential difference,namely

,the abvolt

,volt

,and the

c . g. s. electrostatic unit (see Article93 . General ized Ohm ’

s Law.— The relation between current

strength , electromotive force , potential drop , and resistance in anyportion of an electric c ircuit in which a continuous electric current


in the c ircuit,i .e .

,when E = 0 then I2 1 E_ 1

. Equation ( 18b)R R

is an ex tremely useful one , since it gives a simple and entirely generalformula for calculating the current in any portion of a c ircuit whenthe electromotive force developed in thi s portion of the c ircuit

,the

impressed or term inal electromotive force , and the res istance ofthis portion of the c ircuit are known . In applying this formulacare must be taken in not confusing the electromotive force developed in the given portion of the c ircuit with the electromotiveforce impressed across its term inals .

When the generated electromotive force E and the current Iare in the same direction , i .e .

,in an electric generator or in any

other device in which electric energy is generated, equation ( 18b)may be written

E’E RI ( 180)

In this case the term inal electromotive E ’ force is always less thanthe generated electromotive force E . When the generated electromotive force E and the current I are in opposite directions

,i .e.

,

in an electric motor or in any other device which absorbs electricenergy, equation ( 18b) may be written

E’ =E RI ( l 8d)

In this cas e the impressed electromotive force E ’is greater than

the back electromotive force E .

For example,a continuous current dynamo may be used

either as a generator or a motor . If the electromotive force

generated by the machine is 1 10 volts in each case,and the res ist

ance of its armature is 1 Ohm ,then when a current Of 10 amperes

is supplied by the dynamo running as a generator the term inalelectromotive force is 1 10—( 1 X10) = 100 volts ; while if the dynamo is running as a motor the impressed electromotive forcemust be X 10) = 120 volts .

In employing the spec ial forms ( 18c) and ( 18d) of the generalequation (18b) , the relative directions Of the various quantitiesmust not be lost sight of. In the generator equation ( 18c) the

generated electromotive force E ,the terminal electromotive force

E’ and the current I are all in the same direction . In the motorequation ( l 8d) the generated electromotive force E and the cur

rent I are in Opposite directions ; the impressed electromotiveforce E ’

,considered as localized in the rest Of the c ircuit connecting

the term inals of the portion of the circuit under cons ideration,acts


around the closed c ircuit of. which the given portion is a part inthe same direction as the current

,and therefore in the opposite

direction arOund the c ircuit to the back electromotive force E .

The student should bear in m ind that the various formulas

given in this artic le are all simply different ways Of expressing asingle relation . This relation is that the current in any portion ofa c ircuit formed by one or more conductors In seri es 1s equal to thea lgebra i c difference between the generated and impressed electromotive forces divided by the resistance Of the given portion Of thec ircuit .

94 . Contact E lectromotive Force . It is found by experimentthat whenever an electric current is established in two or moreconductors connected in series

,the current always gains or loses

electric energy at each point of contact between di ssimi lar con

ductors . For any given point of contact , it is found that whetherthere is a gain or loss of energy depends upon the direction of

the current with respect to the two conductors . For example,

when a copper wire is connected to an iron wire and a currentis established across the junction of these two conductors

,the

wires in the vic inity of the j unction become cooled when the current is in the direction from the copper to the iron

,while if the

current is in the oppos ite direction , from the iron to the copper,

the w ires in the vic inity of the junction become heated . Theseeffects as a rule are

'

scarcely apprec iable , but when special precautions are taken they can readily be detected . In the firstcase

,then

,the current gains electric energy at the j unction , and

in the second case it loses electric energy at the junction .

In general , then , at the j unction between any two diss im ilarconductors there is an e lectromotive force

,and the direction of

this electromotive force is independent of the direction of thecurrent . Experiment also shows that the value of this electro

motive force of contact,as it is called , does not depend upon the

strength Of the current or upon the area or shape Of the surfaceof contact between the two conductors , but depends only uponthe nature of the two conductors in contact and upon the tem

perature of the j unction . The value of this contact electromotiveforce between m etallic conductors is quite small

,only a small

fraction of a volt between copper and zinc at 25 degreescentigrade it is volt) , and in practical work it is thereforeusually negligible . However, in case Of a metal conductor in


contact with an electrolyte the contact electrom otive force may

be several volts . In this case , only a very sm all part Of theelectric energy absorbed by or given out by the current manifestsitself as a loss or gain of heat energy at the j unction of the twoconductors , but the gain or loss of electric energy by the currentappears as a loss or gain of chemi ca l energy at the j unction of

the two conductors . That is,the transfer of energy involved

in the chem ical changes which take place at the junction is manytimes greater than the transfer of heat energy from or to thejunction between dissim ilar meta ls . AS in the case of t O m etalsin contact , however , this contact electromotive force betweenconductors and electrolytes is independent of the strength or thedirection of the current through the j unct ion and is also independent of the area and the Shape of the surface Of contact , butdepends only upon the

’nature of the conductors in contact and

the temperature Of the conductors at the j unction .

It is found by experiment that whenever any number of

conductors are connected in series , and there are no electromotiveforces in this chain of conductors other than the electromotiveforces of contact at their j unctions

,then the net electromotive

force between the ends of this series of conductors is the sameas if the two end conductors were connected directly to eachother

, provided a l l the conductors are kept at the same uniformtemperature and the chemi ca l action at the cathode in each electrolyte

in the seri es i s just the reverse of the chemi ca l action at the anode

in this electrolyte. This fact is sometimes called the LawofSuccessive Contacts . For example

,if a copper wire is connected to

an iron wire which in turn is connected to an alum inum w ire,

the net electromotive force between the free ends of the copperand the alum inum wire is the sam e as if the copper wire wereconnected directly to the alum inum wire . Sim ilarly, when twocopper wires are soldered together, the net effect of the solder isnil

,whether the two Copper wires are in actual contact or are

j oined only through the solder . (If the flux used in soldering actschem ically on the wires

,an electromotive force may be produced

at the j unction ; this electromotive force , although sm all,m ay be

sufficiently large to cause‘

considerable trouble when delicate measurements are to be made .) Again , the net electromotive force ofa S ilver voltameter is zero

,S ince at the anode s ilver goes into solu

tion as S ilver nitrate and at the cathode silver is deposited from


and the res istance of the conductors form ing the battery Ohms,

the current in the c ircuit will be—1

amperes . The res istance

of the conductors form ing the battery is usually called the interna lresistance of the battery

,and is never negligible unless the wire

form ing the external circuit has a very large resistance . It

Should also be noted that in the example j ust c ited there is alsoa contact electromotive force between the z inc pole and the copperwire form ing the external c ircuit

, but since this electromotiveforce is but a very small fraction of a volt it is usually negligiblein practical work . For a description of the construction of thecommon forms Of chem ical batteries used in practice

,see any elec

trical engineer’s handbook .

It is found by experiment that the net electromotive forceOf any chem ical battery or cell is constant, provided the chem icalnature Of the electrodes and the electrolyte in contact with themdoes not change . One electrode may waste away and the otherincrease in mass

,due to the chem ical action which takes place

at them, but as long as their chem ical composition and that of

the electrolyte in contact with them does not change and thetemperature remains constant , the electromotive force of thebattery remains constant . In general , however , when an electriccurrent is established through a battery , the electrodes not onlychange in m ass , but som e of the products of the chem ical actionswhich take place collect at the electrodes , and thus change thenature of the substances in contact

,and consequently the

electromotive force of the battery changes . For example,when

a current is established through the copper-sulphuric -ac id-z incbattery

,z inc sulphate is formed at the z inc electrode , and

bubbles of hydrogen gas collect at the copper electrode andthe electromotive force Of the battery falls Off. The battery insuch a case is said to becom e polari sed, and the decrease of its

electromotive force is said to arise from a back e lectromotive forceof polari sation . There are various m ethods for preventing thepolarisation of a battery ; see any electri cal engineer

’s handbook .

The polarisation Of a dry battery is particularly noticeab le , sincethe electrolyte in such a battery

,instead of filling the entire

space between the poles Of the battery , S imply impregnates apractically solid mass between these poles . Consequently , theproducts Of the chem ical actions at the poles of the battery cannot


diffuse rapidly through the battery but collect at the poles wherethey are formed . However

,when such a battery is left open

circuited for a time after it is used,these products gradually dif

fuse through the battery, and its electromotive force graduallyreturns to practically its original value, provided none of theactive materials has been completely destroyed .

96 . Definition of the International V olt. Many of theordinary forms of batteries undergo chem ical changes in theirvarious parts even when left open- c ircuited . This is due chieflyto impurities in the chem icals of which they are made . It is

possible,however

,to cons truct a battery which will rem ain

practically unaltered for several years,provided no current is

taken from it . Hence such a battery,or cell

,makes a very

convenient standard of electromotive force or potential difference .

One of the m ost satisfactory cells of this kind is that known as

the Clark cel l,and in terms of this cell the International Congress

ofE lectric ians defined the volt as follows :As a unit of electromotive force (Shall be taken) the Inter

national V olt,which is the E .M .F . that steadily applied to a

conductor whose resistance is one international ohm,will produce

a current of one international ampere,and which is represented

suffi c iently well for practical use by 11433 of the E .M .F . between

the poles or electrodes of the voltaic cell known as Clark’s cell,

at a temperature of 15° C.,and prepared in the manner described

in the accompanying spec ificationThe spec ification referred to is to be found on page 10 Of

Foster's E lectri ca l Engineer

’s Pocket Book . This definition has

been legal ized by most c ivilized c ountries .

In the practical use of a Clark cell as a standard ofComparisonof potential differences

,an arrangem ent is used which obviates

the necess ity of taking any current from the cell . The princ ipleof this method is to balance the electrom otive force Of the cellagainst an equal res istance drop . See Article 100.

97 . Thermal E lectromotive Forces .

—As noted in Article 94the Law ofSuccess ive Contacts does not hold in case the temperature of the chain of conductors form ing the electric c ircuit isnot the same for all these conductors . For example

,when an

iron and a copper w ire are connected to each other at their twoends in such a manner that they form a closed loop

,the electro

motive force,and therefore the current

,in this loop w ill not be

,158 ELECTRICAL ENGINEERING

zero if the two j unctions between the iron and the copper arekep t at different temperatures , or even if there is a differenceof temperature between any two points of the same wire . Thesetherma l e lectromotive forces

,as they are call ed

,are however o nly

a small fraction of a volt even for a considerable difference of

temperature,and consequently in any c ircuit in which there

are other electromotive forces of the order Of a volt or more,

they may be neglected . One important practical app licationof these thermal electromotive forces , however, is in the thermo

electric couple Or electric pyrometer for m easuring high temperatures . The ordinary form of electric pyrometer cons ists essential ly of a platinum and a platinum - iridium wire fused togetherat one end and connected in series with a m ill ivoltm eter a

.

voltmeter des igned to measure differences Of potential of theorder of a thousandth of a volt) . The j unction between the twow ires

,suitably protected by a porcelain or quartz tube

,is placed

in the furnace the temperature of which it is desired to measure .

The difference of potential indicated by the voltm eter is thenpractically proportional to the difference between the temperatureof the hot j unction in the furnace and the temperature Of theends of the two wires where they are connected to the voltmeter .

98 . Kirchhoff’s Laws .

—We have already had occas ion,in

several instances,to make use of the two experimental facts that

1 . The algebraic sum of the currents com ing up to any junctionin a network of conductors is always zero

,and

2 . The algebraic sum of the potential drops around anyc losed loop in a network of conductors is always zero .

These two experim ental facts are known as Ki rchhofi’s Laws

,

from the nam e of the sc ientist who first c learly enunc iated them .

By making use of these facts one can always predeterm in e (1)the currents in each branch of a network when the res istanceof each branch and the electromotive force in each branch areknown

,or (2) the res istance of each branch of a network when

the current in each branch“and the electromotive force in each

branch are known,or (3) the electromotive force in each branch

when the current in each branch and the resistance of each branchare known . These two laws are therefore of fundamental importance .

It should be carefully borne in m ind in applying these lawsthat a current leaving any point is equivalent to an equal negative


a . Resistances and Electromotive Forces in Series. Con

sider any number of conductors in series (Fig. 58) and let a drop ofpotential V I 2 be established between the two ends Of this series ofconductors . Let the current in these conductors be in the direction from 1 to 2W eb

-

_M fll be the same in eachconductor

,s ince they are in seri es . Let R

’

,R

”

,etc . , be the

res istances of the various conductors and E ’

1 2,E ”

etc .,

the electromotive forces in this portion of the circuit between thepoints 1 and 2 in the direction from 1 to 2 . Then the potentialdrop from 1 to 2 is also E

’

+ R” I E ”

+ R’” I , , E

’”

etc . Hence

V etc .) I _ (El+ E

/l

(20)

Therefore the res istances between the points 1 and 2 a re equivalentto a s ingle resistance

RE E i fi U etc . (20a)

and the electromotive forces between the points 1 and 2 areequivalent to a S ingle electromotive force

E =E’

+ E”

etc . (205)

When the equivalent electromotive force E 1 2 is positive and greaterthan the product of the current I , , and the equivalent res istance R,

k —VL — 9 |12

Fig . 58 .

the drop of potential from 1 to 2 will be negative, that is , there willbe an actual rise of potential from 1 to 2 ; this corresponds to thecondition when the part of the c ircuit from 1 to 2 is supplyingenergy to the rest of the c ircuit which completes the closed loopfrom 2 to 1 . When E 1 2 is negative or is less than the productR I the potential drop from 1 to 2 is positive

,which represents

a transfer of energy to this part of the c ircuit from the portionof the c ircuit closing the loop from 2 to l .

b . Resistances and E lectromotive Forces in Parallel . Whena number of res istances R’

,R

”

,etc .

,are connec ted in

‘ paral lel

between two points 1 and 2,and there are electromotive forces


E” E

’”etc .

, in these respective branches, the d rop of

potential V 12 from the point 1 to the point 2 must be the sam efor each branch, and the totalcurrent entering

‘ the point 1must be equal to the totalcurrent leav ing this point .Hence, call ing I the totalcurrent entering the point 1in the direction from 1 to 2

,

and I ’ I” etc .

,the

currents in the respective Fig . 59

branches in the direction from 1 to 2,we have

+ 1”

+ etc .

112

R!I ! —

.E

’ =Rl l I ” E ” =B

,” Il l/

1 2__ Em =etc . (2 1a)

from which relations the currents in the individual branches may

be calculated when the res istances and the electromotive forcesin these branches are known . Again, if the drop of potentialV I 2 from 11 19 2 is given instead Of the’

tg

otal current we havethe first equation the relationsR

I11

1 2—

1El =R

” 1” El l =R

/l l I !” El l i

(Z l b)whichare also suffic ient for the calculation of the currents in the individual branches . In casethere are noelectromOtive forces in thebranches between the points 1 and 2

,we have from equation (2 1b) ,

V

Iwhere I 1 2 is the total current from 1 to 2 , that

R I Z R, I ! =R

I/I”

RI ” I l l,

putting R

whenceI ’ I I I ] , I I

R R”R R

”’

R R” ”

therefore , adding these equations and substituting for 1 1 , its valuefrom equation gives

1 1+1+1+ etc

R R’R R

’”

That is , the equiva lent resi stance R of any number of branch c ir

cuits in parallel,in which there are no e lectromotive forces

,is the

rec iprocal of the sum of the rec iprocals of the resistances of the

individual branches ; or, the equivalent conductance é Of any

number of branch c ircuits in parallel,in whi ch there are no electro


motive forces,is equal to the sum of the conductances of the In

dividual branches . It a lso follows from equation (21b) that whenthere are any number of branch c ircuits connecting any two points ,and there are no e lectromotive forces in these branches

,the total cur

rent divides among the various branches ih such a manner that theratio of the current in any branch to the current in any otherbranch is equal to the inverse ratio Of the res istances of these twobranches . It shoul d be carefully noted that none of these relationsare true when there are electrom otive forces in any of the branches.In the Spec ial case of two resistances in parallel , but no e. m . f .in ei ther branch

,equation (2 1c) becomes

RIX R”

‘

m99 . The Wheatstone Bridge . A spec ial arrangement of

electric c ircuits which is extensively used in the comparison of

resistances is that known as the Wheatstone Bridge, and is showndiagrammatically in Fig . 60 . B is a battery of any kind

,G a

galvanometer, and R,, R2,R3,and R4 are the res istances of the

branches between the points 1 and 2,2 and 3

,3 and 4

,and 4 and 1

respectively . The currents in each branch of this network can becalculated for any values of these resistances when the electromo

tive force of the battery,and the resistances of the battery and

the galvanom eter ( including the connecting leads) are known .

However,there is a simple relation among the four resistances

R R R3 ,and R4 for which the current in the galvanometer

c ircuit will be zero,independent of the electrom otive force of the

battery and the resistances of the battery and galvanom eter cir

cuits . The condition that there be no current in the galvanom eteris that there be no difference of potential across its term inals .


potential differences . A s imple arrangement of this kind isshown in Fig . 61 . B is a battery or other source Of electro

Fig . 61 .

motive force which is connected to the two ends of a wire of

uniform cross section . To the end 1 of this wire is also connecteda standard cell C

,say a Clark cell

,in series with a galvanometer G

and a high res istance to prevent a large current from flowingthrough the cell while the adjustm ents are being made . Theother end of this c ircuit is an adjustable contact 3 which can bemoved along the wire . The like poles of the two batteries mustbe connected to the same end of the wire

,and the electromotive

force of the battery B must be greater than the electromotiveforce of the standard cel l . Let the contact 3 be moved alongthe wire until the galvanometer shows no deflection (in the finaladj ustment the high resistance is to be short- c ircuited) . Then

,

applying Kirchhoff’s second law to the loop l CG3 1

,we have

,

since there is no current in the galvanometer, that

E =R I

where E is the electromotive force of the standard cell,R,3

theres istance of the wire from 1 to 3

,and I the current in the wire .

E

l

where l ,3 is the length of the wire between 1 and 3 , and the potential

drop across any length of the wire,

say, will then bel— E .

l

Hence,if the standard cell is replaced by any other battery

,the

electromotive force E ’

of which is to be measured,and .v is

the position of the sliding Contact at which the galvanometershows no deflection

,the electromotive force Of this battery will be

The potential drop per unit length of the wire will then be


EE

,provided the current from the battery B

,and therefore

l

the current in the wire 12, does not change .

This potentiometer m ethod of comparing electromotive forcesis extensively employed in calibrating electrical measuring instruments . In practice

,the wire 12 is usually replaced in

whole or in part by a set of res istance coils , the resistances of

which are accurately known .

10 1 . Stream Lines Of Electric Current . So far we have con

fined our attention to conductors which are in the form of wiresor long rods or bars , which we have assumed may be cons ideredas geometrical lines . Certain problems arise in practice

,however

,

when this method is not perm issible ; for example , the calculationOf the ins ulation resistance of a cable , the calculation of themagnetic field intensity within the substance of a wire

,etc . In

such cases it is necessary to look upon the conductor carryingthe current as made up of

“current filaments

,

” j ust as a mag

netised body is conceived to be made up of magnetic filaments .

Experim ent shows that when an insulating gap is cut in aconductor in which an electric current is established

,this gap in

general produces a change in the amount of the effects producedin and around the conductor

,e .g.

,a change in the force produced

on the conductor by a magnetic field,a change in the amount of

heating produced , etc . The amount of these changes produceddepends upon the direction in which the gap is cut . It is possibleto cut a very narrow gap in the conductor in such a direction thatno change whatever is produced in any of these effects , j ust asit is possible to cut a very narrow gap in a magnet without producing any new poles (see Article A conductor of any shapewhatever in which a current is established m ay then be dividedinto filaments separated from each other by insulating walls of

infinitesim al thickness without producing any change in the effectsproduced in and around the conductor . These filaments may becons idered of very sm all cross section and m ay therefore betreated as geom etrical lines . They may be looked upon as thelines along which the electric current flows ; these filam ents arecalled the stream l ines of the electric current . Such stream linesare analogous to hydraulic st ream lines .

Since each of these current filaments m ay be insulated fromthe rest of the conductor without altering the phenom ena prO‘


duced by the electric current,each filament may be treated as an

insulated wire . The definitions of the strength and the directionof an electric current (Articles 66 and 67) are then directly ap

plicable . The direction of the current at any point of a conductoris then the direction of the current fil ament at that point ; thepositive sense of the filament is taken arbitrarily as the direction of the current . The strength of the current in any filamentis the same at every cross section of the filament . Let di be thecurrent in any filam ent and let the cross section of this filamentat any point P be ds then the current densi ty at P is

di

ds,,

The total strength of the current across any surface of area S isthen

(0'

cos a ) ds (23a)

where ds is any element of this surface,and a is the angle between

the direction of the current at ds and the normal to ds . Comparewith equations of Chapter II .

The results of all known experiments Show that the streamlines of an electric current must be considered as c losed loopswithout ends ; in this respect these stream lines are analogousto lines of induction . In the case Of a continuous current

,i .e .

,

a current which does not vary with tim e, these stream lines areconfined entirely to conductors . The stream lines of a variablecurrent

,however

,may be partly or entirely in a dielectric

,even

though this dielectric be a perfect insulator ; the stream lines ina dielectric represent the displacement current (see Chapter V ) .

It should be noted that the direction of the stream lines inany body depends upon the position of the points of connectionof the body to the rest of the c ircuit . These stream lines alwaysrun through the body in the general direction of the line connectingthe two points Of contact or term inals . The stream lines willnot in general be parallel to one another, particularly when thedistance between the term inals is large compared with the lineardimensions Of the cross section of the body perpendicular tothe direction of these lines . However

,as has already been

noted,in the case of a long wire, a long rod or strip , or a long

section of a rail , these lines , excep t for a negligible distance in


between current density and electric intensity is of exactly thesam e form as the relation between magnetic field intensity andflux density (see equation 20 ofChapter II) the reciprocal of theelectric resistance

,i .e.

,the specific conductivity , is exactly anal

ogous to m agnetic perm eability .

The c . g. s . electromagnetic unit of electric intens ity is theabvolt per centimeter ; the practical unit is the volt per centimeteror the volt per inch . E lectric intensity is also som etimes expressedas so many 0 . g. s . electrostatic units per centimeter length .

These units are related to one another as follows1 abvolt per centimeter = 10

' 8 volts per centim eterx volts per inch

1 c . g. s . electrostaticunit per centimeter = 300 volts per centimeter

The resistance drop in an elementary length dl offilament may then be written H

edt and therefore

resistance drop'

between any two points 1 and 2 on

current filament is

V,

Hed l (25)

and the total resis tance drop between any two points 1 and 2 ,whether on the same or different current filaments

,is

( Hecos 9) dl (25a)

where Heis the electric intensity at the elementary length d l

of the path between 1 and 2 and 9 is the angle between thedirection of the electric intensity at dl and the direction of dl .

Compare with equation (25) of Chapter II .

When there is no contact or induced electromotive force inthe path from 1 to 2

,this resistance drop is equal to the total

potential drop along this path ; when there is a contact or inducedelectromotive force e, 2 in the path from 1 to 2 the total drop of

potential from 1 to 2 is

v : ( H,cos 0) dl —e ,, (256)

Since the total electric potential drop around any c losed path iszero

,this reduces to

( H cos 0) dz= 2 6 (25s)

for a closed path ; that is, the total resistance drop around any


c losed path is equal to the total electromotive force in this path,

which is simply another way of stating Kirchhoff’s second law .

Any surface all points Of which are at the same electric potential is called an e lectri c equipotentia l surface. Such a surface isperpendicular at each point to the current filament through thatpoint and is therefore perpendicular to the electric intens ity atthat point . Compare with magnetic equipotential surfaces .

103 . Insulation Resistance of a Single Conductor Cable .

As an example of the use of theabove conceptions

,take the prob

lem of calculating the insulationresistance of a single conductorcable in a lead sheath . Fig. 62

represents the cross section of sucha cable . When all points of thewire are at the same potential andall p oints of the sheath are at thesame potential

,the surface of the

wire and the ins ide surface of thesheath are equipotential surfaces .

Hence the stream lines of the current through the insulation(which , it is to be remembered , is never a perfect insulator butonly an extremely poor conductor) must, from symmetry

,be radial

lines . Let

l = length Of‘ cable in centimetersr ,= radius ofwire in centimeters

r = internal radius of sheath in centimetersI = total current in amperes through the insulation , i .e.

,I

is the leakage current , not the main current through thewire

V =difference of potential in abvolts between the wire andthe sheath

,assumed constant .

p= the spec ific resistance of the insulation in ohms per

centimeter- cube .

Then the current dens ity at any point in the insulation at a distance a; from the center of the wire is

I

2 n x l

whence the electric intensity at this point is


and therefore the difference of potential between the wire and thesheath is *

r

d P Iz df P I m“r, a: 2 7r l r,

The insulation resistance of the given length of cable is thenV P ln

73 (26)

I 2 7T l r ,

Note that the insulation resistance varies inversely as thelength ; this is evidently true when it is remembered that eachelementary length of the insulation of the cable (measured alongthe axis of the wire) Is In parallel with all the other elementarylengths .

Compare equation (26) with the formula , equation (20) ofChapter V

,for the electrostatic capacity of a sheathed cable .

It can be shown that in every case the formula for the insulation conductance the rec iprocal of the insulation resistance)between any two conductors is identical with the formula for the

4electrostat i c capacrty of these two conductors when

77is sub

Pstituted for the dielectric constant K in the formula for capac ity .

See Article 152 for various capac ity formulas .

104 . Field Intensity at any Point Due to a Current in a W ireOfCircular Cross Section . Another application of the conceptionOf current filam ents is the proof of equations (4) and (4a) ofA rticle73

, for the case of a solid wire of c ircular cross section . A rigidproof of these equations requires the use of a geometrical theorem

Fig . 63 .

known as the Theorem of Inverse Points . This theorem may bestated thus : let P be any point either inside or outside a circle

*ln s tands for the na tu ra l logari thm .


in the circumference of the ci rcle representing the cross section of

the tube . When the current is uniform ly di stributed in thewalls of the tube

,the current in each filam ent is

d Itds I ds

2 7TatI

2 rr a

Let P be any point at a perpendi cular distance r from the axisof the tube

,and let a: be the perpendi cular distance of P from a

filament A . Then from equation (34a ) this filament produces afield intensity

dHI ds

a:

in the direction perpendicular to A P . (The direction of the current is assumed up toward the reader . ) This intensity may be

resolved into two components , one parallel to OP and the otherperpendicular to OP . A second fi lament ds ’ of equal cross sec

Fig . 65.

tion symmetri cally located at A ’on the other side of the line OP

wi ll produce an equal field intensity dH ’ at P in the,

'directionperpendicular to A ’ P . Its component parallel to OP will beequal and opposi te to the component Of dH in this direction .

Sim ilarly for any other pair of symm etrically located filam ents .

Hence the resultant field intensity at P due to the entire tube ISequal to the sum of the components perpendicular to OP of thefield intensities due to all the filam ents , and will be in the directionperpendicular to OP .

The component perpendicular to OP of the field intensity atP due to the filament ds IS

dHn

I ds . cos 01.

77 a 23

Let Q be the inverse point of P with respect to the circle representing the cross section of the tube

,and let

,8 represent the angle

P QA and d ,B the angle at Q subtended by the arc ds . Draw a

l ine AB through A perpendicular to QA ; this line will m ake theangle a with ds, Since the angle between OA and QA is a

,and 0 A


is perpendicular to ds and QA to AB . Hence the proj ection of

ds on this line AB is (ds cos Hence,since ds is i nfinitesim al

,

ds

Q

cos ad B : But since Q and P are inverse points QA a

.v r

whence QA Therefore dB :

ds cos a. a

xw _ —d fi

HenceI a I

and therefore the total field intensity at P when P is outside thetube is

77 7.

fiz z o

Therefore the field intensity at any point outside a tube of ci rcularcross section carrying a current I is the sam e as would be produced by a current of the same strength concentrated in a linecoinciding with the center of the tube . Since a solid wire of circular cross section m ay be considered as m ade up of a series ofconcentric tubes

,this form ula is likewise true for any point P

outside a so li d circu lar wire .

When the point P i s inside the tube the inverse point Q is outside . Hence in this case the two lim iting values of the angle ,

8

are identical and equal to r ; therefore the total field intensi ty atP is

That is , the field intensity at any point inside a ci rcular tube inwhich the current is uniform ly distributed is zero .

At any point P inside a sol id circular wire in which the currentis uniform ly distributed the field intensity is therefore due onlyto the current inside the cyl inder the cross section ofwhi ch i s thec ircle through P concentri c wi th the center ofthe wi re. Calling r thedistance of the point from the center of the wire , a the radius of

the wire and I the total current , the current in -this cylinder is then2 2

”

it? I 2—2 I , since-:r r

2 is the area Of the cross section of the cylinder77

and n a’Is the area of the cross section of the whole wire . Hence

the field intensity at the point P inside the wire is

Mg r


That is, at any point inside the solid ci rcular wire the field intensity vari es directly as the distance of the point from the center ofthe wire

,while outside the wire the field intensity varies inversely

as the distance of the point from the center of the wire .

SUMMARY OF IMPORTANT DEFINITIONS ANDPRINCIPLES

1 . AS the measure of the strength of an electric current in awire is taken the ratio of the force per unit length of the wire ,which would be produced on the wire by a magnetic field

,to

the component of the flux density of this field perpendicular tothe wire . The e . g. s . electrom agnetic unit of current strengthis the abampere . The practical unit is the ampere

1 abampere = 10 amperes .

2 . Left-hand Rule . The direction of the current (I) in a wireis the direction in which the m iddle finger of the left hand pointswhen the thumb

,forefinger and m iddle finger of this hand are

held mutually perpendicular and the thumb is pointed in thedirection in which the wire tends to move and the forefinger ispointed in the direction of the component of the flux densityperpendicular to the wire .

3 . A continuous electr ic current is a current the strength of

which does not vary with tim e .

4 . A conductor is a substance in which,when connected to

the poles of a battery,a continuous electric current is established .

An insu la tor or dielectric is a substance in which,when connected

to the poles of a battery,no continuous current

,or only a very

sm all continuous -

current,is established .

5 . Two or mere conductors connected end to end in such amanner that the sam e current flows through each are said to beconnected in series. Two or m ore conductors j oining any twopoints of an electric circuit in such a m anner that the total currententering these conductors at one point leaves these conductorsat the other point are said to be connected in paral lel .

6. The mechan ica l force acting on a wire l centim eters, long

carrying a current of I. abamperes due to the magnetic field inwhich the wire is placed is

I (B sin 9) dl

where B is the flux density in gausses at any elementary length


terminal fiom which the current leaves the electrolyte is calledthe cathode ; the name electrode is used for either term inal .

10 . The relation between the mass m of a substance depositedin tim e t by a continuous current I and this current is

m

t

where k is a constant,called the electrochem ical equ ivalent of the

substance,which constant depends on the nature of the substance

and the_units in which the various quantities are measured . For

a variable current this formula becomesd—m= ki

dt

The electrochem ical equivalent of a substance IS proportional toits chem ical equivalent .

1 1 . The quantity of electric ity Q which flows through any sec

tion of a conductor is defined as the product of the strength I ofthe current in this section by the time t during which the currentflows

,i .e.

,

Q = ItThis applies only to a continuous current ; in the case of a variablecurrent the quantity is

Q

When the current is expressed in abamperes and the tim e in sec

onds the unit of quantity is the abcoulomb; when these quantitiesa re expressed in amperes and seconds respectively

,the unit is the

coulomb .

1 abcoulomb = 10 coulombs12 . The electric resistance of a conductor is defined as the ratio

of the rate at which a continuous current produces heat energy inthe conductor to the square of the strength of this current

, pro

vided the conductor is of uniform structure and is kept at uniformtemperature throughout . The power dissipated in a conductor bya continuous current ofs trength I , due to the resistance R of theconductor is

P, RI2

This formula gives the power in ergs per second when the currentI is expressed in abamperes and the resistance R is expressed inc . g. s . electromagnetic units or abohms ; when the current is ex


pressed in amperes and the resistance in Ohm s (the practical unit)this formula gives the power in watts .

1 Ohm = 10” abohms

13 . The resistance Of a conductor Of length l -and cross sectionA is

l

A

where ' the factor p,called the spec ific resistance of the conductor

,

is a constant for a given material at a given temperature . Thisformula applies

, onl y when the stream lines of the current areparallel and perpendicular to the section A and the current dens ityis uniform over this section . When p is expressed in ohms percentim eter- cube the dimensions of the conductor must be expressedin centim eters ; when p is expressed in ohms per inch- cube thedimens ions must be expressed in inches ; when p is expressed inohms per m il - foot the length must be expressed in feet and thecross section in c ircular m ils .

14 . The rec iprocal of electric resistance is cal led the electricconductance .

15 . The variation of the resistance of a conductor with tem

perature is expressed by the formulaR, R

, ( 1 Bt)where R is the resistance at zero degrees centigrade, R the resistance at any other temperature t degrees centigrade , and B is thetemperature coefficient of resistance . The temperature coefficientof copper is the temperature of other pure

,non-magnetic

conductors has approxim ately the same value . When this valueof the temperature coeffic ient is employed the relation between theresistances R, and R,r at any two temperatures t and t

’ degreescentigrade is given by the formula

238+ t’

238 -H

16 . The electric energy gained by the current in any part of ac ircuit is defined as the am ount of energy lost by this part of thec ircuit or by any other bodies as the result of the existence Of thecurrent in thi s part of the c ircuit ; the electric energy lost by the

current in any part of a circu it/

is defined as the amount Of energy

gained by this portion of the circuit or any other bodies as theresult of the ex istence Of the current in this part Of the circuit .


17. The drop of electric potential from any point 1 to any point2 along a wire carrying an electric current is defined as the ratioof the power developed by the current between these two pointsto the value of the current from 1 to 2 . Hence the power develOped by a current I the rate at which electric energy is lostby the current) in any portion of a c ircuit in which the drop ofpotential in the direction of the current is V is

P VI

This formula gives the power in ergs per second when the currentI is expressed in abamperes and the potential drop V is expressedin c . g. s . electrom agnetic units or abvolts when the current isexpressed in amperes and the potential drop in volts (the practicalunit) this formula gives the power in watts

1 volt = 108 abvolts .

18 . The drop of potential in a wire due solely to the resistanceof the wire

,or the resistance drop, is

V,RI volts

where al l quantities are in practical units .

19 . An electromotive force is that which produces or opposesthe flow of electricity

,other than the Opposition due to the resist

ance of the conductor in which it flows . As the measure of theelectromotive force in any portion of the c ircuit is taken the riseof potential which it would produce in the direction of the currentin this portion of the c ircuit were there no res istance drop in thisportion of the c ircuit . When the electromotive force produces anactual rise of potential in the opposite direction to that of thecurrent it is called a back electrom otive force .

20 . The net rise of potential E ’ from term inal 1 to term inal 2of any c ircuit is called the term ina l electromotive force of the circuit when the current in this portion of the c ircuit is from 1 to 2 ;when the current in this portion of the c ircuit is in the directionfrom 2 to 1 the net rise of potential E ’

,2 from 1 to 2 is called theimpressed electromotive force , due to the rest of the c ircuit . Therelation between the electrom otive force E 1 2 generated in thisportion of the c ircuit

,the current I ,, in this portion of the c ircuit ,

the term inal or impressed electrom otive force E ’

,2 and the resistance R of this portion of the c ircuit is

E — E


vided these filam ents are drawn‘

in the proper directions . Suchfil am ents are called the stream l ines of the electric current .

27. The electric intensity at any point in a conductor coincidesin direction with the stream line through that point and is equalto the product of the specific res istance p of the conductor and thecurrent density 0

' at that point,i .e .

,

He p 0

"

The resistance drop between any two points 1 and 2 in a conductorOf any shape is

( H,cos 9) dl

where H,is the electric intensity at the element d l of the path

between 1 and 2_

and 9 is the angle between the electric intensityat d l and the direction of dl .

28 . E lectric equipoteritial surfaces and stream lines of electriccurrent are mutuall y perpendicular .

29 . The formula for the insulation conductance between anytwo conductors is identical with the formula for the electrostaticcapac ity between these two conductors when 4 77 divided by thespecific resistance p is substituted for the di electric cons tant Kin the formula for capacity . 53

PROBLEMS

l . A straight wire carrying a current of 20 amperes is p lacedIn the gap between the poles of two magnets, the pole- faces of

which are each 1 inch square .

If the flux in the air gap may berepresented by straight lines andthe flux dens ity is 1000 gausses ,calculate the force in dynes act

ing upon the wire, when the wireis perpendicular to the flux lines .

Ans . : 5080 dynes .

2 . Given the electrical circuitshown in the figure . The wireat A is insulated so that all thecurrent flowing at F must passaround the c ircular loop in acounter- clockwise direction be

fore continuing to B . The current may be assum ed to flow In


a geometrical line in all parts of the c ircuit and in the b attery .

Dim ensions : Radius OA = 2 inches,A B = 6 inches

,BC = 4

inches,CE = 8 inches , E F = 4

'

inches,and FA = 2 inches . If

the current from B to C is 50 amperes,find the field magnetic

intensity at 0 due to the entire c ircuit .Ans . : gilberts per cm .

3 . A current Of 100 amperes is established in a long iron rodthe diam eter of which is 1 inch . The rod has a perm eability of300 c . g . 3 . units . What is the m agnetic flux dens ity inchfrom the center of the rod due to this current? What is the totalnumber of lines of induction per foot length ins ide the rod?

Ans.: 2360 gausses . maxwells .

4 . The current established in a solution of copper sulphate(CuSO,) is 100 amperes . Determ ine the weight of copper deposited on a platinum cathode in one hour . What quantity ofelectricity is transferred through the solution?

Ans . : grams . 100 ampere- hours or coulombs .

5 . A copper wire two m iles in length has'

a cross section of

square inch and has a resistance of ohms . What is thelength in feet of a wire of the same material ci rcular m ilsin cross section and having a resistance of ohms?

Ans . : 3 150 feet .6 . The res istance per m il - foot of copper is ohms at 0° cent .

What will be the resistance at this temperature of 40 grams of

copper wire which has a cross section of 2500 ci rcular mils anda spec ific gravity of

Ans . : ohms .

7. The resistance of the armature winding of a given electricm otor at 24° cent . is found to be Ohm s . The armatureresistance is again measured after the motor has been in servicefor several hours and found to be Ohms . What is the average increase in the temperature of the armature winding?

Ans . : 43°cent .

8 . A 100-volt generator and a 50-volt battery are connectedin series . The internal resistance of the generator is 1 ohm andthe internal res istance of the battery 7 Ohms ; the resistance of

each of the two wires connecting the battery and the generator is1 ohm . What is the current in thi s c ircuit and the term inalvoltage of the battery ( 1) when the electromotive forces of the battery and the generator oppose each other , and (2) when these elec


tromotive forces are in the sam e direction around the c ircuit?The electromotive forces are to be assumed constant ; actually,the electromotive force of the battery will depend upon the amountand direction of the current , due to the polarisation which takesplace . (3) How much of the power developed by the enginedriving the generator is converted into electric power in each case ?

(4) At what rate is energy transform ed into chemical energy inthe first case? (5) At what rate is chemical energy transform edinto electric energy in the second case? (6) At whatrate is energytrans formed into heat energy in each case?

Ans . : ( 1) 5 amperes and 85 volts (2) 15 amperes and 55 volts

(3) 500 watts and 1500 watts ; (4) 250 watts ; (5) 750 watts ; (6)250 watts and 2250 watts .

9 . A house service consists of 5 (32 GP .) lamps of 1 10 ohms

resistance each,8 ( 16 GP .) lamps of 220 ohms resistance each and

an electric heater of 10 ohm s resistance,all connected in parallel .

The voltage between the service wires at the entrance to the houseis 1 15

,and each of the two wires leading from the entrance to the

load has a res istance of ohm . What is the energy in kilowatthours delivered to the house during a period of 4 hours?

Ans . : kil owatt-hours .

10 . Three batteries with electromotive forces of 15, 20 and 25volts respectively and with internal resistances of 4

,3 and 2 ohms

respectively,have their positive term inals connected together and

their negative term inals connected together . ( 1) What is the terminal electrom otive force of each battery , and (2) what is thecurrent in each?

Ans . : ( 1) volts ; (2) amperes,

amperes,

amperes .

1 1 . Fig . B represents a so—called three wire system whichis largely used for distributingelectric energy for l ight andpower . A and B are two generators (the field windings areom itted for s implicity) withtheir electromotive forces act

ing in the sam e direction . C'

and D are the loads,whi ch

m ay be either lamps or

F ig . B . motors . a and b are the out


not a perfect c irc le ; in this problem ,however

,the wire m ay be

assumed solid and of c ircular cross section .)Ans . : 1027 megohm s per m ile .

15 . Prove that the resistance of a uniform ly tapered wirea wire such that its di am eter changes by a constant amount perunit distance measured along its axis) is

R=pl

77 r , r2

where p is the specific resistance of the wire per centimeter- cubel is its length in centim eters

,and r , and r2 are the radii in centi

meters oi its cross section at its two ends . The current densityat any cross section of the wire is to be assumed constant overthat cross section . Compare with the resistance of a wire Of uniform cross section .

ELECTROMAGNETISM

105 . Electromotive Force Due to Change in the Number Of

Lines of Induction Link ing a Circu it . Linkages between Flux

and Current. In the last chapter were described two types of

devices for producing an electric current , the chem ical battery andthe thermo- electric couple . In engineering work , however, neitherof these devices is used as a generator of electric energy, exceptin sm all amounts for testing purposes and where only small currents are required

,on account Of the high cost of generating elec

trical energy by such devices . Practical ly all electric generatorsand all electric motors are based upon an entirely different principle, nam ely, that whenever the number of lines ofmagneti c induction l inking a circui t i s changed an electromotive force i s produced inthis

‘

circu i t. This important fact,which i s the basis of electrical

engineering, was discovered by Faraday in the early part of thelast century . E lectrical engineering as an art

,however

,cannot

be said to date back earlier than about 1880,when the develop

ment of the incandescent lamp by Edison first created a dem andfor large am ounts of electric energy .

The electromotive forces produced in an electric c ircuit bychanging the number Of lines of induction linking the circuit arecalled “ induced electromotive forces, and the currents due tothese electromotive forces are called induced currents .

AS has already been noted,every electric c ircuit forms one or

more closed loops and every line of‘

induction forms a closed loop .

A line of induction and an electric c ircuit may therefore link eachother in the sam e way that two links of a chain link each other .When the electric c ircuit is in the form of a coil of a number ofturns

‘

the sam e line Of induction may thread several turns of thecoil ; in this case the line of induction is said to form with thec ircuit a number of

_ ,

linkages equal _to__ the number Of~ turns whichQ "

it threads . The total number of linkages corresponding to anynumber of l iri

'

e'

s of i nduction is the sum of the linkages of all thelines . In the Special case of a coil of N turns and (6 lines of

185


induction each ofwhich threads a l l these turns,the total number

of linkages is K= N (f) if only part of the lines thread all theturns the number of linkages will be less than N (f) . The symbolIx will be used throughout for number of linkages ; linkages areexpressed in the same unit as flux of induction

,i .e.

,in m axwells .

With this understanding the experim entally.

determ ined relation between the value of the induced electrom otive force and thechange in the number of lines of induction linking a c ircuit may

be stated as follows : The va lue of‘

the e lectromotive force induced inan e lectri c circui t i s equa l to the time rate of change of the number

of linkages between the circui t and the lines of induction threading

i t. Or,using the word

“ flux as synonymous with “ lines of

m agnetic induction ,” this law may be stated m ore briefly as

,the

e lectromotive force induced in a circu i t i s equa l to the time rate ofchange of the linkages between the flux and the circu i t.

The direction of this induced electrom otive force is foundto be such that it would of itself set up a current in the c ircuitin such a directiofi as to oppose the change in the flux linking the c ircuit . We have already seen (Article 72) that thedirection of the lines of magnetic induction set up by a cur

rent in any c ircuit is always such that these lines link the cir

cuit in the direction in which a right—handed screw placedalong the wire form ing this c ircuit would have to be turned toadvance it in the direction of the current . Hence the directionof the electromotive force induced in a c ircuit when the flux linkingthis c ircuit is changed is Opposite to the direction in which a righthanded screw placed along the conductor form ing the

x

circu it

would advance if turned in the direction in which the lines ofinduction are increased . Hence

,calling d )x the increase in tim e

dt of the number of linkages between the c ircuit and the fluxthreading it , the induced electromotive force in ,

the right- handedscrew direction with respect to the direction of the lines of induction l irik ing this c ircuit is

d

dt

Where X is expressed in m axwells or c . g . 3 . lines of induction andt in seconds

,this equation gives the value Of the induced electro

m otive force in abvo lts . The value of the induced electrom otiveforce in vo lts

,when X is expressed in maxwells and t in seconds, is

since 1 volt = 108 abvolts


the direction of the current the value e at any instant,then

,in

any infinitesimal interval of tim e dt measured from this instant,

the current in this part of the circuit gains an am ount of electricenergy equal to eidt

,where e is the induced electrom otive force

in the direction of the current. But we have just seen that thevalue of the induced electrom otive force in the direction of thecurrent resulting from an increase d in the number of linkagesof the flux and the c ircuit in the sam e direction as the lines of

d Xdt

'

electric energy ga ined by the current when the linkages of theflux and the c ircuit are increased by an amount d Xis

dW=ei dt= —d - id )\dl

When i is in abamperes and X in maxwells this equation gives thework done in ergs .

Hence,when the number of lines of induction linking any part

of a c ircuit is actually increased, the current in this part of thec ircuit loses electric energy, and when the number of lines ofinduction linking any part of the c ircuit is actually decreased

,

the current in this part of the circuit gains electric energy . Thisagrees with the general princ iple which has already been noted ,that an electrom otive force in the opposite direction to the direction of the current

,that is a back electromotive force

,corresponds

to a loss of electric energy, and an electrom otive force in the direction of the current corresponds to a gain of electric energy .

107. E lectromotive Force Induced by the Cutting of Lines of

Induction . Right-hand Rule .

— When the change in the numberof lines of induction linking a wire is caused by the m otion of

the wire through a magnetic field,the wire m ay be looked upon

as cutting the lines of induction, and the rate of change of thenumber of lines of induction linking the wire may be looked uponas the rate at which the wire cuts the lines of induction . Thedirection of the induced electrom otive force in the wire will thenbe the direction in which the m iddle finger of the right hand pointswhen laid along the wire and the thumb and forefinger Of thishand are held perpendicular to each other and to the m iddlefinger, with the thumb pointing in the direction of the m otionof the wire and the forefinger in the direction of the flux density .

induction due to this current, is Hence the am ount of

ELECTROMAGNETISM 9

This right-hand ru le for the direction of the induced electromotiveforce IS equivalent to the rule given above, but is more convenientin the case of a wire moving across a m agnetic field

. For examp le,

if the li nes of induction are perpendicular to the plane of this pageand are in the downward direction

,a straight wire held parallel

to the sides of the page and moved from left to right will have anelectrom otive force induced in it in the di rection from the bottomto the top Of the page , while if the wire is m oved from right toleft the induced electromotive force w il l be in the direction fromthe top to the bottom of the page . In the above equations forthe induced electromotive force (equations 1) and the electricenergy gained by the current (equation 2) in any part of a circuit

,

d It m ay then be interpreted either as the change in the numberof linkages between the flux and the w ire or as the number oflines Of induction cut by the wire ; in the latter case the electrom otive force is to be taken as positive if its direction as determ inedby the right-hand rule is in the direction of the current in thispart of the circuit , while if in the opposite direction to that of thecurrent it must be taken as negative, i .e .

,a back electromotive

force .

When the change in the number of lines of induction linking awire is due solely to the motion of the wire through a magneticfield

,equations ( 1) and (2) may be

deduced directly from equation (2)of Chapter III

,by the application 3 ll °a m

of the principle of the conservationof energy . Consider the Specialcase of a straight wire of length lcarrying an electric current i

,and

let this wire be in a magnetic fielddue to any other source . To sim

plify the discussion let the lines ofinduction be perpendicular to thewire and let the flux density havethe sam e value at each element ofthe wire . In Fig . 66 the lines ofinduction are taken in the direction downward perpendicular to the plane Of the page, the wireis taken parallel to the edge Of the page and the direction of thecurrent from the bottom to the top of the page . Then from

Motion

Fig . 66.


equation (2b) of Chapter III , the mechanical force produced on

the w1re by the agent producing the field is F =—Bli,and 1s in the

direction from the right to the left (determ ined by the left handrule

,see Article To move the wire a distance dx to the right

then requires that some external agent do an amount of workdW=B lidx . But lda: is the area swept over by the wire andB is the flux density norm al to this area

,and S ince da: is but an

infinitesimal distance,B may be considered constant over this

area . Hence B ldr is the number of lines of induction cut bythe wire in moving the distance dx ,

that is B ldcc =d (5, and therefore the electric energy gained by the current is dW= i dl d) . Hence

the rate at which the current gains electric energy is id—d—Z-Sand

therefore the rise Of potential in the direction of the current isd

d?(see equation 14, Chapter III) which by definition (Article 92)is equal to the electrOmotive force in this direction . The numericalvalues of the electromotive force and the electric energy gainedby the current are identical with those given by equations ( 1)and The direction of the induced e. m . f . in this case is in thedirection of the current

,which agrees with the right

-hand rulestated above . If the wire is allowed to move in the direction ofthe m echanical force acting on it, work is done on the wire or

whatever Opposes its m otion,and the energy for doing this work

com es from a loss of electric energy by the current in the wire .

In this case,the induced electromotive force is in the opposi te

direction to that of the current,which also agrees with the right

hand rule .

From the relationdW i s}: it also follows that the valuedt dt

of the electromotive force in abvolts induced in the wire when itm oves perpendicularly across a magnetic field may also be expressed by the formula

where v is the velocity in centimeters per second at which the wirem oves perpendicular to itself

,l is the length of the wire in centi

m eters and B the flux density in gausses .

When the wire moves through a non-m agnetic medium ,the


coil and the direction of the m agnetic field intensity through thecoil is the sam e as the relation between the direction in whicha , right-handed screw is turned and the direction in which itadvances . The reader m ay prove by a sim ilar argum ent that thecomponent of the field intens ity at any point ins ide the solenoidat right angles to its axis is zero , provided the lines of inductionwhich go out the ends may be neglected . (In proving this therelative directions of the lines of induction and the current on thetwo sides of the solenoid must be taken into account .) Hencethe resultant field intens ity in

, gilberts per centim eter at any pointins ide a long solenoid a considerable distance from the ends of thesolenoid due directly to the current in the solenoid is

H =4 7T N’i (4)

and is parallel to the axis of the solenoid , where i is the currentin abamperes and N’ the

.

number of turns per centim eter length .

Hence the magnetic field ins ide a long solenoid near its centerdue directly to the current in the coil is uniform .

If the solenoid is wound on an iron core,m agnetic poles will

be induced on the ends of the core,and these poles wil l also pro

duce a certain field intensity,which ins ide the iron will be in the

opposite direction to that due directly to the current,but the

field intens ity due directly to the latter will be exactly the sam eas previously existed in the air . When a Slender iron rod

,of a

length cons iderably less than the length of the solenoid,is placed

ins ide the solenoid with its axis parallel to the axis of the solenoid,

these induced poles may be assum ed concentrated in points atthe ends of the rod . Let the strength of these poles be m and—m

,and the length of the rod be l centim eters . Then the field

intensity at the center Of the rod due to these induced poles,or

the so- called dem agnetising force due to the ends of the rod,

is approxim ately

and the resultant field intensity at the center of the rod is therefore

P8m

4H,= H H,

=4 7rN il2

( a )

An exact determ ination Of the pole strength m is extremelydifficult

,and for this reason magnetic tests are seldom made on

ELECTROMAGNETISM 193

rods ; instead a ring or rod and yoke are usual ly employed (seeArticle

109 . Determ ination of the Number of Lines of Induction Linking an Electric Circuit . M easurement of Quantity Of E l ec

tricity . From the fact that an electromotive force is induced in anelectric c ircuit when the number of l ines of induction threading thec ircuit is changed , it is poss ib le by a simple experiment to m easurethe number of l ines of induction through any region in space .

Cons ider first a coil of N turns connected in a c ircuit of whichthe total resistance

,including that of the coil

,is R

,and let the

number of lines of induction threading this c ircuit be changedfrom (p, to 952 , and let t be the tim e required for this change totake place . Then the average e. m . f . induced in the circuit

p rovided each

line of induction links each turn,and therefore the average value

of the induced current is

N

R Rt

N

R

But I t is the quantity of electric ity which flows through the

c ircuit in this interval t. Hence the important relation that,

when the number of lines Of induction linking a coil ofN turnsis changed by an amount qS, (In, a quantity of electricity

Q(152) N

R

is trans ferred across each section of the wire form ing the c ircuit,

where R is the tota l resistance of the c ircuit,where all quantities

are in c . g. 3 . units . Hence,if when a coil which forms part of a

c ircuit which has a tota l resistance of R abohm s is pulled quicklyout of a m agnetic field

,a quantity of Q abcoulombs of electricity

is discharged through the circuit,the number of lines Qf ln

duction in maxwells which threaded the coil when it was in thefield is

R Q95

N_

Or,if the coil is reversed in the field turned 180 degrees about


any axis in the plane of the coil) , or if the direction of the fieldthrough the coil

‘

is reversed, the number of lines of induction inmaxwells original ly threading the coil is

_

R Q

When a momentary current is established in a galvanom eterofwhich the moving elem ent is fairly heavy and has a long periodof vibration

,it can be shown that the m omentary force or impulse

produced on this elem ent causes it to swing out from its positionof equilibrium by an amount which is approxim ately proportionalto the quanti ty of electricity discharged through the galvanometer . * Hence , if the coil we have j ust been cons idering is con

nected in series with such a long period or ba l listic galvanom eter ,and the change in the number of lines of induction threading thecoil is made so rapidly that the current established in the c ircuitlasts for only a sm all fraction of the time required for the movingelem ent of the galvanom eter to make a complete swing, the firstswing of the moving element will be proportional (approxim ately)to the quantity of electricity Q discharged through the circuit

,

(56)

that is,proportional to (Tl

—

Z?) N

From the relation established in the preceding article , thechange in the number of lines of induction , <f) , can be caleulated in the special case when the coil is wound on a non-m agneticSpool which is placed ihside and at the center of a long solenoidwith an air core

,and the current in the solenoid is reversed in

direction . Let

"N’

,= the number of turns per centimeter length of the solenoidor prim ary coil .

I = the strength of the current in abamperes in the solenoidor prim ary coil .

N2= the number of turns on the spool or secondary coil .

R= the total res istance in abohms of the secondary coil,the

galvanom eter and any extra resistance which may be

in the secondary c ircuit .

S ==the area in Square centim eters of the m ean cross sectionof the secondary coil .

Then the field intensity at each point of the area S is H =4 71'N’

, I ,

*J. J. Thom son , Elements ofElectri ci ty and M agnetism , p . 377.


per centimeter,where l is the length in centimeters of the m ean

c ircum ference of the ring . (This formula is an approxim ation ,and applies only when the radi al thickness of the material form ingthe ring is small compared with the radius of the m ean c ircumference of the ring, i .e.

,the dotted line in Fig . A circular

M ean c irc umference

Fig . 68.

ring thus uniform ly wound with a coil ofwire carrying an electriccurrent has no m agnetic poles induced on it , hence this value ofH is the total field intensity inside the ring and is independent ofthe magnetic nature of the ring . When the ring is m ade of amagnetic material such as iron

,the flux dens ity

,i .e .

,the number

of lines of induction per square centimeter,ins ide the ring will

not be equal to the field intensity H but will have som e othervalue B .

. These lines of induction,however

,will coincide in

direction with the lines of force,and therefore across any radial

section A of the ring there will be BA lines of induction . Hence,

if the ring is ' originally unm agnetised* and a current of I abamperes is established in the prim ary winding, the number of linesof induction through the secondary coil will change by an am ount(I) =BA as the result of establ ishing in the ring a field intensity

H4 7T N 1 I

ZThis change of BA lines of induction through the

B A N2

R*The ring can be demagnetised at the start by reversing the current in the

primary winding back and forth and gradually decreasing its strength.

coil will cause a transfer of Q abcoulombs of electricity

ELECTROMAGNETISM l 97

through the galvanometer, produc ing a swing of the m oving element . The quantity of electric ity Q corresponding to this swingcan be read directly from the calibration curve of the instrum ent

,

and therefore the flux density B corresponding to the field inHQ

AN;By increasing the value of the primary current in successive

steps and noting the deflection of the galvanometer correspondingto each change in the primary current

,the B H curve for the

sample is readily determ ined . To determ ine the hysteresis loopcorresponding to any maximum field intens ity H

,the prim ary

current is reversed back and forth a number of tim es betweenthe positive and negative values corresponding to H and H

,

and is then decreased in steps from the value corresponding to Hto the value corresponding to H

,and the galvanom eter deflec

tion corresponding to each step noted . The prim ary current isthen increased in steps from H to H

,and the corresponding

deflections of the galvanom eter again noted . From these obser

vations the value of the flux density corresponding to each valueof H in this cycle of changes may be c alculated

,and hence the

hysteresis loop m ay be plotted .

1 1 1 . The Continuous Current Dynamo . The nam e continu

ous current dynamo is given to any m achine in which a constantor continuous electrom otive force is developed by the rotationof one or m ore conductors in a m agnetic field . Such a machinemay be used as a generator of electric energy when driven bysom e external means such as a steam engine, water Wheel or gasengine ; or when electric energy is supplied to it, it may be used asan electric m otor

,converting electric energy into mechanical

energy . The princ iples involved in the construction of dynamo,

whether it is designed for use as a generator or as a motor, areidentical .The conductors in which the electromotive force is induced

in a continuous current generator are usually insulated copperwires or Copper bars

,called armature conductors

,which are im

bedded in slots in the exterior surface of a hollow iron cylinder

(see Fig . These slots are parallel to the axis of the cylinder .The cylinder itself

,which is called the armature core

,is made up

of sheets of soft iron or sheet steel which has a high perm eability ;the planes of these sheets are perpendicular to the axis of the

tensity H can be calculated from the formula B


cyl inder . In large m achines these fiat rings of sheet steel arefitted on to a cast iron fram e

,called the armature spider, which

resembles the spokes and hub of a wheel . The armature spider

(or , in sm all m achines,the steel sheets themselves) is m ounted on

a shaft or axle which runs through it and proj ects out at each end .

The ends of the shaft are m ounted in suitable bearings , and,in

case the m achine is to be driven by , or is to drive, a belt , a pulleyis mounted on one end of the shaft . The arm ature conductors

,

core and spider taken collectively are called the armature of themachine .

The magnetic field in which the arm ature conductors arerotated is produced by an electric current in two or more coils

Elel d Yok e

F ie l d C

A ir Gap

Fi el d Core

Pole Shoe

F iel d F rame

Ave rage PathLines of Ind uction

Fig . 69 .

of insulated W ire,ca l led fie ld coi ls

,which are wound on stationary

iron or steel cores which are placed symm etrically around thearm ature core

,as shown in the figure . The ends of these field

cores next the arm ature are broadened out so that they coverfrom 50 to 70 per cent of the arm ature surface

,and are made

concave toward the armature so that the end surfaces,called

the pole faces , form part of a cylindrical surface concentric with


electrom otive force in the direction away from the reader ; thee . m . f . induced in those conductors not under a field pole is practically zero . The average paths of the lines of induction areshown by the dotted lines .

In spite of . the fact that the net electrom otive force inducedin the closed arm ature winding is zero , a continuous electromotiveforce can be obtained from this m achine by bringing out suitableconnections or taps from the winding, and m aking contac t withthese connections in a suitable m anner . This can be best understood from Fig . .70

,which shows diagramm atically an armature

winding of a sirnple two—pole machine , with the end connectors on

the end of the arm ature facing the reader in heavy lines and the endconnectors on the other end as dotted lines . The m iddle point ofeach of the front connections is connected to a bar of a device calleda commutator

,which is a cylinder of copper bars insulated from

each other by sheets of m ica . This cylinder is mounted rigidlyon the shaft of the armature

,from which the bars are also insulated .

From the segment m arked 1 there are two paths through the armature winding to the segm entmarked 5

,and the electro

motive force induced in eachof the arm ature conduc torsin each of these paths is inthe direction from 1 to 5 .

Hence the net electrom otiveforce b e t w e e n 1 and 5

through each of these pathsis the arithm etical sum ofthe electrom otive forces induced in a l l the conductorsin each path

,i . e .

,in half the

total number of arm atureconductors . The two halvesof the armature winding

are then sim ilar to two electric batteries in parallel ; the electromotive force of one half the winding Opposes the electrom otiveforce induced in the other half of the winding and consequentlyno current flows in the c losed c ircuit form ed by the entire winding .

However,when the segments 1 and 5 are connected by an external

conductor,the electrom otive force impressed on this c ircuit will be

Fig . 70 .


equal to the electrom otive force induced in each half of the winding, and consequently a current will be established in the externalconductor, one half of the current flowing through each half ofthe arm ature winding ; j ust as in the case of two batteries inparallel and connected to an external c ircuit half the currentflows through each battery , provided the electrom otive forcesand the internal resistances of the two batteries are respectivelyequal . As the arm ature turns in the direction indi cated theelectromotive forces induced in the individual conductors forming each path between 1 and 5 will not all be in the sam e direction,but som e of the electrom otive forces will oppose the other

,and

hence t he net electromotive force between 1 and 5 will decrease .

However,when the arm ature has rotated through an angle

corresponding to one segment of the commutator,another pair

of segm ents, 8 and 4 , come into the position form erly occupiedby 1 and 5 and the electromotive force between 8 and 4 will beexactly the sam e as the electrom otive force which previouslyexisted between 1 and 5

,provided the arm ature rotates with

a constant Speed . Sim ilarly for the next pair of segments , andso on . Cons equently

,if in the positions B and B’ are mounted

two fixed contacts,under which the commutator segm ents slide

as the arm ature rotates,the electromotive force between these

two contacts will rem ain practically cons tant,and the greater

the number of commutator segm ents the more nearly constantwill this electromotive force be . The fixed contacts which rub

against the commutator segments are called brushes , and in mostmodern machines are made of carbon blocks . These brushes areheld in a suitable support

,called the brush-holder

,and the parts

of this brush- holder in contact with the brushes are insulated bysuitable bushings from the brackets which connect them to acommon support called the

”

rocker- arm,which is mounted on

som e part of the stationary struc ture which forms the fram eof the machine . This rocker—arm is so mounted that the positionof the brushes can be adjusted until no sparking occurs betweenthem and the commutator segm ents when a current is establishedthrough the m achine . The brush at the lower electric potentialis called the negative brush , and the brush at the higher electricpotential is called the posi tive brush . The electromotive forceinduced in a dynam o is a lways from the negative to the positivebrush

,whether the dynamo be used as a generator or as a motor .


When the m achine is to be used as a generator it is drivenby som e form of prim e mover

,

”i .e.

,a steam engine , gas engine ,

or water wheel,and the c ircuit which is to be supplied with

electric energy is connected in series with the two brushes B andB

’. The field coils may be connected either in series with this

external c ircuit ; or they m ay be connected di rectly across thebrushes of the machine ; or there m ay be two sets of field coils

,

one set connected across the brushes and the other set in serieswith the external c ircuit . In the first case the m achine is calleda seri es connected generator, in the second a shunt connected

generator, and in the third case a compound connected generator .

These various forms of connections are shown in Fig . 71 . A l l

these types of generators are called se lf- exci ted generators , sincethey produce their own m agnetic field .

There is in general suffi c ient residSe ries Fie l d

ual magneti sm in the iron part ofthe m achine to establish a weak m ag

netic field in the air gap ,which in turn

establishes a small electromotive forcebetween the brushes when the armature is rotated ; this electrom otive forcein turn establishes a sm all current inthe field windings which increases themagnetic field in the gap ; this in turnincreases the induced electrom otiveforce

,and thi s cum ulative process goes

on until the field current reaches asteady value . In the simple shunt

Shuntconnected generator, for example, thefield current increases until the induced electrom otive force in the arm ature establishes a difference ofpotentialbetween the brushes of the m achineequal to the product of the resistanceof the shunt field by the strength ofthe current in this field .

The total current taken from theCompound Connected

Fig . 7 1 .

brushes of any type of continuous cur

rent generator for a given induced , orarm ature , electromotive force

,will depend upon the resist

Series Connected

Se ries F ield


polar m achines may have a single winding of the form describedabove

,or there may be two or more independent windings on

the sam e a rm ature core . The general principles involved inthe construction and operation of such machines are

,however

,

identically the same as in the case of the simple two-pole m achine .

1 1 2 . Calculation of the E lectromotive Force Induced in the

Armature of a Continuous Current Dynamo . The value of theelectrom otive force induced in the arm ature winding of a continuous current dynamo may be readi ly calculated from equationLet

N = the total number of arm ature conductors .

p= the number of field poles .

p’ = the number of parallel conducting paths between the

positive and negative brush sets, that is, fl, i s equal top

the number of armature conductors in series betweenthe positive and negative brush sets .

4) = the total useful magnetic flux per pole in c . g. 3 . lines or

m axwells that is,(f) is the total number of lines of in

duction which pass through the arm ature core from anorth pole to the adj acent south poles of the fieldmagnet .

n = the number of revolutions of the arm ature per second .

The tim e required for each conductor to pass entirely around

the armature is then 1 and therefore the time taken for it to passn

through the m agnetic field under each pole is1

Hence thenp

average value of the electrom otive force induced in each arm atureconductor as it passes under each pole is up 4) abvolts . Sincethe conductors are uniform ly distributed around the surface of

the arm ature,this is a lso the average value at each instant of

the electrom otive force induced in these conduc tors . Since

there are _

7conductors in series between the positive and negative

P

brush sets,the average value of the tota l elec tromotive force

between the brushes is

When the number of commutator segments is large, the in


stantaneous values of the electrom otive force between the brushesare practically equal to this average value . For

,since the brushes

always m ake contact with segm ents in the same posi tion with

respect to the fie ld which produces the flux,the only possible

variation in the electrom otive force between the brushes (when

oS and n remain cons tant) is the variation that m ight occur asthe armature rotates through an angle corresponding to one

commutator segment . But in the distance corresponding tothis sm all displacement of an arm ature conductor

,the flux density

remains practically cons tant (except for the three or four con

ductors which are j ust pass ing under or are j ust leaving a poletip) , and hence the rate at which the conductors are cuttinglines of induction

,and therefore the induced electrom otive force

,

does not change apprec iab ly in this interval .Equation (7) holds whether the dynam o is used as a generator

or as a motor . In the case of a generator, this electromotiveforce is in the direction of the current

,and in case of a m otor

in the opposite di rection from that of the current . It shouldbe noted

,however

,that the flux per pole in a generator or m otor

in'

general depends not onl y upon the current in the field coilsbut also upon the current in the armature ; this is caused by thefact that the arm ature current also sets up lines of induction inthe opposite direction to those due to the field current and theresultant flux per pole is decreased . See under A rm ature Re

action in any text-book on dynam o- electric machinery .

1 13 . Magnetomotive Force . In equation (7) for the electrom otive force of a continuous current dynam o , the only quantitywhich cannot be readily predeterm ined is the flux per pole . Thisquantity can

,however

,be calculated when the dim ensions of

the m agnetic c ircuit,the perm eability of its various parts , the

number of turns on the field coils and the “back am pere- turns ”

of the armature are known . The following considerations willm ake this clear .

Cons ider a wire wound into a coil of any form having N turnsand let a current of I abamperes be established in this wire .

We have already seen that a magnetic field will be establishedaround such a coil

,and that the lines of force representing this

field wil l be c losed loops linking the coil . Consequently , betweenany two points in the field around this coil there will be in generala drop (or rise) of magnetic potential . We wish to find the


relation between the total drop of potential around any closedpath linking this coil and the value of the current establishedin it . By definition (A rtic le the drop of m agnetic potentialalong any path in a m agnetic field is the work done by the agentproduc ing the field when a unit north point-pole moves aroundthis path

,that is

,the drop of potential around any closed path

is the line integral f IU (H cos 0) dl around that path , where dl is

any elem entary length in the path, H the field intensity at d l,(9

the angle between the direction of dl' and the direction of H

,and

the symbol flL l represents the integral around this c losed path .

We have also seen that it is physically impossible to have anorth magnetic pole without at the sam e tim e having an equalsouth magnetic pole on the same piece of m atter . Hence it isphysically impossible to m ove a unit north pole around a c losedpath linking a coil without at the sam e tim e linking the coilwith an equal south pole , un less the magnet at the ends ofwhi ch

these poles exist i s threaded through and bent into a c losed loop

l inking the coi l . Hence to m ove a unit north point-pole arounda c losed coil without at the sam e tim e having the current in thecoil exert a force on the south pole connected to this north pole

,

G) we m ay conceive of this unitnorth point-pole as at the end

Origin.“pos ition of a flexible m agnetic filam ent,

the south pole ofwhich is so farrem oved from the coil that theforce exerted by the latter onthis south pole is negligible .

The north pole end of this filament can then be threadedthrough the coil and backaround to its original position

over the desired path,as shown in Fig . 72 . The only force

exerted by the coil on the filam ent as it is thus bent aroundwill be the force produced by the coil on the north pole ofthe filam ent

,and therefore the work done by the current will

be f|L | (H cos (9)d l . But we have also seen,Article 56

,that 4 77

'

lines of induction exist in a fil am ent which has unit poles ,in the direction from the south to the north pole of the fila

Fig . 72 .


m ean c ircum ference of the ring , the drop of m agnetic potentialaround this c ircum ference is

H l =4 7TNI

When the radius of the ring is large compared w i th the radius ofthe section A

,the field intensity at the m ean c ircum ference

m ay be taken as the average field intensity over this area A .

Hence,calling p. the perm eab ility of the iron form ing the ring,

the average flux density over the area A is,u. H

,and therefore

the total flux or number of lines of induction through A is gb=p. HA .

Hence4 7e

l

uA

The expression ii is analogous to the expression for the re

p.

sistance of a wire of length I, cross section A and Spec ific resi st

ance .

1and it is therefore called the re luctance of the magnetic

,u.

c ircuit form ed by the ring .

The reluctance of any portion of a m agnetic c ircuit has nom eaning unless the two ends of the given portion of the c ircuitwhere the lines of induction enter and leave it are m agneticequipotential surfaces and the total number of lines of inductionthrough each cross section of the given portion of the ci rcuit isthe sam e . When these conditions are satisfied

,the reluctance

R may be defined as the ratio of the difference ‘ of m agneticpotential Vm between the two end faces of the given portion of

the c ircuit to the total flux of induction (fi‘

through the c ircuit , i .e.

,

VR

m(10)

(15

When the difference ofm agnetic potential is expressed in gilbertsand the flux in maxwells

,the unit of reluctance is called the

oersted . The relation expressed by equation ( 10) is of exactlythe sam e form as Ohm ’

s Law for an electric c ircuit ; it is thereforesom etim es called Ohm ’

s Law for a m agnetic c ircuit . Magneticflux

,m agnetom otive force and m agnetic reluctance are strictly

analogous to electric current,electrom otive force and electric

resistance respectively,except that no energy i s required to mainta in

a magneti c flux through a re luctance, while energy is always requiredto m aintain an electric current through a resistance .

ELECTROMAGNETISM 09

Since lines of m agnetic induction are always closed loops,

the flux of m agnetic induction com ing up to any surface mustalways equal

~ the flux of induction leaving that surface . Therefore at any junction in a network of m agnetic c ircuits

2 4; o (1 1)This is the sam e as Kirchhoff ’s first law for an electric c ircuit .Sim ilarly

,since the drop of m agnetic potential Vm around any

c losed loop is equal to the total magnetom otive force in thispath

,

2 ¢ R=2 4 7TN 1 (1 1a)where R is the reluctance of any c losed tube of induction and d)is the flux through this tube . This last relation is the sam e asKirchhoff’s second law for a network of electric c ircuits .

The difficulty in applying these laws to a m agnetic circuitarises from the fact that the m agnetic flux is not confined toapproxim ately geom etrical lines like the currents in a networkof ins ulated wires

,but in general fills all space surrounding the

coils which establ ish the m agnetomotive forces ; also , when thereis iron in the c ircuit the perm eability depends on the flux densityand the previous history of the iron . (The distribution ofm agneticflux in and around an iron c ircuit is analogous to the distributionof current in and around an uninsulated m ass of copper of thesame shape as the iron circuit imm ersed in a liquid having aconductivity about equal to that of carbon .) Ohm ’

s Law for them agnetic c ircuit , however, tells us that in order to obtain a largeflux with the least number of ampere- turns it is necessary toprovide a path of low reluctance for the lines of induction . Hencein nearly all electric m achinery a closed or nearly c losed iron or

steel circuit is provided for the lines of induction,s ince iron and

steel have a high permeability, and therefore for the same dim en

sions a much less reluctance than a non-m agnetic substance .

Onl y in the spec ial case of the uni form ly wound c ircular ringdiscussed above

,however

,are the lines of induction confined

entire ly to an iron c ircuit ; in general a certain number also existin the air and in whatever other substances are in the vicinityof the iron c ircuit . For example , in the case of a dynamo , acertain percentage of the total number of l ines _

of induction establ ished by the field coils “ leak ” around through the air fromone pole to the next without linking the armature conductors .

The predeterm ination of the ratio of the total flux to the useful


flux,i .e .

,the ratio of the total number of lines of induction to the

number which link the arm ature conductors,can be m ade only

very roughly ; this ratio , which is called the “ leakage factor ,may,

however,be determ ined by experim ent ; in modern dynam os

is found to vary from to depending upon the arrangem entof the field m agnets .

1 1 5 . Calculation of Ampere-Turns Required to Establ ish a

Given Flux . The following example will illustrate the Way inwhich the number of ampere- turns required to establish a givenflux may be calculated to a rough degree of approxim ation .

Let it be required to find the number of ampere- turns necessaryto establish a total flux (f) through the arm ature of a two- poledynam o . b et the arm ature be m ade of sheet steel punchingsand the field cores

,pole shoes

,and field yoke be a s ingle piece of

cast iron .

Let

Ag= area of the air gap under each pole .

Aa= the m ean cross section of the path of the lines of induction through the arm ature .

Af= the m ean cross section of the path of the lines of induction through the field .

= the radial depth of the air gap ; usually called the lengthof the air gap .

la= the m ean length of the path of the lines of inductionthrough the armature .

l,= the m ean length of the path of the lines of inductionthrough the field .

k = the leakage factor .

I0

Then the flux density in the air gap is i and this is also equalAg

to the field intens ity in the air gap , that is Hg = if

. Hence thea

drop of magnetic potential across both air gaps is 2 Hg, lg =

The flux density in the armature i s if ; from the B H curvea

for sheet steel punchings find the corresponding value of H;call this value H

a. Then the drop ofm agnetic potential through

the armature is Hala

. The total flux through the field will be


The numerical value of the ratio of the electrom otive forceinduced in a circuit , due to the change of the current in thiscircuit , to the tim e rate of this change, is called the coeffi ci ent ofself induction or the se lf inductance of this c ircuit . That is

,when

the current in the given circuit varies at the rate El}and

,as a

dl

result of this variation,an electrom otive

'

force e is induced inthis circuit, then the self inductance of this c ircuit is

e

di

dl

When the electromotive force is expressed in abvolts , the currentin abamperes and tim e in seconds , the unit of self inductance i s

called the abhenry, when these quantities are expressed in volts,

amperes and seconds respectively, the unit of self inductance iscalled the henry.

1.A mi l li henry is the one—thousandth part of a

henry . These units are therefore related as follows1 henry —10

9 abhenries1 m illihenry 10

6 abhenriesSim ilarly

,the numerical value of the ratio of the electro

motive force el z induced in any c ircuit 1,due to a change in the

current i 2 in any other ci rcuit 2, to the tim e rate of the changeof this current i 2, is called the coeffi ci ent ofmutua l induction or themutua l inductance M 1 2 of c ircuit 2. with respect to circuit 1 that is

6 1 2_

dt,

dt

Mutual inductance is expressed in the sam e unit as self inductance ,The self inductance of a c ircuit m ay also be expressed in term s

of the number of linkages (see Article 105) between the circuitand the flux produced by the current in it . From the fundamental law of electrom agnetic induction , a change in the numberof linkages between a c ircuit and the flux linking it induces inthis c ircuit an electrom otive force equal to the tim e rate of changeof these linkages . Hence

,calling d lx the change in the number

of linkages between the c ircuit and the flux due to a change inthe current in the c ircuit by an am ount di , we have as anotherexpression for the electrom otive force of self induction

ELECTROMAGNETISM 13

dh

dl

Equating this value of the self- induced electrom otive force tothat given by equation ( 13) we have that the self inductance inabhenries is

(15)

where X is expressed in m axwells and i in abamperes . Hence theself inductance of a c ircuit is equal to the change of the linkagesbetween the c ircuit and the flux threading it per unit changein the current in this c ircuit . As will be proved presently

,when

the permeab ility of every body in the m agnetic field is constantand the c ircuit rem ains unaltered in size and shape

,the linkages

between the c ircuit and the flux threading it due to the currentin this circuit are directly proportional to the strength ofthe currentin the c ircuit ; under these conditions

,therefore

,the self induc

tance is a constant ofthe circu it (for a given distribution of current)equa l to the number of linkages between this circui t and the flux

produced by uni t current in i t.

Sim ilarly,the mutual inductance in abhenries of a circuit 2

with respect to any other c ircuit 1 is equal to the change of thelinkages km of c ircuit 1 by the flux due to the current i 2 in 2 perunit change in the current in 2 , i .e.

,

Mdx

.

di 2

where X1 2 is in maxwells and i 2 in abamperes . When the permeabil ity of every body in the m agnetic field is constant and thetwo c ircuits are fixed with respect to each other and rem ain un

altered in size and shape,the mutual inductance of one c ircuit

with respect to another is constant (for a given distribution of thecurrents) and equa l to the number of linkages between one ci rcui t and

the flux produced by uni t circui t in the other ; m oreover, as will beproved presently

,the mutual inductance of one c ircuit with respect

to the other is equal to the mutual inductance of the second c ircuitwith respect to the first .

1 17. Proof of the Relation between Inductance and Linkages .

The intens ity at any point of the magnetic field due to an electric

(sin (9) d lcurrent is given by the equation (see A rtic le


where i is the strength of the current in the circuit,d l any

elem entary length of the circuit,r the distance of the point in

question from the elem entary length al l and (9 the angle betweenthe line drawn from the point to d l and the direction ofdl

,and the

symbol fIL , represents the vector integral of the expression

(sin 9) dl

tion of the c i rcuit rem ain unaltered, the quantity under the integralsign remains unaltered ; hence the field intensity .at any point duedirectly to a current in a given c ircuit is proportional

,to the

strength of this current , no m atter what the shape or size of thec ircuit may be, provided the shape and s ize remain unaltered .

The field intensity at any point in the surrounding region due tothe magnetic poles which m ay be induced by this current on anym agnetic bodies in the vicinity will also be proportional to thestrength of this current

, provided the permeabi l i ti es of these bodi es

are constant. Under these conditions the directions of the resultant lines of force and the resultant lines of induction establishedby the electric current will remain unaltered when the strengthof the current is changed , but their number c rossing any surfacewill vary directly as the strength of the current . The lines of

force and the lines of induction will also coincide in direction .

Hence the number of lines of induction crossing any area in themagnetic field due to a current in a s ingle circuit is directly proportional to the strength of the current in this circuit

,provided

all the surrounding bodies have a constant perm eability .

In particular,the number of lines of induction threading the

circuit itself is proportional to the strength of the current in thec ircuit . Therefore

,calling i the strength of the current in the

c ircuit at any instant , the number of lines of induction threadingthe c ircuit at this instant is proportional to i , and therefore thenumber of linkages X between the circuit and the flux due to thecurrent in it is proportional to i , that is

X=Ai

where A is a constant depending upon the shape and size of thec ircuit and the m agnetic nature of the bodies in the field , but independent of the strength of the current , provided the permeabi li ty

of every body in the field i s constant. Note that this constant Ais equal to the number of linkages between this circuit and the

around the entire c ircuit . As long as the shape and posi


the electric current may be said to be stored in the m agneticfield

,j ust as the kinetic energy of a m oving body m ay be looked

upon as stored in the body itself as a consequence of its inertia,

whi ch may be looked upon as the source of the opposition whichthe body offers to being set in motion . Again, when the strengthof the current in a c ircuit decreases , an electromotive force isinduced in the c ircuit in the same direction as the current

,and

therefore a certain am ount of the energy stored in the m agneticfield of .

the current is converted into som e other form of energyby m eans of the current in the c ircuit heat energy) , j ust aswhen the velocity of a m oving body decre

‘ases a definite amountof its kinetic energy is converted into other forms ofheat energy) .

In general , no matter how a m agnetic field may be formed ,whether by m eans of electric currents or m agnetic poles , adefinite amount of energy is required to establish it and whenthe field is destroyed this energy appears in som e other form .

Every magnetic field therefore represents a certain am ount of

stored,

” or potential,energy . This energy of a m agnetic field

is called magneti c energy; when the m agnetic field is due to electriccurrents this energy is also called the e lectrokinetic energy of thecurrents

,the latter name being due to the analogy between this

energy and the kinetic energy of a m oving body .

The m agnetic or electrokinetic energy of the magnetic fieldset up by a current in a circuit may be expressed in terms of thecurrent strength and the inductance of the c ircuit . Let i be thecurrent in abamperes in the c ircuit at any ins tant and let Lbe the self inductance of the c ircuit

,in abhenries ; then the

linkages X between the c ircuit and the flux due to the cur

rent in it at this instant are A=Li . The back electrom otive

force due to an increase di in the current in tim e dl is then e=Lg,

provided L i s constant,and therefore the energy transferred to

the m agnetic field by the current in tim e dt is

dW eidt Lidi

Therefore the total am ount of energy stored in the m agnetic fieldof the current when the strength of the current increases from 0

to any value i is

ELECTROMAGNETISM 2 17

The energy stored in the field when the current is establishedis equal to - the energy returned to the c ircuit when the current isinterrupted

,for

Lidi

provided L is constant . The energy returned to the circuit fromthe m agnetic field gives rise to an electromotive force whichtends to keep the current flowing in its original direction , j ust aswhen the m otion of a body is opposed in any way

,the kinetic

energy of the body tends to keep it moving in its original di rection .

The instantaneous value of this electrom otive force at the instant

‘

a circuit is opened is generally sufficiently great to establisha m om entary current through the air or whatever else separatesthe ends of the opened circuit

,and consequently the current

continues to flow for a fraction of a second across the space between the open ends of the c ircuit

,produc ing the fam iliar spark

or are which occurs when a c ircuit is opened (unl ess certain specialprecautions are taken) . The energy which produces thi s sparkis the energy which is stored in the magnetic field around the c ircuit when the current is established

,and is generally not suffic ient

to maintain for more than a fraction of a second a current throughthe high resistance of the insulator between the open ends of thec ircuit . However

,whenever a current is established through an

insulator its resistance falls to a comparatively low value,at which

it remains as long as the current through it is m aintained . Hence,

when the gap form ed by Opening the c ircuit is short,the electro

motive force originally in the c ircuit may be suffi c ient to m aintaina comparatively large current through the gap . It is this heavycurrent established across the gap by the electrom otive forceoriginally in the c ircuit which constitutes the so- called arc

,and

which rapidly burns away the conductors at the gap unless asuitable switch or c ircuit-breaker is provided for opening thec ircuit . When the current in the c ircuit is large or the electrom otive force in the c ircuit high , the c ircuit- breaker must open thec ircuit rapidly and m ake a long gap between the opened ends .

It should be noted that the above formula for the energyrequired to establish a magnetic field is deduced on the assumption that the self inductance is constant , which is true only whenthe permeability of every body in the m agnetic field is constant .In the case of iron and other magnetic bodies the permeability


is not constant , and in addition to the transfer of energy to andfrom the magnetic field

,a certain am ount of energy is converted

into heat energy (due to hysteresis , see Article both whenthe current is established and also when the current is interrupted, or whenever there is any change whatever in the currentstrength .

1 1 9 . Analogy between the Magnetic Energy of an ElectricCurrent and the Kinetic Energy of a Moving Column ofWater .

As already noted,the m agnetic energy of an electric current is in

many ways analogous to the kinetic energy of a moving columnof any liquid

,as for example

,a current of water in a pipe . The

kinetic energy of such a water current is the work requiredto set the water in motion ; sim ilarly , the m agnetic energy of anelectric current m ay be looked upon as the work required to set

the electricity in the circuit in motion . When the water com esto rest

,its kinetic energy is converted into som e other form

,

principally heat energy . Sim ilarly,when an electric current is

interrupted,thus causing the electricity to com e to rest

,

” them agnetic energy of the current is converted into som e other form ,

e .g.

,heat energy in the conductors and in the Spark which appears

at the break in the circuit .The kinetic energy of a moving column of water can also be

expressed by a formula sim ilar to the formula W = 5~ L7? for the

m agnetic energy of an electric current . Consider a stream ofwater flowing through a pipe of uniform cross section of A sq .

cm . and let the water completely fill the pipe ; let I be the lengthin cm . ofa given section of the pipe , and let V be the linear velocityof the water in cm . per second . Then

,s ince the m ass of 1 cu .

cm . ofwater is 1 gram ,the kinetic energy of the water in the given

length of pipe is 5 ( lA) V2

. Let J be the volum e or quantityof water that flows across any section of the pipe in one second

,

then J=A V . Hence the kinetic energy of the water column m ay

also be expressed as i J2 KJz,Where K x i is a constant

A A

depending upon the dim ensions of the pipe . The analogy betweenthis last expression and the formula 5 Li

z for the m agnetic energyof an electric current is at once evident

,when an electric current

is considered as the quantity of electric ity per second flowing acrossany section of the wire . Note

,however

,that while the factor K

in the expression for the kinetic energy of the water current


where the X’s and i ’s m ay

'

be positive or negative . Let the twocurrents increase by the am ounts di , and di , in tim e dt; then theback electrom otive forces induced in the two c ircuits are re

spectivelyd A,

[Jig/ 1

+ Mdl z

dl dt dl

d A,L,92 + M

di ,

dl dl dl

and the amounts of energy stored in the m agnetic field by therespective currents are

dW,=e,i ,dt M

dW, e,i ,dt L,i ,di , M

The total amount of energy stored in the m agnetic field bytwo currents when they increase from zero to their final valuesI , and I, may be calculated as follows . First

,let c ircuit 2 be open

,

so that no current can flow in it ; under these conditions i , =0and di , =0 , and therefore the work done by the current in c ircuit1 when the current in this circuit increases from zero to I , is

l di l s lz

x

Now keep the current in c ircuit 1 constant and let the current in2 increase from 0 to I , ; under these conditions i , =I , and di , =0 ,and thereforethe work done by the current in c ircuit 1 is

and the work done by the current in c ircuit 2 is

i 2 = %L2122

Hence the total work done by the two currents in establishingthe m agnetic field corresponding to the final values of the currentsI , and I , is

(20a )

Note that this formula does not contain the coefficientThe explanation of this is the fact

,already noted several tim es

,

that the mutual inductance of one c ircuit with respect to anotheris the sam e as the mutual inductance of the second c ircuit withrespect to the first c ircuit . To prove this

,let the current in 2 now

be kept constant and let the current in 1 be decreased to zero ;


under these conditions i , = I , and di , =0 ; and therefore the workdone on the current in 1 by the magnetic field is

and the work done on the current in 2 by the m agnetic field is

I,di ,

Now open c ircuit 1 and let the current in 2 fall to zero under theseconditions i , =0 and di , =0 , and therefore the work done on thecurrent in 2 by the m agnetic field when this current falls to zero is

L,i ,di , - L,Iz,

Hence the total work done by the m agnetic field when it disappearsis

W=2L1121 M ’

z i l i l z ’ l" %L212

2 (205)From the Princ iple of the Conservation of Energy, the energystored in the field when it is established must be equal to the workdone by the field when it disappears

,hence the expression for the

energy stored in the field must be equal to the expression for theenergy given back by the field ; therefore

M =M

It also follows from the Princ iple of the Conservation of Energythat the total am ount of energy stored in the m agnetic field bythe two currents is independent of the manner in which these twocurrents are established

,and therefore the expression

W= % (2 1)is a perfectly general one for the energy Of the magnetic field dueto the currents i , and i , in two ci rcuits which have constant selfinductances L, and L, and a constant mutual inductance M .

Equation (21) may also be written

715 (L l i l

‘ l' M i zi z) 751 + (Li l/2+ M a i l )which

,in turn, from equation may be written

By exactly s im ilar reasoning it can be shown that the energyof the magnetic field due to the electric currents in any numberof c ircuits is

W :,L2M (22)

where i is the current in any c ircuit and X is the number of linkages between this c ircuit and the total flux which threads it

,and

the summ ation includes all the c ircuits in the field .


It Should be noted that all the formulas in this artic le arebased upon the assumption that every body in the magnetic fieldhas a constant perm eab ility .

1 2 1 . Cal culation of Inductance . Sk in Effect.

— When thepermeability of every body in the magnetic field is constant, it isin general possible to Obtain ~an expression for the inductance Of ac ircuit in terms p f the dimensions i he.

circuit . To . obtain anexact expression for this qu

h

ah tityfhowever, it is necessary toconsider the conductors form ing the circuit as divided into currentfilam ents (see Article 101) and to determ ine the back electrom otiveforce induced in each of these filam ents when the current changes .

The inductance of a circuit therefore depends upon the di stribution of the current in the conductors ,

“ and,as already noted

(Article the distribution of the current in,a conductor

’

depends

upon the rapidity,with which the current i aries with tinie . The

explanation ”

of‘

t li is'

w-

ill now be apparent ; when'

thefl

condu ctor

has a large c ross section or is a m agnetic substance a steelrail) the lines of induction within the substance of . the conductorare an appreciable proportion of the tdtal number of lines of induction which link the conductor ; moreover

, smce_ a lma ef- induc tionwhich l ieS

_ _jwholiy~within -thesubstance of the conductor links only

that portion of the current which threads the loop form ed bythis line

,it fol lqws that j the l number of lines of induc tion whichR “M m-y a ," N

link the iriSidg filaments number which linkthe outs1de

n

or surface filam ents,and cons equently the back elec

tromotive force induced in the inside filam ents is greater thanthat induced in the outsiw . The result of this is thata greater current flows through the outs ide filaments than through the ins ide filam ents

,i .e .

,the current density

is greatest near the surface Of_

the conductor . In certain Simplecases the exact distribution Of the current for a given impressedelectrom otive force can be determ ined by expressing the conditionthat the impressed electrom otive force acting on each filam entmust be equal to the sum Of the resistance drop and the backelectrom otive force induced in that ’ filam ent . This phenomenonof a non—uniform current distribution caused by a rapid variationof the current with respect to tim e is called the skin effect .

”As

noted in Chapter V II,this Skin effect causes an increase in the ap~

parent resistance of the c ircuit ; thos elf inductance of a Ci rcu i t isdim inished by the Skin effect .


field intens ity is perpendicular to this plane and the perm eabilityof the air surrounding the wires is unity

d E d 21dx dx

(IS- l x

x+D—se

Hence,the total number of lines Of induction per centim eter

length of the wires threading the space between the two isD—r x =D - r

D T

q5= d CID= 2I ln x—ln (D—:c)

x =r

(The abbreviation ln is used for the natural logarithm andthe abbreviation “ log for the common logarithm .) Whencethe inductance of one centim eter length of both wires is

L =4 abhenries

Or,since

'

each wire is linked by ha lf the resultant lines of in

duction linking the.space between the two , the inductance Of

each wire per unit length is

L =2 InD _ r

abhenriesr

These formulas,however

,are only approxim ate

,since the lines

of induction inside the wires have been neglected . It can be

Shown (see A lex . Russell,A lternating Currents, V ol . 1 , p . 55) that

the exact formula for the inductance of . each of two non-magneticparallel wires for slowly varying currents is

L 2 lnD

abhenries per cm .

r

DL log m rll ihenrres per 1000 ft . (23b)

r

DL log m i lli henri es per m i le (23c)

r

It should be noted that since D and r occur in these formulas onlyas a ratio

,it is imm aterial in what units D and r be expressed

,

provided they are both expressed in the same unit .The m inimum value of the inductance is when the wires touch .

In this case D = 2r,and therefore L m illihenries per


1000 feet and is independent Of the size of the wire . An ab

solutely non—inductive circuit is impossible,although this condi

tion m ay bes

c losely approxim ated by plac ing the wires form ingthe c ircuit close together, e .g.

,by twisting them together .

123 . Self Inductance of a Concentrated Winding. Let L, be

the self inductance of a single loop of wire of any shape ; then theflux linking this loop due to unit current in it is qS= L , . Cons idera coil m ade of N turns of wire , each turn of exactly the sam edimensions as this single loop and let these N turns be so c losetogether that they m ay all be cons idered as coinc iding exactlywith one another

,i .e.

,let the N turns be considered as coneen

trated in a geometrical line . Then unit current in each of theseturns will set up a flux (f) which will link each Of the N turns ;therefore the total flux linking each turn will be N (f) NL , ,

andsince there are N turns the number of linkages between the totalflux and the entire coil will be N a which, from Article 1 17

,is

equal to the self inductance Lnof the entire coil

,i .e .

,

L,=N2

L , (24)

Hence the self inductance of a concentra ted winding is proportionalto the square of the number of turns in it .

124. Self Inductance of a Long Solenoid . In contradistinc

tion to a concentrated winding, consider the case of a long air- coresolenoid

,which is one form of a distributed winding . Let N be

the number of turns in the solenoid , l its length in centim eters andA its mean cross section in square centim eters . Then

,from

equation the m agnetic field intensity at any point inside thesolenoid at a considerable distance from its ends due to a current

4 77 N

Z

axis of the solenoid . The total number of lines of induction linking each of the central turns (neglecting the lines w ithin the substance of the wire) is therefore

'

<f> =HA

of one abampere in the coil is H and is parallel to the

4 7TNA

l

The flux linking the end turns is less than this , S ince som e of thelines of induction go through the lateral walls of the solenoid ; asan approxim ation

,however

,in the case of a long solenoid all the

lines of induction may be assum ed to link these end turns also .

On this assumption the total number of linkages between the coil


and the flux per unit current in the coil,i .e .

,the self inductance of

the coil,is then

4 7TN2A

l

where all the quantities are in c . g. 3 . units .

The self inductance of a solenoid , therefore , does not vary withthe square of the number Of turns when the variation is made bychanging its length, for in this case the number of turns varies as

the length , and therefore the inductance varies directly as thenumber of turns

,or directly as the length . Hence when a sole

noid is used as a variable inductance , by connecting in the circuita greater or less number of its turns by m eans of a sliding contact

,

for example , the inductance varies directly as the distance between the fixed term inal and the Slider

,provided this distance is

large compared with the diam eter of the solenoid . When theslider is close to the fixed term inal

,the turns between the two

form an approximately concentrated winding, and therefore theinductance when onl y a sm all part Of the winding is used variesapproximately as the square of the distance between the twoterm inals .

It should be noted that when a coil of any kind is used ‘

as apart of an electric c ircuit

,there is always a mutual inductance

between the rest of the circuit and the coil . This mutual inductance

,however

,may be made practically negligible by m aking

the leads to the coil suffic iently long and twisting them together .125. Total Energy of a Magnetic Field in Terms of the Field

Intensity and Flux Density. The total energy of the magneticfield due to any number of electric currents may also be expressedin terms of the field intensity and the flux dens ity . Consider thecase when every body in the magnetic field has a cons tant permeabil ity the flux dens ity and field intensity at every point in thefield will then be proportional to each other and in the sam edirection . Under these conditions the energy corresponding tothe final values of the currents must be independent Of the m annerin which these currents are established ; we may then for con

venience assume that the currents all increase proportionally fromzero to their final values . Underthese conditions

,if we imagine

all space filled with tubes drawn in such a m anner that the wallsof each tube are tangent at each point to the direction of the fluxdensity at that point due to the fina l values of the currents

,then

L =N ¢


When the perm eability of the bodies in the field is not constant

,but varies with the field intens ity

,the reasoning employed

in deduc ing the expression HdB for the work done by

the currents per unit volum e of the field,breaks down ; for in this

case the direction of the tubes which we considered as filling allspace may change as the currents change in value . However

,the

1B

expression HdB for the energy per unit volum e required4 7r

to establ ish the field (only a portion of this , however, is stored inthe field

,see Article l 26a ) is applicable to any case where the

direction of the lines of induction remains unaltered and coinc ideswith

,or is in the oppos ite direction to

,the lines of force . This

condition is realized in the Spec ial case of a c losed anchor ringwhich is magnetised by a coil uniform ly wound around it (asdescribed in Article 1 10) and is also approximately realized in themagnetic c ircuits of most electrical apparatus .

126. Heat Energy Due to Hysteresis . When the perm eabilityof every body in the m agnetic field is constant

,the energy trans

ferred to the magne tic field when the flux density increases fromzero to any value B is exactly equal to the energy transferred fromthe magnetic field to the electric currents when the flux densitydecreases from B to zero ; for, when B is proportional to H

,

1 1

4 77 4 7r

nary magnetic bodies such as iron or steel,we have already seen

(Article 57) that the relation between B and H when the fluxdensity increases is different from the relation between these twoquantities when the flux density decreases . Hence in this case

3 .

However,in the case of ordi

HdB i s not equa l to4

HdB . The first expres7T

sion,however

,represents thework done by the currents in estab

l ishing them selves,while the second expression represents the

work done on the currents when the m agnetic field disappears ;the difference between these two expressions must then representthe production Of som e other form of energy in the m agnetic sub

stance , and experim ent shows that this other form of energy isheat energy . That is

,whenever the flux density in a piece of any


ordinary m agnetic substance is changed, heat energy is alwaysdeveloped . Consequently, when the flux density in a piece Of

iron or steel ” is changed from any value B to any other value andthen brought back again to the original value B ,

an amount ofheat energy per unit volum e of the substance equal to

HdB

B I

for this comp lete cyc le of va lues,is produced in the substance .

Hence,when this cycle of values is plotted (with H and B to the

same scale) in a curve the hysteresis curve) , the area of thiscurve divided by 4 77 gives the amount of heat energy per unit

volume due to hysteresis produced in the substance . WhenH and B are expressed in c . g. s . electrom agnetic units, this heatenergy is m ergs per cub ic centimeter .

126a . Tractive Force ofan E lectromagnet. From the formula,

equation for the energy per unit volum e of a m agnetic field,

can be deduced directly the force of attraction between two electrom agnets or between an elec trom agnet and its keeper . Con

sider the spec ial case where the surface of separation of the two

parits of the m agnetic circuit is perpendicular to the lines of in

duction ; let S be the area of this surface in Square centim etersand let dcc be the infinitesimal am ount by which the two parts ofthe m agnetic circuit are separated . When dx is infinitely sm all

,

the change produced in the reluctance of the c ircuit is negligible ,and therefore the flux density in the air gap formed is the sam eas originally existed in the iron ; let this flux density be B and letit be assum ed constant over the surface of separation . Thevolume of the m agnetic field is changed by an am ount Sdsc , andconsequently the energy of the magnetic field

,from equation

is changed by the am ountB

8 7T

But the work done in separating the two parts of the c ircuit isalso equal to the product of the displacement dx ,

and the m echani

cal force F in dynes required to produce the separation, i .e.,

dW=Fdrc

Equating these two expressions , we have2

F =B S

8 7T


n the flux density is not constant over the surface of separaon

,the total force required is the surface integral


1 . The number of linkages between a line of induction and acoil of wire is equal to the number of turns Of this coil linked bythe line of induction . The total number of linkages between acoil and the flux threading it is the sum of the linkages of all thelines of induction . For a concentrated winding of N turns andeach turn linked by (f) lines , the number of linkages is

=N (bLinkages are expressed in mawel ls .

2 . Whenever the flux threading a circuit changes an electro

motive force is indu ced in this c ircuit equal to the tim e rate ofchange of the linkages between the flux and the ci rcuit , i .e.

,

£2 abvolts EL"dt dt

where X is in maxwells and t in seconds . The m inus S ign in theabove equation indicates that the direction of the electromotiveforce resulting from an increase in the flux in the sam e directionas that of the flux due to the current in the c ircuit produces anelectrom otive force in the opposite direction to that of the current .

3 . The Work done on an electr ic current when the number of

l inkages between the c ircuit and the flux linking it increases byan amount d lx is

dW id Xwhere i is in abamperes and X in maxwells

4 . When the change in the linkages between an electric c ircuitand the flux linking it is caused by the motion of a part of thiscircuit (eg . a wire) through a magnetic field, the induced electrom otive force is equal to the rate at which this part of the c ircuitcuts the lines of induction . The direction of the electromotiveforce is the direction in which the m iddle finger of the right handpoints when laid along the wire and the thumb and forefingerof this hand are held perpendicular to each other and to them iddle finger, w ith the thumb pointing in the direction of the


provided the flux density is uniform and perpendicular to the crosssection .

10 . The ampere-turns required to establish a given flux arefound by equating the drop Of m agnetic potential through them agnetic circuit to the m agnetomotive force, i .e .

,

lL |H dl = NI

where H is_in ampere- turns per inch , l is in inches , I in amperes

and the integral is taken around the closed path formed by themagnetic circuit .

0

1 1 . The value Of the ratio Of the electromotive force inducedin an electric c ircuit , due solely to a change of the current in this circuit

,to the tim e rate of this change , is called the self inductance

L of this circuit ; hence the self- induced electrom otive force is

e= Ldz

dl

The c . g. 3 . unit ofself inductance is the abhenry ; the practical unitis the henry .

1 henry = 109 abhenriesWhen the perm eability of every body in the magnetic field isconstant and the c ircuit rem ains unaltered in Shape

,the self in

ductance is a constant of the c ircuit (for a given distribution ofcurrent) equal to the number Of linkages between this c ircuit andthe flux produced by unit current in it .

12 . The va lue of the ratio of the electrom otive force inducedin an electric c ircuit 1

,due to a change of the current in any other

c ircuit 2,to the tim e rate Of change of this current

,is called the

mutual inductance M of c ircuit 2 with respect to circuit 1 hencethe electromotive force induced in 1 due to a change of the cur

rent i , in 2 is

Mdt

The units ofmutual inductance are the sam e as the units of selfinductance . When the perm eability Of every body in the mag

netic field is constant and the c ircuits are fixed with respect toeach other and rem ain unaltered in S ize and shape

,the mutual

inductance of one c ircuit with respect to the other is constant (fora given distribution of the currents) and equal to the number oflinkages between one c ircuit and the flux produced by unit currentin the other

,and the mutual inductance of the one c ircuit with

ELECTROMAGNETISM 3

respect to the other is the sam e as the mutual inductance of thesecond circuit with respect to the first .

13 . The energy of the magnetic field due to a current i in ac ircuit of constant self inductance L is

W Li‘

When L is in henries and i in amperes this formula gives the energyin j oules ; when L is in abhenries and i in abamperes this formula

gives the energy in ergs .

14 . The energy of the magnetic field due to the currents i , andi , in two circu its which have constant self inductances L, and L,

and a constant mutual inductance M is

W= g. Li z,

When the inductances are in henries and the currents in amperesthis formula gives the energy in j oules ; when these quantities arein abhenries and abamperes this formula gives the energy in ergs .

15 . When the current in a conductor of large cross sectionvaries rapidly with tim e

,the back electrom otive force due to the

variation of the flux linking the conductor is considerably greaterin the inside filam ents of the conductor than in the outside filam ents ; consequently the current density is greatest near the surface of the conductor . This phenom enon is known as the sk ineffect.

16 . The self inductance of each of two long paral lel nonm agnetic W ires, assum ing a uniform current density

,is

logD

m illihenries per m ile7'

where D is the distance between centers and r the radius of eachwire

,both in the sam e units .

17 The self inductance of a concentrated winding variesdirectly as the square of the number Of turns .

18 . The self inductance of a long solenoid is approxim ately

abhenries

where N is the t Of turns in the solenoid,A its cross

section in Square centim eters and l its length in centimeters .

19 . The energy per cubic centim eter of any m agnetic fieldwhen the permeability of every body in the field is constant is

10 2

,“s B

z

8 7T 8 7Tf1.


where H is the field intensity in gilberts per centim eter,B the

flux density in gausses and ,u.the perm eability in c . g. s . electro

magnetic units .

20 . The heat energy dissipated per cycl e per centim eter- cubein a magnetic substance when an alternating magnetic field isestablished in it is

"

equal to the area of the corresponding hysteresisloop divided by 4 7T.

2 1 . The force required to ‘

separate two parts Of a magneticc ircuit is

B2A

8 27

where A is the area in sq. cm . of the surface of separation and Bthe flux density in gausses norm al to this surface, provided B isconstant .

PROBLEMS

1 . A coil which has a concentrated winding of 100 turns isrevolved in a uniform magnetic field about an axis perpendicularto the direction of the field . The intens ity of the field is 100

gilberts per cm .

,the area of each turn of the coil is 10 Sq . cm . ;

the coil is rotated with a uniform veloc ity of 25 revolutions persecond . What is the ins tantaneous value of the e. m . f . inducedin the coil, ( 1) when its plane is perpendicular to the field

, (2).

whenits plane is parallel to the field ; (3) what is the average value ofthe e . m . f . induced in the coil while it is rotating 180

° from thefirst position?

Ans . : ( 1) 0 ; (2) volts ; (3) volts .

2 . How many foot-pounds of work must be done to pull acoil having a concentrated winding of 200 turns out of a magneticfield

,if the flux threading the coil is maxwells and the

current in it is m aintained constant at 100 amperes?Ans . : foot-pounds .

3 . A straight wire 10 inches long is m oved parallel to itselfacross a magnetic field in a plane m aking an angle of 45

° with thedirection of the field . What is the value of the e . m . f . in voltsinduced in the wire if it is moved with a velocity of 5 feet persecond and the field intensity is 200 gilberts per cm . ?

Ans . : volts .

4 . A c oil of 20 turns is wound over the m iddle of a long solenoidwith an air core . The solenoid has a cross- sectional area of 10sq . cm . and has 50 turns per linear inch . If the 20- turn secondary


the total inductance of A and B when connected in series, ( 1) so

that their magnetic fields are in the sam e direction, and (2) so thattheir magnetic fields are in opposite directions . (3) What is themutual inductance Of A and B?

Ans . : ( 1) henries ; (2) henries ; (3) henries .

1 1 . A coil having a concentrated winding of 1000 turns has aself inductance of henries . What is the energy in j oules ofthe m agnetic field due to the current in the coil when this currentproduces a flux Of 2000 m axwells?

Ans . : j oule .

0

12 . When a current of 4 amperes is established in the fieldwinding of a 4—pole shunt generator, the flux Of induction througheach field coil is maxwells . The four field coils are con

nected in series electrically and each has 800 turns . Assum ingcons tant perm eability and no m agnetic leakage , determ ine ( 1)the self inductance of the entire field winding, and (2) the totalenergy of the magnetic field of the generator .

Ans . : ( 1) 24 he’nries (2) 384 j oules .

13 . A cast- iron ring 10 inches in mean diam eter and 4 sq . in .

in cross section is divided into two halves . Upon each half of thering are wound uniform ly 100 turns of wire . If the two halves ofthe ring are placed together to form a complete ring and the twocoils are connected in series so that their m agnetic fields add

,

determ ine ( 1) the energy of the m agnetic field per cubic centim eterwhen the current established in the winding is 2 amperes , assum inga constant perm eability of 300 c . g . 3 . units

,and (2) the initial force

in pounds requ i red to pull the two halves of the ring apart .Ans . : ( 1) 473 ergs ; (2) pounds .

ELECTROSTATICS

127. Electric Charges .

— We have already seen (Article 64)that it is possible to have a difference Of electric potential betweentwo conductors without. having an electric current in these con

ductors . For example,the difference of electric potential between

the two poles of a battery when there is no conductor connectingits poles

,that is

,when the battery is open- ci rcuited

,

”is equal

to the electrom otive force of the battery . It is found by ex

perim ent that when a conductor A of any kind is connected to oneof the poles of a battery but not to the other, a m om entary electriccurrent is estab lished in this conductor

,but after a sm all fraction

of a second this current ceases . By Ohm’s Law

,every point of

this conductor must com e to the sam e potential as the term inalof the battery to which it is connected (neglecting any Slightcontact electromotive force between the conductor and theterm inal) . Sim ilarly

,in any other conductor B connected to

the other term inal of the battery but completely insulated fromthe first term inal

,a mom entary electric current will be established ,

but when this current ceases every point of the conductor Bwill be at the sam e potential as that of the term inal to whichit is connected . Hence between the two conductors A and B adifference of potential is established equal to that of the electrom otive force of the battery . It is also found by experim ent,that the wires connecting the two conductors A and B to thebattery may be rem oved without changing this difference ofelectric potential between them ,

provided the wires are sm alland the position of the conductors A and B relative to eachother and all other bodies in their vicinity rem ains fixed and theconductors remain perfectly insulated from each other and allother conductors . This can be tested by again c onnecting theconductors A and B to the sam e term inals of the battery ; it willbe found that no current , not even a m om entary current

,is estab

l ished in the wires,thus showing that each conductor rem ains

at the potential of the term inal to which it was originally con

237‘


nected . A difference of electric potential m ay then exist betweentwo conductors which are entirely insulated from each otherand all other conductors .

It is found by experim ent that whenever a difference of

electric potential is established between any two conductorsthese conductors exert a force upon each other and in genera lalso upon all other conductors and dielectrics in their vicinity

,

even when there are no currents in the conductors . These forcesare due solely to the difference of electric potential establishedbetween the various bodies , and as long as these potential differences rem ain constant in value the forces are found to rema inconstant . Just as the forces produced by magnets on one anotherand upon m agnetic substances may be expressed in term s of asom ething associated with the surfaces of the m agnets and mag

netic bodies, so m ay these forces due to differences of electricpotential be expressed in terms of a som ething assoc iated withthe surfaces of the cpnductors and the dielectrics in their vicinity .

The som ething that produces these forces which are due to differences of electric potential is called an electri c charge . We wishnow to see what are the properties of these electric charges andwhat is their relation to an electric current .

In the first place,it is found that forces Of exactly the sam e

nature are produced between any two dielectrics,such as glass

and S ilk,when these two bodies are rubbed together and then

separated ; also when an insulated conductor is rubbed with anydielectric

,such as a piece of silk

,and the conductor and the

dielectric are separated,it is

’

found that the two bodies attracteach other . Bodies which are charged in this way are said to be“charged by friction . In fact

,this m ethod of produc ing the

phenom ena which are ascribed to electric charges was known tothe anc ients ; whereas batteries and generators are comparativelyrecent inventions .

128 . Positive and Negative Charges . Attraction and Re

pulsion of Charged Bodies . It is found by experiment thatbodies which are charged in exactly the sam e m anner

,as for

example,two insulated conductors placed m om entarily in con

tact with the sam e term inal Of a generator, always repel eachother

,while they may repel or attract a body which has been

charged in som e other m anner . For example,an insulated

conductor A placed m om entarily in contact with the positive


It is also found by experim ent that a conductor can be chargedwithout placing it in actual contact with a charged conductor ;m erely plac ing the originally uncharged conductor U in thevicinity of a charged body C will cause the originally unchargedconductor U to becom e charged . In this case

,however

,when

the originally uncharged conductor U is perfectly insulated,it is

found that the portion of this conductor U nearer the chargedbody 0 receives a charge of the opposite S ign to that on C whilethe more rem ote portion of the conductor U receives a charge ofthe same S ign as that of C. In other words

,the conductor U

becom es charged by induction in a manner sim ilar to thatby which a piece of soft iron placed in a magnetic field becomesmagnetised by induction . There is

,however

,an important

difference between the phenom enon of electrostatic induction ina conductor and the phenom enon of magnetic induction in iron .

When a piece of iron which is m agnetised by induction is dividedinto two parts and the two parts are separated slightly, SO that anarrow gap is formed between them ,

magnetic poles are foundto exist on the surfaces of the iron form ing the walls of this gap .

When a conductor U (which may be made in two parts originallyin contact) is charged by induction and the two parts U, andU,are separated a slight distance , the surfaces of U, and U,

forming the wa l ls of thi s gap are not charged . Again, when thepiece of iron which is m agnetised by induction is separated intotwo parts and either part is removed from the magnetic field

,

this portion of the original piece of iron is found to have equaland Opposite poles which disappear alm ost entirely when theiron is j arred ; however , when a conductor which is charged byinduction is separated into two parts U, and U, ,

it is found thatthe two portions of the conductor retain their charges , one parta positive charge and the other part a negative charge , as long as

the two parts are kept‘

perfectly insulated from each other andfrom all other conductors . The charges on the two portions ofthe conductor also rem ain unaltered in amount (but not in distribution

,as we Shall see later) even when “these two portions

are removed from - the vicinity of the conductor which inducedthe charges and are separated from each other by any distance .

Hence another important difference between the phenom enonof electrostatic induction in a conductor and the phenom enonof magnetic induction ; the total strength of the poles on a mag

ELECTROSTATICS

netic substance is always zero, no matter into how many piecesthe substance may be broken, but the positive and negativecharges induced on a conductor may be separated from each otherby dividing the conductor .

Experim ent also shows that electric charges are induced on adielectric which is placed in the vicinity of a charged body, inthe same manner that charges are induced on a conductor . In

the case Of a solid or liquid dielectric separated from the chargedbody by air the portion Of the dielectric nearer the charged bodyshows a charge of the Opposite S ign to that on the charged body ,while the portion Of the dielectric m ore rem ote from the chargedbody shows a charge of the sam e S ign as that on the charged body .

However,when a dielectric which is thus charged by induction

is separated into two parts , it is found that charges are produced

onthe walls of the gap between the two parts, a negative chargeon the side of the gap nearer the positive induced charge on theoriginal surface Of the dielectric , and a positive charge on the s ideof the gap nearer the negative

‘

charge induced on the original surface of the dielectric . A lso

,when the dielectric as a whole

,or

any portion of it,is removed from the vicinity of the charged

body which induces the charges on it, it is found that these induced charges entirely disappear . In other words

,the phenome

non of electrostatic induction in a dielectric is”

of a s im ilar natureto the phenom enon ofmagnetic induction in a magnetic substance ,except that it is impossible to produce by electrostatic inductiona perm anently electrified dielectric . The above facts aretrue only Of a di electric which is an absolute non- conductor .Every dielectric , however, is a conductor to a certain extent

,and

the resultant effect produced in a dielectric when it is placedin the vic inity of a charged body is therefore a comb ination ofthe effect due to its true dielectric or insulating property andthe effect due to its conducting property . We shall see laterhow both these effects m ay be taken into account .

An electric charge can be induced only at the surface of

separation of two dissim ilar substances . There is no way of

determ ining experim entally on which of the substances incontact the charge is induced ; for example , when a piece of

glass is placed in air in the vicinity of a charged body, thereis no way of determ ining whether the induced charge at thesurface of separation between the air and the glass is “

on”


the surface of the air or on the surface Of the glass . How

ever,j ust as in the discussion

(

of the phenom ena of magnetismit was found convenient to assum e that air Was non-magnetic (andtherefore at the surface of separation between air’ and any othersubstance the i nduced poles are entirely on the other substance) ,so in the discussion Of the phenomena of electrostatics

,it is found

convenient to assum e that at the surface of separation betweenair* and any other substance , there is no charge induced on thea ir . In general , at the surface of separation between any othertwo substances

,whether they ‘be dielectrics or conductors

,it will

then be necessary,in order to account for the Observed phenomena

,

to assum e that a charge is i nduced on the surfac e of both thesesubstances .

130 . Point-Charges . It is found to be imposs ible to producea finite electric charge at a point in space

,but j ust as in the dis

cussion'

Ofmagnetic phenom ena it was found convenient ‘ to makeuse Of the conception of a point-pole

,so in the discuss ion of

electrostatic phenomena it is convenient to cons ider a chargeof finite amount concentrated in a point . Such a charge may

be called a pointcharge. A physical approximation to a pointcharge is the charge on a small area .

131 . Properties of Electr ic Charges .

—When all the electriccharges produced in any manner whatever, inc luding

‘

the charges

induced on“

di electri cs,are taken into account , it is found that the

forces produced on one another by any number Of charged bodiesmay be accounted for by attributing to these electric charges thefollowing properties

1 . Like charges repel each other and unlike charges attracteach other .

2 . When a charge of one S ign is produced on any body anequal and opposite charge is produced

‘

either on the sam e or onsom e other body

3 . Two point- charges q and q’ located at a distance r apart

repe l each other with a force proportional to the products of thequantities q and q

’ of these charges . and inversely proportionalto the square Of the distance between them ,

independent of the

nature ofthe"

medium between them; that is, with a force

*Strictly ,this assumption is perm issible only for one definite pressure

a pressureof 760 mm . Of m ercury is the adopted standard .


fact that equal and Opposite poles always exist on the sam ebody .

*

132 . Electrostatic Field of Force . Electrostatic Intensity.

Any region of Space in which an electric charge would be actedupon by a mechanical force is called an electrostati c fie ld of force.

The intensi ty of the e lectrostati c fie ld at any point is defined as

the force in dynes which would be exerted by the agents produc ing the field on a unit pointcharge placed at that point .

The nam es e lectrostati c intensi ty and e lectrostati c force are alsoused for this quantity . The electrostatic intensity at any pointdue to a pointcharge q at a distance r centim eters away is then

q (2)12

and the total field intens ity at any point P due to any numberof point- charges is the vector sum

H =§fi

where q is the value of the pointcharge at any point and r is thedistance in centimeters of this charge from the point P ,

and thesummation inc ludes all the charges , induced or otherwise

,on all

the conductors and dielectrics in the field .

The total field intensity at any point P due to any chargedsurface is the vector integral

where ds is any elem eritary area of the surface,or is the charge

per unit area,or the surface destiny of the charge at ds , and r is

the distance of ds from P .

133 . Lines of E lectrostatic Force . Flux of E lectrostaticForce . Lines of electrostatic force may be drawn in the sam emanner as lines Of magnetic force , that is , lines of electrostaticforce are lines drawn in such a manner that the direction of eachline at each point comc l des with the direction of the electrostaticintensity at that point and the number Of these lines per unitarea at any point

,norm al to their direction , is equal to the value

of the electrostatic intens ity at that point . The number Of*The sam e symbols will as a ru le be used throughout for the corresponding

quantities ; in any problem in which both electrostatic and magnetic quantitiesmust be used , the form er m ay be distinguished by the subscript “ e .

ELECTROSTATICS 245

these lines of electrostatic force crossing any area is defined as

the flux of electrostati c force across that area . The m athematicalexpression for the flux Of electrostatic force across any surfaceS is then

it fS (H COS a ) ds (3)

where ds is any elementary area of this surface,H the elec

trostatic intensity at ds and a the angle between the directionof this electrostatic intens ity and the nor mal to ds . Gauss ’sTheorem

,which is simply a consequence of the inverse square

law,also holds for electric charges ; that is , the number of lines

of electrostatic force outward ac ross any closed surface is equalto the

”

algebraic sum of the charges ins ide the surface . Themathem atical expression Of this fact is

f|s I(H cos a ) ds =4 n 2 q

where fiSI represents the integral over the c losed surface .

134 . Lines of E lectrisation .— We have already seen that

when a charged conductor is separated into two parts by a narrowgap ,

there is no electric charge produced on the walls of this gap .

It is also found that when a closed cavity of any kind is form edinside a charged conductor

,no electric charge is produced on the

walls of this cavity no matter how the external surface of theconductor is charged . A charge can be produced on the wallsof such a cavity only by introducing a charged body intO ' this

cavity . Hence in a charged conductor there i s nothing ana logous

to l ines ofmagnetisati on inside a magneti sed body.

In the case of a dielectric charged by induction, however, wehave seen that in general charges do appear on the walls Of a

gap cut in the dielectric,j ust as m agnetic poles appear upon the

walls of a gap cut in a m agnetised body . It is possible,however

,

j ust as in the case of a magnetised body, to cut in a dielectricwhich is charged by induction a gap in such a direction thatno charge will appear on the walls Of the gap . Hence we may

cons ider a dielectric which is charged by induction to be m adeup of filam ents the walls of each of which have such a directionthat were this filam ent separated from the rest of the dielec tric bya gap of infinitesim al width , there would be no charges inducedon the lateral walls of this filament . The two ends of such afilament in the original surface of the dielectric will then have


equal and opposite charges . We’

m ay then take each filament ofsuch a size that it term inates at each end in a charge which has

the numerical value1

The pos itive sense of such a fila4 n

m ent is taken as the direction along it from its negative to itspositive end ; that is , these filam ents are considered as runningthrough the dielectric from its negatively charged end to itspositively charged end . These filam ents are called l ines ofe lectri sation . The intensity of electrisation at any point in abody is defined as the charge per unit . area which would appearon a gap cut in the body at this point perpendicular to the line

.

Of electrisation through this point . The direction Of the intens ityof electrisation at any point is taken as the direction Of the lineOf electrisation through this point . Lines Of electrisation

,lines

of electrostatic force,and the electric charge induced on the surface

of a dielectric are then related to one another in exactly the sam emanner as lines ofmagnetisation, lines of m agnetic force and themagnetic poles induced on the surface of a magnetic body . Thatis

,the number of lines of force ending on a negative charge in

duced on a dielectric is equal to the number of lines of electrisationoriginating from that charge , and the number of lines of electrisation ending on a positive charge induced on the surface Of a dielectric is equal to the number of lines Of electrostatic force originatingfrom that charge . The relation between the intens ity of electrisa

tion at any point in the surface of a dielectric and the chargeinduced on that surface is then

(T J cos a (4)where 0

' is the surface density Of this 1nduced charge at thispoint

,J the intens ity of electrisation of the dielectric at this

point,and a the angle between the direction Of the intens ity of

electrisation at this point and the direction of the normal drawnoutward from the surface of the dielectric at this point . (SeeArtic leThe number of lines of electrisation crossing an elem entary

surface ds the normal to '

which m akes an angle a. with thedirection of the intensity of electrisation is

dN =4 7T( J cos ds (5)Since we have assum ed (Article 1 29) that no charges are in

duced on the Surface of the air in contact with any other substance , the intensity of electrisation in air is zero .


are identical,and therefore the electrostatic intensity and the

electrostatic flux dens ity in air are equal .136 . Dielectric Constant . - Experim ent shows that the elec

trostatic flux density produced in any dielectric when placedin an electrostatic field is directly proportional to the resultantfield intens ity

,and except in the special case of certain kinds of

crystals,the direction of the electrostatic flux density coinc ides

with the direction of the field intensity . The ratio of theelectrostatic flux density to the field intensity is called thedielectric constant of the dielectric . That is

,calling B the elec

trostatic flux density at any point in the dielectric,and H the

resultant electrostatic intensity at this point,the dielectric con

stant of the dielectric at this point isB

H

The dielectric constant is exactly analogous to m agnetic perme

ability . The values of the dielectric constants for a few sub

stances areSolid paraffinParaffin oil

EboniteMicaGlass toDistilled water 76

Al cohol 26

Ordinary gasesPerfect vacuumCombining equations (6) and (8) we have that H + 4 7rJ

KH and therefore that

(K— l ) H

4 77

Hence in any dielectric for which the dielectric constant is unitythe intens ity of elec trisation is zero

,and therefore there are no

lines of electrisation,and hence there are no charges induced on

the surface of such a dielec tric . In a dielectric for which thedielectric constant is greater than unity , the intens ity of electrisation is not zero but a positive quantity , and therefore the fieldintensity is less than the flux density ; hence the number of linesOf electrostatic force ins ide a dielectric of which the dielectric

ELECTROSTATICS 249

constant is greater than unity is always less than the numberof lines of electrostatic induction .

137. A Closed Hollow Conductor an Electrostatic Screen .

It is found by experim ent that an electric charge'

cannot be produced on the inside surface Of a hollow c losed conductor bycharging the conduc tor either by contact with

,or by induction

from,any agent whatever outside the conductor ; nor can an

electrostatic field be produced inside the space enclosed by theconductor by any agent whatever exterior to the conductor

,

provided 1n each case there is noelectric current in the conductor .These two important facts were discovered by Faraday

,who

built a_

large insulated m etal chamber in which he set up them ost delicate m easuring instrum ents he could devise ; he wasunable to detect the slightest ,

electrostatic effect inside thischamber no m atter how highly the chamber was electrified onthe outside . Al l deductions from this discovery of Faraday’shave been found to be in accord with experim ent

,and we there

fore accept as a fundam ental law Of electrostatics that anyregioncompletely enclosed by a conductor is absolutely protected fromexternal electrostatic effects . It is also found that a metal gauzeor cage forms a practically perfect screen .

138 . The Resul tant E lectrostatic Intensity With in the Substanceofa Conductor in which there is no E lectric Current is Always Z ero .

An important deduction from Faraday ’s discovery is thatthe resultant electrostatic intensity within the substance of acharged conductor i s zero , provided there 1s no current in the conductor

,no m atter how the conductormay be charged . For

,should

a c losed cavity be made in this conductor, no charge would appearon the walls of this cavity

,and therefore the original electro

static field would not be altered ; therefore the electrostatic fieldinside this cavity must be the sam e as originally existed in

‘ theconducting material which filled this cavity

,s ince the electro

static intens ity at any point due to any given distribution Ofelectric charges is independent o f the m aterial at the point inquestion . But the electrostatic intensity in the cavity is zeroand since the charges produc ing the field are not altered byform ing the cavity

,the electrostatic intensity in the conducting

material which originally filled this cavity must a lso have beenzero .

Since the resultant intens ity in a conductor is zero,it follows


that when a conductor is placed in an electrostatic field,the induced

charges which appear must be distributed in such a manner thatthe

,field within t he substance

j

of the c onductor due to these induced charges is j ust equal and opposite to the original field inthe space occupied by ' the conductor .

139 . The Total Charge With in any Region Completely Eu

closed by a Conductor,is Always Z ero .

4 We have j ust seen thatthe electrostatic intensity is zero within the substance of a con

ductor in which there is no electric current . Hence when anyregi on b f space is completely enclosed by such a conductor thetotal flux -of electrostatic intensi ty out through any surfacedrawn within the substance of this conducting shell is zero

,since

there are no lines of electrostatic force crossing this surface .

Hence,by Gauss ’s Theorem ,

the total charge inside such a surfaceis zero . A lso

,

- since it is one of the fundamental properties ofelectric charges that a charge of one sign cannot be producedwithout at the sam e time produc ing an equal charge Of the op

posite sign , the total charge outside such a c losed surface mustalso be zero . Hence a

,c losed conducting surface divides all

Space into two parts in.,

each of which parts the total charge iszero . For example

,let A be an insulated conductor which has

Fig . 74

a charge of Q units then on som e other conductor or conductors Bthere must be an equal and opposite charge Of Q units . Let A

be completely surrounded by an insulated conducting shell C;there must then be induced on the inside of this shell a chargeof Q units and on the outside of this shell a charge of Q units .

140. The Electrostatic Intensity Just Outside a Charged Con

ductor is Normal to the Surface of the Conductor .

—By the sam eprocess of reasoning as that employed in Article '

50,by which

it was shown that the tangential components of the m agneticfield intens ities j ust inside and j ust outside a magnetically charged


the surface,the elec trostatic intens ity outside the conductor

will be norm al to ds . Hence the total flux of electrostaticforce across the outside end of the cylinder wil l be Hds .

A cross the lateral walls of the cylinder outside the conduc tingsurface the flux of force will be zero

,since these lateral walls

are parallel to the direction of the field intens ity . A cross thewalls of the portion of the cylinder ins ide the surface the fluxof force will likewise be zero , since there is no electrostatic fieldins ide the conductor . Hence the total flux Of force outwardthrough the walls of this cylinder is Hds . By Gauss

"s Theorem

thismust be equal to 4 7T tim es the total charge ins ide this cylinder,that is

,equal to 4 7 ds . Hence

H 4 7r (0'

But since the lines of electrisation (which coinc ide in directionwith the lines Of force) are perpendicular to the surface at P andare into the dielectric from its surface

,the intens ity of electrisa

tion of the dielectric j ust outside the conductor at P isJ=—d

see equation From equation (6) the electrostatic flux densityat P is B = H+ 4 7TJ Therefore substituting for H and J theirvalues in terms of 0 ' and a

"

,we have

That is,the electrostatic flux dens ity at any point j ust outside a

conductor isB = 4 7T 0

'

(10)where 0

' is the surface dens ity of the charge on the conductor at

this point,and i s independent ofthe nature ofthe di e lectri c in con

tact wi th the conductor at thi s point.* The resultant electrostatic

intensity j ust outside a charged conductor,on the other hand

,

does depend upon the nature of the dielectric in contact with theconductor

,and is equal to

4 7T0”

K (1 1)

where K is the dielectric constant of the dielectric in contactwith the conductor at the point in question .

An important difference between lines of magnetic induction

*In the-deduction ofthis relation the contact charge , ifany , produced on thesurface of, the dielectric

,is neglected . In case there is a contact charge on the

d ielectric this sam e relation holds when 0 '

is taken to represent the surfacedensity of.the charge on the conductor plus the contact charge on the dielectric .

ELECTROSTATICS 253

and lines of electrostatic induction should be noted in this con

nec tion . Lines of magnetic induction are always closed curves .

On the other hand,lines of electrostatic induction which are

due entirely to electric charges originate from and end on con

ductors (or contact charges on dielectrics) ; 4 77 of them o riginatefrom each unit positive charge on a conductor and 4 77 of themend on each unit negative charge on a conductor . They do not

,

however,end on charges induced on dielectrics

,but pass through

these charges (see Article general m aking an abrupt changein direction at the surface of a dielectric on which a chargeis induced . Lines of electrostatic force

,however

,are strictly

analogous to lines Of magnetic force ; 4 rr of them originate fromevery unit positive charge and 4 7T Of them end on every unitnegative charge , whether these charges be on a conductor or areinduced charges on a dielectric . Hence in any dielectric whichhas a dielectric cons tant greater than unity there are fewer linesof force than lines of induction

,for the lines of induction all pass

through the dielectric while some of the lines of force end on itssurface .

142 . Conditions which Must be Satisfied at any Surface in an

Electrostatic Field. By identically the sam e process of reasoningas that employed in Artic les 49 and 50 it can be shown that atevery surface in an electrostatic field other than a conductingsurface the following conditions must be satisfied

1 . The normal components of the electrostatic induction onthe two sides of any surface at any given point in this surfacemust be equal .

2 . The tangential components Of the electrostatic intensityon the two sides Of any surface at any given point in this surfacemust be equal .

At the surface of a conductor the surface conditions are,as

we have j ust seen,that

1 . Both the electrostatic intens ity and the electrostatic induction inside the surface are zero .

2 . Both the electrostatic intens ity and the electrostatic induction j us t outside the surface are normal to the surface .

From these surface conditions the distribution of the chargeson conductors and those

-

induced on dielectrics may be calculatedin certain simple cases .

143 . Electrostatic Potential . The work that would be done


by the agents producing an electrostatic field on a unit positivepoint- charge, were such a pointcharge moved from any pointP in the field to infinity

,is ca lled the e lectrostati c potentia l of the

field at the point P . The electrostatic potential at any point Pdue to a point-charge g at a distance r from P is then

V 51.r

and is independent of the path over which the charge is m ovedfrom P to infinity

,and therefore depends only upon the relative

pos ition of the charge and the point P . (See Artic le Sinceelectrostatic potential is the ratio of work to charge , and sinceboth work and charge are scalar quantities

,electrostatic poten

tial is also a scalar quantity . Hence the resultant elec trostaticpotential at any point due to any number of electric charges isthe algebraic sum of the potentials due to each charge separately .

Therefore the electrostatic potential at any point P due to aCharged surface of any kind is

where”

ds represents any elem entary area of the surface,0

" thesurface density of the charge at ds , and r the distance of the pointP from ds .

144 . Difi erence of E lectrostatic Potential . The difference ofelectrostatic potential between any two points 1 and 2

,or specifi

cally,the drop in electrostatic potential from the point 1 to the

point 2,is the work that would be done by the agents produc ing

the electrostatic field on a unit positive pointcharge were sucha charge moved from the point 1 to the point 2 . Calling dl

an elem entary length in any path from the point 1 to the point2,H the electrostatic intensity at dl

,and (9 the angle between

the direction of the electrostatic intens ity H and the directionof the length dl , the drop of electrostatic potential from 1 to 2along this path is then

(H cos d ) al l

When there is no contact or induced electrom otive force in thepath from 1 to 2

,the drop of electrostatic potential from any point

1 to any pomt 2 is'

independent of the path over which the charge


When a conductor in which there is no electromotive forceis placed in the electrostatic field

,the conductor must become

an equipotential surface,for we have already seen that the electro

static intens ity is always norm al to the surf ace of a conductor inwhich there is no electric current

,and therefore the drop of

potential along any path in the surface of the condu ctor is zero .

146. Para llel Plate E lectrometer.— The following example

will illustrate the principles stated above,and will at the same

tim e show how electric charge and difference of electrostaticpotential may be measured .

,

By the sam e process of reasoningas that employed in Article 37 it can be shown that the electrostatic intensity due to a uniformly charged plane surface is inthe direction norm al to this surface at any point at a perpendicular distance from the surface small compared with the distanceof the point from the perim eter of the surface

,and that its value is

H = 2 7T (Tn (14)where O'

nis the surface density of the net charge at this surface .

By net charge is m eant the algebra ic sum of the charge on theconductor

,the contact charge ( if any) on the dielectric in contac t

with the conductor,and the charge induced on the dielectric .

Consider two equal plane m etal discs A and B placed paral lel

A a , and opposite to each other and ima‘

bm ersed in a dielectric of which the

bfdielectric constant is K,

and let thisFig . 76. dielectric c ompletely fill all the sur

rounding Space . Let the plates A and B be given charges OfQ and Q units respectively . The only charges induced on

the dielectric will then be those at the surface of separationbetween the metal plates and the dielectric . The charges Qand Q and the charges induced -on the dielectric must be distributed in such a manner that ( 1) the electrostatic intensity atany point inside the plates is zero

,and (2) the electrostatic in

tensity at each point in the dielectric infinitely close to the surfaceof the plates is norm al to the surface at this point . A uniformdistribution of the net charge at the two surfaces a and b whichface each other and no charge on the back ”

surfaces a ’ and b’ willsatisfy these conditions for all points except those near the edgesof the plates . For

,ca lling O n

and (Tn the surface densities of

the net charge at the surfaces a and b respectively,the resul

tant intensity (from equation 14) inside each plate will be

ELECTROSTATICS 57

2 7T O'

n2 7T 0

,and the intens ity at any point in the space

between the plates will be 2 Tr O'

n + 2 7T=4 7T O n and will be

perpendicu lar~

to these surfaces,while at each point infinitely close

to the surfaces a’ and b’ the intens ity will be 2 77 (Tn 2 77 U

,,=0

,

which relations satisfy the Surface conditions 1 and 2 . It can alsobe shown m that there 1s onl y one possible distribution Of cha rgeswhich can satisfy these surface condi tions ; hence at all points at aconsiderable distance from the edges of the plates the charges atthe surfaces of separation between the dielectric and the platesmust be distributed uniform ly over the surfaces a and b and therecan be no charge on the surfaces a

’ and b’ (except near theiredges) .

Let O’ be the value of the total charge on the conductor andthe contact charge , if any, on the dielectric at each point Of thesurface a and O " the value of the charge induced on the dielectric atthe surface Then O'

n=O

" J,where J is the intens ity

of electrisation of the dielectric in the direction from a to b. But

J (K 1) H

4 7r

at any point in the Space between the plates,not too c lose to their

edges , is

equation (9a) , whence the resultant field intensity

4 7r 0‘

K

where 0 ' i s the surface densi ty ofthe charge on the conductor . (Com

pare with equation This electrostatic intens ity is thereforeperpendicular to the surfaces of the plates

,and the electro

static field between the two plates is a uniform field except forpoints not too close to the edges of the plates .

The force of attra ction between t he two plates may be readily

_

4 n a

Kbe due to each plate

,that is the resultant field intensity due to each

plate is2 7

120This

,then

,is the value of the force due to the p late

A,say,

on each unit charge on the plate B (or, vice versa , the forcedue to the plate B on each unit charge on the plate A ) . Hence

the pull on each unit area of the plate A is2

linear dim ensions of the plates are equal and are large compared

andwhen the


to their distance apart, so that the non-unnormity of the distribution Of the charge at their edges m ay be neglected, the total forceOf attraction between the two plates is

2 7r 0' 2 S 2 77

'

Q2

K K S

and thereforeKSF

where S is the area of each plate and Q the total charge on eachplate . The force F may be readily m easured by suspendingone of the plates from one arm of a suitable chem ical balance .

When the plates are separated by air, the force of attraction2 7T Q

2

S

stant of air is unity . In this'

case the charge on each plate isnumerically equal to

between the two p lates is F : s ince the di elect ric con

Q (15b)

Hence,by measurmg the force of attraction between the two

plates the charge on each p ate may be calculated . Again ,from the fact that the electrostatic field between the two platesis uniform ,

the difference of electrostatic potential between theplates may also be calculated . For

, calling D the distancebetween the two plates

,we have

,from equation that this

difference of electrostatic potential is4 7r O

' D 4 7r Q D 8 7T F

K K S K S (16)When the plates are separated by air, the dielectric constant isunity

,and therefore the difference of electrostatic potential

between the two plates is8 7r F

S

The arrangem ent of parallel plates above described , whicharrangement is called a para l le l p late e lectrometer

, gives us ameans of m easuring both electrostatic charge and electrostaticpotential difference . The above calculations are all based uponthe assumption that the non-uniform ity Of the charge near theedges of the plates may be neglected . The ~ condition of uni

V ==D


the wire connecting the two plates of the electrom eter and con

nect the other ends of the wires W and W’ to the plates A and Brespectively . The following phenomena will be Observed :

1 . A momentary electric current i s established in the twogalvanometers , as Shown by their deflections, and therefore a

certain quantity of electricity as defined in A rticle 80 is transferred across each sec tion of the wire form ing the galvanom eterwindings and the wires W and W’

. If the galvanometers havebeen calibrated (see Article 109) the quantity of electricity transferred through each can be measured and the direction of its

trans fer can be observed . It will be found that equal quantitiesof electric ity are transferred through each galvanometer (providedthe connecting wires are small) and that the di rection of transferor flow of. electricity is from the plate B which is connectedto the negative term inal of the generator and towards the plateA which is connected to the positive term inal of the generator .

2 . The plates A and B will become charged , as shown by theirmutual attraction , the plate A positively and the plate B negatively. The charges given the two plates will also be num ericallyequal

,as can be tested by Observing the force which each produces

on som e other charged conductor when placed successively in thesame relative position with respect to the latter .When the value Of the impressed electromotive force

,or

the distance apart of the plates,or the s ize of the plates

,or the

dielectric between them ,is altered

,it is found that the charge

given each plate also changes ; the quantity of electricity transferred through each galvanometer likewise changes . In eachcase , however, it is found that the charge given each plate isdirectly proportional to the quantity Of electric ity transferredthrough each galvanom eter . When the charge given each plateis m easured in electrostatic units , as defined in Article 13 1 , andthe quantity Of electric ity trans ferred through each galvanometeris measured in electromagnetic units as defined in Article 80 , it isfound that the num erical value of the charge given each plateis equal (approximately) to 3 x 101

° times the numerical valueOf the quantity of electric ity transferred through each galvanometer . This figure 3 x 10

10 is approximately equal to thevelocity Of light in a ir . Cons equently, if as the unit of chargeis taken a unit 3 x 1010 times the s ize Of the electrostatic unitas defined in Artic le 1 3 1

,the num erical value Of the electric

ELECTROSTATICS 261

charge given each plate w ill be equal to the number of electrom agnet unitso f quantity of electricity transferred to it ; sim ilarly,if as the uni t of charge is taken a unit 3 x 10

° times the size ofan electrostatic unit the num erical value Of the electric charge

given each plate will be equal to the number of coulombs transferred to it . E lectric charge and quantity of electric ity may

then be measured in the same units ; these units are related as

follows1 c . g. s . electromagnetic unit or abcoulomb

—3 x 1010

c . g. s . electrostatic units1 coulomb - 3 x 109 c . g . s . electrostatic units

(The figure x 101° is about as accurate as can be determ ined

by a direct comparison Of electrom agnetic and electrostatic units .

The electrom agnetic theory of light requires that ratio betweenthe electrom agnetic and electrostatic units be exactly equal tothe velocity of light in air , and the velocity of light by directoptical methods i s found to be 2 .998X 10

10centim eters per second .

The experim ental fact that the ratio of the two kinds of unitsis

,as c losely as can be measured , equal to the velocity of light

in air,is strong evidence for our belief in the truth of the electro

magnetic theory .)The phenomena just described are revers ible ; that is , when

the charge on a conductor disappears a current is established inthe conductor which trans fers from each element of the surfaceof the conductor a quantity of electric ity equal to the chargeoriginally on this surface . For example

,when the two plates

of the electrom eter described above are disconnected from the

generator and then connected to each other_

through a ballistic

galvanom eter, it is found that the quantity of electricity transferred through the galvanom eter is exactly equal to the quantityof electricity originally transferred through the galvanom eterswhen the plates were charged (provided the plates are otherwiseperfectly insulated) , and the transfer of charge is in the directionfrom the positive to the negatively charged plate ; the plates alsono longer attract each other, i .e .

,the charges on the two plates

disappear .148 . Relation between Electrostatic Potential and E lectric

Potential . The drop of e lectrostati c potential between anytwo points has been defined as the work done by the agents produc ing the electrostatic field when a unit posi tive point

-charge i s


moved from one point to the other , and the drop Of e lectri c potentialbetween any two points as the work done by a continuous currentin an insulated conductor connecting the two points per uni t

quanti ty ofel ectri city transferred across each secti on ofthe conductor .

Hence,when as the unit of charge is taken the electrom agnetic

unit as above defined and the erg is taken as the unit of work,

the unit of electrostatic potential drop is the number of ergs Ofwork done when one electromagnetic unit of charge is m ovedfrom one point to the other ; this unit is called the electrom ag

netic unit of .electrostatic potential difference . Sim ilarly,when

the electrostatic unit is taken as the unit of quantity Of elec tricityand the erg as the unit of work

,the unit of electric potential

drop is the number of ergs Of work done when one electrostaticunit of quantity is transferred across each section of the con

ductor this unit is called the electrostatic unit Of electri c potentialdifference . The relations between these various units Of potentialdifference are given in Article 87 .

In general , every source of electromotive force,either contact

or induced,produces an electrostatic field and electric charges

appear on any conductors in the vic inity . The results of allknown experim ents show that the drop of e lectrostati c potentia l

through the dielectri c between any two points whi ch are connected

by conductors i s exactly equa l to the drop ofelectri c potentia l through

the conductors between these points , provided both potentia l drops

are expressed in the same units . For example,in the experim ent

considered above,when the momentary currents in the galvanom

eters cease,the plate connected to the positive term inal of the

generator must be at the sam e electric potential as this ter minal

(this follows from Ohm ’s Law) ; sim ilarly , the plate connected to

the negative term inal must be at the sam e electric potential as thislatter term inal . Hence the difference of electric potential betweenthe two plates is equal to the electromotive force of the generator ,which electrom otive force is readily m easured by any of themethods described in Chapter V . The difference of electrostaticpotential between the two plates can be determ ined by the m ethodgiven in Article 146. When the values of the potential

‘

drops

Obtained from these two m easurements are expressed in the sam eunits they are found to be equal .When the drop of potential through the conductor is due solely

to its resistance , i .e., when there is no induced or contact electro


conductor. There are two important differences between the

properties of an electric current which can exist in a perfectdielectric and the properties of a current in a conductor . ( 1)The current in a conductor may be

“continuous

,

” that is,not

varying with tim e , or it may be variable ; in a perfect dielectric ,however

,only a variable current can exist . (2) The current in

a perfect dielectric does not develop heat energy, while in a con

ductor heat energy is always produced . In an imperfect insulatorboth kinds of currents may exist ; that part of the total currentwhich develops heat energy due to the resistance of the dielectricis called the conduction current, while that part which dependsupon the tim e variation of the electrostatic field

,and therefore

upon the time variation of the charges on the conductors incontact with the dielectric, is called the disp lacement current .

In the case of a conductor in contact with a perfect insulator,

the charge given any elem ent of the surface of the conduc tor inany sm all interval of tim e dl is equal to the quantity of electricitydq= idt trans ferred to that surface by the current i in the con

ductor provided charge and current are m easured in the samesystem of units ; hence , s ince in this case the onl y current leavingthat surface is the displacement current in the dielectric

,it fol lows

that the displacem ent current leaving that surface i s

dl

That is,the disp lacement current leaving the surface ofa conductor

in contact wi th a perfect di electric i s equa l to the time rate ofchangeofthe charge on that surface . When the dielectric in contact withthe conductor is not a perfect insulator, part of the conductioncurrent com ing up through the conductor to its surface leavesthat surface as a conduction current through the dielectric .

Hence calling id the displacem ent current in the dielectric and i 0the conduction current in the dielectric , we have that the totalcurrent com i ng up to the surface at any instant is

i = i d+ i c 18)

that is,the total current com ing up to a surface at any instant is

equal to the total current leaving that surface at that ins tant .

dq,where Q is the time rate of change of the charge

dl dl

on the surface , we have that

ELECTROSTATICS 265

dq (i—i c ) dt

whence q=f: (

”i—lo) dt (19)

where q 18 the total charge given to the surface in any intervalof tim e t. That is

,the tota l charge given the surface ofa conductor

in any interva l oftime i s equa l to the a lgebra ic sum ofthe quanti ti esof e lectri city transferred to that surface in thi s interva l by the con

duction currents flowing toward this surface.

Since the electrostatic flux dens ity at a point j ust outside acharged conductor is B r—4 7 where O

' is the surface densityof the charge on the conductor, it follows that the current dens ityof the displacem ent current j ust outside a charge conduc tor

,

which must be equal to the rate of change of the surface densityof the charge on the conductor, is

d o 1 dB

dl 4 77 dl

1 dBThe results of al l known expenments Show that

4 dt77

perfectly general expression for the current density Of the displacement current at any point in a dielectric

,when B represents the

electrostatic flux density at that point . The quantity 3 is4 77

sometim es called the electri c displacement; it represents the quantity Of electric ity transferred by the displacem ent current, or di s

p laced , across unit area perpendicular to the direction of theelectrostatic flux density .

The reader m ay find it diffi cult at first to grasp the idea Ofdisplacem ent current in an insulator , but a re- reading Of Article64 will make evident how such a variable current might occurin a dielectric . If electric ity has the property of an incompressible fluid filling all space , then what is called a positive chargeon a conductor represents a displacem ent of electricity in onedirection across the surface of separation between a conductorand a dielectric

,and what is called a negative charge on a con

ductor represents a displacem ent of electric ity across this’

surface

in the opposite direction . A corresponding displacem ent of

electricity occurs in the dielectric and also in the conductor,

but in the case Of a dielectric the am ount of displacement is lim itedby a property of the dielectric analogous to the elastic propertyof a solid

,such as that form ing the walls of the cellular structure

266 ELECTRICAL '

ENGINBERING

described in A rticle 64,while a conductor has no such elastic

property . What is called a charged surface is then analogousto the surface of separation between two bodies which are differently strained .

150 . Dielectric Strength . Discharge from Points . Corona .

Since the quantity of electric ity displaced across unit areain a dielectric is proportional to the electrostatic flux density atthis area

,which in turn is proportional to the electrostatic inten

sity at this point , the strain produced in the dielectric at any pointm ay be looked upon as proportional to the field intens ity at thatpoint . It is found by experim ent that when the electrostaticintensity at any point in a dielectric is increased up to a certaindefinite value

,the dielectric is ruptured at this point

,as is evi

denced by the Spark which occurs,which in general extends

through the dielectric from one Of the charged conductors to theother . This is analogous to rupture of any substance when it ism echanically strained up to its breaking point . The value Of theelectrostatic intensity which causes the rupture of a dielectricis called the di e lectri c strength Of the dielectric .

Recent experim ents Show that the dielectric strength of or

dinary insulators , such as rubber,m ica

,porcelain

,cloth

,etc .

,

depends not onl y upon the chem ical nature of the insulator butalso upon its thickness . The dielectric strength Of a thin piece

,

of insulation is greater than the dielectric strength of a thickpiece ; also , several thin layers making a given thickness have a

greater dielectric strength than a single layer of the sam e totalthickness . In engineering work dielectric strength is usuallyexpressed as so m any volts per inch ; see Article 102 .

Since the electrostat ic intensity just outside a conductor isproportional to the surface density of the charge on the conductor

(see equation 1 1 ) and since the surface density in turn is in general

greatest at sharp points in the surface Of a conductor,the di

electric as a rule breaks down first at the sharpest point Of theconductor . The protective effect of a lightning rod is basedupon this fact .

In the case of a non-uniform electrostatic field,such as the

field about the wires of a transm ission line or in the insulationof a cable

,the field intens ity under certain conditions can exceed

the dielectric strength Of the insulation in the imm ediatevicinity of the wire without exceeding the dielectric strength in


1 abfarad = 109 farads1 abfarad = 10” m icrofarads1 abfarad = 9 >< 1020 c . g. s . electrostatic units

When the equipotential surfaces between which the diSplacem ent current flows are the surfaces of conductors

,the dielectric

and the conductors form ing these surfaces are said to form anelectric condenser . The capac ity Of a condenser form ed by two

conducting surfaces and a dielectric m ay be expressed in terms

Of the charge on either conductor and the difference of potentialbetween them . When the conducting surfaces are such that thedisplacement current leaving the first surface is the sam e as thatentering the second, the charges on the two surfaces must be equaland opposite

,see equation and therefore the displacem ent

current is dq’ where q is the num erical value of the charge at anydt

instant on either conductor,see equation Hence the capac

ity of the condenser form ed by the dielectric and the two con

ductors is

i dq

dv dv

dt

Therefore the capac ity of the condenser is equal to the numeri ca l

va lue of the charge on ei ther surface per unit difference ofpotentia l

between the two.

The capac ity ofsuch a condenser is constant , i .e .

,$1 is constant

,

v

and equal to the ratio Of the charge q at any instant to the potentialdifference v at this instant

,provided the dielectric cons tants Of

every body in the field are constant . This may be proved as

follows : The electrostatic intens ity H produced at any point Pwhen a conductor is charged with q, units of electricity dependsin general upon the size and shape of the conductor, the nature ofthe surrounding bodies , and the size and shape of these bodiesand their position with respect to the charged conductor . For

these conditions fixed,the field intens ity due to the charge q,

and whatever other charges it may induce is proportional to this

charge q, , and therefore the drop of potential due to this charge q,between any two points in the field is proportional to q, , that is

viz A iqi

ELECTROSTATICS 9

where A , is a cons tant depend ing on the size and shape of thecharged conductor

,the size

,shape and position of any other

conductors inthe vic ini ty , and the nature, S ize , shape and positionof

‘

the surrounding dielectric bodies , but is independent of thevalue Of q, . Sim ilarly the drop of potential between these sam etwo points due to any other charge q, is proportional to q, , thatis

Hence the total drop of potential between 1 and 2 due to bothcharges is

By d efinition , however, in the case of a condenser q, q, , andtherefore the difference of potential between the two conductorsmay be written

2) z vi" = A iqz A 2Q2

: (A l_ A 2) (I:

Whence,from the capac ity of the condens er is a constant

equal to1

A 1_ A 2

152 . Simple Forms of Condensers . a . Para llel Plate Con

denser .

— The parallel plate electrom eter described in Artic le146 is one of the S implest form s of electric condensers . Its

capac ity is calculated directly from equations ( 16) which givethe relation between the charge on each plate and the differenceof potential between them . That is

,the c apac ity of such a

parallel plate condens er isQ KS

V 4 77 D

where K is the dielectric cons tant ofthe dielectric between the plates

,S

the area of each plate and D the distance between the plates .

b. Spherical Condenser .

—AnotherS imple form of condenser consists oftwo concentric spherical shells

,the

space between which is filled with auniform dielectric . Let the shells havethe radii r, and r, and let the dielectricconstant of the dielectric between thetwo shells be K. A charge Of Q c . g. s . electrostatic units given theins ide sphere will induce a charge Of Q units on the ins ide surface

c . g. s . electrostatic units (23)

Fig . 78 .


of the outer shell and a charge of Q units on the outside surfaceof this shell and whatever other conductors m ay be connectedto it . This outside charge Q ,

however,will have no effect on

the field intensity inside this Shell (see ArticleFrom symmetry

,the lines of electrostatic induction are radial

lines norm al to the surfaces of two spheres ; the total numbercom ing out from the charge Q is 4 77 Q (see Article Hencethe electrostatic flux density at any point P in the dielectric is

B47TQ = Q- where r is the distance of P from the center of the4 77 r

2r2

B Qspheres . Hence the field intens ity at P is H

K Krz

therefore the difference of potential between the two spheres is

Q

Kr ,

Hence the capacity of this condenser is

CQ r , T2 K c . g. s . electrostatic units (24)V r ,

—r ,

c . Coaxial Cylinders. Applying exactly sim ilar reasoning tothe case of two coaxial cylinders which are suffic iently long incomparison with their diameters so that the lines of electrostaticinduction going out the ends may be neglected, we have fromsymmetry that the electrostatic flux dens ity at any point P in

4 77 Q 2 Q‘

2 77 r r

where Q is the charge per uni t length of the condenser,and r the

distance of P from the center Of the cylinders . Hence the field

the dielectric between the two cylinders is B

intensity at P is H =_2

K

Q and therefore the difference of potentialr

between the two cylinders is

2 Q 2 QKr K r ,

where r, is the outside radius of the ins ide cylinder and r, theins ide radius of the outside cylinder . Hence the capac ity percentim eter of the condenser form ed by two coaxial cylinders is

C 52:

Kc . g. s . electrostatic units (25)

V2 ln


c . g. s . electrostatic units (26)

For practical calculations this formula m ay be written00 036

27 K

microfarads per 1000 feet (26a)

r

m icrofarads per mi le

[Neither Pe rl ): is correct when the wires are - close together ,

r r

since the charge on the two wires i s no longer uniform ly dis

tributed (see Alex . Russell,A lternating Currents , Vol . I , p .

When the wires are far apart r is n egligible in comparison withD

,hence formula (26) is sufficiently accurate ]153 . Spec ific Inductive Capac ity. In all the formulas for

capac ity it is seen that the capac ity varies direc tly as the dielectric cons tant of the medium between the two conductors . Henceby m easuring the capacity C,, of a given condenser when theplates are separated by air and then m easuring the capac ity C,Cwhen the plates are separated by any other dielectric , the dielectric cons tant may be readily determ ined experim entally , S ince

a s.

Ca

is equal to the ratio of two capacities, this cons tant is frequentlycalled the spec ific inductivecapacity ”

of the dielec tric .

1 54 . Condensers in Series

and in Parall el . When twoor more condensers are con

nected in series,equal and

Q +Q_

Q opposite charges will be induced on each pair of plates

connected by a conductor . Hence if the condensers have capacities C, ,

C, , O, , etc .

,the total potential drop through all the cen

densers is

Since the numerical value of the dielectric constant

ELECTROSTATICS 273

0 0 0 1 1

+CH+C

Hence the equivalent capacity of any number of condensers inseries is C

,where

1

C Q C, C, C,

When the condensers are connected in parallel,the drop of

potential across each condenser is the sam e,and therefore the

Fig . 8 1 .

total positive charge given all the condense rs (equal to the totalnegative charge) is 0

Q = <Cl + 0 2+ 0 3) V

Hence the equivalent capacity of any number Of condensers in‘parallel is

C,+ C,+ 0 ,

Compare with resistances in series and parallel,Artic le , 98 .

155. Absorption in a Condenser . Residual Charge . Ex

perim ent shows that when a given difference of potential is established between the plates of a condens er the dielectric of whichis a solid

,the charge taken by the condenser depends upon the

length of tim e this potential difference is maintained . Again,when a charged condenser is discharged by connecting its platesmom entarily with a conductor

,and this connection is then broken

,

the plates at first appear to be entirely discharged, but after afew seconds a charge again appears on them resulting in the re

estab lishm ent of a differenc e of potential ( less than in the firstcase) between the plates . In Short

,a solid dielectric apparently

absorbs a certain am ount of charge which it gives up only aftera cons iderable lapse Of tim e . The charge which appears on theplates of the condens er after the first discharge i s called theres idual ” charge ; this phenom enon is called e lectri c absorption .

The absorption of a dielec tric is apparently due to impurities in


it or to lack of homogeneity . It is greatest in substance like m icaand glass ; it is doubtful if absorption occurs at all in absolutelyhomogeneous substances .

156. Energy of an E lectrosta tic Field . An electrostatic fieldrepresents a certa in am ount of stored energy j ust as a m agneticfield represents a certain am ount of stored energy . Let a con

denser be charged by being connected to a battery or other sourceof electromotive force ; the amount of energy stored in the electrostatic field can be readily calculated . Let the capacity ofthecondens er be C and let the drop of electrostatic potential fromits positive to its negative plate at any instant be Then thecharge on each plate of the condenser at this ins tant is num ericallyequal to To change the charge by an amount dq= Cdv in

d dtim e dt requ i res a current of uni ts for dt seconds . In

the battery and the wires connecting the battery to the plates of

the condenser this current will be in“the direction from the nega

tive to the positive plate of the condenser and in the dielectricseparating the plates of the condenser this current (as a displacem ent current) will be in the direction from positive to the negative plate of the condenser, that is in the direction of the drop of

potential through the condenser . Hence an am ount of energy

equal to dt= Cvdv will be lost by the current in the con

denser . Hence , when the drop of potential between the twoplates of the condenser changes by an amount dv an am ount ofenergy equal to Ordo is gained by the electrostatic field in thedielectric separating the two plates . Therefore

,when the differ

ence of potential between the two plates is increased from zero tov a total am ount of energy

1 q2 1

_

2 C_

2

is stored in the electrostatic field . In using these formulas , caremust be i taken to express all the quantities in the sam e systemof units .

The energy which is stored in the electrostatic field com esfrom the source of electromotive force to which the condenser isconnected . When the plates are disconnected from the sourceof electromotive force and are connected to each other by a wire

,


but is rather due to the existence in the dielectric of small conducting particles , insulated from one another

,which form m inute

condens ers and in which the varying electrostatic field sets upalternating currents of greater magnitude than would be set upby a cons tant impressed electrom otive force. The increase ofheat energy dissipated when an alternating potential difference is .

established across the condenser over that dissipated when a constant potential difference is established, is then due to the excessof heat energy developed by the alternating currents in theseconducting particles .


(Note : Al l formulas are in c . g. s . electrostatic units unlessotherwise specified .)

1 . An electrostatic unit point-charge is a charge which repelswith a force of one dyne an equal pointcharge one centim eteraway .

2 . Two point- charges of q and q' units at a distance r centi

meters apart repel each other with a force of

qq’

f73

3 . The electrostatic field intensityH at any point in an electrostatic field is the force in dynes which would act on a unit positivepoint- charge place at that point due solely to the agents produc ingthe original field .

4 . The field intensity at any point due to a pointcharge g ata distance r centimeters away is

T2

~ 5 . The mechanical force exerted on a pointcharge q isF =qH

where H is the field intens ity due to every agent in the fieldexcept the charge q.

6. Lines of electrostatic force are lines drawn in an electrostatic field in such a m anner that they coincide in direction ateach point

'

P with the field intens ity at P and their number perunit area at each point P across a surface at right angles to theirdirection is v equal to the field intens ity at P .

ELECTROSTATICS 77

7. Gauss’

s Theorem .

—The algebraic sum of the number oflines of electrostatic force outward from any closed surface is equalto 4 7T times the algebraic sum of the charges ins ide this surface .

8 . A dielectric in an electrostatic field is cons idered to be made

up of filaments such that were the lateral walls of any one of thesefilaments separated from the rest of the di electric by a narrowair gap ,

no charges would be induced on these walls .

9 . The intensity of electrisa tion Jat any point in a body is defined as the charge per unit area which would appear on a gapcut in the body at this point perpendicular to the line of electrisation through thi s point . The direction of the intens ity of electrisation is the direction of the line of electrisation

,the positive sense

of which is the sens e of the line drawn into the gap from the wallon which the positive charge is form ed .

10 . A l ine of electrisation,or unit filam ent

,is a filament of such

a size that were it broken by a narrow gap at any point , thenum erical value of the charge form ed on either wall of the gap

would be —1

4 77

1 1 . The number of lines of electrostatic induction ,or flux of

electrostatic induction,cross ing any surface is defined as the alge

braic sum of the number of lines of electrisation and the lines ofelectrostatic force crossing that surface .

12 . The electrostatic flux density B at any point is the numberof lines of electrostatic induction per square centim eter c rossingan elem entary surface drawn at this point norm al to their direction . The electrostatic flux density is the vector sum

B = 4 n J+ H

13 . In an electrostatic field due solely to electric charges , aline of electrostatic force originates at a positive charge and endsat a negative charge , and may exist in any dielectric but not in aconductor in which no current is flowing . A line of electrisationoriginates at a negative induced charge on a dielectric and runsthrough the dielectric to a positive induced charge on the surfaceof the dielec tric . A line of electrostatic induction is a continuousline which originates at a positively charged conductor (or at apositive contact charge on a dielectric) and ends at a negativelycharged conductor (or at a negative contact charge on a dielectric) ,but passes through the induced charges on dielectrics .

14 . The dielectric constant, or specific inductive capacity, K, of


a dielectric is the ratio of t he electrostatic flux density B in thedielectric to the resultant electrostatic intens ity H when thedielectric is placed in an electrostatic field ; that is

B

H

15 . A closed hollow conductor completely screens everythingins ide from external electrostatic influences .

16 . The electrostatic intensity with in the substance of a con

ductor in which there is no electric current is zero .

17 The total charge within any region completely enclosed bya conductor is zero .

18 . The electrostatic intensity at any point P just outside a

charged conductor in which there is no electric current is normalto the surface and equal to

4 7T 0'

K

where K is the dielectric cons tant of the dielectric in contact withthe conductor at P and 0 ' is the surfacedens ity of the charge onthe conductor j ust opposite P .

19 . The electrostatic flux density at any point P just outsidea charged conductor in which there is no electric current is normalto the surface and equal to

B =4 n d

independent of the dielectric in contac t with the conductor ; 0 ‘ is

the surface dens ity of the charge on the conductor j ust opposite P .

20 . The electrostatic potential V at any point in an electrostaticfield is the work which would be done by the agents produc ingthe field in m oving a unit positive pointcharge from this point toinfinity . The electrostatic potential at any point due to a pointcharge q at a distance r centim eters away is

q

r

The resultant electrostatic potential due to any number of chargesis the algebrai c sum of the potential due to all the individual poles .

2 1 . The drop of electrostatic potential along any path fromany point 1 to any point 2 is

V , V 2: (H cos 0) dl

where d l is the length of any element of the path from 1 to 2 ,H

the field intensity at al l and 9 the angle between the direction of


where B is the electrostatic flux density and all quantities are inthe sam e system of units .

27. The value of the ratio of the displacem ent current betw een “

two equipotential surfaces in a dielectric to the time rate of

change of the difference of potential between these two surfacesis called the capacity C of the portion of the di electric between thetwo surfaces . The displacement current through a given portionof a dielectric between two equipotential surfaces is then

where v is the difference of potential between the two surf aces .

The c . g. s . elec trostatic unit of capac ity, the electrom agnetic uni tor abfarad and the practical unit or farad are related as f ollows :

1 abfarad = 109 farads 1020c . g. s . electrostatic units .

When the two equipotential surfaces are conducting surf aces,the

dielectric and the two conducting surfaces are,

said to form anelectric condenser. The capacity of a condenser is also equal tothe num erical value of the ratio of the charge on either surfaceto the difference of potential between the two .

28 . The capac ity of the condenser formed by two long paral lelW ires of c ircular cross section is

K

Dlog7

where K is the dielectric constant of the surrounding m edium Dthe distance between centers of wires and r the radius of each w i re .

29 . The resultant capac ity of two or more condensers in ser ies

is the reciprocal of the sum of the rec iprocals of the variouscapac ities

,i .e .

,

mic rofarads per mile

1 1 1

C 0 , C2

The resultant capac ity of two or more condensers in paral lelis the sum of the various capac ities

,i .e .

,

C=C,+ (72+

30 . The energy stored in the dielectric of a condenser is

where C’ is the capac ity of the condenser , q the numerical value ofthe charge pn either conductor, and v the difference of potential

ELECTROSTATICS 281

between the conductors . These formulas hold for any system ofunits provided all quantities are expressed in the same system .

PROBLEMS

l . A c ircular disc 20 inches in diam eter and of negligible thickness is charged uniform ly on both sides with a total charge of

electrostatic units . ( 1) What is the intens ity of the electrostatic field at a point on a normal to the disc at its center 10inches from the plane of the disc? (2) What would be the intens ityat this point if the disc had an infinite radius and the surfacedensi ty of the charge rem ains the sam e? (3) Under the latter conditions what would be the difference in electrostatic potential between the point and the disc?

Ans . : ( l ) 1 358 c . g. s . electrostatic units (2) 4650 c . g. s . electrostatic units ; (3) c . g. s . electrostatic units .

2 . Two plates A and B ,the areas of which may be considered

to be infinite , are placed parallel to each other and 10 cm . apart .A is charged positively with 50 electrostatic units of charge persq . cm . while B is charged negatively with the sam e density ; thecharges are confined to the surfaces which face each other . ( 1)Determ ine the pull per sq . cm . which A exerts upon B.

A slab of glass of infinite area and 8 cm . in thickness is nowplaced between the two plates and parallel to them

,but not touch

ing either . The dielectric constant of the glass is 5 . (2) Determ ine again the pull per sq . cm . which A exerts upon B ; (3) theintensity of electrisation in the glass ; and (4) the electrostatic fluxdensity in the glass .

Ans . : ( 1) dynes ; (2) dynes ; (3) 40 c . g. s . electrostatic units ; (4) 629 c . g. s . electrostatic units .

3 . Determ ine the difference of electrostatic potential betweenthe two plates in Problem 2 when the plates are charged positivelyand negatively respectively , with c . g. s . electrostatic unitsper sq . cm . ( 1) when the m edium between the plates is air ; (2) whena slab of glass of infinite area , 6 cm . in thickness and with a dielectric constant equal to 5, is placed between the plates and parallelto them . The charge on the plates and the distance between theplates rem ain constant in the two cases .

If the difference of potential between the two plates is keptconstant at electrostatic unit , what will be the numerical valueof the charge per sq . cm . on each plate (3) when the medium be


tween the plates is air ; (4) when the slab of glass 6 cm . thick isplaced between the plates?

Ans . : ( 1) c . g. s . electrostatic units (2) c . g . s . electrostatic units ; (3) c . g. s . electrostatic units ; (4)0 . g. s . electrostatic units .

4 . Two parallel m etallic plates each 100 sq . cm . in area areseparated by a distance of cm . If the difference of potentialbetween the two plates is maintained at 1000 volts

,determ ine

( 1) the charge in microcoulombs on each plate when the platesare separated by air ; (2) when the dielectric between the twoplates is glass ; (3) the work required to pull the glass out of theelectrostatic field . The dielectric constant of glass is 5. Assum ea uniform distribution of charge .

Ans . : ( 1) microcoulombs ; (2) microcoulombs ;(3) 354 ergs .

5 . Two metallic plates each 100 sq . cm . in area are chargeduntil the difference of potential between the plates is 300 voltsand the source of potential is then removed . The p lates arecm . apart and the medium between them is air . ( 1) Determ inethe work done in separating the plates until they are one centimeter apart . (2) What is the difference of potential between theplates if a sheet of glass cm . thick and having a dielectricconstant of 5 is then pushed in between them ?

Ans ( 1) ergs ; (2) 360 volts .

6 . A transm ission line 10 miles in length cons ists of two No .

0000 wires (B . S. gauge) Spaced 3 feet between centers . If apotential difference of 1000 volts is es tablished between the wiresand the line is open at the far end, determ ine the energy in theelectrostatic field surrounding the line . The diameter of a No .

0000 wire is inch .

Ans . : j oules .

7. The total inductance of a two-wire transm ission line 25m iles in length is m illihenries . What is its capacity in m icrofarads ?

Ans . : m icrofarads .

8 . Three condens ers,A

,B and C

,the capacities of which are

25,20 and 15 m icrofarads respectively, are connected in series.

If the potential drop across B is 100 volts, determ ine ( 1) the potential drop across A ; (2) the potential drop across 0 ; and (3) thepotential drop across the three condensers in series . (4) What is

VARIABLE CURRENTS

158 . Total Energy Associated with an E lectric Circuit .From the foregoing discussion of the m agnetic and electrostaticfield it is evident that the energy assoc iated with an electricc ircuit may m anifest itself in the following ways :

1 . As m agnetic energy , due to the magnetic field producedby the current in the c ircuit .

2 . As electrostatic energy , due to the electrostatic field produced by the differences of electric potential between the variousparts of the c ircuit .

3 . As m echanical energy , due to the relative motion of thevarious parts of the c ircuit or to the m otion of conductors or ofmagnetic or dielectric bodies in the surrounding field .

4 . As chem ical energy , due to chem ical changes which may

take place in various parts of the c ircuit .5 . As reversible heat energy , due to thermal electrom otive “

forces .

6. As heat energy dissipated in the conductors (and to aslight extent in the surrounding dielectric) , due to their electricresistance .

7 . As heat energy dissipated in surrounding m agnetic bodiesdue to the variation of the m agnetic flux through them ; i .e .

,

magnetic hysteresis .

8 . As heat energy dissipated in the surrounding dielectricbodies due to the variation of the electrostatic flux in them ;

i . e .

,dielectric hysteresis .

The effects produced by the first five forms of energy are allrevers ible ; i .e.

,energy is required to establish a magnetic or an

electrostatic field,but the same amount of energy is given back

in som e other form when these fields disappear ; energy is requiredto move a conductor or a magnetic or a dielectric body in the field

,

but the sam e am ount of energy is given back in som e other formwhen the motion of the body is reversed ; energy is required toproduce any chem ical action but the sam e am ount of energy is

284

VARIABLE CURRENTS 85

given back in some other form when this action is reversed ; theheat energy given out at a junction of two diss im ilar substanceswhen the current flows through the j unction in one di rection , isgiven back as some other form of energy when the current isreversed .

The effec ts produced by the last three forms of energy are

not reversible ; i .e .

,energy is always dissipated in a conductor

through which a current is flowing, independent of the directionof the current ; energy is always dissipated when the phenom enonofmagnetic or dielectric hysteresis occurs , independent ofwhetherthe flux is inc reasing or decreas ing .

The first five effects are sometimes said to be due to conserv

ative forces,the last three to di ssipative forces . The word forces

is here used in a general sense , meaning the som ething that produces or opposes the effects .

159 . General Equations of the Simple E lectr ic Circu it .

The trans fer of energy from one region to another in space bymeans of an electric current is in general accompanied by production of energy in all these various forms

,except in the Spec ial

case of a continuous current,i .e .

,a current which does not vary

with tirri e there i s also in general a transfer of energy from one

form to another all along the c ircuit . To understand c learly theeffects produced by a variable current it is necessary to confineone’s attention at first to comparatively s imple c ircuits .

Cons ider a c ircuit cons isting of a coil having a resistance r

and an inductance L m seri es w i th a condenser having a capac ityC and a conductance g. The conductance of a condenser

,that

Fig . 82 .

is,the reciprocal of the resistance of the dielectric between its

plates,is som etimes called the leakance of the condenser . This

c ircuit is typical of the c ircuits one has to deal with in practice ;


in fact the most general form of circuit m ay be looked upon as

made up of such elem entary c ircuits . This simple c ircuit may

be represented diagramm atically as Shown in Fig . 82,the res ist

ance and inductance and the leakance and capac ity being shownseparately S imply for convenience . Let an electrom otive forcee be impressed across the term inals

.

of this circuit . This im

pressed electromotive force will establish a current in the c ircu itand therefore a potential drop across the various parts of thec ircuit . When there is originally no current in the circuit and

no charge on the condenser, this current will be around the c ircu itin the direction of the impressed electrom otive force

,and will

produce a drop of potential through the coil and through thecondenser . The current in the coil will have the sam e value ateach part of its winding at any given ins tant (provided the capacityand leakance of the coil may be neglected) and this current must

in turn be equal to the total current through the condenser .Let i be the current in the coil , id the displacement current throughthe condenser and i6 the conduction current through the con

denser . Then from Article 149

dq dq7/ = l d + l

c

but id =dt ’where

dtrepresents the time rate of change of the

charge on the condenser . Let 1) be the potential drop through thecondenser in the direction of the current through it , then fromArticle 149

dv

di

The conduction current through the condenser is , by Ohm’s Law

i,=gv. Hence

7:=gv+ cd”

dt

From Ohm ’s Law

,the drop of potential through the coil

due to its resistance is ri . From A rticle 1 1 6 the drop of potential through the coil due to its inductance the back electro

motive force of induction) is L3; Hence the total drop of po

tential through the coil is


apply directly only to a single circuit and when there is no dissipa

tion of energy due to ei ther magnetic or di electri c hysteresis or to

currents induced in surroundi ng bodies . When there is any

dissipation of energy due to hysteres is or eddy currents * a term

should be added to take into account this dissipation of energy .

In case there is iron or any other magnetic substance in the

magnetic field produced by the current there are both magnetichysteresis and eddy currents ; in such a case equations ( 1) and (2)

give only a first approximation to the true relations between

current , potential difference and time ; this approximation, however

,is usually sufficiently c lose for practical work . We Shall

see in Article 1 90 how a still c loser approximation can be madein the case of alternating currents .

In this chapter wil l be given the solutions of equations ( 1)and (2) for a fewsimple cases . In every case the inductancewill be assumed constant ; when there is iron in the m agnetic fieldof the current this assumption will give only a roughly approx i

m ate s olution .

160 . Establishment of a Steady Current in a Circu it Conta iningResistance and Inductance . Thec ircuit is represented diagrammatically in Fig . 83 ; E represents aconstant e. m . f. Let the time t bereckoned from the instant theswitch S is c losed ; i .e .

,let the

switch be c losed at tim e t=0 . At

any ins tant an interval of tim e tFig . 83 . after closing the c ircuit .

diE = ri L

dl

whence

d (E —ri) f i“E ri

Integrating ,rt —ln (E—ri ) + GL

where G is a cons tant of integration . The value of G is foun d by

The name eddy current is applied to the currents induced in the

magnetic circuit or in any metal form ing the frame of an ele ctric ma chine .


the condition that at tim e t=0 (that is , at the instant the c ircuitis closed) i Substituting these values in the above equation

givesG = ln E

hence at any instant t seconds after closing the c ircuit

L

Writing each side of this equation as the exponent of the base 6of the natural system of logarithms

,we get

- 1'

-

t E —riL

E6

whence~

l —e L

r

The physica l interpretation of this equation is that the current

reaches its steady value I =Eonly after the tim e t measured

T

from the instant of c losing the c ircuit has becom e suffic ientlyrt

great to make the term 6 L sensibly equal to'

zero . Since,

rt

however, the ratio is usually quite large , this term 6 L as a

ru le becom es practically zero for t equal to a small fraction ofa second

,and therefore the current reaches its steady value

_

1? alm ost immediately after the c ircuit is closed . Whenr

the self- induc tion is large , as in the case of a coil wound on a

c losed iron core,the ratio %may be relatively small

,in which

case several seconds may elapse before the current reaches itssteady value .

In any case,after an interval of time T=£

J

,the current is

1

That is after an interval of tim e T—é ,the current reaches

r

*Otherwise a t the instant the switch is closed energy would be transferred to the magnetic field at an i nfini te

’

rate , which is impossible .


per cent of its final value . The tim e required for the current insuch a c ircuit to reach 100 ( 1 per cent of its final steadyvalue when a steady e . m . f. is impressed upon the c ircuit is calledthe time- cons tant of the circuit . The time- c ons tant of such ac ircuit may be looked upon as a measure of the slowness withwhi ch the current reaches its steady value

,the greater the time

cons tant the longer the interval before the steady value is reached .

For a'

circu it cons isting of an inductance L and a resistance r the

time-c ons tant is equal to 9. It should be noted that this L andr

r refer to the entire circuit ; hence when the impressed e. m. f . isproduced by a generator developing a steady armature ‘

e. m. f.E

,the res istance and inductance of the generator must a lso

be included . In addition, the above formulas hold only in casethe inductance of the entire circuit is a constant

,and there is

no energy dissipated in hysteres is , which conditions never hold

F ig . 84 .

when there is iron or any other m agnetic substance in the mag

netic field produced by the current . In such cases,however

,

the formulas may be looked upon as a first approxim ation to thetru e relation between current and tim e .

The relation between current and tim e given by equation (3)may be represented graphically by plotting the current i as theordinates against the tim e t as absc issas. The curve has thegeneral Shape Shown in figure 84 ; i .e .

,the current rises rapidly

at firs t and becomes asymptotic to the line corresponding to


is large compared with its resistance . The current falls to 100 e“

per cent of its original value in the time T -

I_

J= the tim e

cons tant of the circuit . The relation between the current i and

Fig . 86.

time t is shown graphically in Fig . 86. The current falls rapidlyat first and becom es asymptotic to the axis of t

,i .e.

,to the line

corresponding to i =0 .

162 . Charging a Condenser through a Resistance —The cir

cuit is represented diagrammatically in Fig . 87 ; E is a constant

e . m . f . Let the charge on the condens er be zero at the instant theswitch is c losed

,and let time be

measured from this ins tant ; i .e.,

at time t=0 let go = 0 and i =0 .

Let g and i be the charge andcurrent respectively and v the p .d .

through the condenser at any instant t seconds after closing the switch . Then at this ins tant

,

assuming the idea l condi tions of no leakance in the condenser and

no inductance ofthe circu i t

Fig . 87.

E =m+ v

dvi

dl

From the first equation we have that i which substitutedin the second equation gives


which integrated givest

rC—ln (E—v) + G

where - G is a constant of integration. For t=0,v=0

,* which

substituted in this equation gives G In E . Substituting thisvalue of G in the last equation gives the equation

t—= lnrC

whence

v=E 1 6 TC

which substituted in the second equation above givesE

r

and since q Cv at any instant we get imm ediately from

q=C

’E ( 1

— 6 F0 ) .

Fig 88 .

Equations (5) then give the values of the p .d . through the condenser

,the current

,and the charge at any instant . The variation

of v,i,and q with tim e is shown graphically in Fig . 88 . Note

that both the p .d. through the condenser and the charge start at*Otherwise at the instant the switch is closed energy would be transferred

to the electrostatic field at an infinite ra te, which is impossible .


zero rise rapidly at first and then approach more slowly theircons tant values E and CE respectively . The current

,on the

other hand,has 1ts m aximu mvalue at the ins tant the swi tch

T

is closed and then falls, first rapidly and thenm ore slowly, to zero .

*

As a rule rC is extremely sm all,so that onl y a sm all fraction of

a second is required for these steady conditions to becom eestablished .

The product rC measures the slowness w ith which the con

denser becomes charged ; it is called the tim e- cons tant of thec ircuit . Compare with the time~ constant of a circuit form ed bya resistance and an inductance in series .

163 . Discharging a Condenser

through a Resistance . Let the condenser be charged to a p .d .v

0at

the instant t= 0 at which it is shortcircuited through a resistance r , Fig .

89 . Assuming as before the idea l condi tions of no leakance and no induc

tance,we have

E liminating i from

whence

whence

ln v+ G

But at t=0,v=v

o ,whence G= lnv

o.

Hence at any ins tant t seconds after the condenser is shortc ircuited

,

*Actually the current always starts from zero and rises to a maximumvalue

,due to the inductance of the circuit

,which

,though it m ay be small

,

is never absolutely zero as assum ed in the above di scussion ; see Article 201 .

0 =m+ v

dv

a

two equations

dv

dt


div 1

— ’

U (6)dt

2 LC

This equation is of exactly the sam e form as the equation forthe pendulum ; see Article 23 . Com pare also with the equationfor the vibration ofa magnet in a uniform m agnetic field

,Art icle 42 .

Its solution ist

\/LO

which substituted in equation (b) gives

v=A sin

(e)

The cons tants A and 0 are determ ined from the condition thatat tim e t=0

,v=v

0 ,and i =0 . Substituting these values in (d)

and (c) we get

D= A cos (9

Whence and A = vo

. The equations for the p .d . and cur

rent at any instant are then

v= vocos

t =

These two equations are plotted in Fig . 92 That is,both v and i

are harmonic functions,or S ine waves

,having a period equal

to 2 7rV LC. The maximum value of the p .d . is v,and the maxi

mum value oi the current is0

That is,the charge on the

condenser osc illates from q=Cv

0to q

= Ovo ,and the current

oscillates between the values and v0

the charge

reaching the maximum when the current is zero and vi ce versa .

If the current in the inductance had been equal to i,and the

p .d . across the condenser zero at tim e t=o,then the equations

for p .d . and current would have been


5 1/ LC

i = i0cos

A circuit of this kind is approximately realized in the case ofa transm ission line

,except that the inductance and capac ity in

stead of being lumped in a single coil and a single condenser are

Fig . 93 .

distributed uniformly along the line , and the resistance of the lineis not negligib le . As a first approxim ation we may cons ider theline as equivalent to a lumped inductance and a lumped capac ityas shown in Fig . 93 and neglect the resistance . If the line, open


at the receiving end, becom es shortcircuited at any time P,bya wire fal ling across it , for example, we have the conditions j ustdiscussed .

Take the case of a line cons isting of two number 0000 B . S.

wires (solid) Spaced 6 feet apart ; let the line be 50 m iles long( 100 m iles of wire) .

The inductance of the line will then be henries and thecapacity x farads . Let 500 amperes be flowing in theline at the instant of short- c ircuit at P and let the p .d . acrossthe end of the line across the co ndenser) be zero at thisinstant . Then this p .d . will reach a maximum of

volts

and the current and p .d . will begin to osc illate with a period of

2 7r\/LC=2 X 10 seconds or 580 com

plete swings or cycles per second . This abnorm al rise in voltagewould of course puncture the insulators carrying the wires werenot som e sort of safety valve provided. Such a safety deviceis the lightning arrester, which in its S implest form consists of one

or more Spark gaps so arranged that when the voltage on the linerises a predeterm ined amount above normal

,a Spark j umps across

the gap and thus reduces the voltage . Protecting a line againstlightning discharges is only one of the functions of a lightningarrester ; such a device is l ikewise necessary to protect the line

(and the apparatus connected therewith) agains t the abnorm alvoltage which may be produced when heavy loads are switchedon or off the line .

Note that the ideal case of a condenser without leakance andan inductance without resistance , which we have been considering ,

cannot be completely realized in practice, S ince the dielectricof every condens er is a conductor to a certain extent (thoughusually an extrem ely poor conductor) and every c ircuit m ade of

conductors has a certain resistance . The effect of r es istance andleakance is to damp out the osc illations of the electric current inthe sam e way that friction damps out the osc illation of a vibratingpendulum . (See Article


6. When a condenser of capacity C is charged through aresistance r by a constant impressed e. m . f . E

,the p . d . across the

condenser, the current in the c ircuit and the charge on the condenser t seconds after the e. m is impressed

,are respectively

q=CE l —e r C

provided there is originally no charge on the condenser and nocurrent in the c ircuit and there is no inductance in any part ofthe c ircuit .

7. The tim e- constant of a c ircuit form ed by a res istance r anda capac ity C in series is rC.

8 . When a condenser of capac ity C,through which the p .d .

is v0 . is discharged through a resistance r

,the p .d . through the

condenser, the current in the c ircuit and the charge on the con

denser t seconds after the term inals of the c ircuit are Short- c ircuitedare respectively

t

q= Cv

o6 7 0

provided there is no current in the c ircuit at the instant of Shortc ircuit and there is no e. m . f . and no inductance in the circuit .

9 . When a condenser of capac ity C,through which the p .d . is

v0 ,is discharged through an inductance L ,

the p .d . through thecondenser and the current in the c ircuit t seconds after the term inals of the c ircuit are short- c ircuited are respectively

7) 1)I

I

0

provided there is no current in the circuit at the instant of Shortc ircuit and there is no e . m . f . and no res istance in the c ircuit .

When the term inals of the c ircuit are shortcircuited at the instant


when the current is i , and there is no p ld . through the condenser,the p .d . and current t seconds after the term inals are short- circuitedare respectively

i = tocos

provided there is no e . m . f . and no resistance in the c ircuit .

PROBLEMS

1 . An air- core solenoid of 5000 turns is 20 cm . long and has adiameter of 5 cm . the resistance of the solenoid is 2 ohms . Whena cons tant e . m . f . of 50 volts is impressed across the term inals of

this coil at what rate does the current begin to increase ; (2)what is the final value of the current ; and (3) what is the timeconstant of the coil?

Ans . : ( 1) amperes per second ; (2) 25 amperes (3)second .

2 . In the preceding problem , ( 1) at what rate is energy beingstored in the m agnetic field of the current when the current hasreached a value of 10 amperes (2) when the energy of the electrom agnetic field is j oules at what rate is energy dissipated inheating the coil ; (3) atwhat rate is the current changing when thetotal power supplied to the coil is 250 watts?

Ans . : ( 1) 300 watts ; (2) 400 watts ; (3) amperes persecond .

3 . At the sam e instant that the e . m . f . impressed upon a coilis removed

,the ends of the coil are connected by a resistance of

3 ohms . At the instant imm ed iately after this change in con

nections is made the current in the coil is 12 amperes and is decreasing at a rate of 480 amperes per second . ( 1) What is thetim e- cons tant of the c ircuit? (2) If the energy of the electromagnetic field is j oules at the instant that the change inconnections is m ade

,at what rate is energy dissipated in the

c ircuit in the form of heat energy at that instant? (3) What isthe res istance and the inductance of the coil?

Ans . : ( 1) second ; (2) 576 watts ; (3) resistance 1 ohm,

inductance henry .

4 . An e . m. f . of 20 volts is impressed upon a coil of ohm


res istance and henry inductance . What is the value of thecurrent . in the coil second after the e. m . f . is impressed?

Ans . : am peres .

5 . An e. m. f . of 30 volts is impressed upon a coil of ohm

resistance and henry inductance . One tenth of a second afterthe e. m . f . is impressed, the source of the e . m . i . is rem oved andthe coil is Short- circuited by a resistance of ohm . Determ inethe value of the current in the c ircuit second later .

Ans . : amperes .

6. A battery which has an e . m . i . of 1 0 volts and a resistanceof ohm

,a coil which has an inductance of 3 henries and a

resistance of ohm,and a non- inductive resistance of 4 ohm s

are all connected in series to form a c losed electric c ircuit . Findthe value of the current flowing in a non- inductive resistance of7? of an ohm 2 seconds after it is connected in parallel with the

4 ohm res istance .

Ans . : ampe res .

7 . An e. m . f . of . 250 volts is impressed upon a c ircuit con

form ed by a non- inductive resistance of 1000 ohms in series witha condenser of 50 m icrofarads capac ity . Find (1 ) the initialrate at which the condenser is charged ; (2) the current in thec ircuit when the charge on the condenser is 5000 m icrocoulombs ;(3) the energy in the electrostatic field when energy is dissipatedas heat energy in the c ircuit at the rate of 1 0 watts ; and (4) thecharge in microcou lombs on the condens er when the current ischanging at the rate of 4 amperes per second .

Ans . : (1 ) ampere ; (2) ampere ; (3) j oule ;(4) 2500 microcou lombs .

8 . A 1 00 microfarad condens er charged to a potential difference of 600 volts and is dis charged through a resistance of 500

ohms . Find (1 ) the current when the current is decreasing atthe rate of 1 6 amperes per second ; (2) the charge on the con

dens er when the energy of the electrostatic field is 2 j oules ; (3)the potential drop through the condens er when the condenser isdischarging at the rate of coulomb per second ; and (4) the rateat which the condenser is losing energy when the charge on thecondenser is m icrocoulombs .

Ans . : (1 ) ampere ; (2) microcou lombs ; (3) 250 volts

(4) 20 watts .

9 . One-half a second after an e . m . f . of 1 000 volts is im

ALTERNATING CURRENTS

165 . Introduction .

—We have just seen that when a con

dens er is discharged through a coil having an inductance L but

no resi stance that the current in the c ircuit and the p .d . acrossthe condens er vary harmonically , i .e .

,the instantaneous value

of the current at any instant is i =Iosin wt and the instantaneousvalue of the p .d . across the condens er at any instant is v V

ocos wt

where I, and V 0are the maximum values of the current and p .d.

and to is equal to the factor 2 7T divided by the period of osc il

lation of the current and p .d . Both the current and p .d . varyperiodically from fixed m aximum values (10 and V ,

respectively)in one direction to equal m aximum values - I

,,and V

0respec

tively) in the othei' direction and back again to the maximumvalues in the first direction . The current is therefore called ana lternating current and the p .d . an a lternating p .d . This idealc ircuit we have been cons idering is a very special one ; alternating currents employed in practical work are produced by anentirely different m ethod

,nam ely

,by the rotation of one or more

coils in a m agnetic field . The machine for producing alternatingcurrents in this m anner is called an a l terna tor .

1 66 . The Simple Al ternator . The s implest form of alternatorconsists of a coil of wire ofone turn rotating in a uniform m agnetic

Fig . 940

field,see Fig . 94 . The ends of the wire form ing the coil are

connected to two rings called s lip rings ; m aking contact with

304

ALTERNATING CURRENTS 305

these rings are the brushes 1 and 2 . Let the coil be rotated withan angular veloc ity co ; let tim e be counted from the instant whenthe wire a is

_

above b and the plane of the coil coincides with avertical plane drawn through the axis of rotation

,and let gbo be

the va lue of the magnetic flux of induction threading the coilwhen in this position . Then (f) = rf>o cos wt is the va lue of the fluxthreading the coil at any ins tant t; hence the e. m . f . induced inthe coil at this instant is ; see Art icle 1 05,

—d ¢ —m gbo sin w tdt

and is in the direction from 1 to 2 . The m aximum value of the

e. m . f . is then m (f) , and occurs when w t 3_

9

11'

Calling E0

a w

this m aximum value of the e . m . f. , the induced e .m . f . in the direction from 1 to 2 may be written

e = Eosin w t

The electrom otive force in this case is a harmoni c or Si ne function2 77

of the tim e,the

pperiod is T= and the number of complete

w

periods per second,or the frequency, is f

2

a)

7,

In alternators as actually cons tructed the coils of wire arenot made of a single turn, nor do they rotate in uniform magneticfields ; the field of the alternator also has a large number of poles .

By properly designing the coils and the pole faces of such a ma

chine it is however poss ible to produce in the c oils when rotatedat constant Speed an alternating e. m . f . which is practically aharmonic function of the tim e . In any case

,the e . m . f . will be

a periodic function of the tim e and can therefore be representedby a series of harmonic terms

,called a Fouri er ’s Seri es, of the

forme= E 13 i n w t+ E zsin (2 m l -I 92) E 3sin (3 w t+ 03)

where (0 depends Solely upon the number of pairs of field polesand the angular veloc ity of the rotating part of the machine

(which rotating part may be either the field or the armature) andthe E ’

s and 9’s are constants , in general different for each term .

The first term E 1 sin out is called the fundamenta l or first harmoni c,

the remaining terms are called the second , third, etc .

,harmoni cs .

The frequency of the nth harm onic is equal to n tim es the fre


quency of the fundamental . In most practical cases it is neces

sary to cons ider only the first harm onic or fundam ental .In order to understand the properties of alternating electro

m otive forces and currents it is necessary first to get clearly inm ind certain fundam ental definitions .

167. Definition of Alternating Current and Al ternating Electromotive Force . An alternating current is defined as a currentwhich varies continuously with time from a constant maximum

value in one direction to an equal maximum value in the oppositedirection and back again to the same maximum in the first direction

,repeating this cycle of values over and over again in equa l

F ig . 9 5

intervals of tim e T,in such a manner that the instantaneous

value of the current at any instant t is identical ly the same as

at any other instant t+ kTwhere T is the time, cons tant in value ,required for the current to pass through cyc le of

values,and k is any integer , positive or ne ilarly,

an

alternating e. m . f . is defined as an e . m . f . continuously with tim e from a constant maximum value in one directionto an equal max imum value in the opposite direction and backagain to the sam e maximum in the first di rection

,repeating this

cyc le over and over again in equa l intervals of time T,in such a

manner that the instantaneous value of the e . m . f . at any instantt is identically the sam e as at any other instant t+ lcT where T isthe time

,constant in value

,required for the current to pass through

a complete cycle of values,and k is any integer, positive or nega

tive . In Fig . 95 the success ive values of an alternating currentare p lotted as Ordinates against tim e as the absc issa . Such a

'


The equation of a harm onic or S ine-wave current isi = I

osin (ci t—I 9) (3)2

where w IS a constant equal to TI and (9 IS a constant such thatI,sin 0 gives the value of the current at time t=0 . The cons tant

to is called the peri odi city of the current,and the constant 0 is

called the phase of the current . The relation between periodic ity ,period and frequency is

m :

V169 . Difference in Phase .

—In general , when a harm onic orsine-wave electromotive force is impressed on a c ircuit the resulting current is likewise a harmonic function of time (after a verybrief interval) having the sam e frequency

,but the e. m. f . and

current do not reach their maximum values S imultaneously . In

other words,when e=E

0sin co t represents the e . m . f. , the current

is represented by i =Iosin (wt where E,and l

oare the maxi

mum values of the e. m. f . and current respectively ; w= 2 7r fwhere f is the frequency ; t is the

‘

time measured from the instantwhen e=0 and is increasing in the posi tive direction and 0 is anangle which measures the interval between the instants when thee . m . f . and current reach successive maximum values . The e . m . f .

77reaches 1ts maxrmum value at tim e t wh i le the current

2 a),

reaches its maximum value when7r

(0 t+ 02

Hence when 6’ is positive the current reaches its maximum value0—'

seconds before the e.m .f. reaches its maximum ; when (9 isw

0negative the current reaches its maximum value seconds after

to

the e . m . f . reaches its maximum . In the first case the current issaid to lead the e . m . f. , and in the second case the current is saidto lag behind the e . m . f . The angle 0 is called the difi

’

erence in

phase between the current and e. m . f.' A careful re- reading of

the latter part of Article 23 will assist the reader in obtaining aclear physical conception of phase difference .

When the phase difference is zero the current and e . m. f .

ALTERNATING CURRENTS

71'

radrans or 90°2

the current and e. m. f. are said to be in quadrature; when thephase difference is rr radians or 1 80

° the current and e .m .f. aresaid to be in opposi ti on .

In general , when we have any two harmonic functions of

tim e and of the sam e frequency i,such as

2 77

said to be in phase; when the phase difference is

x ,=X1 sin (a) H! (9)

x2=X2 sin (w t+ 92)

the first function reaches its m aximum value at tim e t1

and the second reaches its first maximum value for t2

Hence the first function x , reaches its first m aximum an interval

of time t2 t1 —1(0, 92) ahead of the second function x, ; the first

0)

function x, is therefore said to lead the second function x, by theangle 9,—92 . Note the order of the subscripts : 1c, leads a) , by

9,—02 ; or x, leads us, by the angle 02—01 . A negative lead isof course equivalent to an actual lag, and a negative lag is equ iva lent to an actual lead .

170 . Instantaneous, Maximum , and Average Values. Theinstantaneous value of an alternating current is the value of thecurrent at any instant .The maximum value of an alternating current is the greatest

instantaneous value during any cycle . For a harmonic a lternating current the value of the current at any instant is given by anequation of the form i =I

osin (a) t+ 9) ; in this case the constant

I,is the maximum value .

The average value of an alternating current is defined as thenumerical value of the average of its instantaneous values betweensuccessive zero values ; hence , when the instantaneous va lues areplotted as ordinates against tim e as absc issa , the average value isthe average ordinate for any positive half cycle of ins tantaneous values . In the case of a harm onic current of the form

the average value l aw ,

is equal to 2 over 77

times the max imum value Io , that is,


2(5)

7T

For, put w t+ then the average value of E0

sin (w t+ t9)between the lim its ont+ 9 =0 and w t+ 9 = 7r is equal to the average value of [ 0 sin as between the lim its a: = 0 and 3: or

sin a: dx

The average value over a comp lete period of an alternating currenthaving symm etrical positive and negative values is zero

,S ince the

average of the positive instantaneous values over half a period isequal to the average of the negative instantaneous values overhalf a period .

The above definitions and deductions also apply to alternating electrom otive forces and alternating potential differences .

In the case of any harmonic function of the form x =X0sin (a ) t+ 9)

the sam e relation exists between its m aximum and average valueas between the maximum and average value of a harmonic cur

rent,that is

,

77

171 . Effective Va lues .

— The total amount of heat energydeveloped during a complete period T in a resistance r throughwhich an alternating current is flowing is equal to

r i2 dl

A steady current I in the sam e interva l of tim e will develop inthis sam e resistance an am ount of heat energy equal to rPT

Hence the alternating current will develop in any given resistancein any given tim e the sam e amount of heat energy as a directcurrent I provided

r12 7

’ = r

The right-hand side of this equation represents the mean of thesquares of the instantaneous values of the alternating current ;


sin2wt

and thereforet sin 2 wt

t =T T

Sinzwtdt z

5“

4 a) 2

Hence the effective value of the current is

10

x/2

Sim ilarly the effective value of a harmonic electromotive forcewhich has the maximum value E0

is

E0

and the effective value of a harmonic potential drop which hasthe maximum value V

0is

V0

2

Note that the effective value in each case is independent of thefrequency but depends onl y upon the maximum value . The

above relations hold on ly for harmoni c functions; when the current ,electromotive force or potential drop is not a harmonic functionof the tim e equations (6) do not hold . (See Article

1 72 . The Use of Alternating Currents . When an alternatingelectrom otive force is impressed acrossthe term inals of an electricc ircuit of any kind an alternating current of the same frequencyis established in this c ircuit . * When the strength of the currentbetween these two points at any instant is i and the potentialdrop in the di rection of the current is v

,the input of electric energy

between any two points 1 and 2 of the c ircuit in any infinitesimal

(66)

*The current established for the first second or two after the alternatingelectromotive force is impressed is not strictly an a lternating current, sinceits amplitude builds up gradually to a constant max imum , just as the value ofthe current established in a c ircuit by a continuous electromotive force buildsup gradually from zero to a constant max imum ,

depending upon the resistance and back electromotive forces in the circuit. The current for the firstsecond or two is then an osci l lating current . The building up ofan a lternating

current is discussed in Article 20 1 . For the presentwe shall confine our attention to what takes place in the c ircuit after the current has become a truealternating current, alternating between constant positive and negative max i

mum values .


interval of time dt is equal to vidt,and therefore the rate at which

energy is transferred to this part of the c ircuit at this instant , orthe power input between 1 and 2 at this ins tant, is p =vi . A lthoughthe average values of an alternating current and an alternatingp .d. over a complete period are both zero

,the average value of

the power input (or output , when the potential drop is in theopposite direction to the current) when the current and p .d . havethe sam e frequency is not zero except in certain spec ial cases .

Hence an alternating current may be used to transm it electricenergy just as a continuous current is used for this purpose . In

fact,the transm ission of electric energy over long distances can

be accomplished much m ore econom ica lly by the use of alternating than by the use of continuous currents . A lternating current

generators are much less expens ive for the same power output,

and certain forms of alternating current m otors,particularly the

induction motor, are cheaper to build and require less care inoperation than a direct current motor of the Same power output .An additional advantage of alternating currents com es from thefact that by means of a device called an alternating currenttransformer

,power may be readily transferred from a c ircuit in

which the current is small and the voltage high to a circuit inwhich the voltage is low and the current large. Since the powerlost in a transm iss ion line depends uponthe square of the current ,it is obvious that for econom ical transmrssmn the current shouldbe kept sm all

,and therefore the voltage high . On the other

hand,a high voltage is dangerous , particularly inside of buildings .

Hence electric power is usually generated at a comparatively lowvoltage , stepped up ”

. by means of a transformer to a high voltage for transm iss ion, and then

“stepped down ” by m eans of

another trans form er to a comparatively low voltage for localdistribution and use .

173 . Al ternating Current Transformer . An alternating cur

rent transform er cons ists essentially of two independent windingswound on the same closed iron core . When a current is establ ished in either winding a m agnetic flux is established throughthe iron core, and when this current varies with tim e the fluxvaries with tim e and therefore an electrom otive force is inducedin each winding . Let N and N2 be the number of turns in seriesin the two windings respectively , and let (f) be the number of linesof magnetic induction established in the core at any instan t .


When all the lines of i nduction link all the turns of both windings(that is , when there is no magnetic leakage) the electrom otive

forces induced in the two windings at this ins tant are e,=N,

d

d

d)

t

respectively . Hence these electromotive forces are

directly proportional to the number of turns in series in the twowindings

'

respectively . When the term inals of the first or primarywinding are connected to the term inals of an alternating current

generator and the term inals of the second or secondary windingare connected to an alternating current motor or other receivingdevice , energy will be transm itted through the transform er to thereceiving device . Neglecting the dissipation of heat energy inthe connecting wires and in the windings and core of the transform er

,the power input into the primary winding of the trans

former is equal to the power output of the secondary winding .

Hence calling i 1 and i 2 the instantaneous values of the currentsin the two windings , we have that

3 1= 6 2

e N i NA :

1

; therefore 7

1:

2

. That IS,the electromoti ve forces

62 N2 1 2 N 1

induced in the two windings are to each other as the number ofturns in the respective windings , and the currents in the twowindings are to each other inversely as the number of turns inthe respective windings . These relations are only approximate

,

since the assumed conditions are only approximately realized inpractice . For a full discussion of the alternating current transform er see any text-book on alternating current machinery .

174. Average Power Corresponding to a Harmonic P D. and

a Harmonic Current . Power-Factor . The average power inputinto a c ircuit when a harm onic current is

,establ ished in the circuit

and a harm onic p .d . is established between the term inals of thec ircuit can be readily expressed in terms of the effective valuesof the current and the p .d . and the difference in phase betweenthe current and p .d . Let

i = IOsin (wt+ 9)

be the value of the current at any instant t; and letv V

0sin (wt+ 02)

be the value of the potential drop in the direction of the current i


The average power transferred to the c ircuit during eachcomplete period T is

1 TV,1

Therefore

(01 02) Sin (t 01 02)

Whence

cos 0,—022

and therefore the average power input is

P V I cos (0, 92) (9)where V and I are the effective va lues respectively of the p .d . and

current,that is V =

V0 and I and 9, (92 is the difference2 x/2

in phase between the current and p .d .

Hence,when the current and p .d . differ in phase

,the average

power input into the c ircuit is less than the product of the effectivecurrent and the effective p .d . The ratio of the true power inputinto a c ircuit to the product of the effective current and effective

p .d. is called the power factor of the c ircuit . In the case of aharmonic current and a harmonic p .d . the power factor is therefore equal to .the cosine of the angle which expresses the differencein phase between the current and the p .d . Hence this angleis frequently called the power - factor angle of the circuit .

Equation (9) may then be expressed in words as the averagepower input into any c ircuit

,when there is established a harmoni c

current in the c ircuit and a - harmoni c p .d . of the sam e frequencyacross the term inals of the c ircuit

,is equal to the product of the

effective values of the current and the p .d . times the power factorof the c ircuit . ”

When current and p .d . are in quadrature,i .e.

,when01 d, i f;

then cos (9, 02) = 0, that is , the power factor is zero , and theaverage power input into the c ircuit is also zero . The Sine curverepresenting the instantaneous power is then symmetrical withrespect to the axis of tim e . In this case as much work is d oneon the current during one half of the cycle of the power curve as

ALTERNATING CURRENTS 3 17

is done by the current during the other half of this cyc le ; the totalwork done by the current in a whole cycle is zero . When c urrentand p .d . are in phase

,i .e .

,when 9, 0, = 0 then cos (0, 92) = l

that is,the power factor is unity and the average electric power

input for the given current and p .d . is a maximum and equal toVI

,where V and I are the effective values of p .d . and current .

In this case the entire curve of instantaneous power lies abovethe axis of time . When current and p .d . are in opposition

,i .e .

,

when 0, 7r then cos (HI—0 —1 ; that is, the powerfactor is again num erically equal to unity , but the average electricpower input is negative, t.e.

,work is done on the current at the

average rate VI . In this case the entire curve for instantaneouspower lies be low the axis of tim e .

In general , whenever the powerfactor angle 0, (92 is greater

than + 3 or less than E, that is , whenever the current leads or

2 2

lags behind the potential drop in the di recti on of the current bymore than work is done on the current at the average rateequal to the num erical value of the expression V I cos (0,This . part of the c ircuit then acts like a generator . Let the instantaneous values of the net rise of potential in the direction ofthe current in this part of the c ircuit

,i .e .

,the term inal e . m . f.

,

be represented by the equation e =E0sin (wt -l Then E

0is

num erically equal to V 0and 0' is equal to (92 4—77 , for the rise of

potential at any instant is opposite to the drop of potential atthis instant . The average power output of this part of the c i rcu i tis then

P0=EI cos (9, (9a)

where E = V is the effective value of the term inal electromotiveforce . This output P0

is positive when (0, is greater than

—Z -r and less than 71. Equation (9) gives the average electric

2 2

power input into the circuit and equation (9a) gives the averageelectric power output of the circuit .

175. Power Corresponding to a Non-Harmonic P D. and a Non

Harmonic Current . When the p .d . and current are not harm onicfunctions of tim e the above expressions for power, equationsdo not apply

. An alternating p .d. or current , however , may

always be represented by a Fourier’s series (see Article no


matter what may be the shape of the curve or wave representing it . For example , let the current and p .d . be

i =I , sin (wt + 1 , sin (2 ml—l (92) + 1 , sin (3 wt ,+ 93)v V

, sin V, sin (2 V, sin (3 ml +

In this case the ins tantaneous power is (see equation 7)vi [V J cos (9, VJ, cos (0, V J , cos (0, 03)VJ, cos (2wt+ 91 + 02) VJ, cos (4 ml + 92+V J , cos (6wt+ (93+ V J , cos (wtVJ, cos (2ml + VJ, cos (mH 92 02)

V J , cos (wt+ 02) V J , cos (2wt 9,

V J , cos (wt+ 9, VJ, cos (3wt+ 9,V ,I , cos (4m (9, V J , cos (3 cot 0, 05)VJ, cos (5wt -l (92+ V J , cos (4wt+VJ, cos (5cot—l 93+

The average of this instantaneous power for a complete cyc le of

the fundamental period T -

2f( 1)

Hence the average power is

VJ,V 11 1 I V 21 2 r l

2(0,

2(0,

2

_ cos (0, (10)

since the integral between the limits 0 and T for each of the harmonic terms containing t in the equation for instantaneous poweris zero . Hence when there ex ists in a circuit an alterna tingcurrent and an alternating p .d . of any kind whatever

,the average

power is equal to the sum of the values of the average powercorresponding to each pair of harm onics of the same frequencywhich exist in the Fourier’s series for the current and p .d . Thatis

,the average power Input corresponding to each pair of har

monics of the same frequency is independent of what other harmonics may be present . Note that when any harmonic is absentin ei ther the p .d . or the current this harmonic contributes nothing

[ 3 V 3

(93to the average power . For example the term2

equals z‘

ero for either I or V ,=0 .

It should a lso be remembered that in m ost practical cases thecurves representing p .d . and current are each symm etrical with


sine wave having an effective value equal to the eflective valueof the actual wave and differing in phase by an angle 9 such thatV I cos 9 represents the average power, where V and I are theeffective values of the true p .d . and current waves respectively .

In certain Special cases , however, it is necessary to analyse thewaves into their cons tituent harm onics .

178 . D etermination of the Maximum Val ue and Phase of the

Harmonics in a Wave of Any Shape . When the wave shape ofa p .d . or current is known , the effective value and the phase ofthe harm onic of any order may be readily determ ined . For

example,the ins tantaneous value of the current may be written

i =I , sin sin (233+ sin (333+ (a)

where x =2 77 ft and f is the frequency of the wave . The instantaneous value of the current i corresponding to any value ofa: is also given by the corresponding ordinate of the curve representing the current wave . We wish to determ ine the cons tants

I , , I , , etc .,and 9 etc . which will make the curve rep

resented by (a) coincide with the actual current wave . Con

sider,for example

,the third harn ionic . Multiply equation (a)

by sin 33: and integrate with respect to at over an entire period ofthe wave

,i .e .

,between the lim its x = 0 and x =2 77 . We then have

i sin 3x da: [I sin 55sin 3x+ I , sin (2x+ sin 33:

+ 1 , sin (31 + sin 3r + etc ] dxBut the integral of each term in the right hand member of thisequation between the lim its 0 and 2 71 is zero

,except for the

particular term I sin (3134—9) sin 3x , the integral of which from2 7rI

0 to 2 7T is2

3

cos 9, 77 1 , cos (See Artic le Hence

i sin 3x dx = 7r I , cos 9,

The integral i sin 32: da: may be determ ined graphically

by plotting the expression i sin 3 i : or ordinates agains t x as

absc issas,and determ ining the area of this curve by means of a

planim eter . Call this area A , ,then

A,= 7TI , cos 9,


Next,multiply the equation (a) by cos 3x=sin 3r +

exactly the sam e manner we then have that

i cos 3x dx = 77 1 , cos 7r I , sin 9,

and the value of ’

t cos 3r dx may be determ ined graphically

as in the first case by plotting the curve i cos 3r and finding itsarea by means of a planimeter . Call this area B, ,

thenB

3 77 1 3 Si ” 03From equations (b) and (c) we then have

1I3:

7T

9, = tan

Note that in case the wave is symm etrical with respect to theaxis of tim e it is unnecessary to look for the even harmonics .

Also,when the wave is symmetrical

,the curves i sin 33: and

i cos 3a: need be'

plotted for only a ha lf period of the wave . Calling a , and ,

B, the areas of these curves for half a period of thewave

,we then have

22 2

Wa s

0, tan"

This method of determ ining the harm onics present in thewave is of course applicable to the determ ination of the harmonicof any order .

Fig . 98 .

Examp le : Suppose the current is a symmetrical rectangularwave a s shown in Fig . 98 . Let I be the value of the current


during the positive half of the wave . Since the wave is sym

metrical,only the odd harm onics can be present . Cons ider the

nth harmonic , where n is any odd positive integer . Then" 2I

I sin nx dx cos na:

0 n

I cos nx dx -

sin nx

S ince i t is odd . Hence from equations (1 3)

0

21

n

Hence all the odd harmonics ex ist in a symmetrical rectangularwave

,the max 1mrim values of the harmonics varying inversely

as their order n . A symmetri cal rectangular wave having a m aximum value I may then be represented by the infinite series

tan

sin (2 7r nft.)77 Z n

n 1

where n has all the odd values from 1 to co and f is the frequencyof the wave .

1 79 . The Fisher-Hinnen Method for Analysing a Non-HarmonicWave .

- (See E lectri c Journa l,V ol . 5

,p . 386 and E lektrotech

n ische Z ei tschrift,May 9 , Thi s m ethod is much simpler

than that described above,except in the rare cases where the

resultant wave may be represented by a simple integrable function . The method is based on the following facts :

1 . The a lgebrai c sum of any n equally spaced ordinates of ak

sine wave,when these ordinates are spaced

5th of a wave length

apart,where k is any integer which is not a multiple of n,

is zero .

2 . The algebraic sum of n ordinates of a sine wave when theseIt

ordinates are Spacedawave lengths apart , where k is a mu ltip le

ofn,is equal to n times the ordinate of this wave at any one of

these points .

3 . The maximum ordinate of any sine wave is equal to thesquare root of the sum of the squares of any two ordinates Spaceda quarter of a wave length apart .


right of the first set of ordinates,their sum will be equal to n

times the ordinate of the nth harmonic a quarter of a wave lengthof this harmonic from y, call this ordinate then

1En

"2

;(92+ 94+ 96+ 9m)

Then from Proposition 3, the maximum value of the ordinateof the nth harmonic is

Y,

From Proposition 4,the angular distance to the left of y, at which

this harmoni c cuts the x axis in the positive direction 1s

ta' ln(15

, —n

Bn

when 360 degrees are taken equivalent to a wave length of thenth harmonic . When 360 degrees are taken equivalent to a wavelength of the given wave this angular dis tance is

An

(fin : —tan‘ 1

-

l

Consider next the mth harmonic , and erect two sets of mordinates

,the ordinates of each set being spaced 360

°apart

(cons idering a wave length of the given wave as equivalent toand the second set a quarter of a wave length of this har

monic to the right of the first set . Then,if the harmonics ofhigher

order are not mul tiples ofm,we have as before that the ordinate

of the mth harmonic at the point 1 is1

Am (y, ya y, yzms )m

and the ordinate of the mth harmonic at the point 2, which is aquarter wave length of this harmonic to the right of 1 , is

1Em =

ht"

yzm)

Whence the maximum value of this harmonic is

Ym V Amz Bm

2

and it cuts the x axis at the angul ar distance

=1 tan“ 1

m

to the left of the first ordina te when 360 degrees are taken equiv?

alent to the wave length of the original wave .

If there also ex ists in the given wave a harmonic of the nthorder

,where n is a multiple of m

,that is if n =—km,

where lc is aninteger , then from Proposition 2

,S ince each set of these m ordinates

is spaced5 k wave lengths of the nth harmonic apart , we have

that the sum of the first set ofm ordinates also contains m times


the ordinate of the nth harmonic at the point 1 . Call ing An

'

the ordinate of the nth harmoni c at the point 1 we then havethat

1Am= y5+ + ym_ 1)

m

Sim ilarly,

1Em

-

2

Eu

’

where is the ordinate of the nth harm onic at the point 2 .

A sim ilar correction must be applied for a ll other harmonicsof higher order than the mth if the orders of these harmonicsare multiples ofm.

The waves of current and electromotive force with whichone has to deal in practice usually contain only the odd harmonics ; als o , as a rul e

,the harmonics of higher orders than

the seventh are negligible . In this case the three harmoni cs ,the third

,fifth

,and seventh

,are not multiples of each other

and cons equently no correction term has to be app lied . Moreover

,it is sufli cient to cons ider the ordinates of only half a wave

length . To determ ine the third harmonic divide the base lineof this half wave into 2n =6 equal parts and measure the ordinatesat the beginning of each of these six segments . Let these ordinates be y, , y, , y, . Let the beginning of the first segmentbe taken where the given wave cuts the x axis and call this pointthe origin, then y,=0 and we have

1A 3=

5_

3

3(y y )

1Ba (9 2+ 96 94)

Then the maximum value of the third harmonic isY,= A ,

2

and it cuts the x axis at the angular distance

¢3 -

1tan

-1

3

to the left of the origin . The equation of the third harmonicis then

y, =Y, sin 3 (33+

Sim ilarly,starting at the sam e point and dividing the wave

into 2n : 10 segments , we have for the 5th harmonrc

1A 5=5(y5+ y9 y3

_

y?)

18 5=3(9 2+ ye+ yro—y4—y,)


Y,

15=— t

“ I(f)

5an

and the equation of this harm onic isy, Y, sin 5

The determ ination of the seventh harmonic is carried out by an

exactly s im ilar process , dividing the base line into 2n = 1 4 equalparts .

The ordinates of the fundamental at the origin and at thepoint corresponding to one quarter of a wave length from theorigin are then respectively

A ,= —A 3

—A 5—A 7

and B l=ym+ B,

where ym is the ordinate of the given wave corresponding to thepoint a quarter of a wave length from the origin the m idordinate of the given wave) and A A, ,

A , ,B, ,

B, ,and B7 a re

the quantities above determ ined . The maximum value of thefundamental is then

Y, A ,

2B

,

2

and it cuts the axis of x at the angular distance

to the left of the origin . Its equation is therefore,

y, Y, sin (33+The equation of the given wave is then

y=Y, sin (St -Hp) Y, sin 3 Y, sin 5 (113+Y7 sin 7 (x+ 937)

The effective value of the given wave is

l + Y3

2+ Y5

2+ Y7

2

2

and the average value is

Ya ,

-

:

1

g, cos 3 gb, +éY,

( 1 40)In emp loying the above formu las stri ct attention must be paid

to a lgebra i c signs .

Examp le. In the curve Shown in the figure the ordinates at6 equally Spaced points

,starting from the point where the curve

cuts the base line are

yr 2 0

y,= 676

y,= 660


The equation of the given wave is then

y sin (x 150 sin 3 (33+100 sin 5 (x

Its effective value isY= 718

and its average valueY

am . 673

1 80 . Form Factor . The form factor of a wave is definedas the ratio of the effective value to the average value . For

I,

210

77

a s1ne wave the form factor IS—z The7r

form factor of a flat- topped wave is less ; of a peaked wave

greater . The form factor of the wave shown in Fig. 99 is 11.—8

673

1 81 . Ampl itude Factor .—The amplitude factor of a wave is

defined as the ratio of the max imum value to the effective value .

For a sine wave the amplitude factor is The amplitudefactor of the wave shown in Fig . 99 is

1 82 . Power and Reactive Components of P D. and Current .Whenever we have to deal with two or more harmonic func tionsof tim e

,we may as a matter of convenience begin counting tim e

at an instant when one of these functions is zero and is increasingin the positive direction . For example

,when the current is

i = Iosin and the p .d . v= V

0sin we may begin

counting tim e at the ins tant when v=0 and is increas ing in thepositive direction ; 92 is then equal to zero . Hence

,dropping the

subscript from 9, and also writing I, 2 I and V,

2 V,

where I and V are the effective values of i and v respectively,

we have

V sin wt

t 1 sin (wt+ 0)The average power is then

P VI cos 9

From the trigonometric formula thatsin (a + b) = sin a cos b+ cos a sin a

we may write the current

\/2 I cos 9 sin I sin 9 00 3 cut

that is,we may consider i as made up

,of two components


cos 9 sin wt

V 2 1 sin 0 cos aim s/2 1 sin 0 sin (m l-Pg)

The first component i , 15 In phase with the p .d . and has the effectivevalue I cos 9 ; the second component is in quadrature ahead ofthe p .d . and has the effective value I sin 9 . The average powercorresponding to the firs t component i , is

V (I cos 9) cos 0 = V I cos 9= P

and the average power corresponding to the second component i , is

V (I sin 9 ) cosg=0

Hence the average power ofa harmonic p .d . and current is equa l

to the effective va lue of the p .d . times the effective va lue of the com

ponent ofthe current in phase wi th the p .d . The component of thecurrent in phase with the p .d . is therefore called the power com

ponent of the current . The component of the current in quadrature with the p .d . is called the reactive component of the current

,

S ince the power input corresponding to this component of thecurrent during each quarter cyc le is exactly equal to the powergiven back during the following quarter cyc le .

Sim ilarly,we may consider the p .d . as made up of two corn

ponents

cos 9 sin (wt+ 9)

sin 9 cos (wt+ 9) sin 9 sin (ml + 9)

The first component v, is in phase with the current and has theeffective value V cos 9 the second component v2 is in quadraturebehind the current and has the effective value V sin 9 . Theaverage power corresponding to the first component v, is

I (V cos 9) cos O= I V ecs 9= P

and the average power corresponding to the second component”

0 2 18

I (V sin 0)7

2

1 :

Hence the average power of a harmoni c p .d . and current i s a lso

equa l to the product of the effective va lue of the current times the

effective va lue ofthe component ofthe p .d . in phase wi th the current.

The effective value of the component of the p .d . in phase with


the current is therefore called the power component of the p .d . ;

and the effective value of the component of the p .d . in quadraturewith the current is called the reactive component of the p .d . Thereactive component of the p .d . is in quadrature behind the currentwhen the resultant current leads the resultant p .d . and is inquadrature ahead of the current when the resultant current lagsbehind the resultant p .d .

The name wattless component is som etim es used for

reac tive component,S ince the average watts corresponding to

this component of current or p .d . is zero ; the instantaneous wattscorresponding to this component

,however

,are not zero . Hence

the adj ective wattless is m isleading .

183 . Apparent Power. The product of the effective valueV of the resultant p .d . and the effective value I of the resultantcurrent is called the apparent power , that is

Apparent Power =V I

From equation we have

Average Power = V I cos 9

where 9 is the difference in phase between p .d . and current .Hence

Avera e PowerPower Factor g

Apparent Power

The terms volt- amperes and apparent watts are also frequently usedfor apparent power ; i .e.

,volt- amperes or apparent watts (effec

tive p .d . in volts) X (effective current in amperes) .1 84. Reactive Power. The expression reactive power is

used for the product of the effective current and the effectivevalue of the p .d . in quadrature with it ; or , what am ounts to thesame thing, the product of the effective p .d . and the effectivevalue of the component of the current in quadrature with the p .d .

That isReactive Power V I sin 9 (17)

where V and I are the effective values of p .d . and current reSpect ively,

and 9 is the power-factor angle . From equations

(1 6) and (1 7) it follows that the apparent power is equal to thesquare root of the sum of the . squares of the average power andthe reactive power

,i .e .

,

V I =V (V I cos (VI sin


motive forces then differ in phase by 90 . If 0,the two act

in the sam e direction at each instant ; if 9 1 800 the two oppose

each other at each instant ; if 9 has any other value the twoelec tromotive forces act together during part of each cyc le andoppose each other during the rest of the cycle . In any case

,the

net or resultant electromotive force at any ins tant issin wt—l—E , sin (ml -H9)

The expression E , sin wt+ E , sin (wt+ 9) may be put equal toE,sin (wt -l where the constants E, and 9, may be deter

m ined from the condition that the relation

E , sin wt+ E , sin (m l - l sin

must hold at all tim es . For t= 0,we then have that

E , sin 9 E,s in 9

,

and for t

E ,+ E , cos 9= E ,cos 9

,

since sin cos 9 and sin

Squaring the equations (a) and (b) and ad

E ,

2+ E ,

2cos

2 9+ 2E ,E , cos 9+ E ,

2sin

2 9 - E,

2

whenceE,

2 = E ,

2+ E ,

2+ 2E ,E , cos 9

Dividing the equation (a) by the equation (b) , we get

E , sin 9tan 9

0

E ,+ .E , cos 9

Thereforesin (mH

where E,is the maximum value of the resultant e.m.f. and is

given by equation ( l 8a) and 9, is the angle by which the re

sultant e . m . f . leads e, and is given by equation (18b) .

Exactly sim ilar relations hold for two harm onic currents ofthe sam e frequency

,and in fact for any two harmonic functions

of the sam e frequency in the same independent variable .

1 86 . Representation of a Harmonic Function by a RotatingVector . The resu lts just deduced may be arrived at graphicallyby a very simple m ethod . Let OQ in Fig . 1 00 be any line. fixedin the plane of the paper and let the line OP , be equal in lengthto E , , and be pivoted at 0 and rotate about 0 with an angular


veloc ity to let this rotating line coinc ide with the line of referenceOQ at tim e t= 0 . Then at any ins tant of tim e t

,this line will

make an angle cu t with OQ and the instantaneous value of thee . m . f . e, = E , sin cut at any instant will be equal to the verticaldistance from P , to OQ . Sim ilarly

,if OP , is a second line

equal in length to E , rotating about 0 with the sam e angularveloc ity co

,and making the angle 9 with OQ at tim e t= 0 , then

the instantaneous value of the e. m . f . e,= E , sin (m l + 9 ) at any

instant w ill be equal to the vertical distance from P , to OQ .

Since both lines rotate with the same veloc ity the angle betweenthe two will be equal to 9 at all times ; that is the phase angle9 m easures the difference in direction between the two rotatingvectors representing the two e . m . i .

’s .

Fig . 100 .

Let OP,= E

,be the vector sum of OP , and OP, . It is then

evident from the diagram that the instantaneous value ofat any instant is the vertical distance from P

, to OQ; also thatthis line OP

,remains fixed in length and fixed in position with

respect to OP ,and OP, and is equal numerically to the maximum

value of and at any instant makes the angle (wt+ P,OP ,)

with the line of reference . Hence the resultant e . m. f.

sin (co t+

where


E,

E 2

,+ cos 9

“ l9,—tan

which are identical with equationsThe above discussion of course applies to any two harmoni c

functions of the same variable . For example,the resultant of

two harmonic currents

i , z ] , sin wt

i , = I , sin (ml 9)i ,+ i , = I , sin (ml +

cos 0 (20a)

9,= tan

”

(20b)

A re- reading of Articles 8 and 9 will be found helpful inunderstanding the vector method of representing alternating cur

rents and potential differences .

1 87 . The Vectors Representing Any Number of HarmonicCurrents and P.D.

’s of the Same Frequency are Stationary with

Respect to One Another. This is immediately evident from thefact that the analytical express ion for any harm onic func tion of

the time is A sin where A and 9 are cons tants for eachsuch function

,and a)

,which is equal to 2 77 tim es the frequency

,

is the angular veloc ity at which the vector representing thisfunction rotates . Hence the vectors representing any numberof such fun ctions of the sam e frequency all rotate through equalangles in any interval of time, i . e .

,their relative positions with

respect to each other remain unchanged . Consequently in problems which involve only harmoni c currents and harmoni c p .d .

’s

and their phase displacements with respect to each other,the

vectors representing the currents and p .d .

’s may be considered

as stationary .

It should be carefully noted that non-harmonic currents or

or currents or p .d .

’s of di fferent frequenc ies cannot be

represented on the‘

same diagram by stati onary vectors .

1 88 . The Lengths of the V ectors Representing HarmonicFunctions Taken Equal to their Effective Values . In any problem which has to do only with the effective values of harmonic


is ri and the back e . m . f . (due to self induc tance) is Lgfsee A rtic leHence the total ins tantaneous potential drop is

v= ri Lilli

dl

Substituting for i its value IO sin (2 77ft) in the above equation we get

v= rI,sin (2 77 fl) (2 77 fL) I , cos (2 77 ft)

rI,sin (2 77 ft) (2 77 fL) I , sin (2 77 ft 3)

Hence when the resistance and inductance are constant the p .d .

from 1 to 2 is also a harm onic function of the tim e,has the sam e

frequency,and cons ists of two components having respectively

the effective values r ] and (2 77fL) I where II”is the effective

value of the current .The first component of the p .d .

,nam ely rI

,is in phase with

the current and the second component of the p .d .

,namely (2 77fL) I ,

leads the current by Hence the effective value of theresultant p .d . is (see the

‘vector diagram and also equations 1 9)

V =I 77fL)

and this resultant p .d . leads the current by the angle

9=tan" 1

This angle 9 is the powerfactor angle of the c ircuit . The powerfactor of the c 1rcuit is then

60 8 0

Note that the potential drop in a given portion of a circuit isequal to the e . m . f . impressed across the term inals of this portionof the c ircuit , which e . m. f . is in the sam e direction around thec losed c ircuit as the potential drop through the given portion ofthe c ircuit ; hence in the above formulas V may be taken to represent either the potential drop in the c ircuit or the e. m. f. impressed ou it .


Since the first component of the p .d . is in phase with thecurrent

,and the second component of the p .d . is in quadrature

with the current,it is the first component alone which determ ines

the average power put into the c ircuit ; r] is therefore the powercomponent of the p .d . The electric energy ri i dt which is con

verted into heat energy in the resistance r during each intervalof time dt is not returned when the current reverses

,s ince this

energy is proportional to the square of the current i and is therefore independent of the direction of i . The average rate at whichthi s energy

'

is suppl ied to the c ircuit is represented by the constantterm in the expression for instantaneous power

,equation

AS j ust noted,the second component of v

,namely ( 2 77fL)I ,

is in quadrature with the current and therefore contributes nothing to the average power, i . e.

, this is the reactive componentof the p .d . This is also evident from the fact that the energystored in the magnetic field while the current is rising from zeroto a maximum is equal to the work done on the current by thefield when the current decreases from its maximum value to zero .

Also note that this cyclic transfer of energy to and from them agnetic field occurs twice during each cyc le of the current

,i .e .

,

it is represented by the double frequency term in the expressionfor ins tantaneous power

,equation

1 90 . Efiective Resistance,Reactance and Impedance of an

Alternating Current Circuit . In general when an alternatingcurrent is established in an electric circuit

,secondary currents are

established in the surrounding conductors (for example , as eddycurrents in the iron c ores of the m agnetic c ircuits of electric ma

chinery) a certain am ount of energy in addition to that dissipatedin the main or primary c ircuit is therefore also dissipated in theconductors in which these secondary currents are established .

A lso,due to the fact that the number of lines ofmagnetic induc

tion linking the center of a wire is greater than the number linkingthe outside of the wire (see Artic le 1 21) the induced back electromotivefins ide the wire is in general greater than tha t induced inthe outside filam ents of the wire

,and therefore the current dens ity

is not unif ormover the cross section of the wire,as is the case

with a continuous current ; hence the effective resistance of aconductor to an alternating current is greater than its resistanceto a continuous current . This skin effect , however, is not as arule serious for the frequencies and sizes of conductors used in


practice , except in the case of steel rails used for conductors mra ilway work . Also

,when there is iron in the magnetic field

established by the current,a certain amount of energy is diss i

pated in the iron,due to . hysteresis . Hence the average rate

at which heat energy is dissipated when an a lternating currentis established in a given, portion of a c ircuit is not equal to theproduct of the square of the effective value of the current by theresistance Of the conductor in which the current is established

,

as determ ined by a continuous current measurement,but is in

general greater than this . This portion of the c ircuit may ,how

ever,be considered as having an effective or apparent res istance r .

“

such that this resi stance mu ltip li ed by the square ofthe effective va lueofthe current gives the true average rate at which heat energy

“

i s dis

sipated when the current i s establi shed in thi s portion of the circui t.

This effective resistance is in m ost practical cases approxim atelyindependent of the current strength (as m easured by the effectivevalue) but does depend upon the frequency of the current

,and

,

in case there is a loss of energy due to hysteresis , upon the effectivevalue and the wave shape of the current also . (The part of theeffective resistance which takes into account the hysteresis lossin iron is not strictly constant , but varies approxim ately as thepower of the maximum flux density

,and therefore as the

power of the maximum value of the current .)The average rate at which heat energy is dissipated when an

alternating current is established in any portion of a c ircuit may

then be written

Ph= T12

where r is the effective resistance of this portion of the c ircuitand I is the effective value of the current .The effective value of the resultant potential drop in any

portion of a c ircuit,however

,is in general greater than rI . Let

V be the vector representing the resultant p .d . in any portion ofthe circuit and let E be the vec tor representing any externally induced e. m . f . in thi s portion of the c ircuit , both in the di rectionof the current if the portion of the c ircuit cons idered is thearm ature of an alternator

,E is the induced e . m . f . due to the

relative motion of the armature and the magnetic field) . Then

the ratio of the numerical value‘of the vector difference E V

to the effective value of the current I in this portion of the c ircuit is


19 1 . Reactance and Impedance of a Coil of Constant Resistance and Inductance * to a Harmonic Current . From equations

(21 ) and the above definitions , it is evident that the impedanceof a circuit of constant res istance ( independent of the value ofthe current at any instant) and a cons tant inductance ( independent of the . value of the current a t each instant) , to a harmonic current is

z (2 77 iL)2

and the reactance is

(264 )

Note that these relations are deduced on the assumptions thatthe coil has no electrosta tic capac ity and that the current is asine wave .

192 . Reactance and Impedance of a Coil of Constant Resistance and Inductance to a Non-Harmonic Current. The reactanceof such a circuit to a non-harm onic current is readily foundwhen the equation of the current is known . Consider the Spec ia lcase of a current which contains the third harm onic . Thiscurrent m ay be represented by the equation

i =I , sin (2 77 ft) + I , sin (6 77 ft+ 9)

where I and I, are the m aximum values of the fundamental andthird harmonic respectively . When

'

the wave shape rem ainsI

constant for all values of the current -

I

-1 is a constant . ThenI

the instantaneous value of the drop of potential through theresistance and inductance is

v=ri + LZ—Z= rI , sin (2 77 ft) + rI , sin (6 77 ft+ 9)

77 fL) I , cos (2 77 fL) I , cos (6 7rit+ 9)

Comb ining the term s of the sam e frequency in the m anner described in Article 1 85, we get

I , sin (2 77 ft -l—a i)

1 , sin

where

Such a coil is frequently called an impedance coil.


a ,=tan

“ l and a , 9 tan“ 1

(see equation 18b) .

The effective value of V is then (see Artic le 1 76)

(2 77 fL) 73

-1—(6 77 fL)2

and the effective value of i is

whence

r2+ (2 77 fL)

2+ 8

2 2

i f is constant,

13

is constant ; put a2[ 3

1 ,

the impedance of the circuit to this current is equal to

r2+ (2 77 fL)

2

( l + 8az

)


a=2°

7r f / l + 8az (27a)

The constant a is the ratio of the effective value of the thirdharm onic to the effective value of the resultant current .Note that the effective value of the first harmonic in the p .d .

I ,wave IS 77 fL)2

,while the effective value of the third

harmonrc IS 77 iL)2

. When the inductance L is large

compared with the resistance , the ratio of the third harmonicin the p .d . wave to its fundamental must then be three tim esas great as the ratio of the third harmonic to the fundamentalin the current wave . V i ce versa

,the third harmonic in the

current wave resulting from a given p .d . will be relatively onlyone third as great in the current wave as in the p .d . wave . Therefore an inductance in a c ircuit tends to dampen out the harmonicsin the current wave when a non-harm onic e. m. f. is impressed onit

,and to ~make this wave approach m ore nearly to a sine wave .

The higher the order of the harmonic the greater the damping.


1 93 . Current through a Condenser when a

.

Harmonic PD . is

Establ ished across It . Capacity Reactance . Let the p .d . acrossthe term inals of the condenser be

-> v- V, s in (an t i

i gv+ C

Fig . 102 .

Thereforei =g V,

sin (2 7r fC) V ,cos (2 77 ft)

—g V ,

sin (2 77 fC) V, sin (2 77 ft+ 25)Hence the component of the current through the condenser dueto the conductance of the dielectric

,i .e.

,the leakage current

,

is in phase with the p .d . across the condenser,and the component

of the current through the condenser due to its capacity,i .e .

,

the displacem ent or charging current,leads the p .d . by

The effective value I of the tota l current is

where V is the effective value of the p .d . Hence the effectiveresistance of a leaky condenser to an alternating current of fre

queney f is (see Article 1 90)9 172

g

[ 2 gz

and its impedance isV 1

I

and therefore its reactance is

v= V,sin 2 77 ft

and let the total current through thecondenser in the direction of the potential drop be i . Let C be the capacityof the condens er and g the conductanceof the dielectric between its plates , i . e .

,

the leakance of the condenser . Then,

S ince the total current through the condenser is the sum of the conduction cur

rent, gv, through the dielectric plus the

displacem ent current Cfi through thedt

dielectric (see Article we have

dv

dl


harmonics in the p .d . wave directly in proportion to their order .For example

,when the seventh harm onic in the p .d . wave has

an amplitude equal to 5 per cent of the fundam ental,the seventh

harm onic in the current wave has an amp litude equal to 35 percent of the amplitude of the fundamental in the current wave .

Compare with the effect produced by an inductance , Artic le 1 92 .

1 95 . Impedance of a Resistance , Inductance and Capac ity inSeries to a Harmonic Current. When a harm onic current ofeffec tive value I is established in such a c ircuit (see Fig . the

p .d . across the resistance is V, =rI and is in phase with I . The p .d .

across the inductance is VL (2 7TfL) I and leads I by The

p .d . across the capac ity (a condenser with negligible leakance) is

Vc

Iand lags behind I by Hence the resultant p .d . is

2 7r fC'

V = Vr+ V L + V 0 r

2+ faw

i == Ios in ( 2

(2 1r iL ) I

Fig . 103 .

The impedance of such a c ircuit is therefore

V 1 27r fL

The reactance is therefore

2 7r fL

The angle by which the p .d . leads the current is

0= tan“ 1 f


Note that equations (30) apply only in case the condenser hasno leakance .

1 96 . Resonance .

—When the inductance L and capacity Cin the case j ust considered are of such values that

1

or when 2 ” (C

1

2m/LC

the reactance of the circuit is zero and the impedance is equal tothe resistance

,and therefore the current corresponding to a

Vgiven p.d . V is I that is

,the current is a maximum and

r

depends onl y upon the resistance of the c ircuit . The frequencycorrespondi ng to this condi tion is the sam e as the frequencywith which the current and p .d . would osc illate were the con

denser short- c ircuited by the inductance ; i .e .

,this frequency

corresponds to the free period of such a c ircuit . (See Artic leNote the analogy with the motion of a body , which is free

to vibrate,produced by a periodic force having a period equal to

the period of vibration of the body ; for example , a heavy churchbell may be caused to swing with large amplitude (correspondingto the maximum value of the current I ) when a comparativelysmall force (corresponding to the p .d . V ) is applied by the m an

pulling the rope , provided the successive applications of thisforce are in tim e with the swinging of the bell .When the frequency of the electromotive force impressed

across the term inals of the c ircuit is equal to the natural or freefrequency of the c ircuit , the c ircuit is said to be in resonance

with this impressed electromotive force .

Note also that,although the resultant p .d . across the resist

ance,inductance and capacity in series

,is equal to T]

,i .e.

,is the

sam e as the p .d . across the resistance , the p .d . ac ross the inductance or across the condenser may be many tim es this . For

example,when the inductance L is 1 henry and the capac ity C

is m icrofarads,and the frequency is 60 cyc les

,the inductive

reactance is xL = 2 77 X 60 X 1 = 377, and the capac ity reactance is


of the circuit is v - l- xc=0

, and the c ircuit is in resonance with theimpressed e. m . f . When the resistance r is 1 ohm and the impressed 6 . m. f. across the entire c ircuit is 100 volts, the current is

100 amperes

The p .d. across the resistance is then volts . The

p .d . across the inductance,however

,is 2 7T 100

volts and the p .d . across the condenser is

2

This brings out in a striking manner the fundamental fact thata lternating p .d .

’s cannot be added a lgebrai ca l ly; they must be added

vectoria l ly.

1 97 . Impedances in Series .

— In the case of continuous cur

rents we have seen that several resistances r , , r2 , r3 , etc .

,in series

are equivalent to a single resistance R which is equal to the arithmetical sum of r, , r2 , r3 , etc .

,that is In

case of alternating currents this same relation -also holds for theresistance of any number of impedances in series

,such as a num

ber of coils of wire in series . For,when the sam e current I

flows through each coil the average power dissipated in all thecoils is —r2 - r3

- etc .)I2 where r 1 , r2 , r3 ,

etc .

,are the effective resistances of the respective coils . Hence

,

calling R the effective resistance of all the coils in series,we have

from Article 1 90 thatR=r 1 r 2+ r

, etc . (32a)

In the caseof a harmonic current the total potential drop throughall the coils in phasewith the current is then RI —

(r l -i—r2 r3 etc .)I .

Sim ilarly,calling cc, x2 , 5133 , etc .

,the reactances of the respec

tive coils to the harmonic current I,the p .d .

’s across the separate

reactances are respectively x 11 , s , x31 , etc .

,and are all in quad

rature wi th the current and are therefore either in the sam e or op

posite direction . Hence the algebraic sum of these p .d .

’s gives the

total p .d . in quadrature ahead of the current . Whence , callingX the equivalent reactance of the c ircuit

,we have from Article

1 90 thatXI —etc .

X etc .


>< z2 = 100 volts,which is 79 per cent greater than

the true difference between the potential drops across the generator and across the receiver

“

term inals .

Fig . 104 will m ake these relations clear . The reason the p .d .

in the line is not equa l to the difference of the p .d .

’s at the

generator and the receiver is that the p .d . in the line and the

p .d . at the receiver are.

not in phase.

198 . Impedances in Para llel . In the case of continuous cur

rents we have seen (Article 98) that when several resistancesr, , r2 , r3 , etc .

,are connected in parallel , the currents in the various

V Vresi stances are respect1vely I , I 3 =E,

etc .,and therefore

T2 T3

that the total current between the j unction points of the several

resistances is etc .,

V + l + 1 41 T3

Whence the equivalent resistance R must be such that1 11 I 1

R V r , r 2 r3

In the case of alternating currents, the currents in any number

Fig . 1 05

of impedances connected in parallel between two points A and B(Fig . 1 05) are respectively

but these currents are not in phase, hence they cannot be added


algebraically ; they must be added vectorially . The componentsof the various currents in phase with the potential drop V betweenA and B are respectively

V rI , cos 9,

1

3 1 3 1

V rI 2 cos 92

2

V etc .

2 2 2 2 Z 2

and the components of these currents in quadrature behind V arerespectively

I , sin 9,

Hence the total current I is the vector sum of these two com

ponents , i .e .

,

I = V

and therefore the equivalent impedance is Z,where

1 I

Z V (33)

1 99 . Conductance, Susceptance and Adm ittance . Note that

for any of the impedances the factor22multip lied by the p .d.

across the impedance gives the component of the current in phase

with the p .d . and therefore this factorT

;multiplied by the square

of the p .d .

,i .e.

, gives the average power dissipated in this

impedance . In general , the ratio of the average rate Ph at whichheat energy is dissipated in any part of a circuit to the square ofthe effective value of the p .d . V across this part of the circuit, is


called the effective conductance g of this part of the circuitthat is

,

Ph

V 2 (34)

The ratio of the effective value of the current in any part of ac ircuit to the num erical value of the vector difference of the external ly induced e. m . f . E and the resultant p .d . V in this portionof the c ircuit is called the admittance y of this part of thec ircuit, that is,

I

E V (35)The square root of the difference between the square of the adm ittance of any part of a ci rcuit and the square of the effec tiveconductance of this part of the c ircuit is called the susceptance

b of this part of the c ircuit, that is ,—g2

(36)In the case of a harmonic p .d .

,established across the term inals

of a c ircuit in which there is no externally induced e. m . f.

, theproduct of the susceptance and the p .d . gives the component ofthe current in quadrature with the p .d .

,s ince the total current

is yV and the component in phase with V is gV ,whence the

component in quadrature with V is —gz

.

The sign of the susceptance b is taken positive when the p .d .

leads the current,negative when the p .d . lags behind the current ;

that is,the susceptance and reactance of a c ircuit always have

the sam e sign . Also,from the definition of reactive power (Artic le

it follows that the reactive power is s .

From the above definitions and the discussion in Article 1 93,

it follows that the effective conductance , or as it is also called,

the leakance,of a condenser

,is equal to the reciprocal of the

resistance'

oi the dielectric between its plates,as m easured by

m eans of a continuous current (provided there is no diss ipationof energy due to dielectric hysteresis and the susceptance ofa condenser of capacity C to a harm onic current of frequency f is

bc= —2 rr fC.

The susceptance of a condenser is negative , since the chargingcurrent corresponding to the capacity 0 leads the p .d . The ad

m ittance of a condenser having a leakance g and capacity C to aharmonic current of frequency f is then

9


negligible leakance, and f the frequency of the current . Let

the coil and condenser be connected in parallel as shown in Fig . 1 06.

Then,the susceptance of the

1

2 7rfL

ceptance of the condenser isbc—2 rriC. Hence the ad

m ittance of the inductanceand capacity in parallel is

coil is b and the sus

Fig . 106 .

TY=— 2 wfl7

1

2 7TfL

Hence the total current from A to B when a drop of potential Vis established from A to B is

I = YV = V1

2 7r/‘L

When1

= 2 7r j'

C,that is

,when f

1_

the total2 7TfL 2 77 1/ LC

adm ittance is zero and therefore the total current fromA to B is zero . This is also evident from the fact that the

current in the inductance isVand lags behind the p .d . by

2 7TfL

the current in the condenser is 2 77 f0 V and leads the p .d . by

Hence,when 2 7rfC —1 the current in the inductance

2 7TfC

at each instant is exactly equal and opposite to the current inthe condenser

,and therefore the total current is zero . Note

2 7fl / LC

c losed c ircuit formed by the condenser and the inductance .

(Compare with ArticleThe above deductions are based on the assumption that a

harmoni c or sine-wave current is estab lished in both the capac ityand in the inductance . In the ideal case of a coil of no resistance ,a harmonic p.d. impressed across its term inals does not necessarily


estab lish a harmonic - current , but the expression for the currentmay contain a constant

'

term . In the case of any actual coil,

however,the resistance of the coil (which can never be actually

zero though it may be quite sm all)‘

causes the constant term inthe expression for the current to becom e zero after a short tim e

,

and the current becom es a true harm onic current,see Artic le 201 .

A lso,due to the resistance of the coil

,and the leakance of the

Condenser as well , the resultant adm ittance of a coil and condenserin parallel can never be zero and hence for a given p .d . across theircomm on term inals the total current can never be absolutely zero .

However,when the resistance of the coil and the leakance of the

condenser are sm all compared with the reactance of the coil andthe susceptance of the condenser

,the total current will be a

m inimum when - the impressed e .

'

m . f . has the frequency1

fzv c

- Examp le : When L = I henry and C = 7 .04 m ic rofarads and

the frequency f is 60 cycles per second,

1 1and

2w 377

2 7TfC and therefore the total adm ittance,neglecting the

377

resistance and leakance,is zero . Hence

,when a coil of inductance

L and a condenser of capac ity C,connected in parallel, are

connected to a 60- cycle generator and a sufficient tim e is allowedfor the current in both the condenser and the coil to becometru e harmonic currents

,a potential drop of 100 volts across

the common term inals A and B will establish a current of100

377

provided the resistance of the coi l and the leakance of the con

denser are small compared with the reactance of the coi l and thesusceptance of the condenser respectively .

201 . Transient Effects Produced when a Harm onic E . M . F. is

Impressed on a Circu it. So far,we have considered the relation

between'

current and p .d . in various types of c ircuits when a harmonic alternating current is established in the c ircuit . However ,when a harmon ic e . m . f . is impressed across the term inals of acircuit

,tim e is required for a harm onic current to becom e estab

l ished j ust as in the case of a cons tant e . m . i . impressed upon a

amperes in the condenser and also in the inductance,


c ircuit time is required for a continuous current to become establ ished . (See Article That is

,when an alternating e . m. f . is

impressed across the terminals of a c ircuit,the current at the

start is an osci l lating current (see Article 1 67) and becom es an

alternating current with fixed maximum positive and negativevalues only after the lapse of an appreciable

,though usually

Fig . 107.

small,interval of time . Let the c ircuit be as shown in Fig . 107 ;

r and L represent the res istance and inductance of an impedancecoil

,and C and g the capacity and the leakance of a condenser .

Let the impressed e. m . f . be e=E sin The generaldifferential equations of this c ircuit have already been given

(Article 159) and are

i =gv+ C

dic = c—ri —L b

dt

where i is the ins tantaneous current in the impedance coil (equalto the . total displacem ent and leakage current of the condenser)and v is the p .d . across the condenser in the direction of the cur~

rent . Substituting in (a) the value of u from (b) we getdz i de

—t

i =ge+ Cdt

(c)

This is a differential equation of the second order . It is integrated by finding first the complem entary function

,i .e.

,the solu

tion corresponding to the right—hand number equal to zero , andthen adding tothis the particular integral . The solution corresponding to the right-hand number zero is i =A e

at,where the

value of a is found by substituting this value of i in the equationwhich gives


A , e + A 2 6 z sin 2 7rft + ,B —tan

where

R= r+g

777072

2 7rfC

92+ (2 M C)

2

The constants of integration A , and A 2 are determined fromthe values of the current i

oand the p .d . v

0across the condenser

at tim e t=0 .

Since the constant u is always greater than the cons tant m,

the transient term A z e'mt] in equation (39) becomes

smaller and smaller as tim e increas es . In most practical cases,

only a fraction of a second is necessary for this term to becomenegligible in comparison with the term

E

R2+ X

2

ing current . Hence the trans ient term is usually neglected, as mostproblems which arise in ordinary practical work have to deal onlywith the steady state produced in a c ircuit when a given e.m .f. is

impressed upon it . However,there are cases in which the tran

sient term becom es of paramount importance ; ia'

particu lar,in

predeterm in ing the effect of switching on or off a heavy loadfrom a transm ission line

,the short- circuiting e or grounding of a

line,and the effect of lightning discharges .

In most practical problems the leakance of the condenser isnegligible , that is , g=0 . When this condition holds

,the con

stants in equation (39) have the values

7T it ,8 - t representing the true alternat


l

1

2 7TfC

202 . Discharge of a Condenser having Negligible Leakancethrough a Resistance and Inductance . As an example of theapp l ication of equationthe manner in which a con

denser of capaci ty C (Fig . 108)charged to a p .d . V dischargesthrough a c ircuit containing aresistance r and inductance Lwill be determ ined . (Comparewith the ideal case of the discharge of a condenser through an inductance having no resistance,Article At tim e t=0 , the switch S is closed . The impressede . m. i . is therefore zero , hence the equation for the current is

i = 6' ut

[A , emt—l—A 2 6

'mt] (a)

where A , and A , are constants to be determ ined from the initialcurrent and p .d . across the condens er .

If there is originally no current in the c ircuit, the current muststart from zero (otherwise there would be an instantaneous trans ferof a finite amount of energy , hLi

z

,to the m agnetic field set up by

the current,which would m ean an infinite rate of transfer

,or

infinite power, which is impossible) and therefore at tim e t=0

i - 0 . We also have from the relation v+ ri + L§3=0

,that at

tim e t=0,

dt

X=x

Fig . 108 .

7

di di VL ordt dt t=0 L

Hence,substituting these conditions in equation (a) , we have

0 A 1 A 2

V—

E= (m—u) A t

—(e uM z

Whence , solving these two equations , we getVand —

i

2mL 2mL

These values of A , and A , substituted in (a ) give

V

2mL


Since the leakance g of the condens er is assum ed to be zero,u

and m have the values

1m :

LC

u is therefore a lways real,but m may be either rea l or imaginary,

1 1de endm u on whether a t or whether u2 that 1s

,u onP g p

LC LCP

whether r2 -

4—C

I-

J

or r2

4LFor r

2

C,the constant m 1s real and equat ion (40) may

therefore be writtenV

mL

This equation tells us that the current starts at zero,rises to

e”w

sinh (mt)

Fig . 10 9 .

a maximum value in the negative direction (corresponding to

Z:-0) and then decreases to zero . The shape of the curve repre

sented by equation (40) in this case is as shown in Fig . 109 .

633—6 1C

*The symbol sinh x is used to represent the expression2

whichis called the hyperboli c sine ofx .


ence in this case equation (40) becomesV te

' “t

L

which likewise represents a non—osc illating discharge such as shown

in Fig. 109 . When rz

Equation (40C) may therefore be looked upon as the lim iting formof the oscillating current given by (40b) .

the current j ust ceases to be oscillatory .

SUMMARY OF IMPORTANT DEFINITIONSAND PRINCIPLES

Note : The definit i ons given In paragraphs 1 to 9 inclusive arein terms of current ; they also apply to electrom otive forces andpotential differences .

1 . An a lternating current is a current which varies con

tinuously with tim e from a constant maximum in one directionto an equal max imum in the opposite direction and back againto sam e maximum in the first direction , repeating this cycle of

values over and over again in equal interva ls of time .

2 . The period T of an alternating current is the tim e takenfor the current to pass through a complete cycle of pos itive andnegative values .

3 . The frequency f of an alternating current is the numberof complete cycles ofvalues which it passes through in one second

1

T

4 . The number ofal ternations a per m inute is the total numberof times per minute that the current changes in direction

120a = 120f

5 . The equation of a harm onic current of maximum value IO isi = I0 sin (wt+ 9)

2 7where w, called the periodicity of the current

,is equal to

7'

=2 7r f,

and 9, called the phase of the current, is a constant such thatI0sin 9 gives the value of the current at time t=0 .

6. The difference in phase between a harmonic current and aharmonic p .d. is the angle corresponding to the time between


successive maxim a values of the current and p .d . respectively7 . The equation of a non-harmonic current of frequency i

m ay be written

i = I , sin sin + 1, sin + etc .

where w=2 7Tfand the I’s and 9’

s are constants . The first termin this expression is called the fundam ental or firs t harm onic ,the succeeding terms the second

,third

,etc .

,harmonics .

8 . The instantaneous va lue of an alternating current is itsvalue at any instant ; the maximum value of an a lternatingcurrent is its greatest instantaneous value during any cyc le ;the average value of an alternating current is the numerical valueof the average of its ins tantaneous values between success ivezero values . For a harmonic current

2

7T

9 . The effective value of an alternating current isthe squareroot of the m ean of the squares of its ins tantaneous values over acomplete period

,that is

,

i

where T is a complete period of the current and i its instantaneousvalue at any instant . When an a lternating current is expressedas so many amperes this effective value is always meant unlessSpecifically stated otherwise . The effective value of a harmoniccurrent is

_

Imax

10 . When a harm onic current of effective value I and a harmonic potential drop of the same frequency and of effective valueV exist in a c ircuit

,the average electr ic power input into this

c ircuit during each cycle isP = V I cos 0

where 9 is the difference in phase between the current and p .d .

1 1 . The power factor of a c ircuit is the ratio offi

the averagepower input P to the product of the effective value

‘

V of the p .d .

by the effective value I of the current , i .e .

,

Pf tPower ac or

VI


12 . The average electric power input into a c ircuit when the

current and p.d. are not harmonic functions of tim e,or sine waves

,

is equal sum of the values of the average power correspondingto each pair of harmonics of the same frequency

,i .e.

,

B = V ,I, cos —V ,I2 cos cos 93 + etc .

where the V ’s and I ’s are the efiective values of the successive

harmonics of the p .d .

’s and currents

,and the the difference

in phase between the corresponding harm onics of the samefrequency .

1 3 . The effective val ue of a non-harmonic current is equal tothe square root of the sum of the squares of the effective valuesof all the harm onics present in the current wave

,i .e.

,is equa l to

where I I , , I, , etc .,are the effective values of the harmonics .

Sim ilarly the effective value of a non-harmon l c p .d . is

where V V , ,V , ,

etc.

,are the effective values of the harmonics

in the p .d . wave .

14 . The equ ivalent sine-wave p.d . and current are the harmonic

p .d . and the harm onic current which have respectively the sameeffective values as the actual p .d . and current and differ in phaseby the angle whose cosme 18 equal to the power factor .

1 5. The form factor of a wave is the rat io of the effectivevalue to the average value .

1 6. The amplitude factor of a wave is the ratio of the m aximumvalue to the effective value .

17 The component of the current in any portion of a c ircuitIn phase with the potential drop through this portion of thec ircuit is called the power component of the current in this portionof the c ircuit . The component of the p .d . in phase with thecurrent is called the power component of the p.d . When thecurrent and p .d . are sine waves having the effective va lues V and

I respectively and differing in phase by the ang le 9, the effectivevalue of the power component of the current is I cos 9 and theeffective value of the power component of the p .d . is V cos 9 .

The component of the current in quadrature with the p .d .

is called the reactive component of the current and the componentof the p .d . in quadrature with the current is called the reactivecomponent of the p . d . For sine-wave current and p . d . the


E—VI

When there is no externally induced e . m. f.V

I

25 . The reactance a: of any portion of a circuit is the squareroot of the difference between

3

the squares of the impedanceand the reactance of the given portion of the c ircuit , i .e.

,

x \/z2 72

The reactance is taken positive when the current lags behindthe potential drop in the direction of the current

,negative when

the current leads this p .d .

26 . In the case of a sine-wave current and no externallyinduced e . m . i . the angle by which the current lags behind the

x9= tan

' 1

r

27. The impedance of a c ircuit of cons tant res istance r andinductance L to a sine-wave current of frequency f is

and its reactance ism=2 u i L

28 . The impedance of a condens er of capac ity C and leakance

g to a sine-wave current of frequency f is1

x/g2+ (2 7r fC)

2

and its reactance is2 q

92

(2 77 f C)2

29 . The impedance of a c ircuit formed by a resistance r,an

inductance L and a capac ity C in series to a sine-wave current is

1F+ w

2 n fC


x =2 q1

2wprovided the condens er of capac ity C has no leakance .


30 . A c ircuit is said to be in resonance with the impressedelectromotive force when the frequency of the impressed e. m. f .is the sam e as the ’natural or free frequency of the c1rcuit . Thefree frequency of a c losed circuit form ed by a capacity C and aninductance L,

when the resistance and leakance are negl igible, is1

2 7T\/L C

31 . In a c ircuit formed of two or more impedances in ser1es,

the resultant effective resistance is the arithmetical Sum of theseparate effective resistances

,the resultant reactance is the

algebra ic sum of the separate reactances,and the resultant

impedance is the square root of the sum of the squares of theresultant res istance and reactance, provided the current andp .d . are sine waves ; that is

R= r , r,+ etc .

Z —l—X2

32 . The effective conductance g of any part of a c ircuit to analternating current is the ratio of the average rate B,, at whichheat energy is developed in this portion of the c ircuit to thesquare of the effective value V of the p .d . through it , i .e .

,

Ph

V2

33 . The adm ittance y of any part of a c ircuit is the ratio of theeffective value of the current I in this portion of the c ircuit to thenumerical value of the vector difference of the externally inducede . m . f . E and the resultant p .d . V in this portion of the c ircuit ,both in the direction of the current , i .e.

,

I

E V

When there is no externally induced e . m . f.I

V

34 . The susceptance of any portion of a c ircuit is square rootof the difference between the squares of the impedance and thereactance of this portion of the c ircuit , i .e.

,

b=vrThe susceptance is taken positive when the current lags behind


the potential drop in the direction of the current,negative when

the current leads this p .d .

35 . In the case of a harmoni c current'

and no externallyinduced e. m . f . the angle by which the current lags behind the p .d .

i s

b

9

36. The resistance r , the conductance g, the reactance x,the

susceptance b, the impedance 2 and the adm ittance y of a givenportion of a c ircuit when the current and p .d . are sine wavesare related as follows :

9 = tan‘ l

z=VI

72

x2

I

.7/

g r

y2

z2

b a:

22

z2

37 The adm ittance of a condenser of capacity C and leakanceg to a sine-wave current of frequency f is

77 f C)2

and its susceptance is—2w

38 . In a c ircuit formed of two or more impedances in paral lel ,the resultant effective conductance is the arithm etical sum ofthe separate effective conductances

,the resultant susceptance

is the algebraic sum of the separate susceptances,and the resultant

adm ittance is the square root of the sum of the squares of theresultant effective conductance and susceptance

,provided the

current and p .d . are s ine waves,that is

etc.

B =-: b1 b,+ etc .

Z —z x/GH- B2

39 . The discharge of a condenser of capac ity C but no leakance through a resistance r and an inductance L in series isoscillatory when


v3=50 sin (w t Find (1 ) the equation of the potent ialdrop through the entire c ircuit, and (2) the effective value of thetotal potential drop .

Ans . : (1) o= l l o.6 sin (0) (2) volts .

An e. m . i . of 1 50 volts , the frequency ofwhich is 60 cycles ,is impressed upon a series c ircuit cons isting of a res istance of 5

ohms,an impedance c oil of 2 ohms res istance and henry induc

tance and a condenser of 50 microfarads capac ity . Find (1) thevalue of the current estab lished in the c ircuit and the potentia ldrops

, (2) through the 5 ohm res istance, (3) through the impedfl

ance,and (4) through the condenser. Find the angle by which

the current lags behind the potential drop (5) through the 5 ohmres istance, (6) through the impedance c oil , (7) through the c

r

ondenser and (8) through the entire c ircuit . Draw a completevector diagram .

Ans . : (1) amperes ; (2) volts ; (3) 336 yplts ; (4)472 volts ; (5) (6) (7) (8)

7. ( 1) What is the max imum 60-cycle e. m. i . which may be

impressed upon a circuit formed by an impedance of ohm

res istance and henry inductance and a c ondenser of

m icrofarads capac ity connected in series,if the effective punctur

ing voltage of the condenser dielectric is volts? (2) If a25- cycle e . m; f . of the same value as this 60-cycle e. m. f. determ inedabove is impressed upon this c ircuit, what is the potential dropac ross the condenser?

Ans . : ( l ) volts ; (2) volts .

8 . When an e. m . f. of 220 volts is impressed upon a c ircuitformed by two impedances A and B connected in series

,the power

absorbed by A is 300 ,watts and the power absorbed by B is

1 200 watts . If the potential drop across A is 200 volts and thecurrent in A is 10 amperes

,find (1) the potential drop across

B, (2) the resistances, and (3) the reactances of A and B re

spectively .

Ans . : ( 1) volts ; (2) 378 volts ; (3) 3 and 1 2 ohms ; (4)i ohms for A and : l: or ziz ohms for B .

9 . The current established in a c ircuit cons isting of a resistance of 3 ohms in series with an impedance of unknown res istance

C

The values of the e . m . f. , potential difference , and current in Problem6 and in al l problems following are effective values and the wave forms

s inusoidal.


and reactance is 1 2 amperes . If the power absorbed by thec ircuit is 600 watts and the potential drop across the impedanceis 100 volts

,find ( 1 ) the impressed voltage on the c ircuit , (2) the

power factor of the c ircuit, and (3) the angle by which the dropthrough the impedance leads the current .

Ans . : (1) 1 1 1 volts ; (2) (3)10 . N o impedances A and B are connected in parallel . A

has a resistance of 5 ohms and an inductance of henry,while

B is formed by a resistance of 10 ohm s and a capacity of 100

m icrofarads in series . If a 25- cyc le e.m .f. of 220 volts is impressedupon this parall el c ircuit find (1 ) the total current, (2) the currentin each branch , and (3) the power factor of the c ircuit .

Ans . : (1) amperes ; (2) and amperes reSpec

tively ; (3)1 1 . At 60 cycles the impedances of two coils A and B are

each 10 ohms . At 25 cycles the impedance of A is ohms

and the impedance of B is ohms . Find the j oint impedanceof A and B at 60 cycles when connected (1) in parallel, and (2) inseries .

Ans . : (1) ohms ; (2) ohm s .

12 . A parallel c ircuit consis ts of a coil (A) of negligible resistance and henry inductance connected in parallel with acondens er (B) of m icrofarads capac ity and negligible leakance . This parallel c ircuit is conn ected in series with an im

pedance (C) of 2 ohm s resistance and henry inductance .

Find (1 ) the current, and (2) the potential drop in A ,B and

C respectively,when an e. m . f . of 220 volts and 60 cycles is im

pressed upon this c ircuit . If a non- inductive resistance (D) of10 ohms is connected in parallel with (A ) , find (3) the current and(4) the potential drop in A ,

B,C.and D respectively . Draw a

vector diagram for each case .

Ans . : ( 1) and 0 amperes (2) 220, 220 and 0 volts ;(3) and amperes (4) and

volts .

13 . When a 50-kw., 60- cyc le, a .c . generator is delivering 30 kw .

at 80% power factor the term inal voltage is 230 volts .

‘

Theeffective res istance of the armature between term inals isohm and the reactance ohm . Neglecting the effect ofarmaturereaction , ( 1 ) what would be the term inal voltage of this machineif the external c ircuit were opened? (2) What is the maximum


load under which this generator could operate continuously at80% power factor?

Ans . : (1 ) volts ; (2) 40 kw .

500 kw . are delivered to a substation from a 6600—voltpower station over a transm ission line of 6 ohms resistance . If

the current taken by the load is 100 amperes , find (1 ) the powerfactor at the power station, and (2) the effic iency of the transm ission line .

Ans . : (1 ) (2)1 5. The potential differences at the load and generator ends of

a transm ission line are each 6600 volts . The resistance of theline is 6 ohm s and the reac tance 8 ohms . If the line current is100 amperes

,find (1) the power factor at the load, and (2) the

effic iency of the transm ission line .

Ans . : (1) (2)1 6. A transm ission line 25 m iles in length consists of two

No . 0 B . S. wires (diam eter inch) spaced 3 feet apart .

At the end of this line is connected a 60-cyc le indubtion motorload ( lagging current) of 1 500 kw . operating at volts andat a power factor of

‘ What must be the potential differenceat the power station supplying this load? The leakage and ca

pacity of the line are to be neglected .

Ans . : volts .

1 7. Two 220 alternators A and B connected in parallel supplyan inductive load of 50 kw . at 90% power factor . If A suppliesone- third of this load at a power factor of find (1 ) the powerfactor at which B supplies power to the load

,and (2) the armature

current of each alternator,if both currents lag behind the gen

erated e . m . i .

’s .

Ans ( 1 ) (2) and 1 60 amperes .


A = 14 13+ A ; ( l a)

Therefore,the length of a vector is equal to the square root of

the sum of the squares of the two components entering into itssymbolic express ion. Taking the ratio of two equations (1a) we get

Atan 9=

2

1dA

Whence,the angle which a vector makes with the line of r,eference

is equal to the angle whose tangent is the ratio of the second (or jcomponent) to the first component in its symbolic expression .

Sim ilarly, any other vector A’OP ’ making an angle 9’

with the same line of reference may be resolved into the twocomponents A ,

’ =A ’cos 9’ and A ,

’ =A ’sin and we may write

A, =A 1

,

+ 70

44;

where j as before indicates that the vector A ,

’ leads the line ofreference by 90

° while A ,

’coincides in direction with the line of

reference .

204 . Addition of Vectors .

—From Fig . 1 12 it follows that theresultant of two vectors A and A ’ is the vector which has the com

Fig ,1 12 .

ponent A cos 9+ A ’cos 9’ =A ,+ A ,

’ parallel to the ax is of referenceand the component A sin 9+ A ’

sin 9’ =A ,+ A ,

’ leading the lineof reference by The resultant of the two vectors A and A ’

is then represented symbol ically as

S=A A ’

+ j (A ,+

The length of th i s vector is then

(A ,+

and 1t m akes an angleA , A ,

’

9,= tan

‘ 1

A r l A I

,

SYMBOLIC METHOD 373

with the axis of reference . Sim ilarly,the sum of any number of

vectors is, in the symbolic notation ,S=A + A

'+ A

"+ etc . (A .+ etc .) +704. A ; +

O

etc .)The length of this resultant vector is

S= V )2

and the angle which it makes with the axis of reference isA ,+

A I +

205 . Subtraction of V ectors . The vector A—A ’

(Fig . 1 13)

is sim ilarly the vector which has the component A cos 9 A ’cos 0’

9, =tan” 1

Fig . 1 13 .

A ,—A ,

’ in the direction of the line of reference and the component A sin 9 A ’

sin 9’ =A , A ,

’ leading the line of referenceby That is

,the vector difference A A ’ is represented sym

bol ical ly as

D =A _ A , = (A 1—A 2

I

)The length of this vector is then

D = l/ (A 1_ A 1

,

)2+ (A 2

_ A 2

,

)2

And it makes the angleA , A ,

’

A , A ,

’

with the axis of reference . When this angle comes out negative,the resultant vector makes a negative angle with the axis of

0d = tan


reference or lags behind this axis ; Fig. 1 13 illustrates such a case .

206. The Symbol “

3 as a Mu ltiplier Signifying Rotation .

So far, we have cons idered the symbol j s imply as a means ofindicating the component of a vector leading the line of reference .

As we shall presently see,it is also frequently convenient to look

upon j as a multiplier as well as a symbol representing direction

,and to define the operation j XA ,

where A i s the symbol i c

expression for any vector, as equiva l ent to turni ng the vector A

through an angle of900 in the posi tive direction . This convention

leads to a useful m athematical equivalent for the symbol f‘

j .

”

Let A (Fig . 1 14) be a vector coinc idingin direction with the line of reference .

Then from the above convention 3A

represents a vector of length A leading5x “ the line of reference by Multiply

Fig 1 14' ing the symbolic expression iA for

the second vector by j , we have from the above convention that

y'

XjA represents a vector of length A making an angle of 90°

with the vector iA ,that is

,a vector equal in length to A ,

bu t

in the opposite direction to A . Therefore

= _ A

j 1 . (4)It should be borne in m ind, however, that although the sym

bol j is mathematically equal to the imag inary quantity 1,

the phys ical m eaning of j is not imaginary at all ; it is s implyan abbreviation placed before the expression for a vector signifying that the vector in question is to be turned through 90

° in thepositive direction; turning a vector through 90

° twice in thesame direction is mathemati

cally equivalent to writing anegative sign before the vector . For example

,in the

express ion

{1

for a single vector, A , and Fig- 1 15

A ,both represent vectors coinciding in direction with the

line of reference, but in the expression for the resultant vec


9’ = tan' liA ,

’

Hence the angle by which A’leads A is

9’ 9= tan’ 1A 2 —tanA ,

’

I I

tan(9’ 9)

—A ’A 2 A ’ A

A IA I

, A 2A 2

For example,let A =8 + j 6 and A

’ = 5+ j 4 . Then A ’ leadsA by the angle 9

’ —9 where

tan (0' 0)

9’ = tan 1

Therefore A ’ leads A by 5

0' 0

209 . Symbolic Representation of a Harmonic Function .

We have seen (Articles 182 and 186) that a harmonic alternatingcurrent sin (mH where I is the effective value of thecurrent , may be written

i =V 2 1 cos 9 sin sin 9 cos wt

and that the terms V 2—I cos 9 sin wtmay be represented by a vector

equal in length to I , = I cos 9

rotating about a fixed point 0with the angular veloc ity to

Sing and the term V 2 I sin 9 cos out

by a second vector of lengthI =I sin 9

,leading the first vec

tor by 90° and rotating aboutthis same point w ith the sam e

Fig 1 17 angular veloc ity . Hence the

function i 2 I sin (wt+ 9) may be represented by a vector

rotating about this point with the angular velocity (0,where I ,

is a vector of length I cos 9,and ‘

jI , a vector of length I sin 9

leading I , by both rotating about the sam e point with thesam e angular veloc ity (0 . Therefore we may look upon equation

(7) as the symbolic expression for the function i V 2 I sin (wt+Note

,however

,that the two vectors I , and j ] , though cons tant

in length are continua l ly changing in direction; therefore in theexpression 0

SYMBOLIC METHOD 77

both I , and I , and therefore I a lso,must be looked upon as functions

oftime and not as constants, since a vector by definition is a quantityhaving direction as well as m agnitude .

Note the analogy with a body moving in a c ircle of radius a

with a constant l inear speed s ; the veloc ity of such a body isconstant in magnitude , but is continually changing in direction .

2

Hence the rate of change of the velocity is not zero , but is i ,

the direction of this rate of change, i .e.,of the acceleration

,is

perpendicular to the direction of the veloc ity .

21 0 . Symbolic Expression for the Derivative of a HarmonicFunction . Since a harmonic function of the form

i = l/ z I sin (m i + 9)may be written in the form

i = V/

2 I cos 9 sin mt+ l/ 2 I sin 9 cos cu t

the derivative of i with respect to time isdi

dt

But 2 ( 0 1 cos 9 cos out is represented by a vector equal in lengthto 00 1 cos 9=wI , leading by 90

° the vector I , which represents

I cos 9 sin wt, and ml sin 9 sin out is represented by avector of length (0 1 sin 9= to ] , leading by 90

° the vector i ,I

which represents 1/ 2 I sin 9 cos wt. Hence the derivative of

the function represented by the vector

1/ 2_

wI cos 9 coswt m l sin 9 sin w'

t

d]

d,+ 71 2) =Jw1 (8)

That is,the derivative with respect to time of a vector rotating

with a veloc ity to results in a vector equal in length to the produc tof

“

w by the length of the original vector, which vector has adirection 90° ahead ofthe original vector .

V i ce versa,from equation (8) we also have that

1 d]

w dt

that i s,mu ltip lying a rotating vector by j i s equiva lent to difier

entiating the vector wi th respect to time and divi ding by w.

(8a)


From this signification of the multiplier j it is at once evidentthat to consider the multiplication of the expressions representingtwo rotating vectors at right angles to each other, such as I and

i V as equivalent to j (I V ) iS' inconsistent with j as being equ iva

lent to the operation1 i . For 1

1(I V ) =I

d—I-f

-l VCE

,and there

to dt dt.

di dt

fo re to be consistent with equation (8a) , the expression j (I V )must equal Ij V—l V iI .

Consequently, although the product of the two expressionsI ,+71 , andV , j V, representing the current and voltage in a c ircuitis I ,V , I ,V ,+ j when j is considered simply as a

multip lier equal to l/ 1,this expression , however, is inconsistent

w ith the m eaning of j , as defined by equation (8a) , since I I V ,

and V , are all rota ting vectors . It is important to bear thisc learly in m ind, s ince one is l ikely to m ake the m istake of takingthis product as the symboli c expression for the power corresponding to the current I ]

'

I and the p .d . V 1 + i V , , and to takethe “real” part of this product as representing the average power .As a m atter of fact

,the value of the average power correspond

ing to the current I 71 , and the p .d . V ,+ i V , is

P = I V ,+ I V ,

(see Article That is,the sign between the two terms in the

express ion for the average power is j ust the opposite of the signresulting from the multiplication of I ,+ j I and V ,+ j V , and tak

ing 7 as a multiplier equal to V 1 .

21 1 . Symbolic Notation for Impedance . Impedance as a

Complex Number . When the current and p .d . in a circuit of

constant resistance r and constant inductance L are both harmoni c

functions ofthe same frequency they may be represented respectively by the two vectors I andV .

The component of theresultant p .d . in phase with the current is a vector coinciding indirection with I and equal in

1 length to the product ofr byQ r

Fig . 1 18 .the length of the current vector

,that is

,is represented by

the vector rI in Fig . 1 18 . The component of the resultant p .d . 90°

ahead of the vector I is a vector leading the current vector by


mathematical form as the symbolic expressi on for a vector .Its properties in any Operation of multiplication or divisi on are

exactly the sam e as the propert ies ofa “ complex number

,

”

that is , a number which con

i x sists of the sum of two terms ,

one real and the other imaginary . Since the product ofthesymbolic expression for cur

Fig , 1 1 9. rent by this expression r -l—ifv

gives the symbolic expression for the p .d . due to this current inthe impedance having a resistance r and reactance x ,

the expression r+ jx m ay be looked upon as the symbolic expression forimpedance , and may be

’

represented by the single symbol 3 witha dot under it , i .e.

,

z= r+fx (10)

We may then write the symbolic expression for the p .d . produ cedby a current in an impedance z=r+ j zc as

V = eI (1 1)Note that the length of the vector represented by zI

is equal to 12 and that this vecto r leads the vecto r I by the

angle Hence multiplying the vector I b y the imr

pedanee z gives ri se to a vector equal in length to the

product of the length of the vector I by the numerical value of

the impedance, which vector leads I by the power- facto r angleof the im pedance .

212 . Symbol ic Notation for Admittance . The relation between current and p .d . when both are harmoni c functions ofthe same

frequency may also be expressed in terms of the adm ittance of thecircuit . Let g be the conductance and b the susceptance, and letI and V be the current and p .d . vectors . Then the component of

I in phase w ith V is a vector coinc iding in direction with thevecto r V and equal in length to the product of the length of thep .d . vector by the conductance

,that is

,is represented by the

vector of g V in Fig . 120 . The component of the current laggi ngbehi nd the

o

p .d . by 900 is a vector lagging behind the vector

V by 90° and equal in length to the product of the length

SYMBOLIC METHOD 38 1

of the p .d . vector by the susceptance,that is

,is represented

by the vector—jbV ; since +7 represents a lead of 90°and there

Fig . 120 .

fore —j represents a lag of Hence the resultant currentvector is

I =gV—i = (g—jb) V (12)

In the particular case of a leaky condenser this relation alsofollows from the differential equation

I =gV + Cd_i_

/

gV + CjwV = (g —jb) Vdt

for s ince C is a constant CjwV =i V = —jbV . In the cas e of a

condenser the charging current wCV leads the p .d . by andtherefore the susceptance of the condenser is —w C

,s ince the

susceptance of the c ircuit is the factor by which the p .d . mustbe multip lied to give the component of the current lagging 90

°

behind the p .d .

When the p . d . is expressed as V ,+ iV , the operation repre

sented by (g can also be shown by a process ofreasoning sim i lar to. that employed in Article 21 1 to be equiva

lent to the product of g—jb and V ,+ iV , when j is con

sidered as equivalent to 1 . Therefore in symbolic notationthe adm ittance of a c ircuit may be written as a complexnumber

y=g

~ ib (13)

and the current due to a p .d .V across this adm ittance may be

writtenI =yV

21 3 . Division of a Rotating.

V ector by a Complex

Since by equations ( 1 1 ) and (14) we haveV =z I

I =yV = (g—ib) Y


the division of the vector V by (r + jx) must be equivalent to

multiplying the vector V by (g jb) , that is

=yV g/ (15)

Sim ilarly, the divis ion of the vector I by (g jb) must be equivalentto multiplying I by (r -l- ix) , that is

I 1

y y

But multiplying equation (a) by (r— jx) , cons idering j s imply as a

multiplier equal to 1/ 1,we get

(r—jx) I

whence,dividing this equation

.

by 734- 115

2

,we get

a:_ J

r2+ x

2

which to be consisteht with I (g jb) V requires that

g_

7.2+ x

2

which agree with the relations already established in Article 199 .

Sim ilarly, mu ltiplyinge quation (b) by (g+ j b) we getV

whence,dividing this equation by r

2+ x

2

, lLb

79,+ b

,

which to be consistent with V = (r+far) I requires that

g and xb

g2_l__ b

2

which agree with the relations already established in Article 199 .

In general, the division of any rotating vector A by a com

plex number of the form is equivalent to multiplying

the rotating vector byOFF?a , that isa 1

2+ a 2

2

Ao

a 1 ¢ 2a 2 A (17)a ,

z+ a ,

z

Consequently,Impedance and adm ittance in the symbolic notation

may be treated as algebraic multipliers or divisors,the symbol

j in each case being considered as mathematically equivalent to


where the currents, impedances , and electromotive forces are allexpressed in symbolic notation, and the currents and

are a l l referred to the same axi s of reference . That is,the currents

'

are to be expressed as I I’ = I ,

'+ iI etc . and the

e .m .fiflg as E =E 1 + j 1 2 , E,

6 130 . where all the compo

ments of the.

currents and with_the subscript 1 are parallel

to one another and all those with the subscript 2 are parallelto one another and lead the first set of components byThe electromotive forces due to inductance and capac ity are

taken account of by the impedance ; the electromotive forcesrepresented by the E

’s in equation (18b) are the externally

induced electromotive forces such as those due generators or

motors or to the mutual inductance of the two windings of atransformer .

In applying equations (18) to the c alculation of the currentsand p .d .

’s in any network of c ircuits , care must be taken to desig

nate clearly the sense. of the vectors representing the currents and

e . m . i .’s . This ismost con

veniently done by numbering all the junction pointsin the network

,and desig

nating each current and e .

m . f . by a double subscriptwritten in the order corresponding to the assum eddirection Of the current ore . m . f. vector . For example

,in Fig . 1 21

,the e .m . f.

from 0 to 1 is represented byFig 1 2 1 E 0 , while the e. m . f . from 1

to 0,which is equal to—Eo, is represented by In the figure ,

then,the net electromotive force from 1 to 2 is

E.

= E + 1?E

.0 1 + E

.0 2

Each equation of the form (18a ) or ( 18b) is in reality equivaleut to two equations , s ince the sum of all the real ’ i terms

on one s ide must be equal to the sum of al l the real terms on

the other s ide,and s im ilarly

,the sum of all the j terms on one

side must be equal to the sum of all the j terms on the other

SYMBOLIC METHOD

side ; the denom inators of all fractions having been cleared of

j” terms by the transformation given by equation This

is merely another way of stating the fact that the component inany direction of the resul tant of any number of vectors must beequal to the algebraic sum of the components in this directionof all the individual vectors . Applying these two laws to any

network enables one,therefore

,to calculate both components of

every p .d . and every current , when the impedances and theelectromotive forces are known .

It should be c learly borne in m ind that the above equationsare true only when the currents

,the and the impedances

are expressed as vector quantities . These equations are not

true when the numerical values of these quantities are employed .

The indications of alternating current voltmeters and voltm etersgive only the effective values of currents and p . d .

’s respectively,

they do not indicate their phase relations . Hence,the sum ofthe

currents entering a function as measured by ammeters in the vari

ous branches is not necessari ly zero; again , the sum of the potentia l

drops in the vari ous branches ofa c losed loop, as measured by volt

meters in the various branches,is not necessari ly zero.

Again , these equations hold only when the currents andare al l simp le harmoni c functions of the same frequency and the

resistances and reactances are constant. When,however

,the re

sistances,inductances and capac ities are constant, a s im ilar set

of equations holds for each frequency that may be present . Sincethe equations are all l inear in the I ’s and E ’

s,the currents and

of any given frequency will be uninfluenced by the presence of currents or of any other frequency . Hence , whenthe harmonics present in each e .m .f. are known, the harmonicspresent in each current may be calculated by solving the equations corresponding to

"

th e frequency of this particular harmonic ,these equations being exactly the sam e as would hold were all theother harm onics absent .Note particularly that the above equations do not hold for

transient currents ; they apply only after the transient terms

(see Artic le 201) have becom e zero .

21 5 . Expression for Average Power in Symbol ic Notation .

When the current in any c ircuit is i 2 I sin (w t+ 9) and thepotential drop

'

in the c ircuit (in the direction of the current) is


2 V sin (00 t+ the average power input in to the circuit isP = VI cos (9

’ 9)where V and I are the effective values of the current and p .d .

(see equation 9 ofChapter V II) . The current i 2I sin(w t+ 9)may also be written

I sin 9 cos w t

and may therefore be represented by the vector

whereI , = I cos 9

I , = I sin 9

Sim ilarly, the p .d . v 2 V sin (00 t+ may be writtenV ecs 9’ sin V sin cos w t

and may therefore be represented by the vectorV V I + j V 2

whereV , V cos 9,V , V sin 9,

Expanding the power equation we getB = VI cos 9 cos V I sin 9 sin 9’

= V cos 9+ V sin 9

But V cos 9’ = V I cos 9 = I , ; V sin 9’ = V ° I sin 9= I , .

Hence the average power corresponding to I = I , j ] ,

V V ,+ i V , is

V ,I 2 (19)Note that this express1on i s not equal to the real part ofthe productof and I when f is cons idered equiv

alent to 1,which product givesV I = V,I , V ,I , + j

The average power is the real part of this product with the s ignbetween the two terms V ,I , and V ,I , reversed .

21 6. Expression for Reactive Power in Symbol ic Notation .

The reactive power corresponding to the current 2 I sin(wt 9)

and the p .d. v 2 V sin wt+ is (see A rticle 1 84)U= VI sin (9

’ 9)

(Note that the reactive power is taken as p os itive when the

current lags behind the p .d .,i .e .

,when 9

Expanding the equation for U we get


as the axis of reference,and calling the current from a to c I“ ,

we have,that (3+ j4+ 8—j6) Ic,c = 100+ j0 whence

100 1 100+ 12 00

1 1—23'

Hence the effective value of the current is

and it leads the potential drop from a to c by the angle-1

The potential drop across the first impedance is(8 .8+ j 1 c) =2e.4 c.4+ i

20+ j40

which has the effective value

and leads the potential drop Va cby the

- 1 it?20

Fig . 1 23 .

The potential drop across the second impedance is- jc) —j

80


= 1/ volts

and lags behind the potential drop by the angle

SYMBOLIC METHOD 389

_ 140

30The complete vector diagram is given in Fig . 123 . Note that thecurrent is plotted to ten tim es the scale of the p .d .

The power input into the first impedance isWa b X 20 X 40=240 watts

The power input into the second impedance is

Wbc 40 =640 wattsThe total power input is

Wu,

100+ = 880 wattswhich of course is the sum ofW

ab and Wbc .

Problem 2 . Paral lel Circuits .

Next,let these two impedances be connected in parallel

,Fig.

124,and let the total current taken by the two be 100 amperes .

Choosing the vector representing the total current from a tob as the axis of reference

,and calling I ’ab the Current

in No . 1 from a to b and”

a b I ,”+ j I ,

” the current in No . 2

from a to b, we have from equations 18) that

(3+ j4) (I + iI (8 76 ) = 0

Whence= 100 + j 0

31 —4I j (4I ,’ = 8I ,

”6I j (81 61

Equating the real and the“j terms in these two equations we get=100

1 ;4I ,

’


Solving these equations we getI ,

’ =80

I ,’

40

I ,” = 20

I = 4O

Whence the current in No . 1 from a to bI ’ab= 80 740


and lags behind the total current Iab by the angle40

-1

80

The current in No . 2 from a to b isI ”

ab=20+ j40

which has the effective valueI ” amperes

and leads the total current Iab by the angle

4 1920

The potential drop across each impedance in the directiona to b is

(3 34 ) (80 =240 160 j (320

400 72 00


V = 447 voltsand leads the total current by the angle

400

(Note, this problem can be solved m ore s imply by ca lculating


1 dX

0) dl

angular velocity to.

5 . The symbol ic notations for a sine-wave current,a sine-wave

p .d .,and a sine-wave e. m. f . are respectively

when X is a vector rotating with a cons tant

where the two components of each quantity are any two components at right angles to each other and each component is a vecto rrotating with an angular veloc ity w= 2 77 f, where f is the frequency .

6. The symbol ic notation for an impedance of constant res istance r and constant reactance a: to a sine-wave current is thecomplex number

Z = r fat

7. The symbol ic notation.

for an admittance of constant conductance g and cons tant susceptance b to a sine-wave current isthe complex number

y=g—Jb

8 . The symbolic notation for the p . d . V in an impedance z

due ' to a current I is

V =zI

9 . The symbolic notation for the current I in an adm ittance

y due to a p .d . V is

I =yV

10 . The relation between impedance and adm ittance in sym

bol ic nothtion is

1 1 . The symbolic notation for the equivalent impedance Z of

two impedances 2, and Z

, in paral lel is

Zzl + z

2

2 1 422

1 2 . An expression of the form‘

where a , and a, are

constants is qqual to

SYMBOLIC METHOD 93

a .H a t

a ,2

0 2

2

1 3 . Kirchhofi’

s Laws in symbolic notation are2 I =0

2 z?I =2 E'

where the currents I and the electromotive forces E are all referred

to the sam e axis of

o

reference .

14 . The average power corresponding to a p .d . V V , j V ,and

a current I = I , + jI , , both referred to the same ax 1s of reference is

P V I 1 1 " l‘ V 2 [ 2

and the power factor is

60 8 9

PROBLEMS

1 . A vector A 10 units in length makes an angle of 60° in acounter- clockwise direction with another vector B 15 units inlength . Find the symbolic expressions of (1) A and B respec

tively, and (2) A + B ,and (3) A—B referred to B as the axis

of reference in each case .

Ans . : (1) 5+ j and 10 ; (2) 20+ j 8 .66 ; (3)2 . A certain vector is represented symbolically by the expres

s ion 1 2+ j24 . Find the symbolic expression of a vector equalin length ( 1 ) when it m akes an angle of 40

° with the axis of

reference, and (2) when it makes a right angle with the axis ofreference in the c lockwise direction .

Ans . : ( 1) —i17.27 ; (2) —j3 . Three impedances A

,B and C have resistances of 5

,8 and

3 ohm s respectively and reactances of 14,0 and — 10 ohms respec

tively . Find the symbolic expressions of the resultant impedanceof these three impedances

, ( 1 ) when they are connected in series,(2) when they are connected in parallel , and (3) find the symbolicexpression of the resultant adm ittance of these three impedances

”

when they are connected in parallel .Ans . : ( 1) l 6+ i4 ; (2) —j0 .902 ; (3)4 . An e . m . f . of 150 volts and 25 cycles is impressed upon a

series c ircuit cons isting of a resistance of 5 ohms , an inductance ofhenry and a capac ity of 1000 microfarads . If the vector


representing the impressed e. m . f. is taken as the axis of referencefind ( l ) the symbolic expression for the current flowing in the c ircuit,and (2) the average rate at which energy is given to the c ircuit .

Ans . : (1) (2) 4140 watts .

5 . Two impedances A and B Of 4 and 1 2 ohms resistancerespectively and 1 5 and —8 ohms reactance respectively are con

nected in series . If the current estab lished in this c ircuit is 10amperes

,find the symbolic expression for the potential drop

, (1 )through A , (2) through B ,

and (3) through the entire c ircuit, al lreferred to the current as the axis of reference .

Ans . : (1) 40+ j 1 50 ; (2) 1 20 (3) l 60+ j70 .

6. If the two impedances described in Problem 5 are connectedin parallel

,and the current in A is 5 amperes

,find the symbolic

express ions for the currents in A and B respectively referred tothe potential drop across the parallel c ircuit as the axis of reference .

Ans . : —j4 .84 and7. When a given 60- cycle e . m . f . is impressed upon a series

c ircuit of 5 ohm s resistance,henry inductance and 50 m icrofarads

capac ity , the symbolic expression for the fundam ental of the current is 12 and for the third harmonic Of the current 6+ j8referred respectively to the first and third harmonics respectivelyof the impressed e . m . f . Find ( 1) the effective value of the impressed e. m . f . and (2) the average power dissipated as heat energyin the c ircuit .

Ans . : (1) 988 volts ; (2) 1 720 watts .

8 . An impedance A is connected in series with two impedancesB and C connected in parallel . The resistances of A

,B and C

are 3,5 and 6 ohms respectively and the reactances are 5

,—7 and

3 ohms respec tively . If an e . m . f . of 200 volts is impressed acrossthe entire c ircuit

,find the symbolic expression of the currents in

A,B and C referred to the potential drop in A .

Ans . : 1 231—32 75 ; —j 16 .3 .

9 . An e. m . f . of 100 volts is impressed upon a series c ircuitform ed by two impedances A and B in series . The impedance Ahas a res istance of 4 ohms and a reactance of 6 ohms and thepotential drop across A is 100 volts . If the entire c ircuit absorbs1 kilowatt of power

,find the symbolic expression of (1 ) the cur

rent, (2) the potential drop through A ,

and (3) the potential dropthrough B,

all referred to the e . m . f . impressed upon the entirec ircuit .

THREE -PHASE ALTERNATING CURRENTS

21 9 . Polyphase Alternating Currents . A polyphase alternator is an alternator upon the arm ature core of which are woundtwo or m ore independent windings which are arranged with respectto each other in such a manner that the electrom otive forces inthe various windings differ in phase by a cons tant angle . In atwo-phase alternator there are two independent windings , whichare arranged in such a manner that when the e . m . f . induced in onewinding is a m aximum ,

the e. m . f . induced in the other windingis zero

,that is

,the e. m . f . induced in one winding is in quad,

rature with the e . m. f . induced in the other winding ; this effectis obtained by plac ing the two windings on the core in sucha manner that when a group of conductors in one winding isdirec tly under a north pole, the corresponding group of con

ductors in the second winding is m idway between a north poleand a south pole . In the case of a two-pole m achine this meansthat a group of conductors in one winding is placed 90 degreesahead of the corresponding group in the other winding ; sim il arly,in the case of a multipolar machine

,if we call the di stance between

two successive poles of like sign equa l to 360 e lectri ca l”degrees ,

we may describe the relative position of the two windings as suchthat the corresponding conductors in the two windings are Spaced90 e lectri ca l degrees apart . In general the two windings overlapeach other

,or are distributed ” over the armature surface

,but

for each conductor in one winding there is a corresponding c on

ductor in the other winding 90 electrical degrees, or one quarterof the pole pitch

,ahead of the conductor in the first winding .

A two—phase alternator may be provided with a separate pairof slip rings for each winding, that is, four rings in all ; or withthree slip rings , and one of these rings made to serve as a commonreturn for the two windings .

A three-phase alternator is s im ilar in construction to a twophase machine except that three separate windings are employed,the corresponding conduc tors in the three windings being spaced

396

THREE-PHASE ALTERNATING CURRENTS 397

120 electri ca l degrees apart,and the

induced electrom otive forces thereforediffer in phase by These windings are connected electrically in twodifferent ways . The three windingsmay be connected end to end as shownin Fig . 1 26

,and leads from the - junc

tions 1,2 and 3

,brought out to

three slip rings ; or one end of each of 1

the three windings may be connectedto a common junction,

called theneutral point

,as shown in Fig .

127,and the other ends 1 , 2 and 3 brought out to three slip rings .

In som e cases the common junction or neutral point is also con

nected to a fourth slip ring. A three-phase alternator with itswindings connected as

.

shown in Fig . 126 is said to be “ meshor Delta ”

connected,the latter from the Greek capital letter

A .

” When the windings are con

nected as shown in Fig . 1 28 the alternator is said to be star 3’ or Y ”

connected . In large generators the

armature is stationary and the fieldrotates ; in such a case the armaturewindings are connected to fixed term inals and the two ends of the fieldwinding are connected to two slip

Fig 1 27 rings . Continuous current is suppliedto the field through brushes bearing on these two slip rings .

The advantages in using three-phase currents are chiefly1 . The m ore satisfactory operation of three-phase motors as

compared with s ingle—phase machines .

2 . The saving in the amount of copper (25 per cent) as com

pared with a s ingle-phase system for the sam e voltage betweenwires .

3 . The lesser cost of three-phase generators and motors of thesam e power and for the same voltage between term ina ls .

4 . Better voltage regulation of three-phase generators .

220 . Vector Sum of the Induced E . M . F .

’s in the ThreeWind

ings of a Three-Phase Generator Equals Z ero . Let E E ,,

and E Fig. 128, be the three induced in the three wind


ings of a three—phase generator, taken as posi tive in the counter

c lockwise direction around the c losed loop formed by the windings of

Fig . 1 28 .

the generator . In a properly designed machine these haveequal effective values

,but due to the relative position of the wind

ings they differ In phase by that 1s E ,, leads B , , by 120° and

E , , leads E ,, by Hence the symbolic express ions for the

three referred to E ,, as the l ine of reference are, when theseare sine waves,

E l + ;i 0 ) E

.

1_ G l E2 2

1

2 2

and thereforeE + E + E =0

Hence the total e .m .f. at any instant acting around the closedloop formed by the three windings is zero , and therefore when noexternal c ircuit is connected to the three term inal s

,i .e.

,when

the generator is running on open (external) circuit, there wil lbe no flow of current in the arm ature provided the aresine waves .

Sim ilarly,in the case of a three-phas e Y- connected generator,

Fig. 1 29 , the induced 3 and are all equal ineffective value and differ In phase by Hence choosing E , , as

the line of reference and taking these as posi tive in the


between the term inals 1 and 2 ; hence, the equivalent“A

”e

m . i .’s,that is

,the between term inals, are

E 23E 0 1

That is , the equivalent A e .m .f. 1s equal num erically to the squareroot of three tim es the effective value ofthe Y c . m. f . and lags 90

° behind thee .m .f. in the Opposite branch of the Y.

The A are therefore representedby the sides of the triangle form ed byconnecting the ends of the three vectorsrepresenting the Y c . m . i .

’s , the positive

direction being taken in the counterclockwise direction , see Fig . 130 .

222 . Currents from a Three-PhaseFig ' 1 30 Generator . Generator and Load Both

Y-Connected .

—When the three term inals of a three-phase generator

,either A or Y connected , are connected to either a Y or A

connected load,currents equal in effective value and differing in

phase by 120° will flow from each term inal

,provided (1) that the

resistance and reactances of the generator windings are reSpec

tively the same for the three windings , (2) that the resistances andreactances of the w indings of the receiving device or load arerespectively the same for the three windings , (3) that the e. m . f.

’s

in each winding Of the generator are equal in effective value anddiffer In phas e by and (4) that the back e. m . f .

’3,if any

,in

the windings of the load are equal In effect ive value and differ Inphase by When the e . m . f .

’s in each phase of a three-phase

system are equal in effective value and differ in phase by 120°

and the currents in each phase are equal in effective value anddiffer in phase by the system is said to be balanced .

”

Cons ider the simple case of a Y- connected generator to

the three term inals of which are connected three equal impedances in Y

,Fig . 13 1 . Let 2 rep resent the total impedance

of each winding of the load plus,vectorial ly, the impedance of

the generator winding in series wi th it . For the closed loops0 1a0

’b20 and 02b0’c30,we have

,assum ing sine-wave currents and

that


E E 0 ,z eI —zI , ,

E , , E0 ,= zI 3 1

Al so at the j unction 0 or 0’ we have that

+ 1 03 =O (c)

which give three equations in the three unknown quantitiesI , , I , , and Referring the to E , , as the axis of refer

ence,equations (a) and (b) may be written

Fig . 13 1 .

I

I( 14 30) E , , E = z (I (d)

E = z (I —I (6)

Substituting the value I from equation (c) in the second of

these equations, g ives

ix’

/3 E = z (2 1 0 2+ Itl )

which subtracted from (d) gives

- l , ,

2 2

Therefore1-

2E = E , ,

Sim ilarly for the other two phases ; hence


Consequently , since the three are equal in effectivevalue and differ in phase by the currents in the three im

pedances have equal effective values and differ in phase byThese currents , however, are not

‘in phase wi th the corr esponding

except in the special case when the reactance componentof the impedance is zero .

When the load is a motor or other device developing in thethree windings back e .m .f.

’s equal ln effective value and

.

differingin phase by it can readily be shown that equations (4)stil l hold provided the E

’s in these equations are taken to repre

sent the vector difference of the generator e .m .f. per winding andthe back e .m .f. of the load per winding .

From the above discussion it therefore follows that when abalanced Y—connected load is connected to a Y—connected generator

,the current established in each winding of the load is exactly

the sam e as would be produced were this winding connected byitself to a single—phase generator having an e. m . f . equal to thee . m . f . induced in each winding of the generator and an internalimpedance equal to the impedance Of each of these windings . It

also follows from equation (4) that the neutral point of the generator and the neutral point of the load must be at the sam e potential

,since the rise Of potential in the generator from the neutra l

point to any one slip ring due to the induced e . m . f . is equal tothe drop in potential 21 due to the resultant impedance of the

generator, line and load . Consequently,were the neutral points

of the generator and load connected by a wire, no current wouldflow in this “ neutral ” wire

, provided the system i s ba lanced . In

case the load is not balanced,a current would flow in the neutral

wire ; such a wire is som etim es installed to reduce the voltage dropbetween generator and load in case of unbalancing . In general ,however

,a three-phase system is very nearly balanced

,as three

phase motors and rotary converters are designed to take the samecurrent per phase

,and in the case of a lamp load or single-phase

motors used on a three- phase system,c are is taken to distribute the

various lamps and single-phase motors equally on the three- phases .

223 . Currents from a Three-Phase Generator .

— Generator A

Connected and Load Y- Connected.-When the generator is A

connected the induced e .m . f.’s between term inals are E E ,,

and Hence,neglecting the res istance and reactance of the

generator, and assuming s ine-wave and currents , we have


plying a ba lanced load i s therefore equiva lent to a Y- connected genera

EA

v3where EA

i s the e.m .f. per winding of the A- connected generator .

As will be shown presently, when the impedance of the generatoris taken into account , as it must be in all practical problems

,it

is also necessary, in order that the Y- connected and the A - con

nected generator produce the same current in the external c ircuit,

that the impedance perwinding of the equivalentY- connected generator beequal to one- thi rd the im

pedance per win ding of theA- connected generator

224 . Coil Current in a

A-Connected Generator forBalanced Load . In a Aconnected generator thecurrent I taken from eachterm inal of the generator

F ig . 132 . must be equal to the vectorsum of the currents flowing to that term inal in the w indings of the generator . Let I ,, and I , , be the curren ts in

tor developing an e .m.f. per winding Ey numeri ca l ly equa l to

the three windings in the directions mdicated in Fig . 133 , and

as before let and I be the currents taken from the three

term inals . The first set of currents are called the A or coilcurrents and the second set the Y or l ine currents . Whenthe load i s balanced

,the three Y currents are equal in effective

value and differ in phase by see A rticle 223 . Hence ,referring all the currents to I 03 as the axis of reference, and as

suming sine-wave currents and e. m . f.

’s,we have that

1 2 3.—I 1 " l” j 0 ) 1 0 3

1

2_

2

I + 1 + I , , (6)

Hence,substituting in (a) and (b) the value of I from (c) we have

2 + 70) I (d)

THREE- PHASE ALTERNATING CURRENTS 405

+ 1

Multiplying (c) by 2 and adding (d) gives—3 Im= j [

03

whence

7

V 3

Sim ilarly for the other two phases whence

Hence the current in each of the windings of a A—connected generator is equal num erica lly to the effective va lue of the Y or linecurrent divided by the square root of three , and lags 90

° behindthe current in the opposite branch of the Y

,provided the system

is perfectly balanced . Compare with the relation between theA e . m . ifs and the Y e. m . i .

’s given by equation

The Y currents , like the Y e . m . ifs, may be represented bythree equal vectors differing in phaseby The sides of the triangleform ed by j oining the ends of thesevectors are then equal to three times thecurrents

,but their directions

,taken

pos itive in the counterclockwise direc 313 ]

tion, give the proper phase relation be

tween the A and Y currents , see Fig .

133 .

225 . Coil E . M . F . and Impedance

of the Equ ivalent Y-Connected Gen

crator and of the Equ iva lent Y-Con

nected Load.

— A Y- connected and aA—connected generator m ay be looked upon as equivalent to eachother provided the voltage between term inal s is the sam e in eachfor the same current taken from the terminals . Let EA ,

IAand

zA ,and

.

EU’Iyand e be the effective e.m .f. per winding , the effective

Fig . 133 .


current per winding and impedance per w inding for a A and Yconnected generator respectively . On the assumption of sine~wave currents and e. m . ifs, the voltage between term inal s and2 in case of the A- connected generator is the vector differenceE —zA [ 1 2 , and the voltage between term inalsm the case of the Yconnected generator is, when referred to E 1 2 as the line of reference,the vector difference —j 3 (Em provided the system is

balanced , see equations Hence in order that these two expressions be equal for all values of the line or Y current

'

we musthave

I

I

I

I

But from equation (6) I 10 3 ; therefore

z = 3z (7)

Therefore,in order that the voltage between term inals be the same

in a Y- connected as in a A connected generator , the coil e. m . i . m

the Y- connected generator must be equal to L tim es the coil

e. m. f . of the A - connected generator and both the resistance andthe reactance of each winding of the Y- connected generator mustbe one- third the resistance and reactance respectively of theA - connected generator .

By a sim ilar argument it can be shown that a A - connectedload may be cons idered equivalent to a Y—connected load provided the back e . m. i . in each winding of the equivalent Y is

taken equal to i tim es the coil e . m . f . of each winding of the

A—connected load and both -the res istance and reac tance of eachw inding of the equivalent Y are taken equal to one- third the re

sistance and reactance respectively of each winding of the A .

226. Reduction of all Balanced Three-Phase Circu its to Equiv

alent Y’s .—Any problem in regard to a ba lanced three-phase

c ircuit may therefore be solved byreduc ing all parts of the c ircuitto an equivalent Y connection

,provided the currents and e. m . i .

’s

are sine waves . The transformations are made as followsAny A

- connected motor or generator is cons idered as equiv


phase between current and p .d . in thi s winding be Hy, then thepower input into each winding is V ,,Iy cos (9g , and therefore thetotal power input 1s

P =3 Vy1ycos 9

1,

Sim ilarly, in the case of a A- connected load,for . which V A is

the potential drop per winding, IA the current per winding, and(9A the difference in phase between current and p .d . in this winding,the total power input is

P =3YAIA cos 0A

In the case of a generator or any kind of a load,the p .d . that is

m ost easily measured is the p .d . between term inals or the p .d .

between the m ains where they connect with the term inals . In

case of a A- connected generator or load this is the p .d . per wmding, that is , in the case of a Y- connected machine this is theequivalent A p .d . and is equal to the p .d. per w inding V ,, mul

tipl ied by the square root of three, that is

V,,

Since V A is the same as the p .d . between wires,V A is frequently

called the line voltage.

The current that is usually most readily measured is thecurrent in each main , that is, the current leaving each term inalof a generator or entering each term inal of the load . In the

case of a Y connection this is equal to the coil current and inthe case of a A—connected machine it is the equivalent Y currentand therefore is equal to the coil current IA multipl ied by thesquare root of three . That is

,

I ,

Since Iyis the same as the current per main , Iy is frequently called

the line current. Hence in express ing the power input into aY—connected load it is more convenient to substitute for V

,,its

value in terms of the line voltage V which gives

V AI,,cos B

,,3

V A I ,, cos (9a)In the case of a A- connected load

,the substitution for IA its

value in terms of the l ine current Iy, gives

P =3 V,,

-13’V 3


V A I ,, cos B,, (9b)Equations (9 ) are the expressions usually employed for the

three-phase power, and are usually written without the sub

scripts ; nam ely, the total three—phase power is usually writtenV I cos t9

It should be clearly borne in m ind,however

,that the 9 in this

expression is not the phase angle between the voltage V and the

current I but is the phase angle between the voltage to neutraland the l ine current

,which in turn is equal to the phase angle

between the voltage between wires and the A current .228 . Example . Energy is supplied from a generating station

to a substation 50 m iles away at a rate of kil owatts . Thesystem is a balanced“ three-phase system and operates at a frequenoy of 25 cycles . The transm iss ion l ine consists of thr eeNo . 0000 B . S. copper wires spaced six feet between centers .

It is des ired to find (1) what will be the voltage between wiresat the generating station when the voltage between w ires at thesubstation is volts

,and the power factor at the substation

is 80 per cent,with the current lagging, (2) how much power is

lost in the transm ission l ine,and (3) what is the power factor

at the generating station . The electrostatic capac ity of the linemay be neglected .

The current per wire is

—241 amperes3 X X

The voltage from l ine to neutral at the substation is

v3Taking Iy as the l ine of reference

,the symbolic notation for

voltage to neutral isVy=34

,600 (cos 9+ j sin 9)

where 0 is the angle by which the voltage leads the line of

reference I,, (see A rtic le But cos and, since the

current is lagging the angle 0 by whi ch the voltage leads thecurrent is positive , and therefore sin Therefore

V,

+ 1 3’

The res istance per m ile of a No . 0000 wire is ohms ; its

inductance per m il e for a spacing of six feet is m ill ihenries,hence the reactance per m ile at 25 cycles of a No . 0000 wire is


2 7T ohms . Hence the total impedanceof each Wire is

zy=50

Hence the voltage to neutral at the generating station isV,

’+ j 20 ,800 + j 15 .2) X 241

The effective value of which is volts,and therefore the

effective voltage between wires at'

the generating station is

The power lost in the l ine is equal to 3 1531; where R is thetotal resistance of each wire and I

ythe line current . Hence the

power lost in the line is3 X X watts kil owatts

The total power del ivered to the l ine and substation is thenkilowatts . Hence the power factor at the generating

station is

v3229 Rating of Three-Phase Apparatus . The rated voltage

ofa three-phase machine always refers to the volts between term inals or the A voltage EA , the power rating is the total powerfor all three-phases

,and the power factor (usual ly written cos 9)

is the ratio of the total power P to the square root of three timesthe voltage between term inals tirnes the current per term inaland the machine is assumed to carry a balanced load . The

current per term inal is then

which is also the current per w inding for a Y- connected machine .

In the case of a A- connected m achine the current per winding i s

230 . Measurement of Power in a Three - Phase Circuit . Two

Wattmeter Method .

— To measure the power output of a threephase generator or the power input of a thr ee-phase load

,the

m ost obvious m ethod is to m easure the power for each phaseseparately by connecting the current coil in the wattm eter inseries with the winding of that particular phase and the potentialcoil of the wattmeter across the term inals of that winding . The


l ine wire a,and the current coil of B is connected in series w ith

a second line wire b. The potential c ircuit of the m eter A is

connected across from a to the third line wire 0,and the potential

c ircuit of m eter B is connected across from the l ine b to thel ine c . Let the instantaneous value

'

of the currents and p .d .

’s

be represented by sm all letters with subscripts indicating theirdirections . The instantaneous torque -ou the moving el em entof meter A is then proportional to v3 1 i 0 1 and the instantaneoustorque on the moving el ement of meter B in the same relative

directi on is vaz (That is , when the ' instantaneous currents and

p .d .

’s are actually in the directions indicated by the arrows on the

coils , and the two meters are exactly alike , these are the valuesof the ins tantaneous torques in each meter either both to theleft or both to the ri ght .)But v3 2

l

l1.

1 2

i0 2= i 1 2 _ 7:23

Hence the total torque on the moving elements of both meters is” 3 1

l

l:0 1 " l” ” 3 2 =v3 1 ifz) ” 23

val" l” ” 23 ” 23)

vs i + v23_ vi z

(smee the total drop of electric potential at any ins tant arounda c losed path is zero) . Hence the total instantaneous torque onthe moving elements of the two m eters is proportional to

” 1 2 i 1 2+ v23 i 23 + v3 1

which is equal to the total ins tantaneous power .

Since the reading of each wattmeter is proportional to theaverage torque acting on its m oving elem ent , it follows that thea lgebrai c sum (the algebraic sum

,s ince the average values of

v3 1 i 0 1 and v23 i 0 2 may be of Opposite sign) of the two wattmeterreadings gives the tru e average power, independent of whetherthe load is balanced or not and also independent of the waveshape of current and p .d . ; provided the wave shape of the currentin the potential c ircuit is the same as the wave shape of the p .d .

across it . This provis ion is onl y approxim ately fulfilled whenthe potential c ircuit of the meter contains iron .

Since the average value of the instantaneous products v8 1 i 0 1 andv3 2 i 0 2 may be either positive or negative , the deflection of the wattm eter needle may be either to the right or to the left of the zero .


Wattmeters,however

,are usually designed with a scale readi ng

onl y to the right ; consequently when a positive value of the average of the product vi corresponds to a deflection to the right, thenthe

'

needle goes off the scale to the left when the value of thisproduct is negative . However

,by revers ing the leads connecting

the current coil in the line, the di rection of the current in the current coil is reversed with respect to the current in the potentialcoil

,and the needl e will then deflect to the right . (Note that the

term inal of the potential c ircuit connected directly to the sus

pended coil of the m eter should always be connected to the linewire in which the current coil of the meter is connected ; otherwise ,since the impedance of this coil is only a fraction of the totalimpedance of the potential c ircuit

,practically the full voltage

across the potential c ircuit will also exist between the potentialcoil and the current coil of the meter, and may cause the insulationbetween the two to break down .)To m easure

,then

,the total three-phase power by this two

wattmeter m ethod,it is necessary to connect the current coils of

the two wattm eters in the lines a and b in such a manner that theneedles of both meters deflect to the right . It is then necessaryto determine whether the two readings shall be added or sub

tracted . This is determ ined from the fact that if the averagevalue of the two products v3 , i 0 1 and vaz i 0 2 are both of the samesign, then both m eters will also read to the right if they are interchanged , and the connection to the m iddl e wire c is kept unaltered .

If,however

,the average values of these two products are of the

opposite sign , then when the two meters are interchanged and theconnection to c is kept unaltered , the deflection of each instrum entwill reverse . Hence

,the general rule , subtract the two readings if

on substi tuting one meter for the other (the connection to the common

wi re c being kept una ltered) , the deflection reverses; otherwi se the two

read ings are to be added.0

231 . Two Wattmeter Method Applied to a Balanced Three

Phase System . In case the load is balanced and the currentsand p .d .

’s are both harm onic functions or s ine waves

,the phase

relations of the currents and potential drops in the two wattmeters can be readily deduced . Let (9 be the angle by which theline current I

ylags behind the potential drop V

,,cos 9 is

the power factor of the load) , then the vectors representing thevarious currents and p .d.

’s are as shown in Fig . 136.


The average value of the ins tantaneous product v3 1 i 0 1 is thenequal to the product of the lengths of the two vectors representingv3 1 and i 0 1 by the cosine of the angle between them ,

that isP l= average (val i n.) V A

Iy cos 9)Sim ilarly

,the average va lue of the instantaneous product vaz i 02

is equal to the product of the lengths of the two vectors repre

senting vaz and i 0 2 by the cosine of the angle between them ,that is

P 2= average (vaz ioz) V AIy cos (30o 0)

Hence when 0 lies between —60° and + 600 both P I and P2 arepositive and therefore their sum gives the true three-phase power ;when 9 is less than —60° or greater than P I and P 2 areof opposite sign , and hence their difference gives the true threephase power. But cOs hence when the power factor of

Fig . 1 36.

the load is greater than 50 per cent either leading or lagging, thesum of the wattm eter readings gives the true power ; when thepower factor of the load is less than 50 per cent the difference ofthe wattmeter readings gives the tru e power . *

In general , the power factor of the load is not known . How

*Notethat in the above discuss ion a negative value of 0m eans a lead ingcurrent.


SUMMARY OF IMPORTANT DEFINITIONS AND

RELATIONS

1 . The distance between the centers of successive magneticpoles of l ike sign in any electric machine is said to correspond to360 electri cal degrees.

2 . A two-

phase generator has two independent windings on its

armature di splaced 90 elec trical with reference to each other ; athree-

phase generator has three independent windings on its arma

ture,successive windings being Spaced 120 electrical degrees with

reference to each other .3 . A A - connected generator has its three windings conn ected

in series and term inals brought out at the three j unction points ;a Y- connected generator has one end of each of its windings con

nected to a common junction point called the neutral and theother ends brought out to term inals .

4 . The sum of the e . m . i .

’s induced in the three windings of an

alternator when these e. m . ifs are sine waves is zero .

5 . The equ ivalent A e . m . f. of a Y—connected alternator isnum erically equal to the square root of three times the effectivevalue of the Y c . m. i . and lags 90

° behind the e. m. f. in the 0pposite branch of the Y; that is ,

E 23 j\/3 E0 ,

provided the e. m . ifs are sme waves .

6. The equ iva lent Y e . m . f. of a A—connected alternator isnumerically equal to the effective value of the A e . m . f . dividedby the square root of three and leads by 90° the c. m. f . in theopposite side of the delta ; that is ,

provided the e. m . i .

’s are sine waves .

7. A three-phase system . is said to be balanced when the currents and e . m . i .

’s respectively in the three phases are equal in

effective value and difler in phase by8 . The equiva lent A current of a Y- connected alternator (or

load) is numerically equal to the effective value of the Y currentdivided by the square root of three and lags 90

° behind the cur

rent ih the opposite branch of the Y ; that is


provided the currents and e . m . ifs are sine waves and the systemis balanced .

9 . The equivalent Y current of a A - connected alternator (orload) is num erically equal to the square root of thr ee times theeffective value of the A current and leads by 90° the current inthe opposite s ide of the A ; that is ,

I


10 . The equ ivalent A impedance of a Y- connected winding isequal to three times the Y impedance, that is,

Z ,=3 Z

,

1 1 . The equivalent Y impedance of a A - connected winding isequal to the A impedance divided by three

,that is

,

1Z y= _

,

ZA

3


12 . When al l parts of a bal anced three-phase system have beenreduced to equivalent Y’

s,each of the three phases may be

treated as a single phase circu it,each such c ircuit being cons idered

completed by a wire of zero impedance conn ecting all the neutrals,provided the currents and e. m . ifs are sine waves .

1 3 . The total power input into a three-phase load isP =V 3 V l cos 9

where V is the p .d . between term inals (the A I is the linecurrent (the Y current) and 9 is the power—factor angle (thedifference in phase between the Y p .d . and Y current or betweenthe A p .d . and A current) . This relation holds only when thecurrents and p .d .

’s are sine waves .

14 . When two wattmeters,connected in a three-phase c ircuit

as shown in Fig . 1 35,read P , and P 2 watts respectively, the total

power is

To determ ine which sign to use , substitute one meter for the other,keeping the connection to the comm on wire unchanged ; if thedeflection of this meter reverses take the difference, if the deflection does not reverse take the sum .


When the system is balanced and the currents and e. m. i .’s

are s ine waves,the power factor of the load is cos 9 where

P —Pt 0

2 l

a" “3P .+P .

PROBLEMS

1 . The ends of the three coi ls of a three-phase alternator arebrought out to six slip rings . When the three coils on this alternator are connected in Y and the machine is operated at its ratedspeed and field exc itation

,the rated output of the m achine is

30 kw . at 1000 volts . If the speed and field exc itation are keptconstant , what would be (1 ) the rated output of the machine , and

(2) the rated voltage when the three coils are connec ted in A ?Ans . : (1 ) 30 kw. ; (2) 577 volts .

2 . A 25- cycle,three-phase, A - connected generator develops an

e. m . f . per coil of 1000 volts effective value . This 6 . m . f . con

tains the third harmonic and the effective value of this harmonicis 10 per cent of the effective value of the resultant e. m. f . The

resistance per coil of the generator is ohm and the reac tanceper coil at 25 cycles is ohm . (1) What will be the value of thecurrent per coil when the generator is supplying no external l oad?(2) Will this current be a s ine-wave current

,and if so

,what will

be its frequency?Ans (1 ) amperes (2) sine-wave current with frequency

of 75 cycles .

3 . A 500—volt,25- cycle, 100-kilovolt- ampere

,three-phase

,A

connected alternator,while Operating at full load on a c ircuit

of 85% power factor, has one of its coils burnt out . With thiscoil open- c ircuited

, (1 ) what is the m aximum balanced load at85% power factor which the machine can supply continuously?

(2) What proportion of this load is supplied by each of the tworemaining coils?

Ans (1 ) kw. ; (2) and4 . The armature of a 230-volt

,two-phase

,50-kilovolt-ampere

alternator is re-wound so that the end of one of the coils l ,is

connected to the middle of the other c oil,B . If

‘

of theturns on coil ‘B are removed

,what will be the rating of the re

sultant alternator?Ans . : 230 volts

, 3-phase

,kilovolt- amperes .


readings of the two wattmeters when connected to this load tomeasure the power by the two-wattmeter method?Ans . : 3420 watts and 2190 watts .

10 .The two wattmeters connected to a balanced three- phas e

load read 5240 watts and 2380 watts respec tively . ( 1) What is thedifference in phase between the current in the current coil of thefirst wattmeter and the potential drop through the potential coilof the second wattm eter? (2) What is the power factor of theload?Ans . : ( 1) 123 degrees or 57 degrees ; (2)1 1 . The current coil of a single wattmeter is connected in one

of the line wires of a balanced three—phase load which takes 30kilowatts at a lagging power factor of Call this wire Aand the other two line wires B and C respectively . What wouldthe wattmeter read when its potential coil is connected (1) betweenA and B

, (2) between A and C,and (3) between B and C?

Ans . : (1) kilowatts ; (2) kilowatts ; and (3) kilo;watts .

12 . kilowatts is supplied from a generating station to asubstation 50 miles distant over a three—phase line form ed ofthree No . 0000 B . S. copper wires spaced 6 feet between cen

ters . The resistance of each wire per m ile is ohms and theinductance per mile of each wire is mil lihenries . A voltageof is m aintained between w ires at the substation and thepower factor of the substation is 80% lagging current . Thesystem is balanced. Calculate (1) the voltage between wires atthe generating station

, (2) the total number of kilowatts lost inthe transm ission line and (3) the power factor at the generatingstation .

Ans . : (1 ) volts ; (2) 2250 kilowatts ; (3)13 . Energy is supplied from a three-phase Y—connected gen

erator over a three-phase line to a balanced three-phase loadformed by two motors connected in parallel . The m otors takerespectively 50 kilowatts at 80% power factor lagging and 100kilowatts at 90% power factor leading . The potential differencebetween the term inals of the motors is 500 volts and the resistanceand reac tance of each line wire are respectively ohm andohm . ( 1 ) What is the sym bolic expression for the total currentper wire taken by the load referred to the potential drop between the term inal of one term inal of the load and the neutral?

THREE-PHASE ALTERNATING CURRENTS 42 1

(2) What is the numerical value of the potential difference between the generator term inal and the neutral ?

Ans . : ( 1) (2) 308 volts .

14 . A three-phase Y- connected generator delivers power to abalanced delta- connected load over a transm ission line . The resistanc e of each line wire is ohm and the reactance of each linewire is ohm . The potential differences at the term inals of the

generator and the load are respectively 500 volts and 400 volts .

The power factor of the load is 80% lagging . Calculate ( 1 ) theline current , (2) the power delivered to the load , (3) the poweroutput of the generator, (4) the power factor of the generator and

(5) the efficiency of the transm ission line at this load .

Ans . : ( 1) am peres (2) kilowatts (3) kilowatts :

(4) (5)

A P P E N D I X A

THE following abbreviations are recommended by the American InstituteofElectrical Engineers and are used in al l its publications . In general theseabbreviations should be used only when expressing definite numerical values .

NAM EAlternating current

AmperesBrake horse powerBoiler horse powerBritish thermal unitsCandle-powerCentigradeCentim etresCircular m ilsCounter electromotive forceCubicDiameterDirect current

Electric horse powerEl ectromotive forceFahrenheitFeetFoot-poundsGallonsGrainsGrammes

Gramm e- caloriesHigh-pressure cylinderHoursInchesIndicated horse powerKilogramm es

Kilogramm e m etres

Kilogramme—caloriesKilom etresKilowattsKilowatt-hoursMagnetomotive forceMean effective pressureMiles

INSTITUTE STYLEspell out , or a- c . when used as compoundadj ective

spell outb .h .p .

Boiler h .p .

B . t .u .

c-p .

cent .cm .

cir . m ilscounter e .m .f.

cu .

spell outspell out

,or d- c . when used as compound

adj ectivee .h .p .

e .m .f.

fahr.ft.ft- lb .

424 APPENDIX A

9 . Never use the characters or to indi cate either feet and inches ,or m inutes and seconds as period of time .

10 . Use capitals sparingly ; when used as units , do not capitalize volt,

ampere ,watt

,farad

,henry , ohm , coulomb etc .

1 1 . DO not use the expression “ rotary or rotary converter use con

verter or“synchronous converter. ”

12 . DO not use a descri ptive adjective as a synonym for the noun described .

Exam ple : a “spare transform er

,not a “

spare ”

, a “

portable instrument ,not a

“

portable ”

;“automatic apparatus , not automatics ”

; a “short cir

cuit, ” not a “ short .13 . Do not use the expressions a c . current or d c . current ; a c . voltage

and dec . voltage . Their equivalents , “alternating-current current

,

” “directcurrent current , ” “alternatingcurrent voltage ,

“direct-current voltage ”

are absurdities .

14 . DO not use the expressions , raising transformer, ” “ lowering transformer these expressions are am biguous . Use

“step-up transformer

,

”

step-down transformer. "

15 . DO not use the words primary and“secondary in connection with

transformerwindings . Use instead “high- tension ”

and“low- tension.

’

NOTATION .

The following notation forms 11324 of Section V of StandardizationRules of the Institute .

E,e,voltage , potential difference

I, i , current Z

, 2 , impedanceP

, power L, l , inductance

(p,magnetic flux C

,0,capacity

B, B ,magnetic density Y

, y,adm ittance

R, r , resistance b, susceptance

x, reactance G, g , conductance

A P P E N D I X B

IN this country wires for electrical purposes , when less than half an inchin diam eter, are nearly always specified in term s ofa wire gauge introduced bythe Brown and Sharpe Manufacturing Co . Thi s gauge , called briefly the B .

S . gauge ,

* is such that successive sizes differ in diam eter by a constant percentage . A solidwire 460 m ils in diam eter is called a No . 0000 wire and a wire5 m ils in diam eter is called a NO . 36 wire . The nex t smaller size to a NO . 0000

wire is No . 000 , the next sm aller size No . 00,the next NO . 0 , the nex t NO . 1

and SO on up to No . 36 . The ratio of the diam eters Of No . 0000 and No . 36

is i gi r -QZ and the ratio of the diameters of successive sizes is constant ; thi sconstant is therefore equal to the 39th root of 92 . The 39th root Of 92 isapproximately equal to the sixth root of 2 ; hence the following approx im aterelations (since the cross section varies as the square of the diam eter and thecube root of2 is approx imately

1 . The ratio of the cross sections ofwires Of successive sizes on the B .

S . gauge is equal to the larger number on the gauge corresponding tothe smaller cross section. Thi s same relation holds for the weights of successive sizes for a given length.

2 . The ratio of the resistances Of wires of successive sizes on the B . S .

gauge is equal to the larger number on the gauge corresponding to thelarger resistance .

3 . An increase Of 3 in the gauge number halves the cross section and

weight and doubles the resistance .

4 . An increase of 1 0 in the gauge number decreases the cross section and

weight to one—tenth their original values and increases the resistance to 10

tim es its original value .

The cross section of a No . 10 wire is approximately circular m ils .

The resistance Of this size copper wire is approx imately 1 ohm per 1000 feetat 20° cent . and its weight approx imately pounds per 1 000 feet . Fromthe above relations the resistance ,

cross section and weight Of any size Of wiremay be calculated approx imately with but little effort. The resistance of a

No . 1 0 alum inum wire is approximately ohm s per 1000 feet at 20° cent .and its weight approx imately 15 pounds per 1000 feet.The above relations are for solid wire . Stranded wire is numbered on the

B . S gauge in accordance with its cross section ,not its diam eter. The

diam eter Of a concentri c strand is approximately 15% greater than that Of asolid wire of the Sam e number ; its weight and resistance are from 1 to 2 per

cent greater, depending upon the number of twists per unit length‘

and the

number ofwires in the strand .

Below are given the exact relations between gauge numbers , diam eter ,cross section

,weight and resistance for Copper wire of 100% conductivi ty

Matthiessen’

s Standard . For alum inum wire Of 62% conductivi ty multiplythe weights by and the resistances by

This gauge is a lso cal led the American W ire Gauge , abbrevia ted A . W . G .

425

426

Solid Copper

460

000 409 6 1

00

0 324 . 9

9 1 14 . 4

10

1 1

12

APPENDIX B

Wire 100% M atthiessen ’s Standard

508

402 . 8

3 19 . 5

Weight, Pounds Resistance ,20

° C .

68° F .

I N D E X

Summaries Of important definitions and princ iples found on the

following pagesM agnest l smContinuous Electric CurrentsElectromagnetismElectrostaticsVariab le CurrentsAlternating CurrentsSymbolic M ethodThree-Phase AlternatingCurrents

Abbreviations, 422 .

Absolute system Of units, 13 .

See under name Of quantity for

individual units .

Absorption, electric , 273 .

Acceleration,definition and units of, 10.

of rigid body, 20 .

Admittance ,definition Of, 349 .

of an inductance and capacity inparallel, 351 .

symbolic notation for, 380 .

units Of, 351 .

Admittances in paral lel, 351 .

A . I . E . E . style,422 .

Alternating current,

average value, 309 .

defin ition Of, 306 .

effective value , 3 1 1 .

establi shment of, 353 .

instantaneous value , 309 .

Alternating currents, 304 .

addition Of, 33 1 .

polyphase , 396.

use of, 3 12 .

Alternations, definition of,307.

Alternator, 304, 396 . C . G . S . system Of units, 13 .

American wire gauge, 425. Cable, insulation resistance , 169 .

429

Ammeters, 122 .

Ampere,definition Of

, 1 1 1 .

international, 1 26.

Ampere-hour,128 .

Ampere-turns,

calculation Of,2 10 .

definition of,207.

Amplitude factor,328 .

Analysis Of wave, 320, ff.

Angle, 3 .

Anode, definition of, 125.

Apparent power,330 .

Are, electric , 2 17.

Armature,electromotive force, 202 , 204 .

Of a generator, 198 .

Average value, 309 .

B . S . wire gauge, 425 .

B-H . curves , 86 .

experimental determ ination of, 195 .

Balance, current, 122 .

Balanced system ,defini tion Of, 400 .

Ballistic galvanometer, 194 .

Battery, electric , 155.

430 INDEX

Calibration of meters 122,145, 194 .

Capacity,and inductance (see Inductance

and Capacity) .definition and units Of, 267.

ofvarious forms of condensers, 269 .

reactance, 342 .

spec ific inductance , .272 .

Capac ities,

in parallel, 273 .

in series,272 .

Cathode, definition of, 125 .

Center Of mass,definition of

,14 .

Charging by contact and by induction, 239 .

Charging current (see D isplacement Current) .

Charge , electrostatic ,definition Of, 237 .

properties of, 242 .

and quantity Of electric ity, 259 .

residual, 273 .

units of,243

,261 .

within a conductor, 250.

Charge, magnetic , 38 .

Chemical energy, 23 , 154 .

Circuit,electric ,

definition Of, 102 .

energy associated with, 284.

general equations of the simple ,

285 .

C ircular m i], 133 .

Coil current in a delta- connectedgenerator, 402 .

Coil electromotive force of Y-con

nected generator, 399 .

Commutation, 200 .

Complex number,adm ittance as a, 380 .

division Of a rotating vector by ,

38 1 .

impedance as a, 378 .

multiplication of a rotating vectorby, 378 .

Components of a vector,power and reactive

, 328 .

Compound wound dynamo, 202 .

Concentrated winding, 120, 225 .

Condenser,electric ,

capacity of,267, 269 .

charging through a resistance , 292 .

cO-axial cylinders, 270 .

definition Of 268 .

discharging through an inductance ,295.

discharging through a resistance,294 .

discharging through a resistanceand inductance, 357.

harmonic current through, 342 .

in parallel and in Series, 272 .

non-harmonic current through, 343 .

parallel cylinders, 271 .

parallel plate , 269 .

reactance Of, 342 .

spherical, 269 .

Condensers in series and in parallel,

272 .

Conductance,


effective, 349 .

units Of, 136.

Conductivity, 136.

M atthiessen’s standard Of

, 136 .

Conductor, definition Of, 103 .

Conductors, electric , 103 .

in parallel, 1 12 .

in series, 1 12 .

Contacts, l aw Of successive ,Continuous current, establishment of

,

288 .

Copper,

specific resistance of, 135 .

temperature coefficient,137.

wires, 425.

Corona,266 .

Coulomb, definition Of, 128 .

Couple, defini tion of, 21 .

Current, electric,absolute measurement of, 120 .

alternating (see Alternating Cur

rent) .comparison of strengths Of, 122 .

continuous, definition of,108 .

decay Of,291 .

definition of, 101 .

432 INDEX

Electrostatic induction, Field ofmagnetic force—Continued .

flux of, 247. intensity Of (see Intensity ofmag

l ines Of, 247 . netic field) .phenomenon of, 239 . Fisher-Hinnen method of wave analElectrostatic intensity, 244. ysis, 322 .

inside a conductor, 249 .

just outside a conductor, 250.

Electrostatic potential, 253 .

and electrical potential, relationbetween, 261 .

Electrostatic screen, 249 .

Electrostatics, 237.

Energy,associated with a circuit, 284.

chem ical, 23, 154 .

conservation of, 23 .

definition of, 21 .

electric, definition Of, 141 .

electric , measurement of, 147.

electrostatic , 274 .

heat, 30, 228 .

magnetic , 2 16, ff, 226.

units Of, 24 .

Equipotential surfaces,electric , 167electrostatic , 255 .

magnetic , 92 .

Equivalent sine-wave p . d . and cur

rent, 3 19 .

Equivalent Y,reduction of al l bal

anced three—phase circuits to ,

406.

Equivalent Y—connected,generator, 405.

load, 405.

Erg, defini tion Of, 24 .

Externally induced 338, 350 .

Farad,267.

Faraday ’

s Law Of Electrolysis, 125 .

Field Of electrostatic force,definition of, 244 .

energy Of, 274 . Galvanometer,intensity Of (see Intensity Of ballistic , 194 .

electrostatid field) . tangent, 122 .

Field Of magnetic force, Galvanometers, 122 .

definition of, 43 . Gauss definition of, 71 .

energy of, 2 16 ff, 226 . Gauss s theorem , 56.

Flux,

of electrostatic force,244 .

of electrostatic induction, 244.

Of electrisation,245 .

ofmagnetic force, 53 , 57, 67 .

Ofmagneti c induction, 67, 72 .

Ofmagnetisation, 66 .

Flux density, magnetic ,

and intensity Ofmagnetisation, 71 .

defini tion of, 71 .

on the two sides of a surface,73 .

units of, 72 .

Flux density, electrostatic,247 251 .

Force,definition Of, 16 .

due to electric charges, 238, 242on a magnetic pole, due to electric

current, 1 14 .

due to magnetic field On wire carrying a current, 1 13 .

required to separate two magneticpoles, 48 .

units of, 19 .

Force, flux of,

electrostatic, 244 .

magnetic, 53 , 57 67.

Force,lines Of,

electrostatic , 244 .

magnetic, 53, 57, 67Force, moment Of, definition of

,19 .

Form factor, 328 .

Foucault currents (see Eddy cur

rents) .Fourier’s Series

,305.

Free period of a circuit, 296, 345 .

Frequency, definition of,307 .

INDEX

Generator, alternating current,single-phase, 304 .

three-phase, 396.

coil current, 404 .

coil e . m . f., 399 .

equivalent Y-connected,405 .

Generator, continuous cuurrent , 1 97e . m . f . of

, 204 .

Geometrical addition, 7 .

Gilbert, definition of, 92 .

Gradient,potential, 92 , 167.

Gravitation, Newton’s law Of, 32 .

Gravity,acceleration of

,1 1 .

center Of, 14 .

specific, 13 .

Harmomc . current, establishment Of ,353 .

Harmoni c e . m . f. and current, 305 .

Harmoni c function,definition of

,28 .

represented by a rotating vector,332 .

symbolic expression for, 376 .

symbolic expression for derivativeof, 377 .

Harmoni c functions, addition Of, 33 1 .

Harmonic motion,25 .

Heat energy,definition and uni ts Of, 30 .

Henry,2 12 .

Hysteresis,dielectric, 275 .

energy loss due to, 228 .

magnetic, 82 .

Hysteresis loop, experimental determination of, 195 .

Impedance,definition Of, 337 .

symbolic notation for, 378 .

uni ts Of, 340 .

Impedance coil, 340 .

reactance Of,to a harmonic cur

rent, 340 .

reactance Of, to a non-harmoniccurrent, 340 .

433

just outside a closed conductor,

Intensity of magnetic field,and flux density, 71 , 76.

Impedance Of a resistance, inductance and capacity in series

,344.

Impedances, equivalent Y and delta,405, 407 .

in parallel, 348, 333, 339 .

in series, 346, 387.

Impressed e . m . f.

,151 .

Inductance and capacity,and linkages, 2 13 .

calculation Of, 222 .

discharge of through a resistance,291 .

in parallel,351 .

in series,295 .

Of concentrated winding, 225 .

Of a solenoid, 225 .

of two parallel wires, 223 .

self and mutual,2 1 1 .

uni ts, Of 212 .

Inductance and resistance (see re

sistance and inductance ) .Induction,electromagnetic , 185 .

electrostatic, 247.

magnetic , 67, 76 .

Induction, flux Of, 39 .

electrostatic, 244 .

magnetic , 67 .

Induction, lines Of,determination Of, linking a c ircuit

,

193 .

electrostatic,244 .

magnetic, 67.

Instantaneous values, 309 .

Insulation resistance of a cable, ,169 .

Insul ator, 103 .

Insulators, definition of,103 .

Intensity, electric, 167, 263 .

Intensity Of electrisation, 246 .

and electrostatic flux density, 247Intensity of electrostatic field,

and, e leetrostatic flux density, 247


434 INDEX

i n/ tensfimnfm agnetic field— Con .

due to bar magnet, 45.

due to current in circular coil, 1 18 .

due to current in long straightwire, 1 16 .

due to current in solenoid, 19 1 .

due to point-pole, 45, 80.

due to magnetically charged disc ,47.

in closed iron ring, 196, 207.

measurement of, 50 .

tangential components of,74 .

units of, 44 .

Intensity Ofmagnetisation,and magnetic flux density, 73 , 77 .

and pole strength per unit area, 65 .

definition of, 62 .


Inverse points with respect to a

circle, 170 .

U’

H

1, meaning Of symbol, 371 , 374 ,378, 377.

Joules’s Law, 129 .

Kilowatt, definition Of, 146 .

Kilowatt-hour,definition of, 146 .

Kinetic energy and magnetic energy ,analogy between, 2 18 .

Kirchhoff’s Laws,

for direct current circuits, 158 .

for the magnetic circuit, 209 .

in symbolic notation, 383 .

Lag, definition of, 309 .

Lead, definition Of, 309 .

Leakance, definition Of, 285 .

Left-hand rule, 1 1 1 .

Length,equivalent, of a magnet, 49 .

units of, 1 .

Line current, 408 .

voltage, 408 .

Lines Of electrisation, 245 .

Lines Of electrostatic force,and electrostatic intensity, 244 .

Lines of electrostatic force—Con .

and lines of induction, 244 .

due to an electric charge, 244 .

Lines of magnetic'

force,and field intensity

, 53 , 57, 61 .

and lines Of induction, 67.

definition Of, 54 , 57.

due to an electric current, 1 15 .

due to single pole, 53 .

resultant, 57.

Lines Of magnetic induction, 67.

butting Of, 188 .

due to an electric current, 1 16 .

measurement Of, 193 .

refraction of, 78 .

Lines of magnetisation, 65, 67.

Linkages, definition of, 185.

Load, equivalent Y-connected, 406 ,Losses, definition of, 3 1 .

M agnet,definition Of, 37.

equality Of the poles Of, 49 .

equilibrium position of, 50 .

equivalent length Of, 49 .

frequency Of vibration Of, 51 .

M agnetic charge, definition of,38 .

M agnetic energy, 216 ff, 226.

M agnetic field intensity,definition of, 43 .

due to an electric ciirrent,1 13 , if,

170 .

due to bar magnet, 45.

due to magnetically charged disc .

47.

due to the earth, 49 , 50 .

effect Of surrounding medium on,

measurement Of, 50 .

tangential components of, 74 .

units Of,44 .

M agnetic field of force,definition Of, 43 .

energy Of, 226.

M agnetic flux t’

see Flux of magneticforce, Flux ,

Of magnetic induction, Flux cf magnetisation) .

Magnetic induction, 39, 67 76, 79 .

436 INDEX

Potential drop , electric—Contin ued ,

definition of, 139 .

measurement Of,134.

units Of, 143 .

Potential drop , electrostatic,definition of

, 254 .

measurement Of,256 .

units Of, 143 , 262 .

Potential drop , magnetic,and magnetomotive force, 205 .

definition and units Of, 9 1 .

Potentiom eter, 163 .

Power, apparent, 330 .

average, 3 14, ff.

definition Of, 24 .

electric , definition and units of, 146 .

electric, measurement of, 147 .

in a balanced three-phase system ,

413 .

in symbolic notation, 385.

in a three-phase c ircui t, measurement of

,410 .

Of a harmonic p . d . and a harmoniccurrent

,3 14 .

of a non-harmonic p . d . and a nonharmonic current, 3 17 .

reactive , 330 .

uni ts Of,25.

Power component, 328 .

Power-factor,definition of, 3 14 .

in symbol ic notation, 387 .

in a three-phas e system,407, 415 .

Quadrature, definition Of, 309 .

Quantity of electricity,and electric charge, 259 .

definition Of, 127.


units Of, 127.

Rating Of three-phase apparatus, 410 .

Reactance,capacity, 342 .

definition Of, 337.

Of a coil, 340, 341 .

of a condenser, 342 .

units of, 340.

Reactive component, 328 .

Reactive power, 330 .

Reluctance, magnetic, 207 .

Remanent or residual magnetism ,

Residual charge , 273 .

Resistance, electric ,absolute measurement of,and conductance , effective, 351 .

defini tion of, 129 .

effective, 337.

inductance and capacity in series,344, 357.

insulation, Of a cable, 169 .

specific, 132 .

temperature coefficient of, 137.

uni ts of, 129 .

Resistances,comparison Of

, 132 , 162 .

in parall el, 160 .

in series, 160 .

Resistance and capac ity in series,

decay of current in, 294 .

establishment of current in,292 .

Resistance and inductance m ser1es,

alternating current in, 335

decay Of current in, 29 1 .

establishment Of current in,288 .

impedance Of, 340.

Resistivity, 132 .

Resonance , 295, 345, 351 .

Right-hand rule, 188 .

Right-handed screw law, 188 .

Root—mean— square, 3 1 1 .

Saturation, magnetic , 86 .

Screen, electrostatic, 249 .

Self-inductance,definition and units of, 2 12 .

Series arrangement,Of condensers

,272 .

Of conductors, 1 12 , 160, 346 .

Series dynamo, 202 .

Shunt dynamo,202 .

Silver voltameter, 127.

Sine-wave (see harmonic function) .Skin effect, 222 .

INDEX

Solenoid, intensity Of magnetic '

fie ld

inside, 191 .

inductance Of, 225 .

Spark, electric , 2 17 .

Specific gravity, 13 .

inductive capacity,272 .

resistance, 132 .

Speed, definition and units Of, 10 .

Star connection, 397.

Strength, dielectric , 266 .

Successive contacts, law of, 154.

Surface, units Of, 2 .

Surface conditions,in an electrostatic field, 253 .

in a magnetic field, 75.

Suscep tance,definition Of, 349 .

units Of, 351 .

Symbolic expression,rationa lization Of, 382 .

Symbolic method, 371 .

examples of, 387.

Symbol ic notation, meaning of sym

bol“ j ” in, 371 , 374, 377, 378 .

Tangent galvanometer, 122 .

Tangential components, 74, 253 .

Temperature, definition and units Of29 .

Temperature coeffi cient of electricresistance, 137

Term inal electromotive force,definition Of, 151 .

Of a three phase generator, 399 .

Three-phase alternating currents,396 .

Time, units of, 3 .

Torque,and power, 25 .

definition and units of, 19 .

Transformer, alternating3 13 .

Transient effects produced by a har

monic e . m . f., 353 .

Transmission line, 297, 409 .

Turning moment (see Torque) .Two wattmeter method Ofmeasuring

power, 410 .

current,

437

Units,

absolute system Of, 13 .

for units of various quantities (seeunder name of quantity meas

ured ) .

Variable currents, 284.

Vector,addition of, 7, 372 .

components of, 6 .

defini tion Of, 4 .

derivative Ofa rotating, 377.

division of, by a complex number,381 .

multiplication of, by a complexnumber, 378 .

referred to another vector as l ineOf reference, 375 .

representation of a harmonic function

, 332 , ff.

rotating, 332 .

subtraction of,

symbol ic expression for, 203,

375 .

Vectors,composition Of, 6,definition of, 4 .

difference in phase between,Velocity, definition and units Of, 9 .

Volt,definition of, 143, 157.

international, 157.

Voltameter, 127.

Voltmeter, 144.

Volume, units of, 2 .

Watt,definition of, 24 .

Wattless component (see Reactivecomponent) .

Wattmeter, 147measurement Ofthree-phase power,410 .

Wave analysis, 320, ff.

Wheatstone Bridge, 162 .

Wire as a geometrical line, 108.

Wire tables, 425.

438 INDEX

Work,current and magnetic flux, re

lation between, 187.

defini tion Of, 2 1 .

relation between current, flux and ,

187.

required to establish a current,2 15, ff.

required to separate electriccharges, 254.

Work—Continued .

requir ed to separate magneticpoles, 88 .

units of, 24.

Y-connection, 397.

reduction Of a ll balanced threephase circuits, 2 , 406 .

terminal e . m . f . of, 399 .

m THIS BOOK I S DUE ON THE LA

STAMPED BELOW

A N INITIA L F INE OF 25

W ILL B E A SS ES

TH IS BOOK ON“

WILL INC REAS EDAY

electrical engineering - forgotten books

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