reading material for basic electricity

8/12/2019 Reading Material for Basic Electricity

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Reading aterial

for

asic Electricity

Reading Material for Basic Electricity

Electrical

Electronics Laboratory

Centre

for H R D

Durgapur Steel Plant


2/38

BASIC ELECTRICITY

CONTENT

SI.

Topics

Page No.

1

Source of E lectricity 1

2 Ohm's Law

1

3 Laws of Resistance,Resistance in series and Uses of Series Resistance.

2 -3

4

Resistance in Pa rallel and U ses of parallel Resistance.

3 - 4

5

Principle of Inductor, Behavier of L-R Ckt with DC

4

6

Behavier of R - C Ckt with DC

5

7

Charging & Discharging of Capacitor with DC

6

8 Di- Electic Strenth of Capacitor

7

9

Capacitor in Series & P arallel

7

10

Defination & N ature of AC Volages

8

11

RM S V alue of an A C V oltage / Cu rrent, Electromagne tic Induction

9

12

Resistance & C apacitor Colour Code chart 10 - 12

13

Practical Assignment on Resistive Ckt at AC / DC

13 - 15

14

Practical Assignment on Inductive Ckt at AC

15- 16

15

Practical Assignment on Capcitive Reactance ( X c)

16 - 17

16 Practical Assignment on Inductive Reactance ( X L)

17 - 18

17

Practical Assignment on Series RC ckt.

18- 19

18 Practical Assignment onS eries LR Ckt.

19

19

Practical Assignment on impe dence of a Series CR Ckt & app lication.

20 -22

20

Practical Assignme nt on impede nce of a Series LR Ckt & ap plication.

23 - 24

21

Practical Assignment on impedence of a Series LCR Ckt & application.

25 - 26

22

Practical Assignment on Parallel and app lication

27--29

23

Working Principle of a Transformer, E.M.F. Equation

29 - 30

24

Power at AC ( 1 Ph & 3 Ph)

30-31

25

Meggering

31- 33

26

Safety Precautions for Electricians.

34-35

27

Treatme nt of Electical shock for Artificial Respiration

36 -


3/38

Source of E lectricity

Electrical energy m ay be produced in several ways. Basically there are five m ethods

of

providing voltage and current.

Friction

This m ethod allows for the separation of electrons and protons b y

the work

of rubbing tw o dissimilar materials together wh ich, in turn produces static

electricity

because of the separation an d storage of opposite charges.

Chemical energy

this metho d is generally found in dry cells or storage

batteries.

Fund am entally, it involes a chem ical reaction between two dissim ilar metals and

electrolyte in w hich charges of o pposite polarities are produced. The m etals

which

beco m es deficient in electrons bec om es the positive plate, and the m etal with ;:r1

abund ance of electrons becom es the negative plate. The chem ical action

continues

until no further transfer of electrons is possible.

Photoelectricity

thou ght are use of the ph otoelectricity effect, light energy

can t

.4

,

;

con verted into electrical energy . The process involes coating the surfaces

of

m etals w ith pho tosensit ive m aterials such as selenium or caesium .

Where

strikes the m aterials, P

hotoelectrons are em itted and create a current flow

by tho

,

steady movement.

Thermal Energy

as the nam e indicates thermal energy involves the

convers:cn

heat into electrical energy. This can be accom plished by h eating metals

which

turn supp ly energy to the free electrons. The electrons in turn break aw ay

from Olt.:

atoms,

creating a current.

Magnestism

M otion can b e applied to a conductor wh ich, when allowed to

move

a magnetic field create a current. This is basically the principle of

generaro

operation.

OHM'S LAW

Temperature and other physical conditions rem aining constant, the potential

different

across the two terminals of a cu rrent carrying con ductor is directly proportional to

the

current flowing through it.

Frv A B is a conductor and the potential at points A & B are VA and Vg

;.pose VA > VE, So potential difference is VA ..V3 = Say V )

Let us assume that when the conductor is connected to battery B (of emf = V),

current

( I

flows through the condu ctor.

According to Ohms law ; A VB ) cc I

or, V I or V = IR

As VA VB = V )

wh ere R is the resistance of the conductor and its unit is ohm Un it of voltage and that of

I s Am p.

Page No. 1


4/38

V

I V

Law s of resistance

R oc p L / A

w here, P = Speci f ic resistance, L= Len gth of the condu ctor,

A = Cross sectional area of the con ductor.

Resistance is directly proportional to the length of the con ductor.

Resistance is inversely proportional to the area of cross section of the

conductor.

Value of the specific.resistance of the cond uctor depend s upon the properties

of conductor m aterial and the tem perature of the condu ctor. Thus a nichrome

wire

Has about Sixty six times more resistance than a copper wire of the same

dimen sions at sam e temperature.

RESISTANCE IN SERIES

R2

Notice that the sum voltage across each resistor is equal to the applied voltage

(V) i.e. total voltage. V = Vi+ V2+V3

Accord ing to Ohm 's law th is can be expanded as IR=IRi + IR2 + IR 3 A s

resistors are connected in series, the sam e value of current (say I ) must flow in

the circuit. So R1 = R1 +R2+ R 3 This Shows that for resistors in series, the

equivalent resistance is equ al to the sum of the resistors in series.

This formula can be writ ten as: R1 = R1 +R2 +R3

Rn for any num bers of

resistances in se ries.

Page No. 2


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6/38


7/38

Capacitor

C

o

ka.5e

-t.-t.rir 7 C

741-1c.

Behavior of R-- C Circuit with DC.

In our earlier assignment we have known about condenser or capacitors. Now for

knowing its nature with DC voltage we shall see the following figure. Here the action wiii

opposite to that in the circuit using an inductor. The initial current will be and the current

will start decreasing until the current become zero and capacitor gets fully charged . The

current curve will be like the following figure.

R

Page No. 5


8/38

414_

Charging

ischarging

CAPACITOR

Capacitor can be defined as two conductors separated by an insulating dielectric. It is a

device to store electrical energy and to release it when required. In figure a simple

parallel plate capacitor is connected with a battery.

Battery

C

CHARGING OF CAPACITOR

Working Principal :-- Suppose plate A is connected to +ve

pole and plate B to -ye pole

of DC supply. On closing the switch there will be momentary flow of electrons in the

direction indicated some electrons are withdrawn from the plate A leaving it positively

charged and transferred to plate B giving it a negative charge. This flow of electrons

gives current which decreases and finally ceases when the voltage across the capacitor

plate has become equal and opposite the applied voltage. A capacitor then blocks DC

circuit . The charged capacitor is now a store of energy. The practical units

of capacity

is farad. A conductor is said to possess capacity of one farad if its potential is raised by

one volt when one coulomb of charge is given to it.

One farad = 1 Coulomb / 1 voltage

Sub-units of Farad are micro and pico farads.

1p F =1 MF = 1/ 1000000 farad

1pp F or 1 pico farad = 1/ 1000000000000

Factors on which the capacity of capacitor depends :

The area of the plate of the condenser (directly proportional).

The distance between the plate ( inversely proportional).

The nature of dielectric, i.e. dielectric constant.

[ C = K* A/d where A = Area, D = distance , K = dielectric constant ].

Page No. 6


9/38

V3

2

voltage

Di- Electric Strength

The maximum kilovolts per millimeter or Volts per mm which a medium can withstand

without break-down, is known as its

Di-electric

strength .

The insulating materiel which resists the break-down is called its dielectric strength e.g.'

Glass = 200 -250 KV / mm

Mica = 500-1500 KV / mm .

The various factors on which the dielectric strength depends are :

Thickness' of material. Break-down voltage (E) is not proportional to the thickness

(T). E =

A* t

2

/

3

) where A is materiel constant .

Temperture. It decreases with rise of temperature .

Absorbed moisture content affects inversely .

Shape and size of electrodes.

Capacitor in series

Let C1, C2, C3 = Capacitance of the capacitors connected in series (figure )

V1, V2, V3 = P. D. drop of

three Capacitors

respectively. ,

V =The total applied

C = Joint capacity

Now in series, V = V1 + V2+ V3

or Q/C = Q/C1+Q/C2+Q/C3 ( since V = Q/C in condenser)

or 1/C = 1/C

1

+1/C2+1/C3 (Taking Q as common ) .

In series combination, charge of all capacitors is the same, but P.D. across each is

different. In this case the reciprocal of the total capacitance is equal to the sum of the

reciprocals of individual capacitance's.

Capacitors in parallel

The three capacitors C1,C2 and C3 are connected in parallel to supply voltage V.

Let the charged on the capacitors be Q1, Q2,& Q3.

Then Q1 = C1 V ,

2 = C2 V,

3 = C3 V

Let C = Equivalent capacitance, then Q = Q1+Q2+Q3

therefore C = C1+C2+C3 .Thus the total capacity is equal to the sum of the

individual-capacitance's .

ci

2

3

Page No. 7


10/38

Vmax

Vp-p

Triangular wave

Square wave

DEFINITION AND NATURE OF A.0 VOLTAGES

So far we have default with pure D.0 voltage and that too only constant magnitude . But

there are other voltages too e.g. varying D.C. voltages of various shapes. Now let us

look at the various wave from below to understand what A.C. voltage is,

The polarity of an A.G. voltage always changes after the completion of every half cycle

and after every full cycle the full wave shape repeats it self . In case of pure A.C. wave

from the area of 'positive' and 'negative' half cycles are equal. Then. of full cycles which

occur in one second time is called frequency of the A.C. voltage. In fact, all voltage

generated out of rotating machines are basically A.G. voltage. But in case of D.C.

generators it is the commutator that converts the A.C. to D.C. voltage. Moreover voltage

level changing in case of A.G. can be done very easily. So we must have a knowledge

of A.G. voltages.

Page No. 8


11/38

R.M.S. VALUE OF AN A.C. VOLTAGE / CURRENT

As the instantaneous values of any AC. waveform are always changing, when we

measure the magnitude of an A.C. voltage by an AVO meter we actually get its R.M.S.

value ( Root Mean Square value). The R.M.S. value of an alternating current is giving

by that steady ( D.C.) current which when flowing through a the same circuit for the

same time.

For example, RMS ( for a sine wave ) = V peak /

or RMS ( for a sine wave) = I peak / 42

ELECTRO-MAGNETIC INDUCTION.

Whenever there is a change of flux linked with any circuit ,an emf, is induced in

the circuit. This is known as Faraday's 1st law of induction. The amount of emf enduced

will depend on the time rate of change of flux linkage . This refers to the 2nd law of

Faraday. The direction of induced emf will always oppose the cause producing the

induced emf. This is called Lenz's law.

Page No. 9


12/38

R CAPACITOR INDENTIFICATION CHART

Colour

Band

A

Band

B

Band C

and D

( Multiplier ) Tolerance )

Band E

Resistor

Capacitor

Resistor

1%

0pF

Capacitors

Resis-

for

---

---

20%

Poly-

-ester

Capa-

-citor

---

---

Upto

2pF

0.1pF

Over

10pF

1%

Black

---

0

10

10

---1

Brown

1

1

10'

1F

-

10

1

10

-2-

10

3

Red

2

2

2%

---

---

---

2%

2.5%

250w

---

---

---

range

3

3

103

Yellow

4

4

10

0

---

---

---

---

Green

5

5

10

5

---

---

---

0.5pF

---

5%

---

_ _ _

---

---

---

lue

6

6

10

6

---

Violet

7

7

107

---

---

---

---

---

---

Grey

8

8

108

0.01pF

---

0.25pF

--- ---

---

White

9

9

10

9

01 pF

---

10%

1pF

---

10%

---

---

---

---

---

ilver

---

---

0.01

---

Gold ---

---

0.1

---

5%

---

---

--- ---

Pink

---

--- ---

---

--- ---

---

---

Hi-

Stab

None

---

---

---

---

20%

---

---

---

Resistor Capacitor letter And Digit code ( BS1852)

Preferred value

E 12

Series

1.0 1.2

1.5 1.8

2.2 2.7

3.3 3.9 4.7 5.6

6.8

8.2 and their

decades.

E 24

Series

1.0

1.1 1.2

1.3 1.5

1.6 1.8 2.0

2.2

2.4 2.7 3.0

.3

3.6

3.9 4.3 4.7 5.1 5.6 6.2

6.8 7.5

8.2

9.1

and their

decades.

Resistor Capacitor letter And Digit code ( BS1852)

Resistor values are indicated as follows:

0.4752

marked

R47

152

marked

1R0

4.70

Marked

4R7

47Q

Marked

47R

100Q

Marked

100R

1k0

Marked

1K0

10k0

Marked

10K

10MQ

marked

10M

A letter following the value shows the tolerance. F =1%: G =2% : J = 5% :

K = 10% : M = 20% : R33M = 0.33Q20% : 6KSF = 6.81(Q1%

Page No. 10


13/38

Capacitor values are indicated as follows

:

0.68 pF

Marked

p68

6.8 of

Marked

6n8

6.80 pf

Marked

6p8

1000 nF

Marked

1p0

1000 pF '

Marked 1n0 _ _ .8 pF

Marked

6p8

Tolerance is indicated by letter as for resistors. Values up to 999 pF are marked in pF

from 1000pf.to999000pF = 999nF as nF 1000pF = 1nF and from 1000nF( = pF

upwards as pF.

Some capacitors are marked with a code denoting the value in pF (first two

figures)followed by a multiplier as a power of ten (3 =10 ).Letters denote tolerance as for

resistors but C = 0.25pf. E.g.123J = 12pF x 10

3

5% =12000pF (or 0.12pF).

Tantanum Capacitors

Color 1

2

3

4

Color

1

2 3 4

Black

---- 0 x1 10V

Blue

6

6 ----

20

Brown 1 1 x10 ---- Violet

7 7

----

----

Red 2 2

x100

----

Grey

8

8 x0.01 25V

Orange 3

3

----

---- White 9

9 X0.1 3V

Yellow 4 4

----

6.3V

Pink

10

10 ---- 35V

Green

5 5

---- 16V

Reactance of Capacitors at spot frequencies.

50Hz

100Hz

1kHz 10kHz 100kHz 1MHz 10MHz 100MHz

1pF

----

----

----

---- 1.6M

160k

16k

1.6k

10pF

---- ----

---- 1.6M

160k

16k 1.6k 160

50pF

----

----

3.2M 320k 32k

3.2k

320 32

250pF ----

6.4Mm

640k

64k

6.4k

640

64 6.4

1000pF 3.2M

1.6M

160k

16k 1.6k

160 16 1.6

2000pF 1.6M 800k

80k

8k 800

80 8 0.8

0.01pF 320k 160k 16k

1.6k 160 16

1.6 0.16

0.06pF 64k

32k

3.2k

320

32

3.2

0.32

----

0.1pF

32k

16k

1.6k 160 16 1.6

0.16 ----

1pF 32k 1.6k 160 16

1.6

0.16

----

----

2.5pF 1.3k 640

64

6.4

0.64 ---- ---- ----

5pF 640 320

32

3.2 0.32 ---- ---- ----

10pF

320 160

16 1.6 0.16

---- ---- ----

30pF

107 53

5.3

0.53 ---- ---- ---- ----

100pF 32

16 1.6

0.16 ---- ---- ---- ----

1000pF

3.2

1.6

0.16

----

---- ---- ----

----

Page No. 1 1


14/38

Reactance of Inductors at spot frequencies

50Hz

100Hz

1kHz

10kHz 100kHz

1MHz 10MHZ 100MHz

1pH ----

----

----

----

0.63 6.3 63 630

5pH ---- ---- ---- 0.31 3.1

31

310

3.1k

10pH

----

----

----

0.63

6.3

63 630

6.3k

50pH

----

----

0.31

3.1

31

310

3.1k

31k

100pH

----

----

0.63

6.3

63

630

6.3k 63k

250pH

0.16

1.6

16

160

1.6k 16k 160k

1mH

0.31 0.63

6.3 63 630

6.3k 63k 630k

2.5mH

0.8

1.6

16 160

1.6k 16k

160k 1.6M

10mH

3.1

6.3

63

630

6.3k 63k

630k

6.3M

25mH 8

16

160 1.6k

16k

160k

1.6M ----

100mH 31

63 630

6.3k

63k

630k 6.3M ----

1H

310

630

6.3k

63k

630k

6.3M ----

----

5H

1.5k 3.1k

31

310k

3.1M

----

----

----

10H

3.1k 6.3k 63k 630k 6.3M

----

----

----

100H

31k 63k 630k 63M

----

---- ---- ----

Page No. 12


15/38

ASSIGNM ENT ON RE SIST IVE CU RCU IT AT AC / DC

OBJECTIVE To investigate a resistive circuit at AC

PROCEDURE

Connect the power supply unit to the mains supply line. DO NOT

switch on yet.

EXPERIMENTAL PROCEDURE

We wish now to see how a resistor behave at a. c.

Connect up the circuit as shown in the patching diagram of figure.

Ensure the variable d. c. vo cage control knob is turned fully counterclockwise, then

switch on the pus. Very the d.c control slowly, and observe the two meters.

PRACTICAL CONSIDERATIONS AND APPLICATIONS Resistors are used at a. c. in

similar ways to those d.c applications of earlier assignments. There is no phase shift

across a true resistance, and the ratio of voltage to current shown by a resistance is

constant with frequency, so the behavior at a/c is no different from its behavior at d.c.

When calculating power dissipation at a.c, rms. values of voltage and current must

apply.

PRACTICAL CONCIDERATIONS AND APPLICATIONS Resistors are used at a.c

in similar ways to those d.c applications of earlier assignments . There is no phase

shift across a true resistance , and the ratio of voltage to current shown by a resistance

is constant with frequency , so the behaviour at a.c is no different from its behaviour at

d.c. When calculating power dissipation at A.G., rms values of voltage and current must

apply.

Practical Assignment on CAPACITIVE ckt. at AC

OBJECTIVE

To investigate a capacitive circuit at a. c.

PRELIMINARY PROCEDURE

We known, from assignment 8, that the relationship

between charge, voltage and capacitance is : Q = CV also we known :Q = It, where I is

current and t is time. From these we can say that if a capacitor of C farad is charged

from OV to V volts, in t seconds then : charging current, I= charging current = capacitor

x rate of increase of voltage. Let us see what happens when a sinusoidal alternating

voltage is applied to a capacitor. Connect up the circuit as shown in the patching

diagram of figure , corresponding to the circuit diagram of figure

Page No. 13


16/38

2.2

micro

F

AC

CH1

CH2

100 R

orn

Set the function generator to give a 10Vpeak-to-peak sine waveform at 250Hz.Set the

Oscilloscope as follows : Y1 channel 1V/cm. Y2 channel 500mV/cm .Time base to

1ms/cm. Zero both the traces and then observe the two waveforms on the oscilloscope.

Mathematically, if the voltage waveform is denoted by the formula :

V = V

max sin t

then as i = C x rate of change of voltage

I = C / dv /dt = C VmAx d /dt (sin w t)

I =CV Ax cos w t

Thus if v is sinusoidal , i.e the same shape , but leading by 900 .

This is because cos t =sin( t+90

)

If we were to plot the voltage and current waveforms in the capacitor by that method we

would require two vectors . Both vectors would rotate while keeping a constant 90

0

angle between them , as shown in figure. As we go on we shall find it useful to think in

terms of these vector, but rather confusing if they are always rotating. Usually it is the

relationships between them that are important, as for instance the 90

0

angle between

those in figure .These relationships can be studied conveniently in a

Page No. 14


17/38

CH

CH2

orn

Inductor

T

Resistor

FG

diagram where the vectors are shown at rest. The vectors are then said to be

represented by 'phasor' and the diagram is a 'phasor diagram'. Figure is phasor

diagram corresponding to figure.

The voltage phasor is taken as the reference and is drawn horizontally pointing to the

right (3 'o'clock). The current phasor leads the voltage phasor by 900

and is thus drawn

900 counterclockwise from the reference. When an alternating voltage is applied across

a capacitor an alternating current flows. Yet when a.d.c voltage is applied, after an initial

flow of charging current, no d.c current flows. This behaviour is different from that of a

resistance. Never the less if an a.c voltage and an a.c current can exist, the ratio

between them is likely to be interest, and the ratio is therefore given a different name. In

an a.c circuit the ratio of voltage to current is called 'impedance' and is denoted by Z.

Thus in an a.c circuit Z = We shall examine this idea further in other assignment. For

the moment it may be noted that impedance may be taken as the ratio of two phasor,

and therefore has both magnitude and phase. Magnitude = Z = Phase

of the

impedance is the angle between the phasor. The impedance of a capacitor has a phase

of 90 radians Impedance's of 900

phase angle have special properties and

are

given the special name 'reactance'.

PRACTICAL CONSIDERATIONS AND APPLICATIONS

The applications

section of

Assignment gives details of capacitors in general, but there are a few more

points

which

appear

at high frequencies(hf).Due to

the form

of construction' wound capacitors of any

type possess appreciable inductance, thus do not act as pure capacitors. The effects of

this inductance are greater at high frequencies and thus wound from of capacitors is

inferior at h.f. Mica capacitors, ceramic tubular and disc capacitors, and air dielectric

capanitnrs

are

ideal

fnr high frequency

work

Capacitors normally have a stated

maximum voltage rating above which they cannot be used safely. When used at a.c the

peak or crest value of the voltage must be used to determine wher ther the capacitor is

within rating.

PRACTICAL ON ASSIGNMENT INDUCTIVE CIRCUIT AT AC

OBJECTIVE

To investigate an inductive circuit at a.c.

EXPERIMENTAL PROCEDURE From assignment 12 we have found

out

that the

relationship between induced emf, current and inductance in a system is : e = - L or :

(Induced emf ) = - (inductance)(rate of change of current) We will now examine what

happens when a sinusoidal alternating voltage is applied to an inductor. Connect up the

circuit as shown in the patching diagram of figure 16.2, corresponding to the circuit

diagram of figure

Page No 15


18/38

Set the function generator to give a 10V peak-to-peak waveform at 250Hz. Set the

oscilloscope as follows : Y1 channel(current) to 1V/cm Y2 channel(voltage) to50OrnV

/cm Time base to 1ms/cm. Zero both traces, the observe the two wave forms on the

oscilloscope. Draw the two waveforms you see, showing their relative positions with

respect to each other.

PRACTIAL CONSIDERTIONS AND APPLICATIONS

By its nature of operation an

inductance will create a magnetic field around it. With an alternating current flowing

through the inductor the magnetic field around it will be alternating. If the inductor is the

presence of other components or conductors this alternating magnetic field will link

with these conductors and induce emfs in them. These emfs will generally be unwanted,

and give rise to noise, hum, or interfering signals.. For this reason, inductors are often

magnetically screened in cans, or housings made from a non - magnetic material such

as mumetal or aluminium. When designing inductors care must be taken in the selection

of type and gauge of wire used. Obviously the lowest possible resistance is desired and

at low frequencies this means the thickest possible wire gauge consistent with a

reasonable sized winding, however at high frequencies multi-stranded litz wire is often

better to minimise the skin effect.

ASSIGNMENT on CAPACITIVE REACTANCE ( Xc )

OBJECTIVE

To investigate the impedance of a capacitor to an a.c sinusoidal waveform

and see how this varies with frequency.


In a capacitor a sinusoidal current and voltage are

always 90

0

out of phase with one another , so that the phase of the impedance is

constant .The magnitude varies however, in a manner which we shall now discover.

connect up the circuit as shown in the patching diagram of figure . corresponding

to the

circuit diagram of figure.

C

Copy the results table as shown in figure, reproduced at the end of this assignment. Set

the frequency of the function generator to 800Hz. Adjust the output of the generator to

give 1V rms as read on the matter. Take the reading for this voltage. Reset the voltage

output of the generator to 2V rms. Record the resultant current. Repeat this procedure

for voltage of 3V, 4V, 5V and 6V rms. Record your results and calculate the ratio of rms

voltage to rms current.

Page No 16


19/38


capacitors show an infinite

impedance to d.c but their impedance is infinite to a.c and decreases as the frequency

of the a.c increases. Therefore one of their used is in coupling between circuits or parts

of circuit, where it is wanted to block any d.c transmission but to allow a.c signals to

pass. Large value electrolytic capacitors are often used in 'smoothing' circuits in which a

d.c voltage which has an a.c ripple voltage superimposed upon it is applied to the

capacitor. The capacitor has no the effect on d.c component of the wave form, as its

impedance to d.c is infinite, but by passes to earth the a.c. This is shown in figure .

PRACTICAL ASSIGNMENT ON INDUCTIVE REACTANCE ( XL )

OBJECTIVE

To investigate the impedance of an inductor to a sinusoidal a.c

Waveform .

E XP E R IME N T AL P R O C E D UR E

We have said that impedance , Z is given by

Z = V

rms rms

And that , for a capacitor , its impedance termed capacitive reactance,. X c .

Similarly we can determine the INDUCTIVE REACTANCE of an inductor. Inductive

reactive is given the symbol XL .Connect up the circuit as shown on the patching

diagram of figure 18.2, corresponding to the circuit diagram of figure 18.1

L

Page No.17


20/38

R3

Copy the results table as shown in figure reproduced at the end of this assignment.

Adjust the generator to give a frequency of 4kHz and an output of 1V rms sine wave, as

shown on the mater. Record the current reading for this voltage. Readjust the output

amplitude to 2V rms and record the resultant current. Repeat this for voltage steps of 3V

4V ,5V and 6V rms. Record your results and calculate the impedance for each step.

PRACTICAL CONSIDERATIONS AND APPLCATIONS

the inductor is the converse

of the capacitor in that it shows an impedance which is practically zero

to d.c but

increases with frequency .Like a capacitor , an inductor can also be used in smoothing

circuits, but it is used in series with the d.c line instead of across it. See figure . The d.c.

is passed without effect while the a.c is greatly impeded by the inductor. In this

application the inductor is often called a 'choke'.

L

t

v i e . .

When an inductor contains magnetic material such as iron , it can only maintain its

inductance over a limited range of current . Excessive current cause the 'iron' to

saturate (i.e

fail to permit more than a limited amount of flux). The inductance for

therefore decreases. This is why you were advised to omit any reading for which the

current exceeded 60mA.

ASSIGNMENT THE SERIES C R CIRCUIT

OBJECTIVE

To investigate the series CR circuit and determine the relationships

governing amplitude and phase shift .


We know that voltage and current are in phase in a

resistive circuit , and that , for a capacitive circuit , the current leads the

voltage by 90

Lets us investigate what happens in a circuit consisting of both capacitance and

resistance . Connect up the circuit as shown in the patching diagram of figure

corresponding to the circuit diagram of figure .

V

.....

......... ......

....

.......

.........

0

C

l

Page No. 18


21/38

Set the Fun ct ion ge nerator f requen cy to 50Hz , and i ts output to 20V p eak-to-peak ,

Sine wave as sho wn on the Y1 chan nel of the oscil loscope .

Have the o scilloscope set to : Y1 chann el(Vi

n

) to 5V / cm . Y2 channel ( ) to 5V /cm

Time b ase to 5mS / cm . Zero both the oscilloscope traces .

W ith link 1 connec ted as show n in f igure 19.2 , ie by pass ing the resistor chain so that

resistor is zero , observe the waveform for and V

c . They should be superimp osed .

M easure their amp litude . Now conn ected l ink 1 betw een point

A and D to leave the

1 kS1 resistor in series with the capacitor .Measu re the am plitude o f V

c

and the phase

shift m ay be m easured as fol lows :

ASSIGNMENT THE SERIES LR CIRCUIT

OB JECTIVE To invest igate the series LR circuit and determ ine the relationships

governing am plitude and phase sh ift .

EXPER IMEN TAL PR OC EDU RE W e have investigate wh at happens in a series RC

circuit when R is varied , let us now see wh at happens w hen R is varied in an LR circuit .

Con nect up the circuit as show n in the patching diagram of figure

corresponding to the

circuit diagram of figure .

0--

R 1

2

--o-H

*-1

R.3

Page No. 19


22/38

0

PRACTICAL ASSIGNMENT IMPEDANCE OF A SERIES CR CIRCUIT

OBJECTIVE

To investigate the series CR circuit and determine the relationships

governing amplitude and phase shift .


We have investigate the impedance ( or reactance ) of

a purely capacitive and an inductive circuit , and arrived at formulae relating impedance

with frequency and capacitance or inductance . Let us now see if we can arrive at some

formulae for the series CR circuit of figure . Connect up the circuit as shown in the

patching diagram of figure corresponding to the circuit diagram of figure .

Set the frequency of the function generator to 100Hz Adjust the

output of the generator

to give 1V rms as read on the meter . Take the current reading for this voltage .

Reset

the voltage output of the generator to 2V rms .Record the resultant current .Repeat this

procedure for voltages of 3V ,4V and 5V rms . Copy the results table as shown in figure

, reproduced

at the end of this assignment and tabulate your results . Work out the

magn itude of impe dance of the circuit at 100Hz , calculating the value at ea ch voltage

step and taking an average o f these .


As well as a

phase shftng

network , the RC circuit can be used as a frequency dependent

circuit as part of a 'thter

A filter is a circuit which will pass some frequencies but attenuate (reject)

others

Consider the circuit in figure .

R

_

0

By the potential divider relationship

2 = V1 (Xc / Z )

/ WC

1

/ COC

R

2

+ ( 1 / 0.)C )

2 r(co CR)2 + 1

now when co CR

1 V2 V1 I

and when w CR > 1,

V2

1 / CO CR

Page

No. 20

C


23/38

At low frequencies the capacitor takes little current , so the voltage drop in R is small .

At high frequencies the capacitor short-circuits the output , and most of the input voltage

is dropped across R .When oCR = 1, f = 1/

2TrCR

and at this frequency

V2 = V / 12 ie 0.707 x

Around the frequency a transition occurs between these conditions , as shown in figure

v

i

4-

4 2/7 ck

This type of circuit is known as a low pass filter , as it passes all frequencies below the

frequency known as the cut-off frequency , f

, and attenuates all those

above f c.

Now consider the circuit of figure.

C

p4)

In this case :

V2=V RZ)

V

1

/ R

R2 + ( 1 / co

C )

2

/ ( )CR

r(c.0 CR )2

+ 1

now when co CR (c 1,

V2 ^* COC R V1

and when

O

CR 1,

V2

1

At low frequencies the high reactance of the capacitor restricts the current flowing in R ,

and therefore also the output voltage . At high frequencies the capacitor virtually

connects the output and input directly together . Similarly around the frequency 1/

2nCR

there is a transition between these two states , as shown in figure .

Page No.21


24/38

-jc '

.

This type of circuit is know n as a high -filter , as it passes all frequencies abo ve the cut-

off frequen cy , f , and attenua tes all those b elow f c .

Filters are used extensively in electronics- .High-pass and low -pass types are bo th

com m on , as is a com bination of the tw o types , cal led a ban d-pass f i lter , which passes

all frequencies w ithin a c ertain band , cal led the pass ban d , an d attenuates al l other .

Drawings o f a band pass filter and its respon se are given in figures

CA

O

w t

t

t t 4

-

2 _

_

C

V

l i

Page No.22


25/38

PRACTICAL ASSIGNMENT ON IMPEDANCE OF A SERIES LR CIRCUIT

OB JECTIVE To invest igate the im pedance of series LR circuit to an a.c sinusoidal

waveform ,and see how this varies with frequency .

EXPERIMENTAL PROCEDURE Using a similar technique to that employed in

Assignm ent 21 , we now wish to investigate the impedan ce of a series LR circuit .

Connect up the circuit as shown in the patching diagram of f igure corresponding to the

circuit diagram o f figure .

t y o 44,14

L

Set the frequen cy of the function generator to 800Hz , and ad just the sinusoidal outpu t

of the generator to give 1V rms as given on the m eter .

Take the cu rrent reading for this voltage .

Reset the voltage outpu t to 2V rm s and record th e resultant current .

Repeat this procedure for outputs of 3V , 4V and 5V .

Cop y the results table as show n in figure , reprodu ced at the end of this assignmen t ,

and tabu late your results .

W ork out the average imp edan ce for the LC circuit at this frequency .

PRA CTICAL C ONSIDERATIONS A ND APPLICATIONS Like CR c i rcui ts ,

LR circuits

may be used as filter networks .

Consider the circuit figure .

L-

Page

No. 23


26/38

With this circuit :

V2 -

i

/R2 X L

2

R

V2

i

/R

+ (27cfL

)

2

Vi

V2

/ R12

This gives a low - pass respon se sim ilar to that in f igure . The ind uctor has a low

impedance to low frequencies (if) , thus the circuit behaves as a shunt

resistan ce at i.f.

and at h igh frequencies (h.f ) the inductor has a high imp edance an d attenuates

h.f

signals .The converse of this is shown in the circuit of figure .

7

V

Here :

L

V2 -

V,

1R2 X

L

2

D L

V2 -

'1R2 + (COL )

2

V2

This is the equation for a high-pass filter .At low frequencies the inductor

shunts the

output to earth , whilst at high frequencies it has a high impedance and the circuit

appears just as a series resistor .The

band pass form of the LR

circuit is

shown in figure

VI

Page No. 24


27/38

R

. S

PRACTICAL ASSIGNMENT ON IMPEDANCE OF THE LCR CIRCUIT

OBJECTIVE

To investigate the impedance of a series LCR circuit and compare it with

the impedance's of its constituent components .


Let us investigate a series circuit containing

inductance , capacitance and resistance as in figure .

We know , from kirchhoff ' Voltage Law , that the phasor sum of

VR VL

and Vc w ill equal

V

n

. Connect up the circuit as shown in the patching diagram of figure corresponding to

the circuit diagram of figure .

11(0

I 6

-

epi41

H

F C,

Connect the voltmeter across the input to the circuit ( points marked

P

and S on figure)

and adjust the generator output to give 4V rams output at 500Hz .Record the resultant

current . Transfer your voltmeter to read the voltage across the resistor ( measure

across points P and Q) . Record this voltage . Measure and record the inductor voltage

( points Q and R) and the capacitor voltage (points R and S) .Copy the results table

as

shown in figure , reproduced at the end of this assignment , and tabulate your results .

Draw to scale , a phasor diagram showing Vi

n , VR , VL and Vc


The LCR network can be

used as a low-pass , a high-pass , or a band-pass network depending on which

component the output voltage is taken across .

Consider the circuit of figure .

Page No.

25


28/38

V

1

Here :

X c

. 1 R

+ ( X L

c )

2

1 I coL

V2

R

2

coL -- 1 / (.0C )2

This gives a low-pass response , but above the cut-off frequency both L and C influence

the rate of cut-off . At high frequencies the inductor has a high impedance and the

capacitor has a low impedance , thus the attenuation above f c is at a greater rate than

for the CR or LR filter .

For a high-pass filter the connection is as in figure .

c.

U

2-

This gives a high-pass filter response , as at high frequencies the capacitor is virtually a

short circuit , and the inductive reactance is very high . The circuit acts like a series

resistance at h.f .

The band-pass circuit form is shown in figure .

V

i

R .

Page No.26


29/38

ASSIGNMENT PARALLEL IMPEDANCES

OBJECTIVE

To investigate the impedance of parallel connected components to a

sinusoidal alternating current .


So far we have investigated series circuit of RC ,RL

and found formula for their impedance's , and how these impedance's vary with

frequency . Let us now look at parallel connected impedance's .For series connected

impedance's the reference that is common to all components is that of current , and the

phasor sum of the voltage drops around the loop is zero .For parallel connected

impedance's the common reference is voltage , and the phasor sum of the currents at

any node is zero . This is illustrated in figure .

The voltage v is applied to all of the impedance Z1 , Z2 and

Z3

in common , and

considering instantaneous currents ,

+i

2

+i3 i = o

Set up the circuit as shown in the patching diagram of figure corresponding to the

circuit diagram of figure .

I

Set the function generator to give an output voltage of 4V rms at 1600Hz , as shown on

the meter .

Disconnect link 1 and insert the 0 5mA millimeter .

Record the total current , I . Disconnect the meter 0 5mA meter and replace link 1.

Remove link 2 and connect the meter instead . Record the resistor current , IR .

In similar fashion , measure and record the inductor current , IL , and the capacitor

current ,Ic

Page No. 27


30/38

Copy the results table as shown in figure , reproduced at the end of this assignment ,

and tabulate your results .

j

qt

urrehl x

Draw a phasor diagram of the currents IR , IL and lc .Use V as your reference direction .

You can use either the completion of a parallelogram or drawing each phasor from the

tip of the previous one to find the resultant . Figure combines phasor both techniques in

order to show how the reactive current ; if several resistors were connected in parallel

their currents would be numerically added to find the total resistive current . The total

current is than

1 R2

x 2

Where IR is the total resistive current and lx is the total reactive current .

In the circuit of figure the total resistive current is simply IR as marked in the diagram ,

and is given by

IR = V / R

Where R is the resistance , V the voltage .

The reactive current lx is given by

Ix

V/XL+V/Xc=V/X

Where XL = coL is the reactance of the inductor ,

Xc = - 1 / coC is the reactance of the capacitor

And X is the combined reactance of the parallel combination .

Notice that this combined reactance of two parallel reactance's is found by a similar

formula to that used for parallel resistances ,i.e,

/X = 1 /X1 + 1 / X2

This dose not mean that resistances and reactance's can be mixed in the same formula

It dose make it easy to calculate reactance's however , and this is one of the reasons

why capacitive reactance is defined to have a negative value . If it were not so the

formula would be different , and one would have to know whether each of X

1

and X2

were inductive or capacitive


Circuits which include a

number of parallel branches are often more easily analysed using the admittance

formula rather than the impedance formulae . Calculations should be carried out in

terms of the admittance of the branches and then , if a final impedance value is

required, the reciprocal of the final admittance should be taken .When taking the

reciprocal of a phasor quantity it is necessary to find both the magnitude and the angle

Page No.28


31/38

of the reciprocal . The magnitude of Z

is simply

found as the reciprocal of the

magnitude of Y . The phase angle of Z is minus the phase angle of Y . Figure illustrates

this . V and I are the total circuit voltage and current respectively in the two diagrams

shown , but in one diagram V is the reference phasor , and in the other I is the reference

phase. Since the phase angle is reckoned from the reference to the other phasor the

signs must be different in the two cases .

Working Principle of a Transformer

A Transformer is a static device by means of which electrical power in one ckt. is

transformed into electrical power of the same frequency in another ckt. but with a

corresponding decrease or increase in current / Voltage. The physical basis of a

transformer is a mutual induction between two ckts. linked by a common magnetic flux.

It consists of two inductive coils

which are electrically separated but magnetically linked

through a path of low reluctance. One coil is connected to a source of alternating supply

Voltage, an alternating flux is set up in the laminated core, most of which is linked with

the other coil in which it

produces mutually induced e.m.f. ( Faraday's laws of

Electromagnetism induction e = M dl / dt ). The second coil is closed a current flows in it

and so electric energy is transferred from first coil to second coil. The first coil which fed

supply is called Primary winding and second coil which power drawn out is called

secondary winding.

In brief, Transformer is a device that

a) Transfers electrical power from one ckt to another ckt.

b) It works without any change of frequency.

c)

It accomplishes this by electromagnetic induction.

d)

Where the two electrical ckt are mutual induction influences of each other.

E.M.F. Equation of a Transformer:

Let, N1 = No. of turns in primary winding

N2 = No. of turns in secondary winding

(I) = Maxm. flux in core in webers = B * A

B = Maxm. flux density,

= Area of the core

f = Supply frequency in herz ( H)

Average rate of change of flux = Maxm. flux / 1/4f = 4f wb/s or Volt

Average e.m.f. per turn = 4IVolt

Page No.29


32/38

If flux varies sinusoidal, then r.m.s. value of induced e.m.f. is obtained by multiplying

the average value with

FORM FACTOR.

FORM FACTOR. = r.m.s. value / average value = 1.11

Therefore, the RMS value of the induced e.m.f. in the whole of primary winding equal to

induced e.m.f per turn into the number of primary turn.

Voltage Transformation Ratio

:

El = 4.44 f N Bm A

E2 = 4.44 f N Bm A

From equation No.(i) & (ii) we get,

K ( K is the constant )

'K' is the Voltage Transformation Ratio.

Condition :- a) If K > 1, i.e. N2 > N1, then transformer is called step-up

transformer.

b) If K


33/38

POWER

in

A

C 3 Phase System

Active Power in one Phase = Vp Ip Cosa) Watt

For all the three phase power ( algebraic sum ) = 3 Vp Ip Cosh Watt

Vp =Phase Voltage

L = Line Voltage

Ip = Phase Current

L = Line Current

Three Phase Power = VLIL Cos cD

Watt/ KW

For Star ( Y ) Connection : = Vp = VL / 43 and Ip = IL

All the Three Power =

3 Vp Ip Cos (13 Watt

Three Ph Power =

q3 VL IL Cos (1) Watt

For Delta ( D ) Connection : = Vp = VL and Ip =1L /

All the three Phase Power -4 3 Ph Power =

3 Vp Ip Coscb Watt

3 Ph Power =

q3 VL IL Cos (1) Watt

MEGGERING ( Practical Assignment )

A Megger is mainly used for testing Insulation resistance. Insulation resistance is

the resistance offered to the passage of current across the surface of the insulation. In

any electrical m/c I.R. can be measured between two conductors which are separated

by some insulation or it may be measured between a conducting part and the body /

earth of the machine. Normally all electrical conductors / live parts are insulated

with

rubber, PVC, enamel varnish etc. and the insulating material may deteriorate in course

of time. So checking insulation resistance at regular intervals becomes very important

from the view point of safety and reliability. Normally minimum value of insulation

resistance between two electrically separated conductors or between a conductor and

body / earth should be 1 MO it is less than 1 gp,the electrical equipment under test will

have to be sent for repair/ reconditioning.

Insulation resistance is measured with a megger or insu-tester or motor checker

etc. The unit of measurement is

Mega-ohm ( M41-.)

Objectives

The objectives of meggering are to :

Test & Measure the Insulation resistance between Stator winding & Body/Earth.

Test & Measure the insulation resistance between different windings.

Test & Measure of Stator & Rotor windings and Slip-Ring & Rotor.

Page No 3 1


34/38

Procedure :

A.

Insulation resistance test between winding body / Earth.

This test is carried out to determine how much insulation resistance is

offered betw een Stator or Rotor wind ing and earth.

Steps -1 Ensure m ain supply is disconnected to the m otor.

Steps -2 Before measu ring insulation resistance, ensure that the body of the m otor is

properly earthed. To do so, connect the L terminal of megg er to the body of the m otor

the E Term inal to earth and rotate the megger han dle. If the reading beco m es Zero,

the earthing is O.K.

Steps -3 Now take Stator terminals ( m ay be m arked A,B , C / Al ,B1,C1-A2, B2, C2 )

Steps-4 Rotate the megger handle at about 160 r.p.m. and note the reading. It

shou ld not be less than 2

m-n ( two Megaohm ).

Steps - 5 Next remo ve the L terminal of the megger and conn ect stator coil individual

( Al / A2 Bl/B2 , Bl/B2 Cl/C2, C1/C2 Al/A2 )

Steps-6 Rotate the megger handle at about 160 r.p.m. and note the reading. It

should be

2M-ohm ( Tw o Megaohm ) or more.

B.

Insulation resistance test between Stator windings and Rotor Ckt.

This test is done for measuring insulation resistance between armature and f ield wd gs.

Steps -1 Take a AC m otor which is not connected to main supply l ine.

Steps -2 Connect the megger L and E terminals to Stator terminal, say A / B /C

and Rotor terminal,

Steps-3 Rotate the megger handle at about 160 rpm. and note the reading. It

should be 2

M -Q ( Two M egaohm ) or more.

C.

Continuity test

This

test is carried out to determ ine the continuity of a particular winding.ln Insu-tester

or m otor checker, low V oltage (9V) range is used for continuity testing. To determ ine

any loose co ntact in the w inding, testing w ith a series test lamp is preferred,

Steps -1 Take a AC mo tor which is not con nected to m ain supply line.

Steps -2 Connect tw o terminals of the con tinuity tester to Stator coil term inals.

If a beep is heard or very low resistance is found then it is o.k.

Steps -3 Connect tw o term inals of the con tinuity tester to Rotor term inals.

If

a beep is

heard or v ery low resistanc e is found then it is o.k.

Page No. 32


35/38

F : -

olarization Index ( P. I.

I t is the ratio betw een the I .R. value ( m eggering va lue ) in 60 second s and I .R. value

( meggering value ) in 15 seconds. P.I. is a good appraisal of the condition of the

insulation, clean dry w dgs w ill have a m uch higher index than m oist or dirty ones. P.I. of

less than 1.3 ind icates the need for recond itioning. Index values of 4 or m ore indicate

the insulation to be in very good condition. An index of less than 1 indicates the

developm ent of carbon ized paths through an d around the insulation.

Megger Application Grade

(i)

For Low Voltage

50 Volt Insulation Tester ( Battery Operated )

(ii)

Low Voltage Upto 440 Volt : 500 Volt Insulation Tester ( Megger ) with hand

hold and battery operated.

(iii)

For 1.1 KV

1000 Volt hand hold Megger

(iv) For 3.3 KV

2.5 Killo Volt hand hold Megger

(v)

For 11 KV

5 Killo Volt hand hold Megger

Leakage Current must not exceed 1 / 1000 th (0.0001)mA of Full load Current.

Page No .33


36/38

SAFETY PRECAUTIONS FOR ELECTRICIANS

Introduction

Safety precautions for electricians are the guidance to mold the habit of

electricians to be more responsible to alertness and consciousness to the hazards in

their day to day work.

TYPES OF

SAFETY PRECAUTIONS FOR ELECTRICIAL HAZARDS

1

lectric shocks are easily received and easily be avoided , RISK is not apparent , should

be carefull .

Beware

of live conductors , either bars or insulated .

Never disconnect a plug point by pulling the flexible card .

Always connect live wire through a switch .

5.

Always switch OFF before replacing a blown out fuse.

6.

Never touch an overhead line unless you have made sure that the line is

electrically dead.

7.

Put on safety belt when working above ground on a pole .

8.

Never put a switch ON unless you are sure that all the men are not working in

the line .

9.

When mixing sulfuric acid in water , add acid to the water and not water to the

acid.

10.

When cells are charged in a room maintain good ventilation and never bring

maked light near accumulator (battery).

Use correct size and quality of fuse wire when rewiring blown out fuse wire.

If

a ladder is used it must be hold by another person so that it may slip away .

Do :lot use wire with poor insulation .

Always use portable hand lamp of insulated safety type and provided with a

rubber , plastic or wooden and wire guard .

15.

Safety demands good Earthing . Hence always keep earth connections in good

conditions .

16. Beware working on Motor or other rotating machinerise make that it can not be

set in motion without your permission .


37/38

e w a r e

s t a r t i n g

3 work always follow "F OU R" principles rs.f F IP,Trica Safety .

ISOLATE ( i.

e. SWITCH OFF , FUSE AND PUT

DANGER BOARD .

TEST ( i.e. CHEACK THAT THE CIRCUIT IS DEAD) .

LOCK ( i.e. SECURITY AGAINST UNAUTHORISED SWITCHING) .

SHORT ( i.e. EARTHING OR SHORTING AGAINST ACCIDENTAL CHARGE )

18

isconnect the supply immediately in case of fire rlar electrical apparatus .

19

o not use fire extinguisher on Electrical equipment unless it is clearly marked

an suitable for that purpose use sand or blanket instead .

20 Do not throw water on live Electrical equipment's in case of fire. It is dangerous for

you

21 Do not work on energized circuit without taking extra precautions , Such as the use

of rubber glovan and wooden handless .

22 Do not wear loose clothing , metal watch straps, bangles, finger rings while working

on electrical applicances

23 Before supplying current to electrical, equipmen t , it should be ascertained that the

equipment is properly earthed .

24 Before using portable Electrical things , see that these are well earthed .

25 Do not use a plier as a hammer .

26 Do not put a sharp edged tool in your pocket .

27 Do not use tools like file , knife ,screw driver etc. , without handle otherwise it can

injury the hand .

28

Preach and practice safety at all times. good work

e spoiled by an accident.

Page No.

35


38/38

YOURS SAFETY IS IN YOUR HAND

TREATMENT OF ELECTRIC SHOCK FOR ARTIFICAL RESPIRATION

This treatment should be con tinued for at least five hours if necessary .

A person apparently dead m ay be revived by the method (SCH AFER'S) describe below

Rem ove the body from contact with the wire , cable or other conductor .

By

Breaking or Disconn ecting the circuit ,

Dragging the patient away by his coat tails .

The hand being protected by Indian Rubber gloves or

ny dry w oolen m aterials,

such as a cap folded sev eral thickness if possible , wood or any m en cond ucting

m aterial may b e used , If possible without discontinuing the treatment , send for a

doctor .

AFTER REMOVAL

Do not w ait to und er the clothing .

Place the patient on his chest w ith head turned to one side .

Kneel at his side and grasping the lower ribs w ith bo th hand s ,

Gradually throw your w eight on to his body .

Spring quickly back and repeat the mo vem ent fifteen tim e a minute .

Do not leave the patient or stop artificial Resp iration , un til a doc tor arrives

Keep the patient warn .

***

Page No.

36

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