chapter 31 electromagnetic oscillations and alternating current

32
Chapter 31 Electromagnetic Oscillations and Alternating Current In this chapter we will cover the following topics: -Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors - Resonance in RCL circuits -Power in AC-circuits (31 - 1)

Upload: adem

Post on 12-Feb-2016

54 views

Category:

Documents


0 download

DESCRIPTION

Chapter 31 Electromagnetic Oscillations and Alternating Current In this chapter we will cover the following topics: - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Chapter  31 Electromagnetic Oscillations and Alternating Current

Chapter 31

Electromagnetic Oscillations and Alternating CurrentIn this chapter we will cover the following topics:-Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors -Resonance in RCL circuits -Power in AC-circuits -Transformers, AC power transmission

(31 - 1)

Page 2: Chapter  31 Electromagnetic Oscillations and Alternating Current

A horizontal power line carries a current of 5000 A from south to north. Earth's magnetic eld (60 μT) is directed towards north and inclined down-ward 50 degrees to the horizontal. Find the magnitude and direction of the magnetic force on 100 m of the line due to the Earth's field.(1)23 N, west (2) 23 N, east (3) 30N, west(4) 30N, east (5) none of these

50

BN

I

Page 3: Chapter  31 Electromagnetic Oscillations and Alternating Current

Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5Wb. If the page is turned by 30◦ around an edge the flux through it will be:

A. 2.5WbB. 4.3WbC. 5WbD. 5.8WbE. 10Wb

Page 4: Chapter  31 Electromagnetic Oscillations and Alternating Current

A car travels northward at 75 km/h along a straight road in a region where Earth’s magnetic field has a vertical component of 0.50 × 10−4 T. The emf induced between the left and rightside, separated by 1.7m, is:

A. 0B. 1.8mVC. 3.6mVD. 6.4mVE. 13mV

Page 5: Chapter  31 Electromagnetic Oscillations and Alternating Current

LC The circuit shown in the figure consists of a capacitor

and an inductor . We give the capacitor an initialchanrge and then abserve what happens. The capacitorwill discharge th

CL

Q

LC Oscillations

rough the inductor resulting in a timedependent current . i

We will show that the charge on the capacitor plates as well as the current 1 in the inductor oscillate with constant amplitude at an angular frequency

The total energy in the circuit is t

q i

LCU

2 2

he sum of the energy stored in the electric field

of the capacitor and the magnetic field of the inductor. . 2 2

The total energy of the circuit does not change with time. Thus

E Bq LiU U UC

dU

2

2

2

2

0

0. 1 0

dtdU q dq di dq di d qLi idt C d

d qLt dt dt

qdtdt dt C

(31 - 2)

Page 6: Chapter  31 Electromagnetic Oscillations and Alternating Current

LC

2

2

2

2

1 0 ( )

This is a homogeneous, second order, linear differential equationwhich we have encountered previously. We used it to d

1 0

escribethe simple harmonic oscillat

o

d qL qdt C

d q qdt LC

eqs.1

22

2 0

with sol

r (SHO)

( )

ution: ( ) cos( )

d x xdt

x t X t

eqs.2

If we compare eqs.1 with eqs.2 we find that the solution to the differentialequation that describes the LC-circuit (eqs.1) is:

1( ) cos where , and is the phase angle.

The current

q t Q tLC

sindqi Q tdt

( ) co sq t Q t

1 LC

(31 - 3)

Page 7: Chapter  31 Electromagnetic Oscillations and Alternating Current

LC

2 22

2 2 2 22 2

2

The energy stored in the electric field of the capacitor

cos2 2

The energy stored in the magnetic field of the inductor

sin sin2 2 2

The total energy

2

E

B

E B

q QU tC C

Li L Q QU t tC

U U U

QU

2

2 2cos sin2

The total energy is constant;

Qt tC C

energy is conserved

2

2

3The energy of the has a value of at 0, , , ,...2 2 2

3 5The energy of the has a value of at , , ,...2 4 4 4

When is maximum is zeE B

Q T Tt TCQ T T TtC

U U

electric field maximum

magnetic field maximum

Note : ro, and vice versa(31 - 4)

Page 8: Chapter  31 Electromagnetic Oscillations and Alternating Current

0t

1

2

/ 8t T

3

/ 4t T

4

3 / 8t T

5

5/ 2t T

432

1

6

6

5 / 8t T

3 / 4t T

7 / 8t T

7

8

7

8

(31 - 5)

Page 9: Chapter  31 Electromagnetic Oscillations and Alternating Current

2

2

If we add a resistor in an RL cicuit (see figure) we mustmodify the energy equation because now energy is

being dissipated on the resistor.

2E B

dU i Rdt

qU U UC

Damped oscillations in an RCL circuit

22

2Li dU q dq diLi i R

dt C dt dt

2

2

2

/ 2

2

2 2

1 0 This is the same equation as that

of the damped harmonics o 0 which hscillator:

The a

as the solution

( ) co ngul r fs a

:

bt mm

dq di d q d q dqi L R qdt dt dt dt d

d x dxm b kxdt dt

x t x e t

t C

2

2

2

2

/ 2 1 ( )

requency

For the damped RCL circuit the solut

cos

ion is:

The angular fre que4

ncy

4

Rt L Rq

k bm m

t Qe tLC L

(31 - 6)

Page 10: Chapter  31 Electromagnetic Oscillations and Alternating Current

/ 2Rt LQe

/ 2Rt LQe

( )q tQ

Q

( )q t / 2( ) cosRt Lq t Qe t

2

2

14R

LC L

/ 2

2

2

The equations above describe a harmonic oscillator with an exponetially decaying

amplitude . The angular frequency of the damped oscillator

1 is always smaller than the angular f4

Rt LQe

RLC L

2

2

1requency of the

1undamped oscillator. If the term we can use the approximation 4

LCRL LC

(31 - 7)

Page 11: Chapter  31 Electromagnetic Oscillations and Alternating Current

A battery for which the emf is constant generates a current that has a constant direction. This typeof current is known as " " or " "In chapter 30 we encountered a d

Alternating Current

dcdirect current ifferent type

of sourse (see figure) whose emf is:

sin sin where , is the area of the generatorwindings, is the number of the windings, is the angular frequency of therotation of the windings, and is the magnetic field.

m mNAB t t NAB AN

B

E E E

This type of generator is known as " " or " " because the emf as well as the currentchange direction with a frequency 2 . In the US 60 Hz. Almost all commercial electrical

f f acalternating current

power used today is ac even though the analysis of ac circuits is more complicated than that of dc circuits.The reasons why ac power was adapted will be discussed at the end of thischapter.

(31 - 8)

sin m tE E

Page 12: Chapter  31 Electromagnetic Oscillations and Alternating Current

LC

Our objective is to analyze the circuit shown in thefigure ( circuit). The discussion will be greatly simplified if we examine what happens if we connecteach of the three elem

RCL

Three Simple Circuits

ents ( , , and ) separatelyto an ac generator.

R C L

From now on we will use the standard notation for ac circuitanalysis. Lower case letters will be used to indicate the

values of ac quantities. Upper case letterswill be used

A convention

instantaneous

to indicate the constant amplitudes of ac quantities.Example: The capacitor charge in an LC circuit was written as:

cos

The symbol is used for the instantaneous value of the charge The symbol

q Q t

q

is used for the constant amplitude of Q q

(31 - 9)

Page 13: Chapter  31 Electromagnetic Oscillations and Alternating Current

In fig.a we show an ac generator connected to a resistor

From KLR we have: 0 sin

The current amplitude

The voltage across is equal to sinThe voltage

mR R

mR

R m

R

i R i tR R

IR

v R t

A resistive load

EEEE

E amplitude is equal to

The relation between the voltage and current amplitudes is: In fig.b we plot the resistor current and the resistor voltage as function of time t. Both quanti

m

R

R

R RV I Ri

v

E

ties reach their maximum values at the same time. We say that voltage andcurrent are .in phase

(31 - 10)

R RV I R

Page 14: Chapter  31 Electromagnetic Oscillations and Alternating Current

A convenient method for the representation of acquantities is that of phasors

The resistor voltage and the resistor current are representedby rotating vectors known as phasors using the following conventions:

Phasors rotate in the counterclockwise direction with angula

R Rv i

1. r speed The length of each phasor is proportional to tha ac quantity amplitude The projection of the phasor on the vertical axis gives the instantaneous

value of the ac quantity. The rotation

2.3.

4. angle for each phasor is equal to the phase of theac quantity ( in this example)t (31 - 11)

Page 15: Chapter  31 Electromagnetic Oscillations and Alternating Current

2 2

We define the "root mean square" (rms) value of as follows:

The equation looks the same

as in the DC case. This power appears a

s2

heat on

rmsmrms

V

VPR

R

V E

2

0

22

0

22

0

2

1 ( ) sin

1

1

si

1sin

n

22

Tm

Tm

mT

P P t dt P tT R

td

P tdtR

tT

T

PR

Average Power for R

E

E

E

2

2mPR

E 2

2

mrmsV

E

(31 - 12)

Page 16: Chapter  31 Electromagnetic Oscillations and Alternating Current

In fig.a we show an ac generator connected to a capacitor

From KLR we have: 0

sin

cos sin 90

The voltage amplitude equal to

The current am

C

C m

CC m

C m

CqC

q C C tdqi C tdt tdt

V

A capacitive load

EE E

EE

plitude 1/

The quantity is known as the

In fig.b we plot the capacitor current and the capacitorvoltage as function of time t

1

. The current

thev

/

CC

C

C

C

CVI CV

C

iv

X C

leads

capacitive reactance

oltage by a quarter of a period. The voltage andcurrent are .out of phase by 90

O

CX

(31 - 13)

1CX

C

Page 17: Chapter  31 Electromagnetic Oscillations and Alternating Current

2

2

2

0 0

sin cos

sin 22

1 1( ) = sin 2 02

A capacitor does not dissipate any power on the average. In some parts o

2sin cos sin 2

h

f t

mC C

C

m

C

T Tm

C

P V I t tX

P tX

P P t dt tdtT X T

Average Power for C

Note :

E

E

E

e cycle it absorbesenergy from the ac generator but at the rest of the cycleit gives the energy back so that on the average nopower is used!

0CP

(31 - 14)

Page 18: Chapter  31 Electromagnetic Oscillations and Alternating Current

In fig.a we show an ac generator connected to an inductor

From KLR we have: 0 sin

sin cos sin 90

The voltage amplitude equa

mL L

m m mL L

L

Ldi diL tdt dt L L

i di tdt tdt tL L L

V

An inductive load

EEEE E E

l to

The current amplitude

The quantity is known as the

In fig.b we plot the inductor current and the inductor voltage as function of time t. The current

m

LL

L

L

L

VIL

iv

X L

E

inductive reactance

the voltage by a quarter of a period. The voltage andcurrent are .

lags behind

out of phase by 90

O

LX

(31 - 15)

LX L

Page 19: Chapter  31 Electromagnetic Oscillations and Alternating Current

2

2

2

0 0

Power sin cos

sin 22

1 1( ) = sin 2 02

A inductor does not dissipate any power on

2sin cos sin 2

the average. In some p rt

a s

mL L

L

m

L

T Tm

L

P V I t tX

P tX

P P t dt tdtT X T

Note :

E

E

E

Average Power for L

of the cycle it absorbesenergy from the ac generator but at the rest of the cycleit gives the energy back so that on the average nopower is used!

0LP

(31 - 16)

Page 20: Chapter  31 Electromagnetic Oscillations and Alternating Current

Circuit element

Average Power

Reactance Phase of current Voltage amplitude

ResistorR

Current is in phase with the voltage

CapacitorC

Current leads voltage by a quarter of a period

Inductor L

Current lags behind voltage by a quarter of a period

1CX

C

LX L

R RV I R

CC C C

IV I XC

L L L LV I X I L

R

SUMMARY

2

2m

RPR

E

0CP

0LP

(31 - 17)

Page 21: Chapter  31 Electromagnetic Oscillations and Alternating Current

An ac generator with emf is connected toan in series combination of a resistor , a capacitor and an inductor , as shown in the figure. The phasorfor the ac genera

sinm

R CL

tThe series RCL circuit

E E

tor is given in fig.c. The current in

this circuit is described by the equation: sini I t sini I t

The current is for the resistor, the capacitor and the inductorThe phasor for the current is shown in fig.a. In fig.c we show the phasors for thevoltage across , the voltage across R C

i

v R v C

common

, and the voltage across .The voltage is in phase with the current . The voltage lags behindthe current by 90 . The voltage leads ahead of the current by 90 .

L

R C

L

v Lv i v

i v i

(31 - 18)

Page 22: Chapter  31 Electromagnetic Oscillations and Alternating Current

O

A B

Kirchhoff's loop rule (KLR) for the RCL circuit: . This equationis represented in phasor form in fig.d. Because and have opposite directionswe combine the two in a single phasor

R C L

L C

L

v v vV V

V

E

2 2 2 22 2 2 2

22

22

. From triangle OAB we have:

The denominator is known as the " "

of the circuit. The current amplitude

C

m R L C L C L C

m

L C

mL C

V

V V V IR IX IX I R X X

I ZR X X

Z R X X I

EE

E

impedance

22

1m

Z

I

R LC

E

22 L CZ R X X

mIZ

E

sini I t

(31 - 19)

Page 23: Chapter  31 Electromagnetic Oscillations and Alternating Current

O

A B

From triangle OAB we have: tan

We distinguish the following three cases depending on the relative valuesof and .

0 The current phasor lags behind the generat

L C L C L C

R

L L

L C

V V IX IX X XV IR R

X XX X

1. or phasor.The circuit is more inductive than capacitive

0 The current phasor leads ahead of the generator phasorThe circuit is more capacitive than inductive

0 The current phaso

C L

C L

X X

X X

2.

3. r and the generator phasor are in phase

sini I t

22 L CZ R X X

LX L

tan L CX XR

1 CXC

(31 - 20)

Page 24: Chapter  31 Electromagnetic Oscillations and Alternating Current

Fig.a and b: 0 The current phasor lags behind the generator phasor. The circuit is moreinductive than capacitive

L CX X 1.

Fig.c and d: 0 The current phasor leads ahead of the generatorphasor. The circuit is more capacitive than inductive

Fig.e and f: 0 The current phasor and the generator ph

C L

C L

X X

X X

2.

3. asor are in phase (31 - 21)

Page 25: Chapter  31 Electromagnetic Oscillations and Alternating Current

In the RCL circuit shown in the figure assume thatthe angular frequency of the ac generator can be varied continuously. The current amplitude in the circuit is given by the equation:

mI

Resonance

E2

2

The current amplitude1

1has a maximum when the term 0

1This occurs when

R LC

LC

LC

The equation above is the condition for resonance. When its is satisfied

A plot of the current amplitude as function of is shown in the lower figure.This plot is known as a "

mresI

RI

E

resonance c "urve

mresI

RE1

LC

(31 - 22)

Page 26: Chapter  31 Electromagnetic Oscillations and Alternating Current

2

We already have seen that the average power used bya capacitor and an inductor is equal to zero. The power on the average is consumed by the resistor.

The instantaneous power P i

Power in an RCL ciruit

2

0

222 2

0

sin

1The average power

1 sin2

cos

The term cos in the equation above is known asthe "

T

avg

T

avg rms

rmsavg rms rms rms rms rms rms rms

R I t R

P PdtT

I RP I R t dt I RT

RP I RI I R I IZ Z

E E E

power fac " of the circuit. The averagepower consumed by the circuit is maximum when 0

tor

2avg rmsP I R cosavg rms rmsP I E (31 - 23)

Page 27: Chapter  31 Electromagnetic Oscillations and Alternating Current

Power Station

Transmission lines Erms =735 kV , I rms = 500 A home

110 V

T1T2

Step-up transformer

Step-down transformer

R = 220Ω1000 km

2

The resistance of the power line . is fixed (220 in our example)

Heating of power lines This parameter is also fixed ( 55 MW in our exa

heat rms

R RA

P I R

Energy Transmission Requirements

mple)Power transmitted (368 MW in our example)In our example is almost 15 % of and is acceptableTo keep we must keep as low as possible. The only wa

trans rms rms

heat trans

heat rms

P IP P

P I

E

y to accomplish thisis by . In our example 735 kV. To do that we need a device that can change the amplitude of any ac voltage (either increase or decrease)

rms rms increasing E E

(31 - 24)

Page 28: Chapter  31 Electromagnetic Oscillations and Alternating Current

The transformer is a device that can change the voltage amplitude of any ac signal. It consists of two coils with different number of turns wound around a common iron core.

The transformer

The coil on which we apply the voltage to be changed is called the " " andit has turns. The transformer output appears on the second coils which is knownas the "secondary" and has turns

P

S

NN

primary

. The role of the iron core is to insure that the magnetic field lines from one coil also pass through the second. We assume that if voltage equal to is applied across the primary then a voltagPV e appears on the secondary coil. We also assume that the magnetic field through both coilsis equal to and that the iron core has cross sectional area A. The magnetic flux

through the primary

S

P

V

B

( )

The flux through the secondary ( )

PP P P

SS S S S

d dBN BA V N Adt dt

d dBN BA V N Adt dt

eqs.1

eqs.2

(31 - 25)

Page 29: Chapter  31 Electromagnetic Oscillations and Alternating Current

( )

( )

If we divide equation 2 by equation 1 we get:

P

S P

P P P P

SS S S S

SS S

P P S PP

d dBN BA V N Adt dt

d dBN BA V N Adt dt

dBN AV NdtdBV NN A

VN

dt

VN

eqs.1

eqs.2

The voltage on the secondary

If 1 We have what is known a " " transformer

If 1 We have what is known a " " transformer

Both types of tran

SS P

P

SS P S P

P

SS P S P

P

NV VN

NN N V VNNN N V VN

step up

step down

sformers are used in the transport of electric power over large distances.

S P

S P

V VN N

(31 - 26)

Page 30: Chapter  31 Electromagnetic Oscillations and Alternating Current

PI SI

If we close switch S in the figure we have in addition to the primary current a current in the secondary coil. We assume that the transformer is " " i.e. it suffers no losses due to heating

P

S

II ideal

then we have: (eqs.2) If we divide eqs.2 with eqs.1 we get:

In a step-up transformer ( ) we have that In a step-down transformer ( )

P P S

P P S S

S SP P

P S S P

PS P

S

S P S P

S P

S

V I V IV IV I

V N V NNI IN

N

I N I N

N I IN N

we have that S PI I

We have that:

(eqs.1)

S P

S P

S P P S

V VN N

V N V N

S P

S P

V VN N

S S P PI N I N

(31 - 27)

Page 31: Chapter  31 Electromagnetic Oscillations and Alternating Current

Hitt

A generator supplies 100 V to the primary coil of a transformer. The primary has 50 turns and the secondary has 500 turns. The secondary voltage is:A. 1000 VB. 500 VC. 250 VD. 100 VE. 10V

Page 32: Chapter  31 Electromagnetic Oscillations and Alternating Current

hitt

The main reason that alternating current replaced direct current for general use is:A. ac generators do not need slip ringsB. ac voltages may be conveniently transformedC. electric clocks do not work on dcD. a given ac current does not heat a power line as much as the same dc currentE. ac minimizes magnetic effects