Topic 2a ac_circuits_analysis

1

Topic 2a :AC Circuits Analysis

• Capacitors network

• Inductors network

• AC analysis – average power, rms power

• Frequency response of RC and RL circuits.

2

Capacitors

• A capacitor is a passive element designed to store energy in its electric field.

• A capacitor consists of two conducting plates separated by an insulator (or dielectric).

• When a voltage source v is connected to the capacitor, the source deposits a charge q on one plate and a negative charge –q on the other.

• q = Cv, where C is the capacitance of the capacitor.

3

Capacitance

• Capacitance is the ratio of the charge on one plate of a capacitor to the voltage difference between the two plates, measured in Farad (F).

• For a parallel-plate capacitor. The capacitance is given by

where A is the surface area of each plate, d is the distance between the two plates, and is the permitivity of the dielectric material.

d

AC

4

Symbols for capacitors

Variable capacitor

Fixed capacitor

Fixed capacitor

5

Current- voltage relationship of the capacitor

• Since different q = Cv gives

• Voltage current relationship

where v(to) is the voltage across the capacitor at time to.

,dt

dqi .

dt

dvCi

)(1

or 1

o

t

t

t

tvidtC

vidtC

vo

6

Power delivered to capacitor

• The instantaneous power delivered to a capacitor

is

• Energy stored in the capacitor is

• v(-) = 0 then

.dt

dvCvvip

t

tCvvdvCdt

dt

dvvCpdtw

tt2

2

1

.2

or 2

1 22

C

qwCvw

7

Important properties of capacitor1. When the voltage across the capacitor is not

changing with time, the current through the capacitor is zero.

A capacitor is an open circuit to dc.2. The voltage on the capacitor must be

continuous. The voltage on a capacitor cannot change

abruptly (suddenly).2. The ideal capacitor does not dissipate energy.3. A real capacitor has a parallel leakage resistance.

The leakage resistance may be as high as 100 M.

8

Example

(a) Calculate the charge stored on a 3-pF capacitor with 20V across it.

(b) Find the energy stored in the capacitor.

Solution(a)

(b) Energy stored

.6020103 12 pCCvq

.600201032

1

2

1 2122 pJCvw

9

Example The voltage across a 5-F capacitor is

Calculate the current through it.

Solution The current is

V. 6000cos10)( ttv

A. 6000sin3.0

6000sin106000105

6000cos10105)(

6

6

t

t

dt

t)d(

dt

dvCti

10

Example If a 10-F capacitor is connected to a voltage

source with

determine the current through the capacitor.

Solution The current is

V. 2000sin50)( ttv

A. 2000cos

2000cos5020001010

2000sin501010)(

6

6

t

t

dt

t)d(

dt

dvCti

11

ExampleDetermine the voltage across a 2-F capacitor if the

current through it is

Assume that the initial capacitor voltage is zero.

Solution

mA. 6)( 3000teti

V. )1(

)1(0)3000(102

106

106102

11

3000

300030006

3

0

300036

0

t

tt

tt

t

e

et

e

dteidtC

v

12

Example

Obtain the energy in each capacitor under dc conditions

13

Solution• Under dc conditions, replaced each capacitor with

an open circuit.

• Current through 2-k and 4k resistor is

mA

mAi

2

)6(423

3

14

Solution

• Voltage v1 across capacitor 2 mF is

• Voltage v2 across capacitor 4 mF is

• Energy stored in capacitor 2 mF is

• Energy stored in capacitor 4 mF is

mJvCw 1041022

1

2

1 232111

mJvCw 12881042

1

2

1 232222

V.420001 iv

V.840002 iv

15

Series and parallel capacitors

• A equivalent capacitance of N parallel-connected capacitors is the sum of the individual capacitors.

Ceq = C1 + C2 + C3 + … + Cn.

• The equivalent capacitance of series-connected capacitors is the reciprocal of the sum of the reciprocal of the individual capacitors.

• For n = 2,

21

21

21

111

CC

CCC

CCC

eq

eq

16

Example• Find the equivalent capacitance seen

between terminal a and b of the circuit.

Ceq

a

b

20F

5F

6F20F

60F

17

Solution• 20F and 5F are in series, their equivalent

capacitance is

• This 4F is parallel with the 6F and 20F capacitors, their combined capacitance is

4+6+20 = 30F.• This 30 F is in series with 60 F capacitor. Hence

the equivalent capacitance is

.4520

520F

.203060

3060FCeq

18

Inductors

• An inductor is a passive element designed to store energy in its magnetic field.

• An inductor consists of a coil of conducting wire.• If current is allowed to pass through an inductor, it

is found that the voltage across the inductor is directly proportional to the time rate of change of the current.

where L is the inductance. dt

diLv

19

Inductance• The unit of inductance is the Henry (H).• Inductance is the property whereby an inductor

exhibits opposition to the change of current flowing through it measured in Henry (H).

• The inductance of an inductor depends on its physical dimension and construction.

• For solenoid where N is the number of turns, l is the length, A is the cross-

sectional area, and is the permeability of the core.

l

ANL

2

20

Circuit symbol for inductors

(a) air-core (b) Iron-core (c) Variable

21

Current-voltage relationship for inductor

• The current-voltage relationship is obtained from

where i(to) is the current at time to.

t

t

o

t

o

tidttvL

i

dttvL

i

vdtL

di

)()(1

)(1

1

22

Power delivered to the inductor

• Power delivered is

• Energy stored is

idt

diLvip

2

2

1LiidiL

idtdt

diLpdtw

i

tt

23

Properties of inductor

1. Voltage across an inductor is zero when the current is constant.

An inductor acts like a short circuit to dc.2. Its opposition to the change in current flowing

through it. The current through an inductor cannot change

instantaneously.3. Ideal inductor does not dissipate energy.4. Real inductor has winding resistance and

capacitance.

24

ExampleThe current through a 0.1-H inductor is

Find the voltage across the inductor and the energy stored in it.

SolutionVoltage

Energy stored

A. 10)( 5tteti

V. )51(

)5(10

1.0

5

555

t

ttt

et

etedt

tedv

J. 5t) 10)(1.0(2

1 10225 tt etew

25

Example Consider the circuit as shown, under dc

condition, find (a) I, vc and iL,, (b) the energy stored in the capacitor and inductor.

26

Solution

(a) Under dc condition, replace the capacitor with an open circuit and inductor with a short circuit.

i = iL = 12/(1+5) = 2 A.

vC = 5 i = 5x2 = 10V(a) Energy in capacitor is Energy in inductor is

J.50)10)(1(2

1 2 Cw

J.4)2)(2(2

1 2 Lw

27

Series Inductors

• Consider a series of N inductors, the equivalent inductor

• The equivalent inductance of series-connected inductors is the sum of the individual inductors.

Neq LLLLL ...321

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Parallel Inductors

• Consider a parallel–connection of N inductors, the equivalent inductor

• The equivalent inductance of parallel inductors is the reciprocal of the sum of the reciprocal of the individual inductors.

Neq LLLLL

1...

1111

321

29

Two inductors in parallel

• For two inductors in parallel combination,

Or • The combination is the same way as resistors in

parallel.

21

111

LLLeq

.21

21

LL

LLLeq

30

Example

Find the equivalent inductance of the circuit.

31

Solution

• The 10-H, 12-H and 20-H inductors are in series, combining them gives 42-H inductance.

• This 42-H inductor is in parallel to 7-H inductor so that they are combined, to give

• This 6-H inductor is in series with 4-H and 8-H inductors. Hence Leq = 4+6+8 = 18H.

.6427

427H

32

ExampleFor the circiut shown,

If i2(0) = -1 mA,

Find: (a) i1(o), (b) v(t), v1(t) and v2(t); (c) i1(t) and i2(t).

.)2(4)( 10 mAeti t

33

Solution

(a)

(b) Equivalent inductance is

Leq = 2 + 4║12 = 2+3 = 5H

.5)1(4)0()0()0(

.4)12(4)0(

21

2121

mAiii

iiiiii

mAmAi

mV. 200)10)(1(45

))2(4(5)(

1010

10

tt

t

eq

ee

dt

ed

dt

diLtv

34

Solution

Since v = v1 +v2.

v2.= v - v1 = 120e-10t mV.

(c) Current i1 is obtained as

similarly

mV. 80)10)(1)(4(22)( 10101

tt eedt

ditv

.3853350

-3e

mA 54

120)0(

4

1)(

101010t-

0 0

10121

mAeemAt

dteidtvti

tt

t tt

.1110

-e

mA 112

120)0(

12

1)(

101010t-

0 0

10222

mAeemAt

dteidtvti

tt

t tt

35

***AC current

36

)2cos(2

1)cos(

2

1)( ivmmivmm tIVIVtp

• Instantaneous power changes with time and is therefore difficult to measure. The average power is more convenient to measure.

• Wattmeter is measuring average power.

Average power (watts) – average power of the instantaneous power over the period.

The average power is given as:

T

avg dttpT

P0

)(1

Average power

37

Average Power

)cos(2

1ivmmavg IVP

)2cos(2

1)cos(

2

1)( ivmmivmm tIVIVtp

T

avg dttpT

P0

)(1

From the,

It can be derived that the average power:

(constant) (sinusoid)

Since, the average of constant is constant and the average of sinusoid = 0. So,

dttIVT

dtIVT

P ivmm

T

ivmm

T

avg )2cos(2

11)cos(

2

1100

38

Average power of purely resistive load, R

090cos2

1 o

mmavg IVP

For purely resistive load, v = i, so that:

Average power of reactive circuit: C, L

# The resistive load absorb power all the time.

For purely reactive elements, v - i = 90o, so that:

RRIIVP mmmavg22 |I|

2

1

2

1

2

1

# The purely reactive load (L or C) absorbs zero power.

39

• Effective value determine the effectiveness of voltage or current source in delivering power to the load.

• For any periodic function x(t) in general, the rms value is given by

• For a sinusoid i(t) = Im cos t Similarly for v(t) = Vm cos t

T

rms dtxT

X0

21

2cos

10

22 mT

mrms

IdttI

TI

2cos

10

22 mT

mrms

VdttV

TV

)cos()cos(2

1ivrmsrmsivmm IVIVP

R

VRIP rms

rms

22

The average power can be written in terms of rms values.

The average power absorbed by a resistor, R is,

Effective or RMS Power

40

The voltage V produced between the terminals of an ac generator fluctuates

sinusoidally in time. Why?

41

What’s the average voltage in an AC circuit? Average current?

42

Does zero average current and voltage imply zero average power?

# NO, it depends on your load is R, C or L.

43

What power is dissipated in a resistor R connected across an AC voltage source?

V = V 0 sin 2 f t I = (V 0/R) sin 2 f t = I 0 sin 2 f t

P = I0*V 0 sin2 2 f t Pave = I0*V 0/2

44

***In this three-phase circuit there is 266 V rms between line 1 and ground. What is the rms voltage between line 2

and 3?

VVVV

tV

ttVVV

VVV

rms

rms

4603

sin22

1)(

)2(cos)(sin2

)]3/2sin()3/4[sin(

2662/

023

2

1

3

2

2

10

023

0

45

AC vs DC: The battle of the currents

46

Example

A stereo receiver applies a peak ac voltage of 34 V to a speaker. The speaker behaves approximately as if it has a resistance of 8.0 , as shown. Determine (a) the rms voltage, (b) the rms current, and (c) the average power for this circuit.

47

Example

48

Example

49

In a resistive load, the current and voltage are in phase

RC and RL circuit

50

In a purely capacitive load, the current leads the voltage by 90º.

51

In a purely capacitive load, the current leads the voltage by 90º.

)(1

)/1(

2cossin

cos

cos

0

maxmax

maxmax

max

max

reactancecapacitiveC

X

CI

tCtCdt

dQI

tC

Q

tV

V

C

C

C

CCC XIV

52

What is the average power dissipated in the capacitor?

On the average, the power is zero and a capacitor uses no energy in an ac circuit.

V=V0 sin (2ft)

I=I0 sin (2ft + /2) = I0 cos (2ft)

Pave = (I·V)ave = 0

Average power, P = 0 But, Reactive power, Q ≠ 0

cc X

VXIQ

22

53

In a purely inductive circuit, the current lags the voltage by 90º

)(

2cos

2cos

)0(sincos

cos

cos

0

maxmax

maxmax

maxmax

max

max

reactanceinductiveLXL

I

tItL

I

CCtL

dttL

I

tdt

dIL

tV

V

L

L

L

54

In a purely inductive circuit, the current lags the voltage by 90º

cc X

VXIQ

22

For purely inductive load, again

Average power, P = 0 But, Reactive power, Q ≠ 0

55

RC circuit diagram

56

RL circuit diagram

57

RLC circuit diagram

58

Phasors for a series RLC circuit

Max or rms: V02 = VR

2 + (VL - VC)2

Vrms = IrmsZ

VR = IrmsR, VC = IrmsXC, and VL = IrmsXL

59

Power factor for a series RLC circuit

60

Summary of the AC phasor relations

61

Summary of impedance relations

= 1/ωC

= ωL