1

Topic 2b

AC Circuits Analysis

2

Example

An ac generator with rms voltage of 110 V is connected in series with a 35- resistor and 1-F capacitor. At what frequency must the generator operate if it is to maintain a current of 1.2 Amps.

3

Example

Z = V/I = 110/1.2 = 91.7 ΩZ2 = R2 + (1/ ωC)2; f = ω/2π = 1.9 kHz

4

Example

An ac generator emf is ε = ε0∙sinωt, with ε0∙ = 22.8 V and ω = 353 rad/s. It is connected to a 17.3 H inductor. When the current is a maximum, what is the emf of the generator? When the emf of the generator is -11.4 V and increasing in magnitude, what is the current?

5

Example

mAXVI L 73.3)3.17353/(8.22/maxmax

)30180cos(

3sin

3018030360

2/18.22/4.11sin

max

II

quadrantrdmagnitudegincrea

or

t

6

Example

An ac generator emf is ε = ε0∙sin ωt, with ε0∙ = 28.6 V and ω = 333 rad/s. It is connected to a 4.15 µF capacitor. When the current is a maximum, what is the emf of the generator? When the emf of the generator is -14.3 V and increasing in magnitude, what is the current?

7

Example

mAXVI C 5.39)1015.4333/1/(6.28/ 6maxmax

)30180cos(

3sin

3018030360

2/16.28/3.14sin

max

II

quadrantrdmagnitudegincrea

or

t

8

Example

A 0.4-H inductor and a 220-Ω resistor are connected in series to an ac generator with an rms voltage of 30 V and a frequency of 60 Hz. Find the rms values of the current, voltage across the resistor, and voltage across the inductor.

9

Example

Irms = Vrms/Z = 0.112 A

(VL)rms = IrmsXL= 17 V

(VR)rms = IrmsR = 24.6 V

10

Example An ac generator has a frequency of 1200

Hz and a constant rms voltage. When a 470- resistor is connected between the terminals of the generator, an average power of 0.25 W is dissipated in the resistor. Then, a 0.08-H inductor is connected in series with the resistor, and the combination is connected between the generator terminals. What is the average power dissipated in the series combination?

11

Example

12

Example A person is working near the secondary

of a transformer, as shown. The primary voltage is 120 V at 60.0 Hz. The capacitance Cs, which is the stray capacitance between the hand and the secondary winding, is 20.0 pF. Assuming the person has a body resistance to ground Rb = 50.0 kΩ, determine the rms voltage across the body.

13

Example

14

Example A typical "light dimmer" used to dim the

stage lights in a theater consists of a variable inductor L (whose inductance is adjustable between zero and Lmax) connected in series with a lightbulb. The electrical supply is 110 V (rms) at 62.5 Hz; the lightbulb is rated as "110 V, 900 W.“ What Lmax is required if the rate of energy dissipation in the lightbulb is to be varied by a factor of 5 from its upper limit of 900 W? Assume that the resistance of the lightbulb is independent of its temperature. Could a variable resistor be used instead?

15

Example

mHR

L

R

LR

LR

LR

ZV

ZV

RI

RI

P

P

rms

rms

rms

rms

3.71)60(2

900/)110(22

)(

)(

)(5

5/

/

2

max

2

2max

2

2min

2

2max

2

2

max2

2

min2

2min,

2max,

min

max

16

Modern dimmer

17

Phasors for a series RLC circuit

Max or rms: V02 = VR

2 + (VL - VC)2

Vrms = IrmsZ

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

18

Power factor for a series RLC circuit

19

Example

A series RLC circuit contains a 148-Ω resistor, a 1.50-μF capacitor, and a 35.7-mH inductor. The generator has a frequency of 512 Hz and an rms voltage of 35 V. What is (a) the rms voltage across each circuit element and (b) the average electric power consumed by the circuit.

20

Example

21

rms current in RL and RC circuits depends on frequency

22

When the phase difference between current and voltage is zero (Xc=XL), the circuit is in resonance

23

How does the variable air capacitor work?

24

Phasors

• Sinusoids are easily expressed in terms of phasors, which are more convenient to work with than sine and cosine functions.

• A phasors is a complex number that represents the amplitude and phase of a sinusoid.

25

Complex number

• A complex number z can be written in rectangular form as z = x + jy where : x is the real part and y is the imaginary part of z.

• The complex number z can also be written in polar form or exponential form as

where r is the magnitude of z and is the the phasor of z.

1j

jrerz

jrerjyxz

26

Complex number

• Rectangular form• Polar form• Exponential form• Given x, y can get r and

• Know r and , can obtain x and y

jyxz rzjrez

.tan , 122

x

yyxr

.sin ,cos ryrx

27

Important operation

• Addition:• Subtraction:• Multiplication:

• Division:• Reciprocal:

• Square root:

)()( 212121 yyjxxzz )()( 212121 yyjxxzz

)( 212121 rrzz

)( 212

1

2

1 r

r

z

z

rz

11

2/ rz

28

Complex conjugate

• Complex conjugate of z is

Real part of e j.

Imaginary part of ej.

jrerjyxz

jj

1

j

j

j

e

e

je

Imsin

Recos

sincos

29

Real and imaginary part

• can be express as

• V is the phasor representation of the sinusoid v(t).

)cos()( tVtv m

mj

mtj

tjjm

tjmm

VeVetv

eeVtv

eVtVtv

VV whereRe)(

Re)(

Re)cos()( )(

30

Sinusoidal-phasor transformation

Time –domian representation

Phasordomian representation

)cos( tVm

)cos( tIm

)sin( tVm

)sin( tIm

mV

)90( omV

mI

)90( omI

tion)representadomain -(Phasor tion)representadomain -Time(

cos)( mm VtVtv V

31

Derivative and Integral

• The derivative of v(t) is transformed to the phasor as jV.

• The integral of v(t) is transformed to the phasor as V/ j.

domain)(Phasor domian) (Time

j Vdt

dv

domain)(Phasor domian) (Time

j

V

vdt

32

Differences between v(t) and V

1. v(t) is the instantaneous or time domain representation, while V is the frequency or phasor-domain representation.

2. v(t) is time dependent, while V is not

3. v(t) is always real with no complex term, while V is generally complex.

33

Example Transform these sinusoids to phasors

(a)

(b)

Solution(a)

(b)

oti 4050cos6 otv 5030sin4

oI 406

o

o

oo

o

t

t

tv

1404

14030cos4

905030cos4

5030sin4

V

sin (t±180o) = sin t

Cos (t±180o) = cos t

sin (t±90o) = ±cos t

cos (t±90o) = sin t

34

ExampleFind the sinusoids represented by these phasors:

(a) I = -3 +j4

(b)

Solution(a) I = -3 +j4 = 5(126.87o)

(b)

ojej 208 V

otti 87.126cos5)(

o

o

ooj

ttv

ejo

70cos8)(

708

)20(89018 20

V

oj 901

35

Phasor relationships for circuit elements

• For resistor R , V = RI I = V/R

• For inductor L, V = jLI

• For capacitor C, I = jCVCj

IV

LjV

I

36

Example

The voltage is applied to a 0.1-H inductor. Find the steady-state current through the inductor.

Solution = 60, V = 1245o

otv 4560cos12

oo

oo

jLj452

906

4512

)1.0)(60(

4512

V

I

A 4560cos2)( otti

37

Impedance and Admittance of passive elements

• Impedance Z of a circuit is the ratio of the phasor voltage V to the phase current I, measured in ohms.

• The admittace Y is the reciprocal of impedance, measured in siemens.

Element Impedance Admittance

R Z = R

L Z = jL

CCj

1Z

R

1Y

CjY

Lj1

Y

38

Impedance Z

• The complex quantity Z may be represented by rectangular form as Z = R +jX.

where R = ReZ is the resistance and

X = Im Z is the reactance.

sinX ,cos

tan , 1-22

ZZ

Z

ZZ

RR

XXR

jXR

39

Example Find v(t) and i(t) in the circuit shown.

+_

i 5

0.1F v+

_vs=10cos4t

40

Solution = 4. Vs = 100o.

• Voltage across the capacitor

oo

oos

j

jj

57.26789.157.2659.5

010

5.25

010

5.25)1.0)(4(

15

Z

VI

Z

)43.634cos(47.4)(

)57.264cos(769.1)(

)43.63(47.4

904.0

57.26789.1

)1.0)(4(

57.26789.1

o

o

o

o

oo

c

ttv

tti

jCj

I

IZV

41

Impedance combination• Consider N series-connected impedance,

V = V1 + V2 +… +VN = I(Z1+Z2+…+ZN)

• Consider N parallel-connected impedance,

N21eq Z...ZZI

VZ

N21eq

N21eq

N21N21

Y...YYY

Z

1...

Z

1

Z

1

V

I

Z

1

Z

1...

Z

1

Z

1VI...III

42

Example

Find the input impedance in the circuit shown. Assume that the circuit operates at = 50 rad/s.

Zin

2mF0.2H

83

10mF

43

Solution• Impedance of 2mF capacitor

• Impedance of the 3- resistor in series with 10 mF capacitor

• Impedance of the 8- resistor in series with 0.2 H inductor

10)002.0)(50(

111 j

jCjZ

)23()01.0)(50(

13

132 j

jCjZ

)108()2.0)(50(882 jjLjZ

44

Solution

• The input impedance is

07.1122.3

07.122.310811

)811)(1444(10

811

)108)(23(10

)108()23(

)108)(23(10

22

j

jj

jjj

j

jjj

jj

jjj

321in ZZZZ

45

Example

• Determine vo(t) in the circuit.

+_+

_vo5H10mF

60

20sin(4t-15o)

46

Solution

• Vs = 20(-105o)

• Impedance of 10 mF capacitor

• Impedance of 5-H inductor ZL= j(4)(5)=j20• Impedance of the parallel combination of 10-mF

capacitor and 5-H inductor

25)01.0)(4(

1j

jcZ

V )04.744cos(15.17)(

04.7415.17)105(20)(96.308575.0(

)105(20(10060

100

60

1005

500

)20()25(

)20)(25(2025

oo

ooo

oso

ttv

j

j

jjjj

jjjj

VZ

ZV

Z

47

Effective or RMS Value

• The idea of effective value arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load.

• The effective value of a period current is the dc current that deliver the same average power to a resistor as the periodic current.

48

Root Mean Square Value

• Effective value for current

• Effective value for voltage

• The effective value of a periodic signal is them root mean square (rms) value

T

eff dtiT

i0

21

T

eff dtvT

v0

21

T

rms dtxT

X0

21

49

RMS value for Sinusoidal function

• The rms value of a constant is the constant itself.

• For sinusoidal i(t) = Im cos t, the effective value is

• For v(t) = Vm cos t, the effective value is

.2

)2cos1(2

1cos

1 222 m

T

o

mT

o

mrms

Idtt

T

ItdtI

TI

.2m

rms

VV

50

Average power

• The average power absorbed by a resistor R can be written as

• For instance, the 230 V available at every household is the rms value.

• It is convenient in power analysis to express voltage and current in their rms value.

R

VRIP rms

rms

22

51

Example Determine the rms value of the current waveform

as shown. If the current is passed through a 2- resistor, find the average power absorbed by the resistor.

0

10

-10

i(t)

t2 4 6 8 10

52

Solution• The period of the waveform T = 4.• Can write the current waveform as

• The power absorbed by a 2- resistor is

Att

dtdttdtiT

I

t

ttti

T

rms

165.82003

200

4

1

2

4100

0

2

325

4

1

)10()5(4

11

42 ,10

20 ,5)(

3

4

2

22

0

2

0

2

.3.133)2()165.8( 22 WRIP rms

53

Example The waveform shown is a half-wave

rectified sine wave. Find the rms value and the amount of average power dissipated in a 10- resistor.

10

0

v(t)

t 2 3

54

Solution• The period of the voltage is T = 2, and

WR

vP

v

tt

dtttdtv

t

tttv

rms

rms

rms

5.210

5

is absorbedpower Average

5

2502

2sin25

02

2sin

4

100

2cos12

100

2

1sin100

2

1

2 ,0

0 ,sin10)(

22

00

22

55

Maximum Power Transfer

• Why use Thevenin and Norton equivalents?– Very easy to calculate load related quantities

– E.g. Maximum power transfer to the load

• It is often important to design circuits that transfer power from a source to a load. There are two basic types of power transfer– Efficient power transfer: least power loss (e.g.

power utility)

– Maximum power transfer (e.g. communications circuits)

• Want to transfer an electrical signal (data, information etc.) from the source to a destination with the most power reaching the destination. Power is small so efficiency is not a concern.

56

Maximum Power Transfer Theorem

Given a Thévenin equivalent circuit of a network, the maximum power that can be transferred to an external load resistor is obtained if the load resistor equals the Thévenin resistance of the network. Any load resistor smaller or larger than the Thévenin resistance will receive less power.

57

Maximum Power Demonstrated (1/2)

PL is max if

and

VTh

RTh

RL

a

b

I

Th

Th L

VI

R R

2

2L

L Th

Th L

RP V

R R

0L

L

dP

dR

58

Maximum Power Demonstrated (2/2)

• Maximum Power Transfer Theorem applied when matching loads to output resistances of amplifiers

• Efficiency is 50% at maximum power transfer

andVTh

RTh

RL

a

b

I

0L

L

dP

dR

L ThR R