Chap 11 AC Power Analysis

Outline

• Instantaneous and Average Power

• Maximum Average Power Transfer

• Effective or RMS Value

• Apparent Power and Power Factor

• Conservation of AC Power

• Power Factor Correction

Chap 11 AC Power Analysis 2


Instantaneous Power

( ) ( ) ( )p t v t i t

( ) cos( )

( ) cos( )

( ) ( ) cos( )cos( )

1 cos

1cos(2 )

2

cos cos(

1c

) c

os( )

(

( )

)

2

os2

m

m v

m i

m

m m vm v

m v i

ii

v t V t

i t I t

v t i t V I

V I

t t

V

p t

tI

Frequency doubled!

The instantaneous power (in watts) is the power at any

instant of time.

Power Provider Power Consumer


Illustration of Instantaneous Power

1cos

1cos (( ) )

2 2( ) 2m m v i m m v iV Ip tt V I

• p(t) is periodic but with frequency 2ω

• The power is transferred from the circuit to the source.

(Because of the storage elements such as capacitors and

inductors in the circuit)

An Example of Instantaneous Power


1co

1cos( ) s(2 )

2( )

22 60

4 4p t t


Average Power

The average power (in watts) is the average of the

instantaneous power over one period.

0

*

0

*

1( )

1 cos( )cos( )

1

2

1

cos(

Re ;

)

2

m m v

T

T

m

i

m v i

m m v i

p t dtT

V I t t

V

T

P

I

dt

V I

VI VI

( ) cos( )

( ) cos( )

m v

m i

v t V t

i t I t


Two Special Cases of Average Power

2

Purely resistive circuit:

Purely reactive circui

1

1

2

9

cos 2

0:

90 0

t

v i

v i

eq

m mV I

P R

P

I

cos1

(2

)m m v iP V I

The resistive load (R) absorbs power at all times, while a

reactive load (L or C) absorbs zero average power.

Examples of Special Cases


221 1

: 0 || |

2

90 0

1

|2 2

1 :

92

01

: 0

2

m m

m m

m m

RR

j L

j C

P V I R

P V I

P V I

Z I

Z

Z

V

Phasor Domain Interpretation


+

-

V

I

Z

V

I

22

*

*

1Re

2

1 1Re Re

2

||

2

||

P

P

VI Z

Z

V ZI VI


Example 11.1

Q: Given that v(t)=120cos(377t+45o)V, i(t)=10cos(377t-

10o)A, find the instantaneous power and the average

power absorbed by the passive linear network.

344.2 600cos(754

1200cos(377 45 )cos

35 )

34

(377 10 )

600 cos(754 35 ) cos55

1 1cos( ) 1200cos 45 ( 10 )

2 2

600cos55 4

W

W.2

m m v i

p

t

i

t

V IP

v t t


Example 11.2

Q: Calculate the average power absorbed by an

impedance Z = 30 – j70 Ω when a voltage V = 1200°is applied across it.

120 0 A

76.16 66.8

1cos( )

2

1(120)(1.576)cos(

1.576 66.8

30 66.8 )2

7. W24

m m v iV IP

V

ZI


Example 11.3

Q: For the circuit, find the average power supplied by

the source and the average power absorbed by the

resistor.


Example 11.3 (cont.)

1. A

W

115 30 5 30

4 2 4.472 26.57

1Re

2

1 5 1.118cos(30 5

8 56.57

2.

6.57 )2

5

P

j

VI*

I

1.118 56.57 A

4 4.472 56.57

2.5

V

14.4 W72 1.118

2

R

R R

RP

I I

V I


Example 11.4

Q: Determine the average power generated by each

source and the average power absorbed by each

passive element in circuit (a).



1

2

2

1

4 A

10 5 10 60 30 0

By and 12 6

( )

( )

( ) ( ) 10.58 79.1

For mesh 1:

KCL for mesh :

8

2

A0

a

bj

a

j

b

j

I

I

I I

*

2

1 1Re

For the vol

207

tage sourc

60 10.58 cos(30 79.1 )2

:

2

e

8 W.VP VI

V

0. Hence, this average powe absorbed by the r sis o c .ur eVP



1 1 1 2

*

1 1

20 10( ) 80 10(4 2 10.39)

184.984 6.21

1

For

1Re 184.984 4cos(6.21 0)

2 2

the current source:

3 W67.8

I

j j j

P

V

V I I I

V I

1

2 1

*

2 1

4 0 A

20 80 0 V

1 1Re 80

F

42 2

160

or the resis or

W

t :

RP

I

V I

V I

0. Hence, this current source power to thsup e cpli irc .s te uiIP



2

C 2

*

2

L 1 2

L 1 2

10 58 79 1 ,

5 (5 90 )(10.58 79.1 ) 52.9 (79.1 90 )

1 1

For the capacitor:

For the inductor:

0Re (52.9)(10.58)cos( 90 )2 2

2 10.39 10.58 79.1

10( ) 10.58 ( 79.1

C C

. .

j

j

j

P

I

V I

V I

I I I

V I I

*

90 )

1 1Re (105.8)(10.58)cos90

20

2L L LP

V I

Finally,

367.8 160 0 0 20 07.8

V I R C LP P P P P

LVLI

CV


Maximum Average Power Transfer

For maximum average power transfer, the load

impedance ZL must be equal to the complex conjugate

of the Thevenin impedance ZTh.

2

Th

max

Th8P

R

V

*

ThL Z Z


Th Th Th

Th Th

Th Th Th

2

2* * Th

2 2

Th Th

;

( ) ( )

1 1 1 1Re Re

2 2 2 2 ( ) ( )

L L L

L L L

L L

L

L

L

L

R jX R jX

R R j

R

R R

X

P

X

RXX

Z Z

V VI

Z Z

VV I Z II I

Derivation of Maximum Average Power

Transfer

2

Th Th

22 2

Th Th

2 2 2

Th Th Th Th

22 2

Th Th

To find the condition with maximum power,

( )

( )0

0

( )

( ) ( ) 2 ( )

2 ( ) ( )

L L

L L

L L L

L L

L

L

L

P

X

P

R X X

R R X X

R R X X R R R

R R R X X

V

V


2

Th Th

22 2

Th Th

2 2 2

Th T

Th

22

Th

*

Th

2

h Th Th

22 2

Th Th

T

Th

h

Th Th

max

Th

0( ) ( )

( ) ( ) 2 ( )0

2 ( ) ( )

;

8

L L

L L L

L L L

L

L th

L

L L L

L

L

th

L

R X XP

X R R X X

R R X X R R RP

R R R

X X

R R X

X X

X -X

R j

X

R

PR

X

Z Z

V

V

V

Derivation of Maximum Average Power

Transfer

2 2

T

2 2

Th

h h T

Th

T h

0

L

L

L L

X

R R (X X

R R X

)

Z


Example 11.5

Q: Determine the load impedance ZL that maximizes the

average power drawn from the circuit. What is the

maximum average power?



Th

Th

2.933 44(8 6)

From Fig. (a), 5 4 (8 6) 5 4 8 6

8 6From Fig (b),

.467

7.454 1 (10) 4 8 6

0. V3

jj j j

j

j

j

j

Z

V

*

2

x

2

ma

(7.454)

8

2.933 4.

8(2.933

467

2.368 W)

TH

T

H

L

H

TR

j

P

Z Z

V


Example 11.6

Q: In the circuit, find the value of RL that will absorb the

maximum average power. Calculate the power.



Th

Th

20(40 30)(40 30) 20 9.412 22.35

20 40 30

20(150 30 ) 72.76 134 V

20 40 30

j jj j j

j j

j

j j

Z

V

2 2

Th

Th

Th

2

max

9.412 22.35

72.76 1341.8 100.42 A

3

24

3.66 22.35

1 1(1.8)

.

(24

25

39..25)2

292

W

L

L

LR

P

R j

R

Z

VI

Z

I

RL


Effective or RMS Value

The effect value of a period current is the dc current

that delivers the same average power to a resistor as

the period current.


Effective or RMS Value (cont.)2 2

eff0

1 T

R i RdtP I RT

2

0eff rms

1 T

TI i dt I

eff rms

2

0Simi

1larly,

T

v dtT

V V

2

0

The ( ) value of a periodic signal ( )

1

T

rms

x troot - mean - square rms

X x dtT


R

VRIP

2rms2

rms

Effective or RMS Value (cont.)

The effect value of a period signal is its root-mean-square

(rms) value.

rms

22

rms

2

0 0

For the sinusoid ( ) cos , the rms valus is

1 1 cos 1 cos 2

2

Similarly, for ( ) cos ,

2

2

m

T Tm

m

m

m

m

i t I t

II tdt t dt

T T

v t

II

V

V t

V


Example 11.7

Q: Determine the rms value of the current waveform. If

the current is passed through a 2-Ω resistor, find the

average power absorbed by the resistor.



• The period of the waveform is T = 4,

42,10

20,5)(

t

ttti

2 4

2 2 2

0rms

0 2

23

4

2

0

1 1(5 ) ( 10)

4

1 1 20025 100 200

48.165

3 4 3A

T

i dt t dt dtT

t

I

t

2

rm2

2

s (8.165) (2) 1 3 3 W3.R IP R


Example 11.8

Q: The waveform is a half-wave rectified sine wave.

Find the rms value and the amount of average power

dissipated in a 10-Ω resistor.



• The period of the waveform is T = 2,

2,0

0,sin10)(

t

tttv

2

2 2 2 2

rms0 0

2

0

rms

0

1 1( ) (10sin ) 0

2

1 100 50 sin 2 11 cos 2 ; sin (1 cos 2 )

2 2 2 2 2

50 1sin 2 0 25

2 2

V5

T

π

V v t dt t dt dtT

( t)dt t t t

V

π

2 2

rms10 W2.5

5

10R

V

RP

RMS of Sinusoidal Signal


rms

22

r

0

ms

2

0

For the sinusoid ( ) cos , the rms valus is

1 1 cos 1 cos 2 2

2

Similarly, for ( ) cos ,

2

2

m i

T Tm

m im

i

v

m

m

i t I t

II t dt

II t dt

T T

v t V

V

t

V

Power Representation Via RMS

Values


*

*

Represented b

Given ( ) cos and ( ) cos

1 1The average power cos( ) Re[ ]

2 2

y RMS:

Denot

cos( ) cos

e phasors:

1Re[

( )

2

] Re[

2

2

,

2

2

m i m v

m m v i

m mv i rms rms

rms rm

rms rms

v i

i t I t v t V t

P V I

V IP V I

P

s

VV I

V

I

I I

V

V

I

*

22

For resistive load:

]

v i

rmsrms rms rms L

L

VV I I R

RP

Complex Power


Complex power S (in volt-amperes or VA) is the product

of the rms voltage phasor and the complex conjugate of

the rms current phasor. Its real part is real power P and its

imaginary part is reactive power Q.

* 2* 2

*

2

*

2

*

1 | || |

2 22 2

| |

sin

| |

t

c s

1

o

a

2

n

rms rms

rmsrms

jS S

j

SS

P PQ P

e j

j

V I VV I I Z

Z

VS I Z

S

S

Z

S VI

*

rms rmsS V I

Summary of Terminologies


rms rms

2 2

rms rms

Terminology Unit Form

1*

Complex Power VA 2

( )

Apparent Power VA

Real Power Watt Re( ) cos( )

Reactive Power VAR Im( ) sin( )

Power p Factor os( )f c

v i

v i

v i

v i

S

P

Q

P

S

P jQ

V I

V I P Q

S

S

VI

S

S

S

S

Summary of Terminologies (cont.)

• S is called the complex power consumed by

the load Z

• P is the average or real power.

– The power delivered to the load

– The actual power dissipated by the load

• Q is the reactive or quadrature power.

– Unit: volt-ampere reactive (VAR)

– A measure of the energy exchange between the source

and the reactive part of the load.


Summary of Terminologies (cont.)

• S is called the apparent power which denotes

the power the provider provides.

• pf: is the power factor in [0,1]. The power

factor is an index of the load reaching the

satisfaction of the provider; the larger the

power is, the happier the provider is.


Power and Impedance Triangles


Power triangle

2

2

*

** 2

*

2

1 | || |

2 2

| || |

c

1

2

ta

os s n

n

i

rms rms

rmsrms

jS

P

e S S jS

PP jQ j

VV I I Z

Z

VS I Z

S

S

Z

VI

S

rms

rms

( )rmsv i

rms

V

I

VV

I IZ

Impedance triangle

Passive Load


| | ( )

0 ( )

(1)

(2) ; in the

Resistive load

Inductive load

f quadrant and the curr

[ 90 ,90 ]

0 :

0 0 :

ent is

the voltage by

irst

C

(3) ap0 ac0 : it

v i

v i

R

X

X

laggi

R jX

R

ng

Z Z

Z

S

; in the

quadrant and the current is

ive loa

the voltage by

d

our

f th

leading

S


Power Triangle (cont.)

Resistive load

Inductive load

Capacitive load


Example 11.9

Q: A series-connected load draws a current i(t) = 4

cos(100t + 10°) A when the applied voltage is v(t) =

120 cos(100t - 20°) V. Find the apparent power and

the power factor of the load. Determine the element

values that form the series-connected load.

Sol:

rms rmsapparent power: 240

powe

120 4 VA

2 2

cos( ) cos( 20 1r factor: pf 0.86) 60v i

S V I



• The load impedance Z can be modeled by a 25.98-Ω

resistor in series with a capacitor with

The power factor can also be obtained from

120 20 30 30 25.98 15

4 10

pf 0.cos 866( 30 )

j

VZ

I

2

115

1 1 F

1512.

15 12

00

CXC

C


Example 11.10

Q: Determine the power factor of the entire circuit as

seen by the source. Calculate the average power

delivered by the source.



• The average power supplied by the source is

2 46 4 ( 2) 6 6.8 7 13.24

pf 0.9734

1.6 4 2

cos( 13.24)

eq

jj j

j

Z

rrms

ms 30 0A

7 13.24.286 3 24

41 .

I

V

Z

rms rms

2

rms

2

(30)(4.286)0.9734 W

or (4.286) (6.8) W

where is th

p

e resistor part o

f 125

5

f

1

.

2

P V I

R

R

P I

Z


Example 11.11Q: The voltage across a load is v(t) = 60cos(t – 10°) V

and the current through the element in the direction of

the voltage drop is i(t) = 1.5 cos(t + 50°) A. Find: (a)

the complex and apparent powers, (b) the real and

reactive powers, and (c) the power factor and the load

impedance.

*

rms s

r

m

m

r

s

( ) ( ) 60cos( 10 ); ( ) 1.5cos( 50 )

60 1.510 , 50

2 2

The complex power is

60 1.5 10 50 VA

2 2

The ap

45 6

p

0

arent power is

V5 A4

o o

rms

a v t i t

S

t t

S V I

S

V I



(b) Form (a),

S 45 60 45 cos( 60 ) sin( 60 )

22.5 38.97

S

The real powe 22.5

38.97

r W

The reactive power VA R

j

j

P j

P

Q

Q

pf 0.4

(c) cos( 60 )

6 40 60

0 10

1.5 50

V

IZ


Example 11.12

Q: A load Z draw 12 kVA at a power factor of 0.856

lagging from a 120-V rms sinusoidal source. Calculate:

(a) the average and reactive powers delivered to the

load, (b) the peak current, and (c) the load impedance.

Sol:

The apparent power is S = 12,000 VA, the average

or real power is

and the reactive power is

1(a) cos , copf 0.856 s 0.856 31.13 .

cos 12000 0.856 10. kW272P S

sin 12000 0.517 6.2 kVA04Q S



(b) Since the pf is lagging, the complex power is

*

rms rms

*

rms

rms

rms

rms

S V Ι

Ι

1

S 10.272 6.204 VA

S 10272 6.204

V 120 0

85.6 51.7 A A

Thus and the peak c

00 31.13

urrent I 1 is

2 2 100

00 31.13

141 4 A.m

P jQ j

j

II

j



(c) The load impedance

which is an inductive impedance.

rms

rms

120 0

101.2 31.13

0 31.13

Z

V

I


Conservation of AC Power

1 2 N S S S S

• The principle of conservation of power is stillapplicable to the ac circuits.

The principle of conservation of ac power: The

complex, real, and reactive powers of the sources equal

the respective sums of the complex, real and reactive

powers of the individual loads.

1 N 1 2 N

1 1

2 2 * * *S V V S S SI I I


Example 11.13

Q: The following figure shows a load being fed by a

voltage source through a transmission line. The

impedance of the line is represented by the (4 + j2)Ω

impedance and a return path. Find the real power and

reactive power absorbed by; (a) the source, (b) the

line, and (c) the load.



(a)

20.

(4 2) (15 10)

19 8

22

62 22.

0 0=

83

10.67 22.83 A20.62 22.83

eq

rmr s

s

s

m

j j

j

Z

IV

Z

* (220 0 )(10.67 2

(2163.5 910.8)

2.83 )

2347.4 22.83 VA

s

rms rs ms

j

S V I



(b)

The complex power absorbed by the line is

(4 2) (4.472 26.57 )(10.67 22. 47.7283 ) 4 .4 V9line

r s mm r sj IV

*

line

2

line line

2

455.4 j227

(47.72 49.4 )(10.67 22.83 )

509.2 26.57 VA

or

(10.67) (4 2) VA

.7

455.4 227.7

line

rms rms

rms j j

S V I

S I Z



(c)

*

(15 10) (18.03 33.7 )(10.67 22.83 )

V

(192.38 10.87 )(10.67 22.83 )

2053

192.38 1

33.7

0.87

( V1708 11 9) A3

Load

rms

Load

rms

Load

rms rms

j

j

I

V I

V

S

s line Load S S S


Example 11.14

Q:

Calculus the total: (a) apparent power, (b) real power,

(c) reactive power, and (d) pf, supplied by the source

and seen by the source.

1 260 30 and 40 45 . Z Z



• The complex powers absorbed by the impedances are

• The total complex power is

1

1

2

2

120 10A rms2 40

3 3

60 30

120 10A5 rms

40 45

VI

Z

VI

Z

2 2

rms

*

1

2 2

ms

*

2

1

2r

(120)240 30 VA

60 30

(120)360 45 VA

40 45

207.85 120

254.6 254.6

jV

Vj

S

S

Z

Z

2t 1 462.4 134.6 VAj SS S



1 2

t

1 2

t

The total complex power is

VA

(a) The total apparent power is

VA

(b) The total real power is

Re( ) W or

(c) The total r

462 4 134 6

481 6

4

eactiv

4

e

.

p

62

o

t

t t t

. j .

S .

P P PP

S S

S

S

S

1 2134

wer i

.6

s

Im( ) VAR or t tt Q QQ Q S



p

(

f

d) The powe

0.

r f

96

actor

/ 462.4 / 481.6t tP S

1

*

s

2 (1.532 1.286) (2.457 1.721)

(4 0.4 4

Result checking: The complex power

35) A rms.024 6.21

463 135

(120 10 )(4.024 6.21 )

488.88

supplied by the sourc

16.21

VA

e

t

t j j

j

j

I I

VI

I

S


Power Factor Correction

• Most practical loads are inductive since motors

are the most useful driving force in this world.

• Hence, we can use a parallel-connected

capacitor to attain the goal of power factor

correction

Power factor correction: the process of increasing the

power factor without altering the voltage or current to the

original load.


Illustration of Power Factor Correction


Phasor Diagram for Inductive Load

by Capacitor Compensation

2

1

IIII

VI

IV

I

CL

C

LL

Cj

LjR


Capacitor Compensation

1 1

1 2

1 1 1 2 2 2

1 1

2

2

2 2 2

1 2 1

T

tan

tan

(

o increase pf from cosθ to cosθ without

altering the real power,

cos cos

tan tan )

sin

sin

Applying the ac power conservation gives

C

Q P

P S P S P

S

S

Q Q

Q P

Q P

2

rms

*

1 2

2 2

rms r

22rm

ms

srms

1 2

2

Not real poe tha wer

because is

t the

zero

is not affected by

the pf corre

But

(tan tan )

ction .

C

C

C

C

rms

VQ CV

X

Q QC

ωV

P

P

V

Q P

V V

SZ


PF Correction for Capacitor Load

• Similar idea can be used to realize power factor

correction for a capacitive load.

• The required shunt inductance L can be calculated as

2 2

rms r

2

rm1

mss2

LL

L

V VQ L

L Q

VQ Q

X


Example 11.15

Q: When connected to a 120-V (rms), 60-Hz power line,

a load absorbs 4 kW at a lagging power factor of 0.8.

Find the value of capacitance necessary to raise the pf

to 0.95.

Sol:

If pf = 0.8, then

1 1

1

1

1 1 1

cos 0.8 36.87

40005000 VA

cos 0.8

sin 5000sin36.87 3000VAR

PS

Q S



When pf is raised to 0.95,

2 2

2

2

2 2 2

1 2

2 2

rms

cos 0.95 18.19

40004210.5 VA

cos 0.95

sin 1314.4 VAR

3000 1314.4 1685.6 VAR

1685.6 F

2 60 120310.5

C

C

PS

Q S

Q Q Q

Q

VC

Chap 11 AC Power Analysis -Rev

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Transcript of Chap 11 AC Power Analysis -Rev