Chapter 11

AC Steady-State Power

Design the matching network to transfer maximumpower to the load where the load is the model of an antenna of a wireless communication system.

Matching Network for Maximum Power Transfer

Cellular Telephone

George Westinghouse, 1846-1914

Nikola Tesla, 1856-1943

The greatest engineer of his day, George Westinghouse modernized the railroad industry and established the electric power system.

Tesla was responsible for many inventions, including the ac induction motor, and was a contributor to the selection of 60Hz as the standards ac frequency in the United States.

Instantaneous Power and Average Power

( ) ( ) ( )p t v t i t A circuit element

Instantaneous Power

If v(t) is a periodic function

( ) ( )v t v t T Then for a linear circuit i(t) is also a periodic function

( ) ( )i t i t T

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

Average Power0

0

1( )

t T

t

P p t dtT

Arbitrary point in time

Instantaneous Power and Average Power(cont.)

If v(t) is a sinusoidal function

( ) (cos )m Vv t V t For a linear circuit i(t) is also a sinusoidal function

( ) (cos )m Ii t I t

( ) (cos )(cos )m m V Ip t V I t t

( ) cos( ) cos(2 )2

m mV I V I

V Ip t t

0

0 0

0 0

1cos( ) cos(2 )

2

1 1cos( ) cos(2 )

2 2

1cos( ) cos(2 )

2 2

cos( ) 02

cos( )2

Tm m

V I V I

T Tm m m m

V I V I

T Tm m m m

V I V I

m mV I

m mV I

V IP t dt

T

V I V Idt t dt

T T

V I V Idt t dt

T T

V I

V I

average value of the cosinefunction over a completeperiod is zero

Example 11.3-1 P = ?

i(t) through a resistor R

Using the period from t = 0 to t = T

; 0mIi t t T

T

The instantaneous power is2

2 22

; 0mI Rp i R t t T

T

The average power is2

220

1 TmI R

P t dtT T

2 2 23

23 30

W3 3

Tm m mI R I R T I R

P t dtT T

Example 11.3-2 PL = ? PR = ?

( ) 721cos(100 41 ) mAi t t

The element voltages are

( ) 20cos(100 15 ) V

( ) 18cos(100 41 ) V

( ) 8.66cos(100 49 ) V

s

R

L

v t t

v t t

v t t

The average power delivered by the voltage source is

(20)(0.721)cos( 15 ( 41 )) 6.5 W

2sP

The average power delivered to the voltage source is 6.5 W

Example 11.3-2 (cont.)

The average power delivered to the resistor is

(18)(0.721)cos( 41 ( 41 )) 6.5 W

2RP

The average power delivered to the inductor is

(8.66)(0.721)cos(49 ( 41 )) 0 W

2LP

WHY the average power delivered to the inductor = 0 ?

The angle of vL always be larger than the angle of iL and cos(90 ) 0

90

Effective Value of a Periodic Waveform

The goal is to find a dc voltage, Veff (or dc current, Ieff), for a specified vs(t) that will deliver the same average power to R as would be delivered by the ac source.

The energy delivered in a period T isW PT

2

0

1 TP i Rdt

T

The average power delivered to the resistor by a periodic current is

Effective Value of a Periodic Waveform (cont.)

The power delivered by a direct current is

2effP I R

2 2

0

1 T

effP i Rdt I RT

Solve for Ieff 2

0

1 T

eff

rms

I i dtT

I

rms = root-mean-square

The effective value of a current is the steady current (dc)that transfer the same average power as the given time varying current.

Example 11.4-1 Ieff = ?

i(t) = sawtooth waveform

Express the waveform overthe period of t = 0 to t = T

; 0mIi t t T

T

22 2

20 0

2 3 2

3 3

0

1 1

3

T Tm

eff

T

m m

II i dt t dt

T T T

I t I

T T

3m

eff

II

Complex Power

A linear circuit is excited by a sinusoidal input and the circuit has reached steady state.

The element voltage and current can be represented in

(a) the time domain or (b) the frequency domain

Complex Power (cont.)To calculate average power from frequency domain representation of voltage and current i.e. their phasors

( ) and ( )m I m VI V I V

The complex power delivered to the element is defined to be * ( )( )

2 2

( )2

m I m V

m mV I

I V

I V

VIS

*where complex conjugate ofI I

Apparent power

Complex Power (cont.)

The complex power in rectangular form is

cos( ) sin( )2 2

m m m mV I V I

I V I

P

V

Q

j S

P jQ Sor

real or average power reactive power

: , P:Unit , QVs A W R:VASVolt-Amp ReactiveVolt-Amp

Complex Power (cont.)

The impedance of the element can be expressed as

( )( ) ( )

( )m V m

V Im I m

V V

I I

V

ZI

In rectangular form

( ) cos( ) sin( )m mV I V I

m m

R

Vj

I

X

V

I Z

or

( ) R jX Z

resistance reactance

Complex Power (cont.)

The complex power can also be expressed in terms of the impedance

2 2

2 2

cos( ) sin( )2 2

cos( ) sin( )2 2

Re( ) Im( )2 2

m m m mV I V I

m m m mV I V I

m m

m m

I V I Vj

I V I Vj

I I

I I

P Q

j

S

Z Z

Complex Power (cont.)

The impedance triangle The complex power triangle

The complex power is conserved *

02k k

allelements

V I

The sum of complex power absorbed by all elements of a circuit is zero.

Complex Power (cont.)

The complex power is conserved implies that both average power and reactive power are conserved.

* *

* *

*

Re Re 02 2

and Im Im 02

02

2

k k k k

all allelements elements

k k k k

all allelements eleme

k k

allelement

ts

s

n

V I V I V I

V I V I

or 0 and 0k kall all

elements elements

P Q

Example 11.5-1 S is conserved ?

100cos1000 V

=100 0sv t

Solving for the mesh current

( ) 7.07 451

s

R j L jC

VI

Use Ohm’s law to get the element voltage phasors

( ) ( ) 70.7 45

( ) ( ) 141.4 45

( ) ( ) 70.7 135

R

L

C

R

j L

j

C

V I

V I

V I

Example 11.5-1 (cont.)Consider the voltage source

*

2353.5 45 VA

sV

V IS supplied by the source

For the resistor *

2250 0 VA

RR

V IS absorbed by the resistor

For the inductor*

2500 90 VA

LL

V IS delivered to the inductor

Example 11.5-1 (cont.)

For the capacitor

*

2250 90 VA

CC

V IS delivered to the capacitor

250 0 500 90 250 90

353.5 45R L C

V

S S S

S

The total power absorbed by all elements (except source)

*

02k k

allelements

V IFor all elements

Example 11.5-2 P is conserved ?

100cos1000 V

=100 0sv t

The average power for the resistor, inductor, and capacitor is2

Re( )2mI

P

Z2

250 W 02m

R L C

IP R P P

The average power supplied by the voltage source is

*

Re Re2

Re(353.5 45 ) 250 W

sV VP

V IS

Power Factor

The ratio of the average power to the apparent power is called the power factor(pf).

cos( )

2m m

V I

I VP

S

average power

apparent power

cos ( )V I

pf angl

p

e

f

Therefore the average power

2m mI V

P pf

Power Factor (cont.)

cos( ) cos( ) The cosine is an even function

cos( ) cos( )V I I Vpf

Need additional information in order to find the angle

0.8 0 36.87

and

leading

l

Ex

agging0.8 0 36.87V I

V I

pf for

pf for

Ex The transmission of electric power

Time domain

Power Factor (cont.)

Frequency domain

We will adjust the power factor by adding compensating impedance to the load. The objective is to minimize the power loss (i.e. absorbed) in the transmission line.

1 1 1 1

1 1

( )2 2 2 2LINE

R L R Lj j

R j L

Z The lineimpedance

Power Factor (cont.)The average power absorbed by the line is

2 2

1Re( )2 2m m

LINE LINE

I IP R Z

The customer requires average power delivered to the load Pat the load voltage Vm

2m mV I

P pfSolving for Im

2m

m

PI

V pf

2

12LINEm

PP R

V pf

max pf =1

Power Factor (cont.)

compensatingimpedance

A compensating impedance has been attached across the terminals of the customer’s load. 1pf

cos cpfc

correctedThe load impedance is and the compensating impedance isWe want ZC to absorb no average power so

R jX ZC C CR jX Z

C CjXZ

Power Factor (cont.)The impedance of the parallel combination ZP

CP P P P P

C

R jX Z

ZZZ

Z ZThe power factor of the new combination

1cos cos tan PP

P

Xpfc

R

Calculate for RP and XP

2 2

2 2

2 2

2 2 2 2

( )

( )

( )

( )

( )

( ) ( )

C CP

C C

C C C C

C

C C C C

C C

R jX jX

R jX jX

RX j R X X X XX

R X X

RX R X X X XXj

R X X R X X

ZZZ

Z Z

Power Factor (cont.)2 ( )P C

P C

X R X X X

R RX

1cos tan P

P

Xpfc

R

From 1tan(cos )P

P

Xpfc

R

Solving for XC2 2

1tan(cos )C

R XX

R pfc X

Typically the customer’s load is inductive ZC = capacitive

C C

jjX

C

Z

2 2

1

1

tan(cos )

R X

C R pfc X

Power Factor (cont.)

Solving for C1

2 2

12 2

tan(cos )

tan(cos )

X R pfcC

R XR X

pfcR X R

1tanX

R

Let

2 2tan tan C

RC

R X

where

1 1cos and cosCpf pfc

Example 11.6-1 I and pf = ?

Load = 50 kW of heating (resistive) and motor 0.86 lagging pf

1 1 50 kWP S

Load 1 50 kW resistive load

Load 2 motor 0.86 lagging pf 2 0 1 1

2 2cos ( ) cos (0.86) 30.7pf

2 2 2 100 30.7 86 51 kVAj S SP Q

Example 11.6-1 (cont.)

1 2 136 51 145.2 20.6 kVAj S S S

cos(20.6 ) 0.94pf To calculate the current

4

2

145200

1014.52 Arms

m mrms rms

rmsrms

V IV I

IV

S

S

Example 11.6-2 pf ==> 0.95, 1 C = ?

377 rad/s

=100+j100

Z

cos cos45 0.707 laggingpf We wish to correct the pf to be pfc

0.95 laggingpfc 2 2

1tan(cos )

297.9

C

R XX

R pfc X

18.9 μF

C

CX

1pfc Example 11.6-2 (cont.)

2 2

1tan(cos )

200

C

R XX

R pfc X

113.3 μF

C

CX

Or use

2 2tan tan C

RC

R X

13.3 μFC

The Power Superposition Principle

1 2i i i

2 2 2 21 2 1 2 1 2( ) ( 2 )p i R i i R i i i i R

2 21 2 1 20 0

2 21 2 1 20 0 0

1 2 1 20

1( 2 )

2

2

T T

T T T

T

RP pdt i i i i dt

T TR R R

i dt i dt i i dtT T T

RP P i i dt

T

0

1 20

20 ?

TRi i dt

T

The Power Superposition Principle (cont.)

Let the radian frequency of the 1st source = m and the radian frequency of the 2nd source = n

1 1

2 2

cos( )

cos( )

i I m t

i I n t

12 1 20

1 20

2

2cos( )cos( )

T

T

RP i i dt

TR

I I m t n t dtT

integer

1 212 0

1 2

0

1 2

2cos( )cos( )

2(cos(( ) ( )) cos(( ) ( )))

0 ;

cos( );

2

T

T

RI IP m t n t dt

TRI I

m n t m n t dtT

m n

RI Im n

The Power Superposition Principle (cont.)

For the case that m and n are not integer for example m = 1, n = 1.5 0

2 212 1 2

2 2

21 2

2

1 1lim lim 2 cos cos(1.5 )

1lim 2 (cos0.5 cos2.5 ) 0

t t

t tt t

t

tt

P pdt RI I t t dtT T

RI I t t dtT

The Power Superposition Principle (cont.)

The superposition of average power

The average power delivered to a circuit by several sinusoidal sources, acting together, is equal to the sumof the average power delivered to the circuit by each source acting alone, if and only if, no two of the source have the same frequency.

If two or more sources are operating at the same frequencythe principle of power superposition is not valid but theprinciple of superposition remains valid.

1 2 3 N I I I I IFor N sources2

2mI R

P

Example 11.7-1 P = ?

(1) ( ) 12cos3 V and ( ) 2cos4 A

(2) ( ) 12cos4 V and ( ) 2cos4 AA B

A B

v t t i t t

v t t i t t

Example 11.7-1(cont.)Case I 1 2( ) 1.414 45 and ( ) 1.6 143I I These phasors correspond to different frequencies andcannot be added.

1 2( ) 1.414cos(3 45 ) and ( ) 1.6cos(4 143 )i t t i t t Using the superposition

( ) 1.414cos(3 45 ) 1.6cos(4 143 )i t t t

The average power can be calculated as

2

0(1.414cos(3 45 ) 1.6cos(4 143 ))

TRP t t dt

T

Since the two sinusoidal sources have different frequencies2 2

1 2

1.414 1.66 6 13.7 W

2 2P P P

Example 11.7-1(cont.)Case II 1 2( ) 1.2 53.1 and ( ) 1.6 143I I

Both phasors correspond to the same frequency andcan be added.

The sinusoidal current is

( ) 2.0cos(4 106.3 )i t t

The average power can be calculated as

Power superposition cannot be used here becauseBoth sources have same frequencies

22.06 12 W

2P

( ) 1.2 53.1 +1.6 143 2.0 106.3I

The Maximum Power Transfer Theorem

t t t L L LR jX and R jX Z Z

22

2 22 ( ) ( )t Lm

Lt L t L

RIP R

R R X X

V

( ) ( )t

t t L LR jX R jX

V

I

We wish to maximize P set2

2( )t L

t L

RP

R R

V

L tX X

For 0 we get L tL

dPR R

dR

*L t t t Maximum PoweR r Transf rjX e Z Z

(a) both coil currents enter the dotted ends of the coils

(b) one coil current enters the dotted end of the coil, but the other coil current enters the undotted end

Coupled Inductors

1 21 1

2 12 2

di div L M

dt dtdi di

v L Mdt dt

1 21 1

2 12 2

di div L M

dt dtdi di

v L Mdt dt

Summary

Instantaneous Power and Average Power

Effective Value of a Periodic Waveform

Complex Power

Power Factor

The Power Superposition Principle

The Maximum Power Transfer Theorem

Coupled Inductor and Transformer