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Chapter 11AC Power Analysis

Huseyin BilgekulEeng224 Circuit Theory II

Department of Electrical and Electronic Engineering Eastern Mediterranean University

Chapter Objectives: Know the difference between instantaneous power and average

power Learn the AC version of maximum power transfer theorem Learn about the concepts of effective or rms value Learn about the complex power, apparent power and power

factor Understand the principle of conservation of AC power Learn about power factor correction

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Instantenous AC Power

( ) cos( ) ( ) cos( )

( ) ( )11

cos( ) cos )( (22

2) m m m vm

m v m

v i

i

i

v t V t i t I t

p t v t ti V I V It

Instantenous Power p(t) is the power at any instant of time.

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Instantenous AC Power Instantenous Power p(t) is the power at any instant of time.

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

The instantaneous power is composed of two parts.

• A constant part.

• The part which is a function of time.

Assume a sinusoidal voltage with phase , ( ) cos( )

Assume a sinus

1cos( )

2CONSTANT POWER

oidal current with phase , ( ) cos( )

( ) ( ) ( )1

cos(2 )2

SINUSO A( ) + ID L

v m v

i

m m vv i im

m i

m

v t V t

i t I t

p t v t i Vt

p

I I t

t

V

(frequen POW cyER 2 )

1c( ) ( ) ( )

1cos( )

2os(2 )

2m m v m m v iip t v t i t V I tV I

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Instantenous and Average Power The instantaneous power p(t) is composed of a constant part (DC) and a time dependent part having frequency 2ω.

Instantenous Power p(t)

12

Average Po we

c )

r

os(m m v iP V I

1co

( ) ( ) ( )

( ) c

1cos(2 )

2

os( ) ( ) cos( )

( s2

) ( )

m v m i

m m vv i im mV I

p t v t i t

v t V t i t I t

I tp t V

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Instantenous and Average Power 1 1

1 22 2( ) cos( ) cos(2 ) ( ) ( )m m v i m m v ip t V I V I t p t p t

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Average PowerThe average power P is the average of the instantaneous power over one period .

0

1( ) Average

( ) ( ) (

Powe

) Instantaneous Power

( ) cos

r

( ) ( ) cos( )m v

T

m i

p t v t i t

v t V t i

P p t

tT

t

dt

I

1 12 20 0 0

1 1 1( ) cos( ) cos(2 )

T T T

m m v i m m v iP p t dt V I dt V I t dtT T T

12

12

cos( )

1Re cos( )

2

m m v i

m m v i

P V I

P V I

VI

1 12 20 0

12

1 1cos( ) cos(2 )

= co (Integral of a Sinusoidal=0)s( ) 0

T T

m m v i m m v i

m m v i

P V I dt V I t dtT T

V I

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Average PowerThe average power P, is the average of the instantaneous power over one period .

12

12

cos( )

1Re cos( )

2

m m v i

m m v i

P V I

P V I

VI

A resistor has (θv-θi)=0º so the average power becomes:

221 1 12 2 2R m m mP V I I R R I

1. P is not time dependent.

2. When θv = θi , it is a purely resistive load case.

3. When θv– θi = ±90o, it is a purely reactive load case.

4. P = 0 means that the circuit absorbs no average power.

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Instantenous and Average Power

Example 1 Calculate the instantaneous power and average power absorbed by a passive linear network if:

)60 10( sin 15 )(

)20 10( cos80 )(

tti

ttv

1cos( )

2385.7

1cos(2 )

2 600cos(20t 10 )

P= 385.7 W is the average power

(

)

low

=

f

W

m mm iv vi mVI Ip V tt

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Average Power Problem Practice Problem 11.4: Calculate the average power absorbed by each of the five elements in the circuit given.

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Average Power Problem

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Maximum Average Power Transfer

a) Circuit with a load b) Thevenin Equivalent circuit

Finding the maximum average power which can be transferred from a linear circuit to a Load connected.

• Represent the circuit to the left of the load by its Thevenin equiv.• Load ZL represents any element that is absorbing the power generated by the circuit.• Find the load ZL that will absorb the Maximum Average Power from the circuit to which it is connected.

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Maximum Average Power Transfer Condition

• Write the expression for average power associated with ZL: P(ZL). ZTh = RTh + jXTh ZL = RL + jXL

2

2

2 2

2

22 2

2

L

2

L

2

2

Ajust R and X to get maximum P

1 2( ) ( ) 2 ( ) ( )

( )

( ) ( )

( ) ( ) 2 ( )

2 ( )

LThTh Th

LTh L Th Th L L Th L Th L

Th L Th L

L Th L Th L

Th Th L Th L L Th L

L Th L

RVV VI P I R

Z Z R jX R jX R R X X

V R X XP

X R R X X

V R R X X R R RP

R R R

22

2 2

0 0 (

( )

)

L Th L Th Th L Th

Th L

L L

L L L Th Th Th

P PX X R R X X

R

X X

RX

Z R jX R jX Z

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Maximum Average Power Transfer Condition

For Maximum average power transfer to a load impedance ZL we must choose ZL as the complex conjugate of the Thevenin impedance ZTh.

2

max 8

L L L Th Th Th

Th

Th

Z R jX R jX Z

VP

R

• Therefore: ZL = RTh - XTh = ZTh will generate the maximum power transfer.• Maximum power Pmax

22

max 2 8ThL L

Th

VI RP

R

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Maximum Average Power Transfer Practice Problem 11.5: Calculate the load impedance for maximum power transfer and the maximum average power.

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Maximum Average Power Transfer

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Maximum Average Power for Resistive Load

When the load is PURELY RESISTIVE, the condition for maximum power transfer is:

Now the maximum power can not be obtained from the Pmax formula given before.

Maximum power can be calculated by finding the power of RL when XL=0.

2 2 2 20 ( ) L L Th Th L Th Th ThX R R X X R X Z

●

●

RESISTIVE LOAD

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Maximum Average Power for Resistive Load

Practice Problem 11.6: Calculate the resistive load needed for maximum power transfer and the maximum average power.

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Maximum Average Power for Resistive Load

Notice the way that the maximum power is calculated using the Thevenin Equivalent circuit.

RL

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a) AC circuit

Effective or RMS Value The EFFECTIVE Value or the Root Mean Square (RMS) value of a periodic current is the DC value that delivers the same average power to a resistor as the periodic current.

b) DC circuit

2 2 2 2

0 0

1( ) ( )

T T

eff Rms

RP i t Rdt i t dt I R I R

T T

2 2

0 0

1 1( ) ( )

T T

eff Rms eff RmsI I i t dt V V v t dtT T

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Effective or RMS Value of a Sinusoidal The Root Mean Square (RMS) value of a sinusoidal voltage or current is equal

to the maximum value divided by square root of 2.

22 2

0 0

1 1cos (1 cos 2 )

2 2

T Tm m

Rms m

I II I tdt t dt

T T

12 cos( ) cos( )m m v i Rms Rms v iP V I V I

The average power for resistive loads using the (RMS) value is:

22 Rms

R Rms

VP I R

R

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Effective or RMS Value Practice Problem 11.7: Find the RMS value of the current waveform. Calculate the average power if the current is applied to a 9 resistor.

A309.23

16rmsI

W48)9(3

162

RIP rms

4 0 1( )

8 4 1 2

t ti t

t t

1 22 2 2 2

0 0 1

1 1(4 ) (8 4 )

2

T

rmsI i dt t dt t dtT

1 22 2 2

0 1

16 (4 4 )2rmsI t dt t t dt

3

2 2 21

1 168 4 2

3 3 3rms

tI t t

2T 4t

8-4t

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