Andersen CH 11 Final R1

 

EE2003 Circuit Theory Chapter 11 AC Power Analysis Copyright © The McGrawMcGraw-Hill Hill Companies, Inc. Permission required for reproduction or display.

 

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AC Power Analysis   Chapter 11 11.1 11 .1

Ins nsttant nta aneo eou us and and Av Ave era rage ge Po Pow wer

11.2 11.2 11.3 11.4 11 .4 11.5

Maxim Maxi mum Av Aver era age Po Powe werr Tra rans nsfe ferr Effective or RMS Value App Ap par are ent Po Pow wer an and Pow Powe er Fa Facto torr Complex Power

11.6 11.7 11.8

Conservation of AC Power Power Factor Correction Power Measurement

 

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11.1 Instantaneous and Average Power (1)

 

• Th The e ins insta tant ntan aneo eous usly ly pow power er,, p(t) p(t)  p (t )  v(t ) i (t )   V m I m cos ( t    v ) cos (  t    i ) 1

1

2

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 V m I m cos ( v   i )   V m I m cos (2  t    v   i ) Constant power 

Sinusoidal power at 2 t 

p(t) > 0: power is absorbed by the circuit; p(t) < 0: power is absorbed by the source.  

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11.1 Instantaneous and Average Power (2)

 

• The The av ave era rag ge po powe wer, r, P, is the average of the instantaneous  power over one period.

 P  

1 T 

 T 



0

1

 p (t ) dt     V m  I m cos ( v    i ) 2 1. P is not time dependent. 2. When θv = θi , it is a purely resistive load resistive  load case. 3. When θv – θi = ±90o, it is a purely reactive load reactive load case. 4. P = 0 me mean ans s that that th the e cir circu cuit it absorbs no average power.  power. 

 

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11.1 Instantaneous and Average Power (3)

 

Example 1

Calculate the instantaneous power and average power absorbed by a passive linear network if: v(t )   80 cos  (10 t   20) i (t )   15 sin (10 t   60)

)W,, 387.5W    10)W Answer: 385.7  600cos(20t  

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11.1 Instantaneous and Average Power (4)

 

Example 2

A current

I

  10  30 flows through an impedance Z

 .20    the 22Ωaverage power Find

delivered to the impedance.

Answer:  927.2W Answer:  

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11.2 Maximum Average Power Transfer (1)

 

ZTH   R TH   j XTH ZL   R L   j X L The maximum average power can be transferred to the load if XL = –XTH  and RL = RTH

Pmax    

2 TH

If the load is purely real, then R     R     X L

 

2 TH

VTH

2

8 R TH

   ZTH 7

11.2 Maximum Average Power Transfer (2)

 

Example 3 For the circuit shown below, find the load impedance ZL that absorbs maximum average power. Calculate that maximum average the power.

Answer: 3.415 – j0.7317 , 1.429W  

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11.3 Effective or RMS Value (1) The total power dissipated by R is given by:  P  

1 T 



T

0

i  Rdt   2

 R T 



T 

0

i dt    I rms R 2

Hence, Ieff is equal to:  I   eff  

2

 

1 T 

T 



  i 2 dt    I rms  0

The rms value is a constant itself which depending on the shape of the function i(t).

The effective of a periodic current is the dc current that current that delivers the   same average power  to  to a resistor as the periodic current.

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11.3 Effective or RMS Value (2) The rms value of a sinusoid i(t) = Imcos(t) t)   is given by: Im 2 I rms   

2 The average power can be written in terms of the rms values:

I eff   

1 Vm I m cos (θ v   θ i )  Vrms I rms cos (θ v  θ i ) 2

Note: If you express amplitude of a phasor source(s) in rms, then all the answer as a result of this phasor source(s) must also be in rms value.  

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11.4 Apparent Power and Power Factor (1)

 

• Apparent Power, S, Power, S, is the product of the r.m.s. values of voltage and current. • It is measured in volt-amperes volt-amperes or  or VA to distinguish it from the average or real power which is measured in watts.

P  Vrms I rm coss (θv  θ  i )   S co coss (θv  θi ) rmss co

Apparent Power, S

Power Factor, pf  

• Powe Powerr fac facto torr is is the the co cosi sine ne of th the e phase difference between the voltage and current. It current. It is also the cosine of the angle of the load impedance.  

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11.4 Apparent Power and Power Factor (2)

 

Purely resistive load (R) Purely reactive load (L or C)

P/S = 1, all power are

θv – θi = 0, Pf = 1 θv – θi = ±90o,

consumed P = 0, no real power consumption

pf

=0 •

Resistive and reactive load (R and L/C)

 

θv – θi > 0 θv – θi < 0

Lagging Lagging  - inductive load • Leading Leading  - capacitive load

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11.5 Complex Power (1) Complex power S is the product of the voltage and the complex conjugate of the current: 

V

 Vmθ v  

 

I

 I mθi

1 V I  Vrm rmss I rm rmss  θ v  θ i 2

 

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11.5 Complex Power (2) 1

 



S   V I  V  rm rmss I rm rmss  θ v  θ i 2

 S V

coss (θ  θ )    j V co

I

rms

S  

rms

=

v

P 

i

rms

+ j

sin (θ  θ )

I rms

v

i

Q 

P: is the average power in watts delivered watts delivered to a load and it is the only useful power. Q: is the reactive power exchange between exchange between the source and   the reactive part of the load. It is measured in VAR. •Q = 0 for resistive loads (unity pf).  

•Q < 0 for capacitive loads (leading pf). •Q > 0 for inductive loads (lagging pf).

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11.5 Complex Power (3)  S  Vrms I rmscos (θv  θi )    j Vrms Irms sin (θv  θi )

S  =

P 

+ j

Q 

2 2 Apparent Power Power,, S = |S| = Vrms*Irms = P    Q

Real power, power, 

P = Re(S) = S cos(θv – θi)

Reactive Power, Power,  Q = Im(S) = S sin(θv – θi) Power factor, factor,  

pf = P/S = cos(θv – θi) 15

 

 

11.5 Complex Power (4)  S  Vrms I rms cos (θv  θi )    j Vrms I rms sin (θ v  θi )

S  =

  Power

Triangle Impedance Triangle

P 

+ j

Q 

Power Factor 

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11.6 Conservation of AC Power (1) The complex real, real, and reactive powers of powers of the sources equal the equal  the respective sums of  the  the complex, real, and reactive powers of the individual loads. loads. 

For parallel connection: S  

1 2

V I   *

1 2

V (I1  I )    *

* 2

1 2

V I   * 1

1

V I 2   S1  S2 *

2

  The same results can be obtained for a series connection. 17

 

11.7 Power Factor Correction (1) Power factor correction is correction is the process of  increasing the  increasing the power factor without altering the altering the voltage or current to the original load.

Power factor correction is necessary for economic reason. reason.   

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11.7 Power Factor Correction (2)

Qc = Q1 – Q2  = P (tan θ1 - tan θ2) = ωCV 2rms

Q1 = S1 sin θ1  = P tan θ1   

P = S1 cos θ1 

Q2 = P tan θ2

C  

Qc 2 ωVrms

  P (tan θ1  tan θ 2 )   2 ω Vrms

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11.8 Power Measurement (1) The wattmeter is the instrument for measuring the average power.

The basic structure If  

Equivalent Circuit with load

v(t )  V m cos(      t     v ) and i (t )   I m cos(      t     i )

P    Vrms I rms cos (θ v  θ i )      12 Vm I m cos (θ v  θ i )

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