CHAPTER 11 - AC POWER ANALYSIS

List of topics for this chapter :Instantaneous and Average PowerMaximum Average Power TransferEffective or RMS ValueApparent Power and Power FactorComplex PowerConservation of AC PowerPower Factor CorrectionApplications

INSTANTANEOUS AND AVERAGE POWER

Problem 11.1 [11.3] Refer to the circuit depicted in Figure 11.1. Find the averagepower absorbed by each element.

Figure 11.1

+ 3010)30t2cos(10 , 2=2jLjH1 =

-j2Cj1

F25.0 =

j2 j2

2 4

1030 V +

I1

I2

I

1 H 0.25 F

2 4

+

10 cos(2t + 30) V

2j22

)2j2)(2j()2j2(||2j +==

=++

= 565.11581.12j24

3010I

=== 565.101581.1j22j

1 III

=

= 565.56236.22

2j22 II

For the source,

)565.11-581.1)(3010(21

*== IVS

5.2j5.718.43905.7 +==SThe average power supplied by the source = 7.5 W

For the 4- resistor, the average power absorbed is

=== )4()581.1(21

R21

P 22

I W5

For the inductor,

5j)2j()236.2(21

21 2

L

22 === ZIS

The average power absorbed by the inductor = W0

For the 2- resistor, the average power absorbed is

=== )2()581.1(21

R21

P 22

1I W5.2

For the capacitor,

5.2j-)2j-()581.1(21

21 2

c

21 === ZIS

The average power absorbed by the capacitor = W0

The average power supplied by the source = W5.7

The average power absorbed by the 4- resistor = W5

The average power absorbed by the inductor = W0

The average power absorbed by the 2- resistor = W5.2

The average power absorbed by the capacitor = W0

Problem 11.2 The load for the following circuit is given by the 5 ohm resistor and the0.02297 Henry inductor. In addition, vin(t) = 200 sin(377t) volts. Determine the average powerdelivered to the load.

Using the frequency domain circuit on the right, we can solve for I.

-141.4 + (5 + j8.660)I = 0. Which leads to,

I = 141.4/(1060o) = 14.14-60o amps. But power delivered to the load is equal to,

Pavg = 2I R = (14.14)2 x 5 = 200x5 = 1000 watts.

MAXIMUM AVERAGE POWER TRANSFER

Problem 11.3 Given the circuit in Figure 11.1 and ( )tsin200)t(v = volts, calculate thevalues of LR and X for maximum power transfer to LR .

Figure 11.1

+

v(t)

10 j5

jX

RL

+vin(t)

5

0.02297 H

+

200/ 2 5

j8.660

I

This is a straightforward classical maximum power transfer problem. If you remember that formaximum power transfer, ZL = Zs* or the value of the load is equal to the complex conjugate ofthe source impedance.

RL + jX = 10 j5 Thus the load resistor must be 10 and the reactance mustbe capacitive and equal to [1/(5)] F.

What if you do not remember the maximum power transfer theorem? Well, you can usually workit out just by looking at what you have. It seems reasonable to cancel whatever source reactancethere is. Then if the source resistance is either zero or infinity, there is no power transfer. Theonly thing that makes sense is that the load resistance must equal the source resistance.

Problem 11.4 [11.15] Find the value of LZ in the circuit of Figure 11.1 for maximumpower transfer.

Figure 11.1

We find ThZ at terminals a-b as shown in the figure below.

j1080(80)(-j10)

20j20-j10)(||8040||4020jTh

++=++=Z

154.10j23.21Th +=Z

==*

ThL ZZ 15.10j23.21

40 j10 80 40

j20 Zth

40 j10 80

j20

+

ZL50 A

600 V

40

EFFECTIVE OR RMS VALUE

Problem 11.5 Calculate the RMS value of the signal shown in Figure 11.1. The curve canbe represented by the function PP V)tsin(V + .

Figure 11.1

Since the value of the voltage, v(t), can be expressed as Vp[sin(t) + 1], we can ignore the valueof Vp in our calculations. The true value of the rms voltage will be what we obtain times Vp.Also, since the value of the rms voltage is independent of , we will let T = 1, which meansthat = 2.

2p

2rms

VV

= ++=+= T0 22T0T0 2 dt]1)t2sin(2)t2([sindt]1)t2[sin(dt)t(vT1

= 5.1121

t8

)t4sin(2tdt0dt

2)t4cos(1 1

0

1

0

1

0

1

0=+=+

=++

Vrms = 5.1 = 1.2247Vp

Problem 11.6 Calculate the RMS value of the signal shown in Figure 11.1. The curve canbe represented by the function ),tsin(VP (please note, this is often referred to as the full wave,rectified sine wave).

Figure 11.1

2 VP

VP

We can use the same approach that we used in problem 11.5. In this case, stays the same butnow T = 0.5 sec.

5.0

0

5.0

0

5.0

02

2p

2rms

8)t4sin(

2t2dt

2)t4cos(12dt)t2(sin

5.01

VV

=

==

5.0025.02 =

+= or Vrms = 0.707Vp which is to be expected since

the rms value is taken by squaring the value of a signal. Squaring either wave produces the sameresult.

Problem 11.7 Calculate the RMS value of the signal shown in Figure 11.1.

Figure 11.1

Again, the same approach is used.

5.0

0

5.0

0

5.0

0

1

5.02

2p

2rms

8)t4sin(

2tdt

2)t4cos(1dt0dt)t2(sin

11

VV

=

=

+=

= (0.5)/2 0 = 0.25 thus, Vrms = 25.0 Vp = 0.5Vp

Problem 11.8 [11.21] Find the effective value of the voltage waveform in Figure 11.1.

Figure 11.1

4

v(t)10

t

5

2 6 1080

VP

4T = ,

08.157j30)210( 2

*

2

==

ZV

S

Apparent power === 160)210( 2

S 275.6 VA

)19.79cos(36

08.157tancoscospf 1- =

==

=pf (lagging)1876.0

COMPLEX POWER

Problem 11.11 [11.35] Determine the complex power for the following cases:(a) W269P = , VAR150Q = (capacitive)(b) VAR2000Q = , 9.0pf = (leading)(c) VA600S = , VAR450Q = (inductive)(d) V220Vrms = , kW1P = , = 40Z (inductive)

(a) == jQPS VA150j269

(b) === 84.259.0cospf

31.4588)84.25sin(2000

sinQ

SsinSQ =

===

48.4129cosSP ==

=S VA2000j4129

(c) 75.0600450

SQ

sinsinSQ ====59.48= , 6614.0pf =

86.396)06614)(600(cosSP ===

=S VA450j9.396 ++++

(d) 121040)220(

S2

2

===

ZV

8264.012101000

SP

coscosSP ====

= 26.34

25.681sinSQ ==

=S VA2.681j1000 ++++

CONSERVATION OF AC POWER

Problem 11.12 [11.43] Obtain the power delivered to the 10-k resistor in the circuit ofFigure 11.1.

Figure 11.1

From the left portion of the circuit,

mA4.05002.0

o ==I

mA820 o =I

j1 k

10 k

j3 k

8 mA

Ix

4 k

j1 k

10 k

j3

20 Io 4 k

500

+

Io

0.20 V rms

From the right portion of the circuit,

mAj716)mA8(

3jj1044

x

=

++=I

)1010(50

)1016(RIP 32-3

2x

========

====P mW2.51

POWER FACTOR CORRECTION

Problem 11.13 A small industry operates from 220 volts supplied by a utility. The smallindustry represents a load to the utility that represents 22,000 watts and a power factor of 0.8.Develop an equivalent circuit for the load. Determine the value of a capacitor to correct thecircuit to unity power factor.

power = VI cos = 220xIx0.8 = 22,000 or I = 125 A

Thus, |Z| = 220/125 = 1.76 and cos = 0.8 leads to = 36.87Although it was not specified, most industries, if not all, represent an inductive load, thus thepower factor is lagging and is positive.

Z = 1.7636.87 = (1.408 + j1.056)

Since this represents a resistor in series with an inductor, we place a capacitor in parallel with thecombination in order to correct to unity power factor. The easiest way to do this is to just cancelthe reactive power with the parallel capacitor.

Q = VIsin = 220x125xxin(36.87) = 16881 = 2202/XC

XC = 2202/16881 = 2.867 = 1/(C) with = 377 the C = 925 F

Problem 11.14 For the network in Figure 11.1, determine the value of C that corrects thepower factor to 0.8, 0.85, 0.9, 0.95, 1.0.

Figure 11.1

pf = power/VA = 5

C1

C15

Xwhere

51

X1

5/1|Z|V

RVSP

22

2

2

=

+

==

or C51

5X

=

==++

= 1pf1

251

X1

pf251

251

X1

251

X1

25/1pf 22222

2

1pf125X2

2

= If we let a = X, then 1

pf15

a

2

= and

C515

= a

Solving this for C, 5/a = 1 5C or C = [1 5/a]/(377x5) = [1 5/a]/1885

pf a C

0.707 5 00.8 6.667 132.65 F0.85 8.068 201.7 F0.9 10.324 273.6 F0.95 15.212 356.1 F1.0 530.5 F

5j5C

Problem 11.15 Referring to the results of Problem 11.14, what can you say about the relativecosts of power factor correction?

To correct to 0.8 pf requires a 132.65 F capacitor. (Clearly one would purchase the closest valuecommercially available for the desired use. Taking into account energy requirements, this mightbe a 150 F capacitor.)

To correct to 0.85 requires only a 200 F capacitor. However, to go to 0.95 requires almost twoof these. This would mean that the cost of correcting to 0.95 is twice as much as correcting to0.85. To go to unity costs even more. It would require 4 times the number of capacitors tocorrect to unity as to correct to 0.8. Fortunately, utilities gain little from corrections from 0.85 tounity, which can save a lot since these compensating capacitors are very expensive.

Problem 11.16 [11.53] Refer to the circuit shown in Figure 11.1.

Figure 11.1

(a) What is the power factor?(b) What is the average power dissipated?(c) What is the value of capacitance that will give unity power factor when

connected to the load?

(a) Given that 12j10 +=Z== 19.50

1012

tan

== cospf 6402.0

(b) 09.354j12.295)12j10)(2()120(

2

2

*

2

+=

==

ZV

S

The average power absorbed === )Re(P S W1.295

(c) For unity power factor, = 01 , which implies that the reactive power due to thecapacitor is 09.354Qc =

But 2VC21

X2VQ

c

2

c ==

=

=

= 22

c

)120)(60)(2()09.354)(2(

VQ2

C F4.130

+

Z = 10 + j12 120 V60 Hz C

APPLICATIONS

Problem 11.17 [11.63] The kilowatt-hour-meter of a home is read once a month. For aparticular month, the previous and present readings are as follows:

Previous reading: 3246 kWhPresent reading: 4017 kWh

Calculate the electricity bill for that month based on the following residential rate schedule:Base monthly charge: $12.00First 100 kWh per month at 16 cents/kWhNext 200 kWh per month at 10 cents/kWhOver 300 kWh per month at 6 cents/kWh

kWh consumed kWh77132464017 ==

The electricity bill is calculated as follows :(a) Base charge = $12(b) First 100 kWh at $0.16 per kWh = $16(c) Next 200 kWh at $0.10 per kWh = $20(d) The remaining energy (771 300) = 471 kWh

at $0.06 per kWh = $28.26.

Adding (a) to (d) gives a total of 26.76$

SearchHelpEWB Help PageWe want your feedbacke-Text Main MenuTextbook Table of ContentsProblem Solving WorkbookWeb LinksTextbook WebsiteOLC Student Center WebsiteMcGraw-Hill Website

PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisInstantaneous and Average PowerMaximum Average Power TransferEffective or RMS ValueApparent Power and Power Factor Complex PowerConservation of AC PowerPower Factor CorrectionApplications

Chapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

sctoc:

TOC:

e-text:

forward:

back-last:

background:

back:

forward-last: