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# 328 Chapter 7 AC Power

## CYU 7.18 I* = 3Iy

CYU 7.19 Ia = 1890 A; Ib = 189120 A; Ic = 189120 A
CYU 7.20 Losses for a 120-V circuit are approximately double the losses for a 240-V circuit of the same power rating.

HOMEWORK PROBLEMS
Section 1: efficiency at rated output is 80 percent. [Note: 1
Basic AC Power Calculations horsepower = 0.746 kW.]
7.8 Given the waveform of a voltage source shown in
7.1 The heating element in a soldering iron has a Figure P7.8, find:
resistance of 391 . Find the average power dissipated
a. the average and rms values of the voltage.
in the soldering iron if it is connected to a voltage
source of 117 V rms. b. the average power supplied to a 10- resistor
connected across the voltage source.
7.2 The heating element in an electric heater has a
resistance of 10 . Find the power dissipated in the
vS (t)
heater when it is connected to a voltage source of (V)
240 V rms.
1
7.3 A current source i(t) is connected to a 100-
resistor. Find the average power delivered to the 0 1 2 3 4 5 6 t (s)
resistor, given that i(t) is:
a. 4 cos 100t A 3
b. 4 cos(100t 0.873) A
Figure P7.8
c. 4 cos 100t 3 cos(100t 0.873) A
d. 4 cos 100t 3 A
Section 2: Complex Power
7.4 Find the rms value of each of the following periodic 7.9 For the following numerical values, determine the
currents: average power, P , the reactive power, Q, and the
a. cos 377t + cos 377t complex power, S, of the circuit shown in Figure P7.9.
Note: phasor quantities are rms.
b. cos 2t + sin 2t
a. vS (t) = 650 cos (377t) V
c. cos 377t + 1
iL (t) = 20 cos(377t 0.175) A
d. cos 2t + cos(2t + 3/4)
b. VS = 4600 V
e. cos 2t + cos 3t
IL = 14.14/4 A
7.5 A current of 10 A rms flows when a single-phase c. VS = 1000 V
circuit is placed across a 220-V rms source. The IL = 8.61.5 A
current lags the voltage by /3 rad. Find the power d. VS = 208/6 V
dissipated by the circuit and the power factor. IL = 2.31.1 A
7.6 A single-phase circuit is placed across a 120-V rms,
60-Hz source, with an ammeter, a voltmeter, and a iL(t)
wattmeter connected. The instruments indicate 12 A,
120 V, and 800 W, respectively. Find
a. The power factor. vS (t) +
~ ZL
_
b. The phase angle.
c. The impedance.
d. The resistance. Figure P7.9

7.7 The nameplate on a single-phase induction machine 7.10 For the circuit of Figure P7.9, determine the power
reads 2 horsepower (output), 110 V rms, 60 Hz, and factor for the load and state whether it is leading or
24 A rms. Find the power factor of the machine if the lagging for the following conditions:

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Part I Circuits 329

## a. vS (t) = 540 cos(t + /12) V ~

IS
Line
iL (t) = 20 cos(t + 0.82) A
b. vS (t) = 155 cos(t /12) V R=1
iL (t) = 20 cos(t 0.384) A + ~
~
_ VS = 2300 ZL Load
c. vS (t) = 208 cos(t) V
iL (t) = 1.7 sin(t + 3.054) A
d. ZL = (48 + j 16) 
Figure P7.14
7.11 For the circuit of Figure P7.9, determine whether
the load is capacitive or inductive for the circuit shown
if 7.15 Calculate the apparent power, real power, and
a. pf = 0.87 (leading) reactive power for the circuit shown in Figure P7.15.
Draw the power triangle.
c. vS (t) = 42 cos(t)
~
iL (t) = 4.2 sin(t) IS
d. vS (t) = 10.4 cos(t /15)
iL (t) = 0.4 cos(t /15)
R = 20
+ ~
~
_ VS = 50 V
7.12 Find the real and reactive power supplied by the jXC = j34.6
source in the circuit shown in Figure P7.12.

Figure P7.15

2H 1/18 F

## 7.16 A single-phase motor draws 220 W at a power

vS(t) = 10 cos 3t V +
~ 4 factor of 80 percent (lagging) when connected across a

200-V, 60-Hz source. A capacitor is connected in
parallel with the load to give a unity power factor, as
Figure P7.12
shown in Figure P7.16. Find the required capacitance.

7.13 For the circuit shown in Figure P7.13, find the real I
and reactive power supplied by each source. The
sources are Vs1 = 36/3 V and Vs2 = I1
240.644 V. I2

V R

8 6 j12 C
L

~ ~
VS 1 +
~ j6
+
~ V
S2
Figure P7.16

Figure P7.13
7.17 Suppose that the electricity in your home has gone
out and the power company will not be able to have
7.14 The load ZL in the circuit of Figure P7.14 consists you hooked up again for several days. The freezer in
of a 25- resistor in parallel with a 100-F capacitor. the basement contains several hundred dollars worth
Assume = 377 rad/s. Calculate of food that you cannot afford to let spoil. You have
also been experiencing very hot, humid weather and
a. The apparent power delivered to the load. would like to keep one room air-conditioned with a
b. The apparent power supplied by the source. window air conditioner, as well as run the refrigerator
c. The power factor of the load. in your kitchen. When the appliances are on, they

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330 Chapter 7 AC Power

draw the following currents (all values are rms): c. The power factor angle and the power factor.
Air conditioner: 9.6 A @ 120 V Vs = 70.873 V Is = 13 0.349 A
pf = 0.90 (lagging)
Freezer: 4.2 A @ 120 V 7.22 Determine C so that the plant power factor of
pf = 0.87 (lagging) Figure P7.22 is corrected to 1; i.e., Is is minimized and
in phase with Vo .
Refrigerator: 3.5 A @ 120 V
pf = 0.80 (lagging) vs (t) = 450 cos(t) V = 377 rad/s

## In the worst-case scenario, how much power must an Z = 7 + j1 

emergency generator supply? ZG = 3 + j 0.11 m
7.18 The load on a single-phase three-wire system in a
home is generally not balanced. For the system shown ~
Is
in Figure P7.18, let Vs1 = 1150 Vrms and
Vs2 = 1150 Vrms . Determine:
ZG
a. The total average power delivered to the connected ~
C Z Vo
loads: ZL1 , ZL2 , and ZL3 . + ~
b. The total average power lost in the lines: Zg1 , Zg2 , _ Vs
and Zn .
c. The average power supplied by each source.
Power Plant
plant
Zg1 = 0.1 Figure P7.22
7.23 Determine C so that the plant power factor of
~
VS 1 +
Figure P7.22 is corrected to 1 (or the power factor
~
ZL1 = 60
angle to zero) so that Is is minimized and in phase with
Zn = 2 Vo .
ZL3 = 20
vs (t) = 450 cos(t) V = 377 rad/s
~
VS 2 + ZL2 = 80
~
Z = 70.175 

## 7.24 Without the capacitor connected into the circuit of

Zg2 = 0.1 Figure P7.22,
Figure P7.18 Vo = 4500 V Is = 17 0.175 A
f = 60 Hz C = 17.40 F
7.19 A large consumer of electricity requires 10 kW of
power at 230 Vrms at a pf angle of /3 rad lagging. The The value of C is that which will correct the power
transmission line between the electric utility and the factor angle to zero, i.e., reduces Is to a minimum
consumer has a resistance of 0.1 . If the consumer value in phase with Vo . Determine the reduction of
can increase the pf from 0.5 to 0.9 lagging, determine current which resulted from connecting the capacitor
the change in transmission line losses and load current. into the circuit.
7.20 A 1000-W electric motor is connected to a source 7.25 Without the capacitor connected into the circuit:
of 120 Vrms , 60 Hz, and the result is a lagging pf of 0.8.
vo (t) = 170 cos t V
To correct the pf to 0.95 lagging, a capacitor is placed
in parallel with the motor. Calculate the current drawn is (t) = 130 cos(t 0.192) A
from the source with and without the capacitor
connected. Determine the value of the capacitor f = 60 Hz C = 387 F
required to make the correction. The value of C given is that which will correct the
7.21 If the voltage and current given below are supplied power factor angle to zero, i.e., reduces Is to a
by a source to a circuit or load, determine: minimum value in phase with Vo . Determine by how
a. The power supplied by the source which is much the current supplied to the plant is reduced by
dissipated as heat or work in the circuit (load). connecting the capacitor.
b. The power stored in reactive components in the 7.26 Determine the time-averaged total power, the real
circuit (load). power dissipated, and the reactive power stored in each

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Part I Circuits 331

of the impedances in the circuit shown in Figure P7.26 b. The voltage gain v2 /vg .
if: c. The value to which the 16- load resistance should
Vs1 = 1700 V be changed so it will absorb maximum power from
the given source.
Vs2 = 170 V(/2) V
Z1 = 0.7(/6)  1

Z2 = 1.50.105  1:4 +

Z3 = 0.3 + j 0.4  vg +

_ v2 16
_

+ Z1
_ Vs1 Figure P7.29
Z3
+ Z2
7.30 An ideal transformer is rated to deliver 400 kVA at
_ Vs2 460 V to a customer as shown in Figure P7.30.
a. How much current can the transformer supply to
Figure P7.26 the customer?
b. If the customers load is purely resistive (i.e., if
7.27 If the voltage and current supplied to a circuit or pf = 1), what is the maximum power that the
c. If the customers power factor is 0.8 (lagging),
Vs = 170 0.157 V Is = 130.28 A what is the maximum usable power the customer
a. The power supplied by the source which is d. What is the maximum power if the pf is 0.7
dissipated as heat or work in the circuit (load). (lagging)?
b. The power stored in reactive components in the e. If the customer requires 300 kW to operate, what is
circuit (load). the minimum power factor with the given size
c. The power factor angle and power factor. transformer?

Section 3: Transformers
7.28 A center-tap transformer has the schematic
representation shown in Figure P7.28. The
Z
primary-side voltage is stepped down to a Customers
secondary-side voltage, Vsec , by a ratio of n : 1. On the load
secondary side, Vsec1 = Vsec2 = 12 Vsec .
a. If Vprim = 12032 V and n = 9, find Vsec , Vsec1 , Ideal transformer Customer
and Vsec2 .
Figure P7.30
b. What must n be if Vprim = 2080.175 V and we
desire |Vsec2 | to be 8.7 V?
7.31 For the ideal transformer shown in Figure P7.31,
find vo (t) if vS (t) is 294 cos 377t.
n:1
+
+ +
~ 100 2.5:1
Vsec1
_ +
~ ~
Vprim Vsec
+ vS(t) +

_ 25 vo(t)
~
Vsec2 _
_ _
_
Figure P7.31
Figure P7.28
7.32 If the transformer shown in Figure P7.32 is ideal,
7.29 For the circuit shown in Figure P7.29, find: find the turns ratio N = 1/n that will provide
a. The total resistance seen by the voltage source. maximum power transfer to the load.

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332 Chapter 7 AC Power

RS n:1 leaving the secondary side open. The current into the
+ primary side is measured, as is the power dissipated.
vS(t) +

_ RL vo(t)

_
A W
RS = 1800 RL = 8
~
Vs +

_ V
Figure P7.32

(a)
7.33 Assume the 8- resistor is the load in the circuit
shown in Figure P7.33. Assume a turns ratio of 1 : n.
What value of n will result in the load resistor A W
absorbing maximum power from the source? +
~
Vs V
_

3 (b)

1:n
rw Lw
+
~ ~
Ip IS
vg +

_ 8
rc Lc Vsecondary
4

Figure P7.35

## 7.34 If we knew that the transformer shown in Figure

P7.34 was to deliver 50 A at 110 V rms with a certain The short-circuit test, shown in Figure P7.35(b),
resistive load, what rms phasor voltage source, VS , is performed by increasing the primary voltage until
would provide this voltage and current? rated current is going into the transformer while the
secondary side is short-circuited. The current into the
transformer, the applied voltage, and the power
dissipated are measured.
~ 50 A
The equivalent circuit of a transformer is shown
IS
1 j0.1 VW 2 : 1 in Figure P7.35(c), where rw and Lw represent the
+
~
IL winding resistance and inductance, respectively, and rc
~
IW
and Lc represent the losses in the core of the
~ + transformer and the inductance of the core. The ideal
Vs
_ j200 110 V
transformer is also included in the model.
With the open-circuit test, we may assume that
_
IP = IS = 0. Then all of the current that is measured
Figure P7.34
is directed through the parallel combination of rc and
Lc . We also assume that |rc ||j Lc | is much greater
than rw + j Lw . Using these assumptions and the
open-circuit test data, we can find the resistance rc and
7.35 A method for determining the equivalent circuit of the inductance Lc .
a transformer consists of two tests: the open-circuit test In the short-circuit test, we assume that Vsecondary
and the short-circuit test. The open-circuit test, shown is zero, so that the voltage on the primary side of the
in Figure P7.35(a), is usually done by applying rated ideal transformer is also zero, causing no current flow
voltage to the primary side of the transformer while through the rc Lc parallel combination. Using this

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Part I Circuits 333

~ ~
assumption with the short-circuit test data, we are able VW VR
to find the resistance rw and inductance Lw .
Using the following test data, find the equivalent +

_
+

_
circuit of the transformer: 120 2 3 120 0

## Open-circuit test: V = 241 V

I = 0.95 A
P = 32 W
120 4 3
_
=5V
+
Short-circuit test: V
I = 5.25 A
P = 26 W ~
VB
7.36 Using the methods of Problem 7.35 and the
following data, find the equivalent circuit of the 7.40 For the three-phase circuit shown in Figure P7.40,
transformer tested: find the currents IW , IB , IR , and IN .

## Open-circuit test: VP = 4,600 V ~

IR
~
IOC = 0.7 A VR = 110 0 V
50
~
P = 200 W IW
j 45
~
Short-circuit test: P = 50 W VW = 110 2 3 V

VP = 5.2 V ~
_ j 20
IB
~
The transformer is a 460-kVA transformer, and the VB = 110 4 3 V
tests are performed at 60 Hz. ~
IN

## Section 4: Three-Phase Power

Figure P7.40
7.37 The magnitude of the phase voltage of a balanced
three-phase wye system is 100 V. Express each phase 7.41 For the circuit shown in Figure P7.41, find the
and line voltage in both polar and rectangular currents IR , IW , IB , and IN .
coordinates.
~
IR
7.38 The phase currents in a four-wire wye-connected ~
VR = 220 0
load are as follows: ~ 10
IW
~ 10
Ian = 100, Ibn = 125/6, Icn = 82.88 VW = 220 2 3

## Determine the current in the neutral wire.

~ 10
IB
7.39 For the circuit shown in Figure P7.39, we see that ~
VB = 220 4 3
each voltage source has a phase difference of 2/3 in ~
IN
relation to the others.
a. Find VRW , VW B , and VBR , where
VRW = VR VW , VW B = VW VB , and
VBR = VB VR . Figure P7.41
b. Repeat part a, using the calculations
7.42 In the circuit of Figure P7.42:
VRW = VR 3 /6 vs1 = 170 cos(t) V

VW B = VW 3/6 vs2 = 170 cos(t + 2/3) V
vs3 = 170 cos(t 2/3) V
VBR = VB 3/6
c. Compare the results of part a with the results of f = 60 Hz Z1 = 0.520 

part b. Z2 = 0.350  Z3 = 1.7 90 

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334 Chapter 7 AC Power

~ ~ ~
Determine the current through Z1 using: V1 V2 V3
a. Loop/mesh analysis.
b. Node analysis. R1 R2 R3
c. Superposition. 1st 2nd 3rd
winding winding winding
L1 L2 L3
~
Vs1
_+

Z1
+
_

~ ~ Z2 Z3
Vs3
+_

Vs2
(a)
~ ~ ~
Figure P7.42 V1 V2 V3

## 7.43 Determine the current through R in the circuit of

Figure P7.43: R1 R2 R3

v1 = 170 cos(t) V
L1 L2 L3
v2 = 170 cos(t 2/3) V
v3 = 170 cos(t + 2/3) V
f = 400 Hz R = 100 
C = 0.47 F L = 100 mH (b)

~ ~ ~
V1 VR, IR
~
V3
_+

+ R1
R
416 _30
+
_

~ ~ R2
C L VW, IW
+_

~ _ L1 L2
V2 416 210 +
L3
416 90
Figure P7.43 ~ ~
+ VB, IB
_ R3
7.44 The three sources in the circuit of Figure P7.44 are
connected in wye configuration and the loads in a delta
IN
configuration. Determine the current through each
impedance.
vs1 = 170 cos(t) V
(c)
vs2 = 170 cos(t + 2/3) V
vs3 = 170 cos(t 2/3) V Figure P7.45

## f = 60 Hz Z1 = 30  7.46 With reference to the motor of Problem 7.44,

Z2 = 7/2  Z3 = 0 j 11  a. How much power (in watts) is delivered to the
motor?
b. What is the motors power factor?
Z1
~ c. Why is it common in industrial practice not to
_+

V1
Z2 connect the ground lead to motors of this type?
+
_

~ ~
V3
+_

## V2 Z3 7.47 Find the apparent power and the real power

delivered to the load in the Y-* circuit shown in Figure
Figure P7.44 P7.47. What is the power factor? Assume rms values.
7.45 If we model each winding of a three-phase motor W
like the circuit shown in Figure P7.45(a) and connect R
the windings as shown in Figure P7.45(b), we have the +
+

_ _
three-phase circuit shown in Figure P7.45(c). The 120 0 120 2 3 j60 40
motor can be constructed so that R1 = R2 = R3 and N
L1 = L2 = L3 , as is the usual case. If we connect the 40 j 60
120 4 3
_
motor as shown in Figure P7.45(c), find the currents
+ 40 j 60
IR , IW , IB , and IN , assuming that the resistances are B
40  each and each inductance is 5 mH. The frequency
of each of the sources is 60 Hz. Figure P7.47

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Part I Circuits 335

7.48 The electric power company is concerned with the 7.49 A balanced, three-phase Y-connected source with
loading of its transformers. Since it is responsible to a 230-Vrms line voltages has a balanced Y-connected
large number of customers, it must be certain that it load of 3+j 4  per phase. For the case that the lines
can supply the demands of all customers. The power have zero impedance, find all three line currents and
companys transformers will deliver rated kVA to the the total real power absorbed by the load.
secondary load. However, if the demand were to 7.50 The circuit shown in Figure P7.50 is a Y-*-Y
increase to a point where greater than rated current connected three-phase circuit. The primaries of the
were required, the secondary voltage would have to transformers are wye-connected, the secondaries are
drop below rated value. Also, the current would delta-connected, and the load is wye-connected. Find
increase, and with it the I 2 R losses (due to winding the currents IRP , IW P , IBP , IA , IB , and IC .
resistance), possibly causing the transformer to
overheat. Unreasonable current demand could be ~ ~
caused, for example, by excessively low power factors IRP IA
4:1
460 0 V 10 10
The customer, on the other hand, is not greatly
concerned with an inefficient power factor, provided
that sufficient power reaches the load. To make the _ j7 _ j7
~
customer more aware of power factor considerations, IWP
4:1
the power company may install a penalty on the _ j7
460 2 3 V
customers bill. A typical penalty-power factor chart is
shown in Table 7.3. Power factors below 0.7 are not
permitted. A 25 percent penalty will be applied to any ~ 10
billing after two consecutive months in which the IBP ~
4:1 IB
460 _2 3 V
customers power factor has remained below 0.7.

Table 7.3 ~
IC

## Power factor Penalty Ideal

transformers
0.850 and higher None Figure P7.50
0.8 to 0.849 1%
0.75 to 0.799 2%
7.51 For the circuit shown in Figure P7.51, find the
0.7 to 0.749 3% currents IA , IB , IC , and IN , and the real power
Courtesy of Detroit Edison.

## The Y-Y circuit shown in Figure P7.48 is ~

IA
representative of a three-phase motor load. Assume
rms values. 220 0 V 40 20
a. Find the total power supplied to the motor. j10
_ j5
b. Find the power converted to mechanical energy if
the motor is 80 percent efficient.
c. Find the power factor. ~ j5
IB
d. Does the company risk facing a power factor
penalty on its next bill if all the motors in the 110 2 3 V 20
factory are similar to this one? ~
IC

110 _2 3 V
W
R 5 In
+ + j6

_
_ j6
120 2 3 Figure P7.51
120 0 5
5
120 _2 3
_
+
j6
B

Figure P7.48

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