Solution manual fundamentals of electric circuits 3rd edition chapter12

Chapter 12, Problem 1.
If Vab = 400 V in a balanced Y-connected three-phase generator, find the phase voltages,
assuming the phase sequence is:
(a) abc

(b) acb

Chapter 12, Solution 1.
(a)

If Vab = 400 , then
400
Van =
∠ - 30° = 231∠ - 30° V
3
Vbn = 231∠ - 150° V

Vcn = 231∠ - 270° V
(b)

For the acb sequence,
Vab = Van − Vbn = Vp ∠0° − Vp ∠120°

⎛ 1
3⎞
Vab = Vp ⎜⎜1 + − j ⎟⎟ = Vp 3∠ - 30°
2 ⎠
⎝ 2
i.e. in the acb sequence, Vab lags Van by 30°.
Hence, if Vab = 400 , then
400

Van =
∠30° = 231∠30° V
3
Vbn = 231∠150° V
Vcn = 231∠ - 90° V
Chapter 12, Problem 2.
What is the phase sequence of a balanced three-phase circuit for which Van = 160 ∠30° V
and Vcn = 160 ∠ − 90° V? Find Vbn.

Chapter 12, Solution 2.
Since phase c lags phase a by 120°, this is an acb sequence.
Vbn = 160∠(30° + 120°) = 160∠150° V
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Chapter 12, Problem 3.
Determine the phase sequence of a balanced three-phase circuit in which
Vbn = 208 ∠130° V and Vcn = 208 ∠10° V. Obtain Van .
Chapter 12, Solution 3.
Since Vbn leads Vcn by 120°, this is an abc sequence.
Van = 208∠(130° + 120°) = 208∠ 250° V
Chapter 12, Problem 4.
A three-phase system with abc sequence and VL = 200 V feeds a Y-connected load with
ZL = 40 ∠30°Ω . Find the line currents.
Chapter 12, Solution 4.
VL = 200 = 3V p

200
3
V
200 < 0o
I a = an =
= 2.887 < −30o A
o
ZY
3x 40 < 30
I b = I a < −120o = 2.887 < −150o A
⎯⎯
→ Vp =

I c = I a < +120o = 2.887 < 90o A
Chapter 12, Problem 5.

For a Y-connected load, the time-domain expressions for three line-to-neutral voltages at
the terminals are:

vAN = 150 cos ( ω t + 32º) V
vBN = 150 cos ( ω t – 88º) V
vCN = 150 cos ( ω t + 152º) V
Write the time-domain expressions for the line-to-line voltages vAN, vBC, and vCA .
Chapter 12, Solution 5.
VAB = 3V p < 30o = 3x150 < 32o + 30o = 260 < 62o

Thus,
v AB = 260 cos(ωt + 62o ) V

Using abc sequence,
vBC = 260 cos(ωt − 58o ) V
vCA = 260 cos(ωt + 182o ) V
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Chapter 12, Problem 6.

For the Y-Y circuit of Fig. 12.41, find the line currents, the line voltages, and the load
voltages.

Figure 12.41
For Prob. 12.6.
Chapter 12, Solution 6.
Z Y = 10 + j5 = 11.18∠26.56°

The line currents are
Van
220 ∠0°
Ia =
=
= 19.68∠ - 26.56° A
Z Y 11.18∠26.56°
I b = I a ∠ - 120° = 19.68∠ - 146.56° A
I c = I a ∠120° = 19.68∠93.44° A

The line voltages are
Vab = 220 3 ∠30° = 381∠30° V
Vbc = 381∠ - 90° V
Vca = 381∠ - 210° V

The load voltages are
VAN = I a Z Y = Van = 220∠0° V
VBN = Vbn = 220∠ - 120° V
VCN = Vcn = 220∠120° V

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Chapter 12, Problem 7.

Obtain the line currents in the three-phase circuit of Fig. 12.42 on the next page.

Figure 12.42
For Prob. 12.7.

Chapter 12, Solution 7.

This is a balanced Y-Y system.

440∠0° V

+
−

ZY = 6 − j8 Ω

Using the per-phase circuit shown above,
440∠0°
Ia =
= 44∠53.13° A
6 − j8
I b = I a ∠ - 120° = 44∠ - 66.87° A
I c = I a ∠120° = 44∠173.13° A

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Chapter 12, Problem 8.

In a balanced three-phase Y-Y system, the source is an abc sequence of voltages and Van
= 100 ∠20° V rms. The line impedance per phase is 0.6 + j1.2 Ω , while the per-phase
impedance of the load is 10 + j14 Ω . Calculate the line currents and the load voltages.

Chapter 12, Solution 8.

Consider the per phase equivalent circuit shown below.
Zl

Van

Ia =

+
_

ZL

Van
100 < 20o
5.396∠
=
= 5.3958
< –35.1˚
−35.1o AA
Z L + Z l 10.6 + j15.2

I b = I a < −120o = 5.3958
−155.1oAA
5.396∠<–155.1˚
5.396∠<84.9˚
I c = I a < +120o = 5.3958
84.9oAA
o
VLa = I a Z L = (4.4141 − j 3.1033)(10 + j14) = 92.83
V
92.83<∠19.35
19.35˚

A

92.83<∠−–100.65˚
VLb = VLa < −120o = 92.83
100.65o A
V
o
VLc = VLa < +120o = 92.83
92.83<∠139.35
139.35˚ V
A

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Chapter 12, Problem 9.

A balanced Y-Y four-wire system has phase voltages
Van = 120∠0°

Vbn = 120∠ − 120°

Vcn = 120∠120° V

The load impedance per phase is 19 + j13 Ω , and the line impedance per phase is
1 + j2 Ω . Solve for the line currents and neutral current.

Chapter 12, Solution 9.
Ia =

Van
120 ∠0°
=
= 4.8∠ - 36.87° A
Z L + Z Y 20 + j15

I b = I a ∠ - 120° = 4.8∠ - 156.87° A
I c = I a ∠120° = 4.8∠83.13° A

As a balanced system, I n = 0 A

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Chapter 12, Problem 10.

For the circuit in Fig. 12.43, determine the current in the neutral line.

Figure 12.43
For Prob. 12.10.
Chapter 12, Solution 10.

Since the neutral line is present, we can solve this problem on a per-phase basis.
For phase a,
Ia =

Van
220∠0°
220
=
= 7.642∠20.32°
=
Z A + 2 27 − j10 28.79∠ − 20.32°

Ib =

Vbn
220 ∠ - 120°
=
= 10 ∠ - 120°
ZB + 2
22

Ic =

Vcn
220∠120° 220∠120°
=
=
= 16.923∠97.38°
12 + j5
13∠22.62°

ZC + 2

For phase b,

For phase c,

The current in the neutral line is
I n = -(I a + I b + I c ) or - I n = I a + I b + I c
- I n = (7.166 + j2.654) + (-5 − j8.667) + (-2.173 + j16.783)
I n = 0.007 − j10.77 = 10.77∠90°A
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Chapter 12, Problem 11.

In the Y- ∆ system shown in Fig. 12.44, the source is a positive sequence with
V an = 120 ∠0° V and phase impedance Z p = 2 – j3 Ω . Calculate the line voltage V L and
the line current I L.

Figure 12.44
For Prob. 12.11.

Chapter 12, Solution 11.

VAB = Vab = 3V p < 30o = 3(120) < 30o
VL =| Vab |= 3 x120 = 207.85 V

I AB

3V p < 30o
VAB
=
=
ZA
2 − j3

I a = I AB 3 < −30o =

3V p < 0o
2 − j3

=

3 x120
= 55.385 + j83.07
2 − j3

I L =| I a |= 99.846 A

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Chapter 12, Problem 12.

Solve for the line currents in the Y-∆ circuit of Fig. 12.45. Take Z ∆ = 60∠45°Ω .

Figure 12.45
For Prob. 12.12.

Chapter 12, Solution 12.

Convert the delta-load to a wye-load and apply per-phase analysis.

Ia
110∠0° V

ZY =

+
−

ZY

Z∆
= 20 ∠45° Ω
3

110∠0°
= 5.5∠ - 45° A
20∠45°
I b = I a ∠ - 120° = 5.5∠ - 165° A
Ia =

I c = I a ∠120° = 5.5∠75° A

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Chapter 12, Problem 13.

In the balanced three-phase Y-∆ system in Fig. 12.46, find the line current I L
and the average power delivered to the load.

Figure 12.46
For Prob. 12.13.

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Chapter 12, Solution 13.
Convert the delta load to wye as shown below.
A

110∠0o V rms
2Ω

–+

110∠–120o V rms

ZY
N

2Ω
ZY

–+

110∠120o V rms

ZY

2Ω

–+
1
ZY = Z = 3 − j 2 Ω
3
We consider the single phase equivalent shown below.
2Ω
110∠0˚ V rms

+
_

3 – j2 Ω

110
= 20.4265 < 21.8o
2 + 3 − j2
I L =| I a |= 20.43 A
Ia =

S = 3|Ia|2ZY = 3(20.43)2(3–j2) = 4514∠–33.96˚ = 3744 – j2522
P = Re(S) = 3744 W.
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Chapter 12, Problem 14.

Obtain the line currents in the three-phase circuit of Fig. 12.47 on the next page.

100 –120°

Figure 12.47
For Prob. 12.14.

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Chapter 12, Solution 14.
We apply mesh analysis.

1 + j 2Ω

A

a
+
100∠0 o V
-

ZL
ZL

I3

n
100∠120 o V
+
c

I1

-

100∠120 o V
+

b

I2

B

C
Z L = 12 + j12Ω

1 + j 2Ω

1 + j 2Ω
For mesh,
− 100 + 100∠120 o + I 1 (14 + j16) − (1 + j 2) I 2 − (12 + j12) I 3 = 0
or

(14 + j16) I 1 − (1 + j 2) I 2 − (12 + j12) I 3 = 100 + 50 − j86.6 = 150 − j86.6 (1)
For mesh 2,
100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j12) I 3 + (14 + j16) I 2 = 0
or
− (1 + j 2) I 1 + (14 + j16) I 2 − (12 + j12) I 3 = −50 − j86.6 + 50 − j86.6 = − j173.2 (2)
For mesh 3,
− (12 + j12) I 1 − (12 + j12) I 2 + (36 + j 36) I 3 = 0
(3)
Solving (1) to (3) gives
I 1 = −3.161 − j19.3,

I 2 = −10.098 − j16.749,

I 3 = −4.4197 − j12.016

I aA = I 1 = 19.58∠ − 99.3 A
o

I bB = I 2 − I 1 = 7.392∠159.8 o A
I cC = − I 2 = 19.56∠58.91o A

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Chapter 12, Problem 15.
The circuit in Fig. 12.48 is excited by a balanced three-phase source with a line
voltage of 210 V. If Z l = 1 + j1 Ω , Z ∆ = 24 − j 30Ω , and ZY = 12 + j5 Ω , determine the
magnitude of the line current of the combined loads.

Figure 12.48
For Prob. 12.15.
Chapter 12, Solution 15.
Convert the delta load, Z ∆ , to its equivalent wye load.
Z∆
= 8 − j10
Z Ye =
3
(12 + j5)(8 − j10)
= 8.076 ∠ - 14.68°
20 − j5

Z p = 7.812 − j2.047

Z p = Z Y || Z Ye =

Z T = Z p + Z L = 8.812 − j1.047
Z T = 8.874 ∠ - 6.78°

We now use the per-phase equivalent circuit.
Vp
210
Ia =
,
where Vp =
Zp + ZL
3
Ia =

210
3 (8.874 ∠ - 6.78°)

= 13.66 ∠6.78°

I L = I a = 13.66 A
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Chapter 12, Problem 16.

A balanced delta-connected load has a phase current I AC = 10 ∠ − 30° A.
(a) Determine the three line currents assuming that the circuit operates in the positive
phase sequence.
(b) Calculate the load impedance if the line voltage is V AB = 110 ∠0° V.

Chapter 12, Solution 16.

(a)

I CA = - I AC = 10∠(-30° + 180°) = 10∠150°

This implies that
I AB = 10 ∠30°
I BC = 10∠ - 90°
I a = I AB 3 ∠ - 30° = 17.32∠0° A

I b = 17.32∠ - 120° A
I c = 17.32∠120° A

(b)

Z∆ =

VAB 110 ∠0°
=
= 11∠ - 30° Ω
I AB 10 ∠30°

Chapter 12, Problem 17.

A balanced delta-connected load has line current I a = 10 ∠ − 25° A. Find the phase
currents I AB , I BC , and I CA.
Chapter 12, Solution 17.

I a = I AB 3 < −30o

⎯⎯
→ I AB =

Ia
10
=
< −25o + 30o = 5.773 < 5o A
o
3 < −30
3

I BC = I AB < −120o = 5.775 < −115o A
I CA = I AB < +120o = 5.775 < 125o A

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Chapter 12, Problem 18.

If V an = 440 ∠60° V in the network of Fig. 12.49, find the load phase currents I AB , I BC,
and I CA .

Figure 12.49
For Prob. 12.18.

Chapter 12, Solution 18.
VAB = Van 3 ∠30° = (440 ∠60°)( 3 ∠30°) = 762.1∠90°
Z ∆ = 12 + j9 = 15∠36.87°
I AB =

VAB 762.1∠90°
=
= 50.81∠53.13° A
Z ∆ 15∠36.87°

I BC = I AB ∠ - 120° = 50.81∠ - 66.87° A
I CA = I AB ∠120° = 50.81∠173.13° A

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Chapter 12, Problem 19.

For the ∆ - ∆ circuit of Fig. 12.50, calculate the phase and line currents.

Figure 12.50
For Prob. 12.19.

Chapter 12, Solution 19.
Z ∆ = 30 + j10 = 31.62 ∠18.43°

The phase currents are
Vab
173∠0°
=
= 5.47 ∠ - 18.43° A
I AB =
Z ∆ 31.62 ∠18.43°
I BC = I AB ∠ - 120° = 5.47 ∠ - 138.43° A
I CA = I AB ∠120° = 5.47 ∠101.57° A

The line currents are
I a = I AB − I CA = I AB 3 ∠ - 30°
I a = 5.47 3 ∠ - 48.43° = 9.474∠ - 48.43° A

I b = I a ∠ - 120° = 9.474∠ - 168.43° A
I c = I a ∠120° = 9.474∠71.57° A

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Chapter 12, Problem 20.

Refer to the ∆ - ∆ circuit in Fig. 12.51. Find the line and phase currents. Assume that the
load impedance is ZL = 12 + j9 Ω per phase.

Figure 12.51
For Prob. 12.20.

Chapter 12, Solution 20.
Z ∆ = 12 + j9 = 15∠36.87°

The phase currents are

210∠0°
= 14∠ - 36.87° A
15∠36.87°
= I AB ∠ - 120° = 14∠ - 156.87° A

I AB =
I BC

I CA = I AB ∠120° = 14∠83.13° A

The line currents are
I a = I AB 3 ∠ - 30° = 24.25∠ - 66.87° A
I b = I a ∠ - 120° = 24.25∠ - 186.87° A
I c = I a ∠120° = 24.25∠53.13° A

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Chapter 12, Problem 21.

Three 230-V generators form a delta-connected source that is connected to a balanced
delta-connected load of ZL = 10 + j8 Ω per phase as shown in Fig. 12.52.
(a) Determine the value of IAC.
(b) What is the value of Ib?

Figure 12.52
For Prob. 12.21.

Chapter 12, Solution 21.

(a)

− 230∠120°
− 230∠120°
=
= 17.96∠ − 98.66° A(rms)
10 + j8
12.806∠38.66°
17.96∠–98.66˚ A rms

I AC =

230∠ − 120 230∠0°
−
10 + j8
10 + j8
= 17.96∠ − 158.66° − 17.96∠ − 38.66°
= −16.729 − j6.536 − 14.024 + j11.220 = −30.75 + j4.684
= 31.10∠171.34° A

I bB = I BC + I BA = I BC − I AB =
(b)

31.1∠171.34˚ A rms

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Chapter 12, Problem 22.

Find the line currents Ia, Ib, and Ic in the three-phase network of Fig. 12.53 below.
Take Z ∆ = 12 − j15Ω , ZY = 4 + j6 Ω , and Zl = 2 Ω .

208 0° V

Figure 12.53
For Prob. 12.22.
Chapter 12, Solution 22.

Convert the ∆-connected source to a Y-connected source.
Vp
208
Van =
∠ - 30° =
∠ - 30° = 120 ∠ - 30°
3
3
Convert the ∆-connected load to a Y-connected load.
Z
(4 + j6)(4 − j5)
Z = Z Y || ∆ = (4 + j6) || (4 − j5) =
3
8+ j
Z = 5.723 − j0.2153
ZL

Van

Ia

+
−

Z

Van
120∠ − 30°
= 15.53∠ - 28.4° A
=

Z L + Z 7.723 − j0.2153
I b = I a ∠ - 120° = 15.53∠ - 148.4° A

Ia =

I c = I a ∠120° = 15.53∠91.6° A
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Chapter 12, Problem 23.
A three-phase balanced system with a line voltage of 202 V rms feeds a delta-connected
load with Zp = 25 ∠60°Ω .
(a) Find the line current.
(b) Determine the total power supplied to the load using two wattmeters connected to the
A and C lines.

Chapter 12, Solution 23.

(a)

I AB =

VAB
202
=
Z∆

25∠60 o
o

I a = I AB 3∠ − 30 =

202 3∠ − 30 o
25∠60

o

= 13.995∠ − 90 o

I L =| I a |= 13.995A
(b)
⎛ 202 3 ⎞
⎟ cos 60 o = 2.448 kW
P = P1 + P2 = 3VL I L cos θ = 3 (202)⎜⎜
⎟
25
⎝
⎠

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Chapter 12, Problem 24.

A balanced delta-connected source has phase voltage Vab = 416 ∠30° V and a positive
phase sequence. If this is connected to a balanced delta-connected load, find the line and
phase currents. Take the load impedance per phase as 60 ∠30°Ω and line impedance per
phase as 1 + j1 Ω .
Chapter 12, Solution 24.
Convert both the source and the load to their wye equivalents.
Z∆
= 20 ∠30° = 17.32 + j10
ZY =
3
Vab
∠ - 30° = 240.2∠0°
Van =
3
We now use per-phase analysis.

1+jΩ

Van

Ia =

+
−

Ia

20∠30° Ω

Van

240.2
=
= 11.24∠ - 31° A
(1 + j) + (17.32 + j10) 21.37 ∠31°

I b = I a ∠ - 120° = 11.24∠ - 151° A
I c = I a ∠120° = 11.24∠89° A
But
I AB =

I a = I AB 3 ∠ - 30°
11.24 ∠ - 31°
3 ∠ - 30°

= 6.489∠ - 1° A

I BC = I AB ∠ - 120° = 6.489∠ - 121° A
I CA = I AB ∠120° = 6.489∠119° A
PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
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Chapter 12, Problem 25.
In the circuit of Fig. 12.54, if Vab = 440 ∠10° , Vbc = 440 ∠250° , Vca = 440 ∠130°
V, find the line currents.

Figure 12.54

For Prob. 12.25.

Chapter 12, Solution 25.
Convert the delta-connected source to an equivalent wye-connected source and
consider the single-phase equivalent.
Ia =

where

440 ∠(10° − 30°)

3 ZY
Z Y = 3 + j2 + 10 − j8 = 13 − j6 = 14.32 ∠ - 24°.78°

Ia =

440 ∠ - 20°
3 (14.32 ∠ - 24.78°)

= 17.74∠4.78° A

I b = I a ∠ - 120° = 17.74∠ - 115.22° A
I c = I a ∠120° = 17.74 ∠124.78° A

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
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Chapter 12, Problem 26.
For the balanced circuit in Fig. 12.55, Vab = 125 ∠0° V. Find the line currents IaA, IbB, and
IcC.

Figure 12.55
For Prob. 12.26.

Chapter 12, Solution 26.
Transform the source to its wye equivalent.
Vp
Van =
∠ - 30° = 72.17 ∠ - 30°
3
Now, use the per-phase equivalent circuit.
Van
,
Z = 24 − j15 = 28.3∠ - 32°
I aA =
Z
I aA =

72.17 ∠ - 30°
= 2.55∠2° A
28.3∠ - 32°

I bB = I aA ∠ - 120° = 2.55∠ - 118° A
I cC = I aA ∠120° = 2.55∠122° A

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part

of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.


Chapter 12, Problem 27.
A ∆-connected source supplies power to a Y-connected load in a three-phase
balanced system. Given that the line impedance is 2 + j1 Ω per phase while the load
impedance is 6 + j4 Ω per phase, find the magnitude of the line voltage at the load.
Assume the source phase voltage Vab = 208 ∠0° V rms.
Chapter 12, Solution 27.
Since ZL and Z l are in series, we can lump them together so that
ZY = 2 + j + 6 + j 4 = 8 + j 5
VP
< −30o
208 < −30o
Ia = 3
=
ZY
3(8 + j 5)
208(0.866 − j 0.5)(6 + j 4)
VL = (6 + j 4) I a =
= 80.81 − j 43.54
3(8 + j 5)
|VL| = 91.79 V

Chapter 12, Problem 28.
The line-to-line voltages in a Y-load have a magnitude of 440 V and are in the positive
sequence at 60 Hz. If the loads are balanced with Z1 = Z 2 = Z 3 = 25 ∠30° , find all line

currents and phase voltages.
Chapter 12, Solution 28.
VL = Vab = 440 = 3VP or VP = 440/1.7321 = 254
For reference, let VAN = 254∠0˚ V which leads to
VBN = 254∠–120˚ V and VCN = 254∠120˚ V.
The line currents are found as follows,
Ia = VAN/ZY = 254/25∠30˚ = 10.16∠–30˚ A.
This leads to, Ib = 10.16∠–150˚ A and Ic = 10.16∠90˚ A.

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part
of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior
written permission of the publisher, or used beyond the limited distribution to teachers and educators
permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual,
you are using it without permission.