Solution Manual - Fundamentals of Electric Circuits 3rd Edition Chapter 12 PDF

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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 ∠ - 30° = 231∠ - 30° V Van = 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 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 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. V L = 200 = 3V p

⎯⎯ → Vp = Ia =

Van = ZY

200 3 200 < 0 o = 2.887 < −30 o A o 3x 40 < 30

I b = I a < −120 o = 2.887 < − 150 o A

I c = I a < +120 o = 2.887 < 90 o 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. V AB = 3V p < 30o = 3x150 < 32o + 30o = 260 < 62o Thus, vAB = 260 cos(ω t + 62o ) V

Using abc sequence, vBC = 260 cos(ωt − 58o ) V vCA = 260 cos(ω t + 182o ) V 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 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. ZY = 10 + j5 = 11.18 ∠26.56 °

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

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° = 44∠ 53.13° A Ia = 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 < 20 o = = 5.396∠–35.1˚ A Z L + Z l 10.6 + j15.2

I b = I a < − 120 = 5.396∠ –155.1˚ A o

I c = I a < +120 o = 5.396∠ 84.9˚ A V La = I aZ L = (4.4141− j 3.1033)(10 + j14) = 92.83∠19.35˚ A VLb = VLa < − 120o = 92.83∠ –100.65˚ A VLc = VLa < + 120 o = 92.83∠ 139.35˚ 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 =

220∠0° 220 Van = = = 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° = = ZC + 2 12 + j5 13∠22.62°

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 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 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.

V AB = Vab =

3V p < 30o =

3(120)< 30o

V L = | V ab |= 3x120 = 207.85 V 3V p < 30o VAB I AB = = ZA 2 − j3

I a = I AB 3 < − 30o =

3Vp < 0o 3 x120 = = 55.385 + j 83.07 2 − j3 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. 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 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 + j2 Ω

A

a + 100∠ 0 o V -

ZL ZL

I3 I1

n 100∠120 o V + c

B

C

-

100 ∠120 o V + b

I2

Z L = 12 + j12Ω

1 + j 2Ω

1 + j 2Ω For mesh, − 100 + 100∠120 o + I 1 (14 + j 16) − (1 + j 2)I 2 − (12 + j 12)I 3 = 0 or

(14 + j16)I 1 − (1 + j 2)I 2 − (12 + j12) I 3 = 100 + 50 − j 86.6 = 150 − j 86.6 (1) For mesh 2, 100∠120 o − 100∠ − 120 o − I 1 (1 + j 2) − (12 + j 12)I 3 + (14 + j 16)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 + j 12)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 = I1 = 19.58∠ − 99.3 A o

I bB = I2 − I1 = 7.392 ∠159.8o 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 ZYe = ∆ = 8 − j10 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 ZT = 8.874∠ - 6.78°

We now use the per-phase equivalent circuit. Vp 210 , where Vp = Ia = Z p +Z L 3 Ia =

210 3 (8.874∠ - 6.78°)

= 13.66∠6.78°

I L = I a = 13.66 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 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

Ib = 17.32∠ - 120° A Ic = 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 < −30 o

⎯⎯ → I AB =

Ia 10 = < − 25 o + 30 o = 5.773 < 5 o 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 173 ∠0 ° Vab = = 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)...