Chapter 11 Solutions
11
Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC .
VAB = VAN + VNB = VAN − VBN
Since VAN , VBN , and VCN form a balanced set, and VAN = 240/ − 30◦ V, and the phase sequence is positive,
VBN = |VAN |//VAN − 120◦ = 240/ − 30◦ − 120◦ = 240/ − 150◦ V
Then,
VAB = VAN − VBN = (240/ − 30◦ ) − (240/ − 150◦ ) = 415.46/0◦ V
Since VAB , VBC , and VCA form a balanced set with a positive phase sequence, we can find VBC from VAB :
VBC = |VAB |/(/VAB − 120◦ ) = 415.69/0◦ − 120◦ = 415.69/ − 120◦ V
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
11–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
11–2
CHAPTER 11. Balanced Three-Phase Circuits
Thus,
VBC = 415.69/ − 120◦ V
AP 11.2 Make a sketch:
We know VCN and wish to find VAB . To do this, write a KVL equation to find VBC , and use the known phase angle relationship between VAB and VBC to find VAB .
VBC = VBN + VNC = VBN − VCN
Since VAN , VBN , and VCN form a balanced set, and VCN = 450/ − 25◦ V, and the phase sequence is negative,
VBN = |VCN |//VCN − 120◦ = 450/ − 23◦ − 120◦ = 450/ − 145◦ V
Then,
VBC = VBN − VCN = (450/ − 145◦ ) − (450/ − 25◦ ) = 779.42/ − 175◦ V
Since VAB , VBC , and VCA form a balanced set with a negative phase sequence, we can find VAB from VBC :
VAB = |VBC |//VBC − 120◦ = 779.42/ − 295◦ V
But we normally want phase angle values between +180◦ and −180◦ . We add
360◦ to the phase angle computed above. Thus,
VAB = 779.42/65◦ V
AP
Balanced Three-Phase Circuits
Assessment Problems
AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC .
VAB = VAN + VNB = VAN − VBN
Since VAN , VBN , and VCN form a balanced set, and VAN = 240/ − 30◦ V, and the phase sequence is positive,
VBN = |VAN |//VAN − 120◦ = 240/ − 30◦ − 120◦ = 240/ − 150◦ V
Then,
VAB = VAN − VBN = (240/ − 30◦ ) − (240/ − 150◦ ) = 415.46/0◦ V
Since VAB , VBC , and VCA form a balanced set with a positive phase sequence, we can find VBC from VAB :
VBC = |VAB |/(/VAB − 120◦ ) = 415.69/0◦ − 120◦ = 415.69/ − 120◦ V
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval
11–1 system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department,
Pearson Education, Inc., Upper Saddle River, NJ 07458.
11–2
CHAPTER 11. Balanced Three-Phase Circuits
Thus,
VBC = 415.69/ − 120◦ V
AP 11.2 Make a sketch:
We know VCN and wish to find VAB . To do this, write a KVL equation to find VBC , and use the known phase angle relationship between VAB and VBC to find VAB .
VBC = VBN + VNC = VBN − VCN
Since VAN , VBN , and VCN form a balanced set, and VCN = 450/ − 25◦ V, and the phase sequence is negative,
VBN = |VCN |//VCN − 120◦ = 450/ − 23◦ − 120◦ = 450/ − 145◦ V
Then,
VBC = VBN − VCN = (450/ − 145◦ ) − (450/ − 25◦ ) = 779.42/ − 175◦ V
Since VAB , VBC , and VCA form a balanced set with a negative phase sequence, we can find VAB from VBC :
VAB = |VBC |//VBC − 120◦ = 779.42/ − 295◦ V
But we normally want phase angle values between +180◦ and −180◦ . We add
360◦ to the phase angle computed above. Thus,
VAB = 779.42/65◦ V
AP