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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield $15.99   Add to cart

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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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  • Power System Analysis and Design
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  • Power System Analysis And Design

Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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  • November 11, 2024
  • 414
  • 2024/2025
  • Exam (elaborations)
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  • Power System Analysis and Design
  • Power System Analysis and Design
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Solutions Manual
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Power System Analysis And Design

Seventh Edition




J. Duncan Glover
Mulukutla S. Sarma
Thomas J. Overbye

, Contents
Chapter 2 1

Chapter 3 27

Chapter 4 71

Chapter 5 95

Chapter 6 137

Chapter 7 175

Chapter 8 195

Chapter 9 231

Chapter 10 303

Chapter 11 323

Chapter 12 339

Chapter 13 353

Chapter 14 379

,Chapter 2
Fundamentals

Answers To Multiple-Choice Type Questions
2.1 B 2.19 A
2.2 A 2.20 A. C
2.3 C B. A
2.4 A C. B
2.5 B 2.21 A
2.6 C 2.22 A
2.7 A 2.23 B
2.8 C 2.24 A
2.9 A 2.25 A
2.10 C 2.26 B
2.11 A 2.27 A
2.12 B 2.28 B
2.13 B 2.29 A
2.14 C 2.30 (I) C
2.15 A (Ii) B
2.16 B (Iii) A
2.17 A. A (Iv) D
B. B 2.31 A
C. A 2.32 A
2.18 C




1
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, 2.1 (A A1 = 530 = 5Cos30 + J Sin 30= 4.33 + J 2.5
)
−1 4
(b) A2 = −3 + J4 9 + 16 Tan = 5126.87 = 5ej126.87
= −3
(c) A3 = (4.33 + J2.5) + (−3 + J4 ) = 1.33+ J 6.5 = 6.63578.44
(d) A4 = (530)(5126.87) = 25156.87 = −22.99 + J9.821

A5 = (530) / (5 − 126.87) = 1156.87 = 1ej156.87
(e)

2.2 (A I = 400 − 30 = 346.4 − J200
)
(b) I(T) = 5sin(T + 15) = 5cos(T + 15 − 90) = 5cos(T − 75)

( )
I = 5 2  − 75 = 3.536 − 75 = 0.9151− J3.415

(c) I = (4 2 ) − 30 + 5 − 75 = (2.449 − J1.414) + (1.294 − J4.83)
= 3.743 − J 6.244 = 7.28 − 59.06

2.3 (A Vmax = 359.3 V; Imax = 100 A
)
(b) V = 359.3 2 = 254.1v; I = 100 2 = 70.71a
(c) V = 254.115V; I = 70.71 − 85A
− J6 6 − 90
2.4 (A I1 = 100 = 10 = 7.5 − 90A
) 8 + J6 − J6 8

I2 = I − I1 = 100 − 7.3 − 90 = 10 + J7.5 = 12.536.87A
V = I2 (− J6) = (12.536.87) (6 − 90) = 75 − 53.13V
(B)




2.5 (A) (T) = 277 2 Cos(T + 30) = 391.7 Cos(T + 30)V

(B) I = V / 20 = 13.8530A
I(T) = 19.58cos(T + 30)A




2
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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