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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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Solutions Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield Solutions Manual For Power System Analysis and Design 7th Edition SM Power System Analysis and Design y J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield | 7th Edition | Copyright 2023

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AN INSTRUCTOR’S SOLUTIONS MANUAL TO ACCOMPANY

POWER SYSTEM ANALYSIS AND DESIGN
SEVENTH EDITION

J. DUNCAN GLOVER
THOMAS J. OVERBYE
MULUKUTLA S. SARMA
Adam B. Birchfield

, Contents

Chapter 2............................................................................................................................. 1

Chapter 3........................................................................................................................... 24

Chapter 4........................................................................................................................... 63

Chapter 5........................................................................................................................... 85

Chapter 6......................................................................................................................... 125

Chapter 7......................................................................................................................... 160

Chapter 8......................................................................................................................... 166

Chapter 9.........................................................................................................................185

Chapter 10....................................................................................................................... 219

Chapter 11....................................................................................................................... 293

Chapter 12....................................................................................................................... 314

Chapter 13....................................................................................................................... 333

Chapter 14....................................................................................................................... 346

Chapter 15....................................................................................................................... 369

,Chapter 2
Fundamentals



2.1 (a) 𝐴1̅ = 9∠30° = 9[𝑐𝑜𝑠 30° + 𝑗 𝑠𝑖𝑛 30°] = 7.8 + 𝑗4.5
5
(b) 𝐴̄2 = 4 + 𝑗5 = √16 + 25 ∠ 𝑡𝑎𝑛−1 4 = 6.40∠128.66° = 6.40𝑒 𝑗128.66°

(c) 𝐴̅3 = (7.8 + 𝑗4.5) + (−4 + 𝑗5) = (11.8 + 𝑗9.5) = 15.15∠38.8°


(d) ̅̅̅
𝐴4 = (9∠30°)(6.4∠51.34°) = 57.6∠81.34° = 8.673 + 𝑗56.9
9∠30°
(e) ̅̅̅
𝐴5 = = 1.41∠81.34° = 1.41𝑒 𝑗81.34°
6.4∠−52.34°

2.2 (a) 𝐼̄ = 500∠ − 30° = 433.01 − 𝑗250
(b) 𝑖(𝑡) = 4𝑠𝑖𝑛(𝜔𝑡 + 15°) = 4𝑐𝑜𝑠(𝜔𝑡 + 15° − 90°) = 4co s(𝜔𝑡 − 75°)
4
𝐼̅ = ∠ − 75° = 2.83∠ − 75° = 0.73 − 𝑗2.73
√2

(c) 𝐼̄ = (5/√2)∠ − 15° + 4∠ − 60° = (3.42 − 𝑗0.92) + (2 − 𝑗3.46)
= 5.42 − 𝑗4.38 = 6.964∠ − 38.94°
13
2.3 𝑉̅2 = 2 ∠(−(125 × 10−6 )(2𝜋60)) kV = 9.19∠ − 2.7° kV


− j6 6 − 90
2.4 (a) I1 = 100 = 10 = 7.5 − 90 A
8 + j6 − j6 8
I 2 = I − I1 = 100 − 7.3 − 90 = 10 + j 7.5 = 12.536.87 A
V = I 2 ( − j 6 ) = (12.536.87 ) ( 6 − 90 ) = 75 − 53.13 V

(b)




1
© 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.

, 2.5 (a)  (t ) = 277 2 cos ( t + 30) = 391.7cos ( t + 30 ) V
̅
𝑉
(b) 𝐼 ̅ = 45 = 6.155∠30° A

𝑖(𝑡) = 8.7 cos (𝜔𝑡 + 30°) A


(c) Z = j L = j ( 2 60 ) (10  10 −3 ) = 3.77190 
I = V Z = ( 277 30 ) ( 3.77190 ) = 73.46  − 60 A
i(t ) = 73.46 2 cos (t − 60 ) =103.9cos (t − 60 ) A

(d) Z = − j 25 
I = V Z = ( 27730 ) ( 25 − 90 ) = 11.08120 A
i(t ) = 11.08 2 cos ( t + 120 ) = 15.67cos ( t + 120 ) A
75
2.6 (a) 𝑉̄ = ( 2) ∠ − 15° = 53.03∠ − 15°;  does not appear in the answer.


(b) 𝜐(𝑡) = 50√2 𝑐𝑜𝑠(𝜔𝑡 + 10°); with  = 377,
𝜐(𝑡) = 70.71 𝑐𝑜𝑠(377𝑡 + 10°)
(c) 𝐴̄ = 𝐴∠𝛼; 𝐵̄ = 𝐵∠𝛽; 𝐶̄ = 𝐴̄ + 𝐵̄
c(t ) = a(t ) + b(t ) = 2 Re Ce jt 

The resultant has the same frequency .
2.7 (a) The circuit diagram is shown below:




(b) 𝑍̄ ̶ = 3 + 𝑗8 − 𝑗4 = 3 + 𝑗4 = 5∠53.1° 𝛺
(c) I = (1000 ) ( 553.1 ) = 20 − 53.1 A

The current lags the source voltage by 53.1
Power Factor = cos53.1 = 0.6 Lagging




2
© 2023 Cengage Learning®. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
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