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

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




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

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




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

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




Chapter 2
Fundamentals



2.1 (a) 𝐴̅1 = 9∠30° = 9[𝑐𝑜𝑠 30° + 𝑗 𝑠𝑖𝑛 30°] = 7.8 + 𝑗4.5
5
(b) 𝐴̅2 = 4 + 𝑗5 = √16 + 25 ∠ 𝑡𝑎𝑛−1 4 = 6.40∠128.66° = 6.40𝑒𝑗128.66°

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


(d) 𝐴̅4 = (9∠30°)(6.4∠51.34°) = 57.6∠81.34° = 8.673 + 𝑗56.9
(e) 𝐴̅5=
9∠30° = 1.41∠81.34° = 1.41𝑒𝑗81.34°
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
𝐼̄ = ∠ − 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 = ∠(−(125 × 10−6)(2𝜋60)) kV = 9.19∠ − 2.7° kV
√2
− j6 6 − 90
2.4 (a) I = 100 = 10 = 7.5 − 90 A
1
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)




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

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




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

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) 𝑉̅ = ( ) ∠ − 15° = 53.03∠ − 15°;  does not appear in the answer.
√2

(b) 𝜐(𝑡) = 50√2 𝑐𝑜𝑠(𝜔𝑡 + 10°); with  = 377,
𝜐(𝑡) = 70.71 𝑐𝑜𝑠(377𝑡 + 10°)
(c) 𝐴̅ = 𝐴̅∠𝛼; 𝐵̄ = 𝐵̄∠𝛽; 𝐶̄ = 𝐴̅ + 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) 𝑍̶̄ = 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.