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 (Complete And Verified Study material) (402pages) LEARNEXAMS

Chapter 2 Fundamentals 2.1 (a) ????̅ 1 = 9∠30° = 9[???????????? 30° + ???? ???????????? 30°] = 7.8 + ????4.5 (b) ????̄ 2 = 4 + ????5 = √16 + 25 ∠????????????−1 5 4 = 6.40∠128.66° = 6.40???? ????128.66° (c) ????̅ 3 = (7.8 + ????4.5) + (−4 + ????5) = (11.8 + ????9.5) = 15.15∠38.8° (d) ????4 ̅̅̅ = (9∠30°)(6.4∠51.34°) = 57.6∠81.34° = 8.673 + ????56.9 (e) ????5 ̅̅̅ = 9∠30° 6.4∠−52.34° = 1.41∠81.34° = 1.41???? ????81.34° 2.2 (a) ????̄ = 500∠ − 30° = 433.01 − ????250 (b) ????(????) = 4????????????(???????? + 15°) = 4????????????(???????? + 15° − 90°) = 4co s(???????? − 75°) ????̅= 4 √2 ∠ − 75° = 2.83∠ − 75° = 0.73 − ????2.73 (c) ????̄ = (5/√2)∠ − 15° + 4∠ − 60° = (3.42 − ????0.92) + (2 − ????3.46) = 5.42 − ????4.38 = 6.964∠ − 38.94° 2.3 ????̅ 2 = 13 √2 ∠(−(125 × 10−6 )(2????60)) kV = 9.19∠ − 2.7° kV 2.4 (a) ( ) ( ) ( ) 1 2 1 2 6 6 90 10 0 10 7.5 90 A 8 6 6 8 10 0 7.3 90 10 7.5 12.5 36.87 A 6 12.5 36.87 6 90 75 53.13 V