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3 Pages
IRWIN 9e 11_52
Course: EE 2120, Fall 2008School: LSU
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LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d.I 6 A rmsVR 84.85V rms VL 84.85V rmsRVR IVL I84.85 14.14 684.85 14.14 6L Z load 14.14 j14.14 S 3 3 I Z loadS 3 36 14.14 j14.14 22S 3 2.1645 kVAChapter 11: Po
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a.Z 12 j12 Finding the equivalent per-phase wye:Z Y 4 j 4 5.6645V AB 230 V rmsV AN 230 3 132.79V rmsI aA V AN ZY132.79 23.5 A rms 5.66P3 3 V AN I aA cos Z 3(132.79
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is b.ZY1 ZY 224 j18 8 j 6 3 6 j 4V AB 208V rms 120V rms 3 1200 I aN 1 12 36.87 A 8 j6 1200 I aN 2 16.64 33.69 A 6 j4 I aA I aN 1 I aN 2 28.63 35.02 A rmsV AN 208Chapter
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is c. S 3 2430 kVA 20784.61 j12000 VAP3 20.78kW P1 6.93kWChapter 11: Polyphase CircuitsProblem 11.FE-4
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is d. By use of the power triangle:S 2 P2 Q2Q S 2 P2Q (100k ) 2 (90k ) 2 Q 43.59k varChapter 11: Polyphase CircuitsProblem 11.FE-5
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120