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IRWIN 9e 10_23
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
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
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:M4 j 5 j 824 0VI1j 2I25k=0.5The correct answer is a.M k L1 L2 1H240 I 1 (4 j 2) j 2 I 2 j 2 I 1 (5 j 3) I 2 0 I 1 4.92 19.75 AI 2 1.6939.29 A 240 Z1 4.8819.75 4.92 19.75Chapter 10: Ma
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:48 n 2 j 32n 2 j 2R L 3V2The correct answer is b.Z TH 3 for maximum power transfer. Reflecting the primary into the secondary: Z TH 48n 2 j 32n 2 j 2 3 48n 2 j32n 2 j 2 3 1 if n : 4Z TH 48
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION:1k+ Vs+5kVx -4V x 10010 kVoc-The correct answer is d. 5 4 Voc Vs x10 4 6 1001k+ Vs5kVx -4V x 10010 kI sc 5 4 I sc Vs 6 100 Find Rout of the amplifier: 5 V x Vs 6Chapter 10: Magn
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is c. Z1 I1 ZL 2 2 (1 j1) 4 j 4 n21200 11.7711.31 A 6 j2 4 j4 I I2 1 n I 2 23.5411.31 AChapter 10: Magnetically Coupled NetworksProblem 10.FE-4
LSU - EE - 2120
Irwin, Basic Engineering Circuit Analysis, 9/E1SOLUTION: The correct answer is a.I1 V1 Z1ZL n2 N n 2 2 N1 10 j10 Z1 2.5 j 2.5 22 1200 I1 24 j 24 A 2.5 j 2.5 I 24 j 24 I2 1 n 2 I 2 12 j12 A 16.97 45 A Z1 Chapter 10: Magnetically Coupled NetworksProbl
LSU - EE - 2120
LSU - EE - 2120
LSU - EE - 2120