ECE 201 Spring 2010 Homework 37 Solutions Problem 5 a For Figure a 2 Ief f Ief f 1 T 2 i t dt T Z0 Z 6 3 1 9 dt 16 dt 9 0 3 25 3 5 3 Z For Figure b Z 3 1 1 9 dt 16 dt 4 0 2 25 4 Z 2 Ief f Ief f 2 5 b Current through RL is given by i t 60 30 60 2 3 i t Power absorbed by RL is then given by the square of the effective current through it times the resistance Thus 4 25 30 9 3 1000 W 9 P 1 c 4 25 30 9 4 250 W 3 P Problem 7 In all parts of the problem the following identities will be used Z 2 m Z 2 m cos mnx dx 0 0 sin mnx dx 0 0 Here m and n are integers a 1 T 2 v t dt T 0 1 Z 1 T 102 40 cos 20t 2 cos 40t dt T 0 102 2 V1ef f Z V1ef f 10 0995 b 1 T 2 v t dt T 0 2 Z 1 T 50 1 cos 4t 12 5 1 cos 8t 50 cos 6t cos 2t dt T 0 62 5 2 V2ef f Z V2ef f 7 9057 c 1ZT 2 v t dt T 0 3 1ZT 50 1 cos 4t 36 4277 1 cos 8t 6 2499 1 cos 8t dt T 0 92 6776 2 V3ef f V3ef f 9 6269 2 Problem 9 a 1 j 30 5 10 3 5 0 8 4 j3 ZL 1 VL Iin ZL 4 4 j3 2 vL t 20 cos 30t 36 87 V b Pinst t iin t vL t Paverage 100 cos 30t cos 30t 36 87 W Vm Im cos v i 2 50 cos 36 87 40 W Problem 10 a 506 90 6 j12 j4 56 143 13 magnitude 5 Is b Paverage Vef f Ief f cos v i 5 50 cos 53 13 150 W 3 c Paverage R Is 2 6 25 150 W same as part b d 506 90 30 j50 j10 16 143 13 magnitude 1 Is Paverage Vef f Ief f cos v i 1 50 cos 53 13 30 W 4
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