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ECE 201 Spring 2010 Homework 40 Solutions Problem 30 a Zth R1 R2 j C Voc 20 j10 R2 Vs R1 R2 20 Vrms 20 j10 For maximum power transfer ZL Zth b Voc2 4RL 5W Pavg Problem 37 a Here we cannot apply the maximum power transfer theorem because RL is fixed The expression for average power across RL is given by Pavg Vs2 RL R RL 2 L 1 C 2 Clearly the average power is maximum when the denominator is minimum which happens when the following two conditions hold R 0 L 1 1 C C Pavgmax 1 2L 2 5 mF Vs2 RL 502 5 500 W b Pavg Vs2 RL R RL 2 L 1 C 2 For maximum Pavg dPavg 0 dRL RL2 R2 L 1 C 2 RL 4 011 2
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