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ECE 81_hw13_solutions

# ECE 81_hw13_solutions - ECE 81 Lehigh University HW#13...

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ECE 81 Lehigh University 03/19/2007 HW #13 P5.73: Æ V L = 440 * 3 = 762.1 V RMS Æ I L = 440 /30 = 14.67 A RMS Æ P = 3 * 440 * I L * cos (0) = 19360 W P15.60: ¾ a) Æ Power delivered by the source: P s = V sRMS * I RMS * cos (0) = V sRMS *(V sRMS / R eq ) P s = V² sRMS /20 = 500 W Æ Power dissipated in the line resistance : P line = V lineRMS * I RMS * cos (0) = 10*I² RMS P line = 10*(V sRMS /20)² = 250 W Æ Power delivered to the load : P L = V LRMS * I RMS * cos (0) = 10*I² RMS P L = 10*(V sRMS /20)² = 250 W Æ Efficiency : E = P L /P s = 250/500 = 0.5 = 50% ¾ b) Z’ L = Z L (N 1 / N 2 V 2RMS = (N 2 / N 1 ) * V 1RMS ; I 2RMS = (N 1 / N 2 ) * I 1RMS The impedance seen by the source of “step down” is: R’ L2 = R L *(10)² = 1000 The impedance
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