HW27Sol - Platt, David Homework 27 Due: Nov 11 2005, 4:00...

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Unformatted text preview: Platt, David Homework 27 Due: Nov 11 2005, 4:00 am Inst: Ken Shih 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A lightbulb is connected to a 60 Hz power source having a maximum voltage of 184 V. What is the resistance of the light bulb that uses an average power of 50 . 4 W? Correct answer: 335 . 873 . Explanation: Let : V max = 184 V and P av = 50 . 4 W . The rms voltage is V rms = V max 2 = 184 V 2 = 130 . 108 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (184 V) 2 2(50 . 4 W) = 335 . 873 . 002 (part 1 of 1) 10 points A capacitor is placed parallel across a resis- tive load in the circuit below. E . 371 F 367 What fraction of the current goes through the capacitor when the frequency is 100 s- 1 ? Correct answer: 0 . 0852386 . Explanation: Let : R = 367 , C = 0 . 371 F , and f = 100 s- 1 . The capacitive reactance is X C = 1 2 f C = 1 2 (100 s- 1 )(3 . 71 10- 7 F) = 4289 . 89 . The electric potential across the AC source, the capacitor, and the resistor is V , since they are in parallel. The current through the capacitor is I C = V X C , and the current through the load is I R = V R ....
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This note was uploaded on 07/26/2009 for the course PHY 303L taught by Professor Turner during the Fall '08 term at University of Texas at Austin.

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HW27Sol - Platt, David Homework 27 Due: Nov 11 2005, 4:00...

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