HW27Sol - Platt David – Homework 27 – Due 4:00 am –...

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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 Ω. 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 Ω . 002 (part 1 of 1) 10 points A capacitor is placed parallel across a resis- tive load in the circuit below. E . 371 μ F 367 Ω What fraction of the current goes through the capacitor when the frequency is 100 s- 1 ? Correct answer: 0 . 0852386 . Explanation: Let : R = 367 Ω , C = 0 . 371 μ F , and f = 100 s- 1 . The capacitive reactance is X C = 1 2 π f C = 1 2 π (100 s- 1 )(3 . 71 × 10- 7 F) = 4289 . 89 Ω . 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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HW27Sol - Platt David – Homework 27 – Due 4:00 am –...

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