Homework_16 - Nguyen, Thanh Homework 16 Due: Apr 1 2008,...

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Unformatted text preview: Nguyen, Thanh Homework 16 Due: Apr 1 2008, 7:00 pm Inst: Weathers 1 This print-out should have 10 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 173 V. What is the resistance of the light bulb that uses an average power of 82 . 6 W? Correct answer: 181 . 168 . Explanation: Let : V max = 173 V and P av = 82 . 6 W . The rms voltage is V rms = V max 2 = 173 V 2 = 122 . 329 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (173 V) 2 2(82 . 6 W) = 181 . 168 . keywords: 002 (part 1 of 1) 10 points An inductor has a 54 . 7 reactance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 176 . 9 V rms voltage? Correct answer: 5 . 48829 A. Explanation: Let : X L = 54 . 7 , f = 60 Hz , f = 50 Hz , and V rms = 176 . 9 V . The maximum voltage of the circuit is V max = V rms 2 , and the inductive reactance is X L = L, so X L = X L = f f X L , and the maximum current is I max = V max X L = 2 V rms f f X L = 2(176 . 9 V)(60 Hz) (50 Hz)(54 . 7 ) = 5 . 48829 A . keywords: 003 (part 1 of 3) 10 points A 680 resistor, a 2 . 67 F capacitor, and a 1 . 7 H inductor are connected in series across a 270 Hz AC source for which the maximum voltage is 210 V....
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This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Homework_16 - Nguyen, Thanh Homework 16 Due: Apr 1 2008,...

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