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Unformatted text preview: song (shs546) – Homework 18 – weathers – (22202) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A lightbulb is connected to a 60 Hz power source having a maximum voltage of 116 V. What is the resistance of the light bulb that uses an average power of 92 . 1 W? Correct answer: 73 . 051 Ω. Explanation: Let : V max = 116 V and P av = 92 . 1 W . The rms voltage is V rms = V max √ 2 = 116 V √ 2 = 82 . 0244 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (116 V) 2 2 (92 . 1 W) = 73 . 051 Ω . 002 10.0 points An inductor has a 56 . 4 Ω reactance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 138 . 4 V rms voltage? Correct answer: 4 . 16441 A. Explanation: Let : X L = 56 . 4 Ω , f = 60 Hz , f ′ = 50 Hz , and V rms = 138 . 4 V . The maximum voltage of the circuit is V max = V rms √ 2 , and the inductive reactance is X L = ω L, so X ′ L = ω ′ ω 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 (138 . 4 V) (60 Hz) (50 Hz) (56 . 4 Ω) = 4 . 16441 A . 003 10.0 points A 1 . 5 mF capacitor is connected to a standard outlet ( rms voltage 191 V, frequency 66 Hz ). Determine the magnitude of the current in the capacitor at t = 0 . 00877 s, assuming that at t = 0, the energy stored in the capacitor is zero....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Power, Work, Light

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