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Unformatted text preview: Patel, Kinal – Homework 16 – Due: Apr 1 2008, 7:00 pm – Inst: Weathers 1 This printout should have 10 questions. Multiplechoice 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 158 V. What is the resistance of the light bulb that uses an average power of 96 . 5 W? Correct answer: 129 . 347 Ω. Explanation: Let : V max = 158 V and P av = 96 . 5 W . The rms voltage is V rms = V max √ 2 = 158 V √ 2 = 111 . 723 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (158 V) 2 2(96 . 5 W) = 129 . 347 Ω . keywords: 002 (part 1 of 1) 10 points An inductor has a 51 . 6 Ω reactance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 195 . 3 V rms voltage? Correct answer: 6 . 42316 A. Explanation: Let : X L = 51 . 6 Ω , f = 60 Hz , f = 50 Hz , and V rms = 195 . 3 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(195 . 3 V)(60 Hz) (50 Hz)(51 . 6 Ω) = 6 . 42316 A . keywords: 003 (part 1 of 3) 10 points An 890 Ω resistor, a 1 . 52 μ F capacitor, and a 3 H inductor are connected in series across a 220 Hz AC source for which the maximum voltage is 50 V....
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This note was uploaded on 10/24/2010 for the course PHYS PHYS 2220 taught by Professor Weathers during the Spring '08 term at North Texas.
 Spring '08
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