# 26 - Shie Gary – Homework 26 – Due Nov 5 2004 4:00 am...

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Unformatted text preview: Shie, Gary – Homework 26 – Due: Nov 5 2004, 4:00 am – Inst: Turner 1 This print-out should have 12 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 2) 10 points The alternating voltage of a generator is rep- resented by the equation E = E sin ω t , where E is in volts, E = 332 V, ω = 643 rad / s, and t is in seconds. Find the frequency of the electric potential E . Correct answer: 102 . 337 Hz. Explanation: We compare the given equation E = (332 V) sin(643 rad / s) t to the general form for such an equation, E = E max sin ω t. By comparison, we see that ω = 2 π f = 643 rad / s . Therefore f = 643 rad / s 2 π = 102 . 337 Hz . 002 (part 2 of 2) 10 points Find the voltage output of the source. Correct answer: 234 . 759 V. Explanation: The rms voltage is E rms = E max √ 2 = 332 V √ 2 = 234 . 759 V . 003 (part 1 of 1) 10 points A lightbulb is connected to a 60 Hz power source having a maximum voltage of 138 V. What is the resistance of the light bulb that uses an average power of 40 . 3 W? Correct answer: 236 . 278 Ω. Explanation: The rms voltage is V rms = V max √ 2 = (138 V) √ 2 = 97 . 5807 V . and the average power is P av = V 2 rms R , so the resistance is R = V 2 rms P av = V 2 max 2 P av = (138 V) 2 2 (40 . 3 W) = 236 . 278 Ω . 004 (part 1 of 2) 10 points The output of a generator is given by V = V max sin ω t . If after 0 . 0109 s , the output is 0 . 25 times V max , what is the largest possible angular velocity ω of the generator?...
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## This note was uploaded on 11/28/2010 for the course PHY 56705 taught by Professor Turner during the Spring '10 term at University of Texas.

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26 - Shie Gary – Homework 26 – Due Nov 5 2004 4:00 am...

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