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Unformatted text preview: Shie, Gary Homework 26 Due: Nov 5 2004, 4:00 am Inst: Turner 1 This printout should have 12 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 2) 10 points The alternating voltage of a generator is rep resented by the equation E = E sin t, where E is in volts, E = 332 V, = 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 t. By comparison, we see that = 2 f = 643 rad / s . Therefore f = 643 rad / s 2 = 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 . 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 . 004 (part 1 of 2) 10 points The output of a generator is given by V = V max sin t. If after 0 . 0109 s , the output is 0 . 25 times V max , what is the largest possible angular velocity of the generator? Correct answer: 23 . 1817 rad / s....
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This note was uploaded on 01/20/2010 for the course PHY 1 taught by Professor Erskine during the Spring '10 term at University of Texas at Austin.
 Spring '10
 ERSKINE
 Physics, Work

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