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Unformatted text preview: Rodriguez, Brandon Homework 19 Due: Oct 26 2006, 11:00 am Inst: Andrew J Rader 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Please notice that for your homework to be considered towards your grade, it needs to be submitted one hour before the corresponding recitation starts. Work submitted after this time, but before the DUE DATE on top of this page, will be accepted but not graded. PLEASE REMEMBER THAT YOU MUST CARRY OUT YOUR CALCULA- TIONS TO AT LEAST THREE SIGNIFI- CANT FIGURES. YOUR ANSWER MUST BE WITHIN ONE PERCENT OF THE CORRECT RESULT TO BE MARKED AS CORRECT BY THE SERVER. 001 (part 1 of 1) 5 points A lightbulb is connected to a 60 Hz power source having a maximum voltage of 182 V. What is the resistance of the light bulb that uses an average power of 60 . 1 W? Correct answer: 275 . 574 . Explanation: Let : V max = 182 V and P av = 60 . 1 W . The rms voltage is V rms = V max 2 = 182 V 2 = 128 . 693 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (182 V) 2 2(60 . 1 W) = 275 . 574 . keywords: 002 (part 1 of 2) 4 points When a particular inductor is connected to a sinusoidal voltage with a 264 V amplitude, a peak current of 4 . 7 A appears in the inductor....
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This note was uploaded on 10/18/2010 for the course PHYSICS 251 taught by Professor Pedon during the Spring '08 term at IUPUI.
- Spring '08