This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Nguyen, Thanh 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 173 V. What is the resistance of the light bulb that uses an average power of 82 . 6 W? Correct answer: 181 . 168 . Explanation: Let : V max = 173 V and P av = 82 . 6 W . The rms voltage is V rms = V max 2 = 173 V 2 = 122 . 329 V . The average power is P av = V 2 rms R R = V 2 rms P av = V 2 max 2 P av = (173 V) 2 2(82 . 6 W) = 181 . 168 . keywords: 002 (part 1 of 1) 10 points An inductor has a 54 . 7 reactance at 60 Hz. What will be the maximum current if this inductor is connected to a 50 Hz source that produces a 176 . 9 V rms voltage? Correct answer: 5 . 48829 A. Explanation: Let : X L = 54 . 7 , f = 60 Hz , f = 50 Hz , and V rms = 176 . 9 V . The maximum voltage of the circuit is V max = V rms 2 , and the inductive reactance is X L = L, so X L = 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(176 . 9 V)(60 Hz) (50 Hz)(54 . 7 ) = 5 . 48829 A . keywords: 003 (part 1 of 3) 10 points A 680 resistor, a 2 . 67 F capacitor, and a 1 . 7 H inductor are connected in series across a 270 Hz AC source for which the maximum voltage is 210 V....
View
Full
Document
This note was uploaded on 10/24/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Work

Click to edit the document details