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Unformatted text preview: song (shs546) Homework 9 weathers (22202) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points An electric heater operating at full power draws a current of 13 . 7 A from a 124 V circuit. What is the resistance of the heater? Correct answer: 9 . 05109 . Explanation: Let : I = 13 . 7 A and V = 124 V . From Ohms law, R = V I = 124 V 13 . 7 A = 9 . 05109 . 002 (part 2 of 2) 10.0 points Assuming constant R , how much current should the heater draw in order to dissipate 810 W? Correct answer: 9 . 46002 A. Explanation: Let : P = 810 W . Since P = I 2 R , I 1 = radicalbigg P R = radicalbigg 810 W 9 . 05109 = 9 . 46002 A . 003 (part 1 of 2) 10.0 points Suppose that you want to install a heating coil that will convert electric energy to heat at a rate of 399 W for a current of 0 . 76 A. Determine the resistance of the coil. Correct answer: 690 . 789 . Explanation: Let : P = 399 W and I = 0 . 76 A . Power is P = I 2 R . The resistance of the coil is R = P I 2 = 399 W (0 . 76 A) 2 = 690 . 789 . 004 (part 2 of 2) 10.0 points The resistivity of the coil wire is 1 . 35 10 6 m, and its diameter is 0 . 461 mm. Determine its length. Correct answer: 85 . 409 m. Explanation: Let : = 1 . 35 10 6 m and r = 0 . 2305 mm = 0 . 0002305 m . We use the formula for resistance as a function of resistivity , length and crosssectional area A : R = A = r 2 , where r is the radius of the wire. Solving for , = R r 2 = (690 . 789 ) (0 . 0002305 m) 2 1 . 35 10 6 m = 85 . 409 m ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.
 Spring '00
 Littler
 Power, Work, Heat

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