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HW 9 solution

# HW 9 solution - song(shs546 Homework 9 weathers(22202 This...

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song (shs546) – Homework 9 – weathers – (22202) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points 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 Ω. Explanation: Let : I = 13 . 7 A and V = 124 V . From Ohm’s law, R = V I = 124 V 13 . 7 A = 9 . 05109 Ω . 002(part2of2)10.0points 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 Ω = 9 . 46002 A . 003(part1of2)10.0points 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 Ω. 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 Ω . 004(part2of2)10.0points The resistivity of the coil wire is 1 . 35 × 10 6 Ω · m, and its diameter is 0 . 461 mm. Determine its length. Correct answer: 85 . 409 m. Explanation: Let : ρ = 1 . 35 × 10 6 Ω · m and r = 0 . 2305 mm = 0 . 0002305 m . We use the formula for resistance as a function of resistivity ρ , length and cross-sectional area A : R = ρ A = ρ π r 2 , where r is the radius of the wire. Solving for , = R ρ π r 2 = (690 . 789 Ω) π (0 . 0002305 m) 2 1 . 35 × 10 6 Ω · m = 85 . 409 m . 005 10.0points A 10 . 3 V battery delivers 114 . 8 mA when connected to a 33 . 7 Ω load.

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HW 9 solution - song(shs546 Homework 9 weathers(22202 This...

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