Practice HW 9 solution

Practice HW 9 solution - song (shs546) Practice HW 9...

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Unformatted text preview: song (shs546) Practice HW 9 Solutions weathers (22202) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points At 45 C, the resistance of a segment of gold wire is 79 . When the wire is placed in a liquid bath, the resistance increases to 180 . The temperature coefficient is 0 . 0034 ( C) 1 at 20 C. What is the temperature of the bath? Correct answer: 452 . 986 C. Explanation: Let : R 1 = 79 , R 2 = 180 , T = 20 C , T 1 = 45 C , and = 0 . 0034( C) 1 . Neglecting change in the shape of the wire, we have R 1 = R [1 + ( T 1 T )] R = R 1 1 + ( T 1 T ) and R 2 = R [1 + ( T 2 T )] , where T = 20 C . Thus R 2 = R 1 [1 + ( T 2 T )] 1 + ( T 1 T ) R 2 + R 2 ( T 1 T ) = R 1 + R 1 ( T 2 T ) R 1 T 2 = R 2 R 1 + R 2 ( T 1 T ) + R 1 T so that T 2 = R 2 R 1 + R 2 ( T 1 T ) + R 1 T R 1 . Since R 2 R 1 + R 2 ( T 1 T ) + R 1 T = 180 79 + [0 . 0034( C) 1 ] (180 ) (45 C 20 C) (79 ) [0 . 0034( C) 1 ] (20 C) = 121 . 672 , then T 2 = 121 . 672 (79 ) (0 . 0034( C) 1 ) = 452 . 986 C . 002 10.0 points A variable resistor is connected across a con- stant voltage source. Which of the following graphs represents the power P dissipated by the resistor as a function of its resistance R ? 1. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 2. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) 3. 0 1 2 3 4 5 6 7 8 910 1 2 3 4 5 Resistance () Power(W) correct 4....
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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.

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Practice HW 9 solution - song (shs546) Practice HW 9...

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