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Practice HW 9 solution

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

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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 10.0points At 45 C, the resistance of a segment of gold wire is 79 Ω. When the wire is placed in a liquid bath, the resistance increases to 180 Ω. 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 Ω , R 2 = 180 Ω , T 0 = 20 C , T 1 = 45 C , and α = 0 . 0034( C) 1 . Neglecting change in the shape of the wire, we have R 1 = R 0 [1 + α ( T 1 T 0 )] R 0 = R 1 1 + α ( T 1 T 0 ) and R 2 = R 0 [1 + α ( T 2 T 0 )] , where T 0 = 20 C . Thus R 2 = R 1 [1 + α ( T 2 T 0 )] 1 + α ( T 1 T 0 ) R 2 + α R 2 ( T 1 T 0 ) = R 1 + α R 1 ( T 2 T 0 ) α R 1 T 2 = R 2 R 1 + α R 2 ( T 1 T 0 ) + α R 1 T 0 so that T 2 = R 2 R 1 + α R 2 ( T 1 T 0 ) + α R 1 T 0 α R 1 . Since R 2 R 1 + α R 2 ( T 1 T 0 ) + R 1 α T 0 = 180 Ω 79 Ω + [0 . 0034( C) 1 ] × (180 Ω) (45 C 20 C) (79 Ω) [0 . 0034( C) 1 ] (20 C) = 121 . 672 Ω , then T 2 = 121 . 672 Ω (79 Ω) (0 . 0034( C) 1 ) = 452 . 986 C . 002 10.0points 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 0 1 2 3 4 5 Resistance (Ω) Power (W) 2. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) Power (W) 3. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) Power (W) correct 4. 0 1 2 3 4 5 6 7 8 910 0 1 2 3 4 5 Resistance (Ω) Power (W)

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

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