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14_Chapter 17 HomeworkCH17 Current and Resistance

14_Chapter 17 HomeworkCH17 Current and Resistance - 82...

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82 CHAPTER 17 17.29 (a) From R L A ρ = , the initial resistance of the mercury is ( ) ( ) ( ) 7 2 3 9.4 10 m 1.000 0 m 1.2 1.00 10 m 4 i i i L R A ρ π × Ω⋅ = = = × (b) Since the volume of mercury is constant, f f i L A L i V A = = gives the final cross- sectional area as ( ) f i i f A A L L = . Thus, the final resistance is given by 2 f f f f i i L L R A A L ρ ρ = = . The fractional change in the resistance is then ( ) 2 1 f i L L = 2 1 1 f i f f i i i i i i R R R L A L R R L A ρ ρ ∆ = = = 2 4 100.04 1 8.0 10 100.00 ∆ = = × or a 0.080% increase 17.30 The resistance at 20.0°C is ( ) ( ) ( ) 0 1 -3 0 200.0 217 1 1+ 3.92 10 C 0 C 20.0 C R R T T α = = = + × ° ° ° ( Solving ) 0 0 1 R R T T α = + for T gives the temperature of the melting potassium as
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