14_Chapter 16 HomeworkCH17 Current and Resistance

# 14_Chapter 16 HomeworkCH17 Current and Resistance - 82...

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82 CHAPTER 17 17.29 (a) From RL 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 π ×Ω == =Ω × (b) Since the volume of mercury is constant, ff i L AL i VA = ⋅=⋅ gives the final cross- sectional area as ( ) f ii f A ALL =⋅ . Thus, the final resistance is given by 2 f f LL R A AL ρρ . The fractional change in the resistance is then ( ) 2 1 f i L L  =   2 11 fi f fi i i i RRR LA L RR L A −⋅ ∆= = − = 2 4 100.04 18 . 0 1 0 100.00 × 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 TT α = +−  ×° ° °  ( Solving ) 00 1 =+− for T gives the temperature of the melting potassium as
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## This document was uploaded on 10/20/2011.

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