12_Ch 17 College Physics ProblemCH17 Current and Resistance

12_Ch 17 College Physics ProblemCH17 Current and Resistance...

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80 CHAPTER 17 17.23 At 80°C, () 1 3 00 5.0 V 1 200 1 0.5 10 C 80 C 20 C VV I R RT T α ∆∆ == =  +− Ω+ −× ° ° −°  or 2 2.6 10 A 26 mA I = 17.24 If , then 41.0 at 20 C and 41.4 at 29.0 C =Ω= ° =Ω=° ( ) 1 TT =+− RR gives the temperature coefficient of resistivity of the material making up this wire as ( ) 1 °C 3 0 41.4 41.0 1.1 10 41.0 29.0 C 20°C RTT −Ω = ° × 17.25 ( )( ) 8 2 0 62 1.7 10 m 10.0 m 5.67 10 3.00 10 m L R A ρ ×Ω = × × (a) At ( ) 0 30.0 C, 1 R T =° = + − 0 T TR gives a resistance of 1 32 30.0 C 20.0 C 5.89 10 −− ° ° ° = × 0.056 7 1 3.9 10 C R =Ω + × (b) At ( ) 10.0 C, 1 R T yields T 1 10.0 C 20.0 C 5.45 10 ° = × 0.056 7 1 3.9 10 C R + × 17.26
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

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