P27_027 - 27. (a) Let T be the change in temperature and be...

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27. (a) Let ∆ T be the change in temperature and κ be the coefficient of linear expansion for copper. Then L = κL T and L L = κ T =(1 . 7 × 10 5 / K)(1 . 0 C) = 1 . 7 × 10 5 . This is equivalent to 0 . 0017%. Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the value of κ used in this calculation is not inconsistent with the other units involved. Incorporating a factor of 2 for the two-dimensional nature of A , the fractional change in area is A A =2 κ T =2(1 . 7 × 10 5 / K)(1 . 0 C) = 3 . 4 × 10 5 which is 0 . 0034%. For small changes in the resistivity ρ ,length L ,andarea A of a wire, the change in the resistance is given by R = ∂R ∂ρ ρ + ∂R ∂L L + ∂R ∂A A. Since R = ρL/A , ∂R/∂ρ = L/A = R/ρ , ∂R/∂L = ρ/A = R/L ,and ∂R/∂A = ρL/A 2 = R/A .Fu r the rmo re ,∆
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