ch26_p78 - 78. (a) Let T be the change in temperature and...

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78. (a) Let T be the change in temperature and κ be the coefficient of linear expansion for copper. Then L = L T and L L T == × ° = × −− (. / . . 17 10 17 10 55 K)(1.0 C) This is equivalent to 0.0017%. Since a change in Celsius is equivalent to a change on the Kelvin temperature scale, the value of 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 T × ° = × 2 2 17 10 34 10 / . K)(1.0 C) which is 0.0034%. For small changes in the resistivity ρ , length L , and area A of a wire, the change in the resistance is given by ∆∆ R RR L L R A A = + + . Since R = L / A , R / = L / A = R / , R / L = / A = R / L , and R / A = – L / A 2 –R / A . Furthermore, / = α T , where is the temperature coefficient of resistivity for copper
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