P27_027.PDF

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Chapter 27 - 27.27
27. (a) Let ∆ T be the change in temperature and κ be the coecient 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 .Furthermore,∆ ρ/ρ = α ∆ T ,where α is the temperature coe cient of resistivity for copper
(4 . 3
×
10 − 3 / K=4 . 3
×
10 − 3 / C ◦ , according to Table 27-1). Thus,
∆ R
R
=
ρ + ∆ L
−
∆ A
A
L
= α + κ
−
2 κ )∆ T =( α
−
κ )∆ T
=(4 . 3
×
10 − 3 / C ◦ −
1 . 7
×
10 − 5 / C ◦ )(1 . 0C ◦ )=4 . 3
×
10 − 3 .
This is 0 . 43%, which we note (for the purposes of the next part) is primarily determined by the
∆ ρ/ρ term in the above calculation.
(b) The fractional change in resistivity is much larger than the fractional change in length and area.
Changes in length and area affect the resistance much less than changes in resistivity.
−
∆ ρ

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