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Physics / Chapter 26: Current and Resistance / Section 26-3

Resistance and Resistivity — Question 6

Section 26-3 · Physics pages 4202–4203

Question 6 Thermal expansion slightly alters the resistance temperature coefficient A freely expanding isotropic wire has ρ = ρ0(1 + α∆T) and each linear dimension scales by 1 + β∆T. Both products are small. R(T) + − I Schematic; labels, not drawing lengths, specify the data. (a) Find the exact resistance ratio within these stated laws and its first-order expansion. (b) For α = 4.0 × 10−3 K−1 and β = 2.0 × 10−5 K−1, compare the effective coefficient with alpha.
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Question 6 - Solution Thermal expansion slightly alters the resistance temperature coefficient R(T) + − I (a) Method and calculation R R0 = ρ ρ0 L L0 A0 A = 1 + α∆T 1 + β∆T . R/R0 = 1 + (α −β)∆T + O(∆T 2). (b) Interpretation and further calculation αR = α −β = 3.98 × 10−3 K−1. (α −αR)/α = β/α = .00500 = 0.500%. Physical check Expansion increases area faster than length, partially offsetting the material resistivity increase.

Question 6

Thermal expansion slightly alters the resistance temperature coefficient

A freely expanding isotropic wire has ρ=ρ0(1+αΔT)\rho=\rho_0(1+\alpha\Delta T) and each linear dimension scales by 1+βΔT1+\beta\Delta T. Both products are small.

Diagram shown in the original SVG worksheet.

  1. Find the exact resistance ratio within these stated laws and its first-order expansion.

  2. For α=4.0×10−3\alpha=4.0\times10^{-3} K−1^{-1} and β=2.0×10−5\beta=2.0\times10^{-5} K−1^{-1}, compare the effective coefficient with alpha.

Question 6 - Solution

Thermal expansion slightly alters the resistance temperature coefficient

Diagram shown in the original SVG worksheet.

(a) Method and calculation

RR0=ρρ0LL0A0A=1+αΔT1+βΔT.\frac R{R_0}=\frac{\rho}{\rho_0}\frac{L}{L_0}\frac{A_0}A=\frac{1+\alpha\Delta T}{1+\beta\Delta T}.R/R0=1+(α−β)ΔT+O(ΔT2).R/R_0=1+(\alpha-\beta)\Delta T+O(\Delta T^2).

(b) Interpretation and further calculation

αR=α−β=3.98×10−3K−1.\alpha_R=\alpha-\beta=3.98\times10^{-3}\,\mathrm{K^{-1}}.(α−αR)/α=β/α=.00500=0.500%.(\alpha-\alpha_R)/\alpha=\beta/\alpha=.00500=0.500\%.

Physical check

Expansion increases area faster than length, partially offsetting the material resistivity increase.

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