Resistance and Resistivity — Question 10

Section 26-3 · Physics pages 4210–4211

Question 10 Resistance uncertainty is especially sensitive to a measured diameter A uniform round wire has ρ = Rπd2/(4L). Independent hard bounds are R = 10.0 ± .1 Ω, d = 1.00 ± .02 mm and L = 2.00 ± .01 m. d +x Wire cross-section and reference direction; dimensions are schematic. (a) Find nominal resistivity and its exact hard-bound interval. (b) Find the first-order fractional bound and identify the dominant measurement.
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Question 10 - Solution Resistance uncertainty is especially sensitive to a measured diameter d +x (a) Method and calculation ρ0 = 10π(10−3)2/(4 · 2) = 3.92699 × 10−6 Ωm. ρ ∈ [ 9.9π(.98 × 10−3)2 4(2.01) , 10.1π(1.02 × 10−3)2 4(1.99) ] . (b) Interpretation and further calculation ρ ∈[3.71519, 4.14723] × 10−6 Ωm. |δρ|/ρ ≲.01 + 2(.02) + .005 = .0550. The diameter contributes 4 percent, the largest term. Physical check Hard bounds combine extreme values; the squared diameter doubles its first-order fractional contribution.

Question 10

Resistance uncertainty is especially sensitive to a measured diameter

A uniform round wire has ρ=Rπd2/(4L)\rho=R\pi d^2/(4L). Independent hard bounds are R=10.0±.1ΩR=10.0\pm.1\,\Omega, d=1.00±.02d=1.00\pm.02 mm and L=2.00±.01L=2.00\pm.01 m.

Diagram shown in the original SVG worksheet.

  1. Find nominal resistivity and its exact hard-bound interval.

  2. Find the first-order fractional bound and identify the dominant measurement.

Question 10 - Solution

Resistance uncertainty is especially sensitive to a measured diameter

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ρ0=10π(10−3)2/(4⋅2)=3.92699×10−6Ωm.\rho_0=10\pi(10^{-3})^2/(4\cdot2)=3.92699\times10^{-6}\,\Omega\,\mathrm m.ρ∈[9.9π(.98×10−3)24(2.01),10.1π(1.02×10−3)24(1.99)].\rho\in\left[\frac{9.9\pi(.98\times10^{-3})^2}{4(2.01)},\frac{10.1\pi(1.02\times10^{-3})^2}{4(1.99)}\right].

(b) Interpretation and further calculation

ρ∈[3.71519,4.14723]×10−6Ωm.\rho\in[3.71519,4.14723]\times10^{-6}\,\Omega\,\mathrm m.|δρ|/ρ≲.01+2(.02)+.005=.0550.|\delta\rho|/\rho\lesssim .01+2(.02)+.005=.0550.The diameter contributes 4 percent, the largest term.

Physical check

Hard bounds combine extreme values; the squared diameter doubles its first-order fractional contribution.