The Electrical Properties of Solids — Question 1

Section 41-1 · Physics pages 6792–6793

Question 1 Separating intrinsic resistivity from specimen geometry A uniform bar has length L = 12.0 mm, cross section A = .300 mm2 and resistance R = 8.00 Ω. Contacts are ideal and the response is ohmic. I L A Schematic; labels, not drawing lengths, specify the data. (a) Find resistivity and conductivity. (b) Find the resistance of a same-material bar twice as long with twice the diameter.
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Question 1 - Solution Separating intrinsic resistivity from specimen geometry I L A (a) Method and calculation R = ρL/A ⇒ρ = RA/L = 2.00 × 10−4 Ωm . σ = 1/ρ = 5.00 × 103 S/m . (b) Interpretation and further calculation L′ = 2L, d′ = 2d ⇒A′ = 4A. R′/R = (L′/L)(A/A′) = 1/2, R′ = 4.00 Ω. Physical check Resistivity is a material parameter at the specified conditions; resistance also depends on shape. Doubling a diameter quadruples area.

Question 1

Separating intrinsic resistivity from specimen geometry

A uniform bar has length L=12.0mmL=12.0\,\mathrm{mm}, cross section A=.300mm2A=.300\,\mathrm{mm^2} and resistance R=8.00ΩR=8.00\,\Omega. Contacts are ideal and the response is ohmic.

Diagram shown in the original SVG worksheet.

  1. Find resistivity and conductivity.

  2. Find the resistance of a same-material bar twice as long with twice the diameter.

Question 1 - Solution

Separating intrinsic resistivity from specimen geometry

Diagram shown in the original SVG worksheet.

(a) Method and calculation

R=ρL/A⇒ρ=RA/L=2.00×10−4Ωm.R=\rho L/A\Rightarrow\boxed{\rho=RA/L=2.00\times10^{-4}\,\Omega\,\mathrm m}.σ=1/ρ=5.00×103S/m.\boxed{\sigma=1/\rho=5.00\times10^3\,\mathrm{S/m}}.

(b) Interpretation and further calculation

L′=2L,d′=2d⇒A′=4A.L\prime=2L,\quad d\prime=2d\Rightarrow A\prime=4A.R′/R=(L′/L)(A/A′)=1/2,R′=4.00Ω.\boxed{R\prime/R=(L\prime/L)(A/A\prime)=1/2,\quad R\prime=4.00\,\Omega}.

Physical check

Resistivity is a material parameter at the specified conditions; resistance also depends on shape. Doubling a diameter quadruples area.