Resistance and Resistivity — Question 9

Section 26-3 · Physics pages 4208–4209

Question 9 A small neck can dominate the resistance of a nearly uniform wire A wire of total length L and resistivity rho has area A over fraction 1 −f of its length and area ηA over fraction f, with 0 < f, η < 1. Transitions are idealized as short. ηA Schematic; labels, not drawing lengths, specify the data. (a) Find the total resistance and fraction of the voltage drop across the neck. (b) For f = .0200, find eta that makes the neck account for half the voltage drop.
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Question 9 - Solution A small neck can dominate the resistance of a nearly uniform wire ηA (a) Method and calculation R = ρL A [(1 −f) + f/η]. Vneck V = f/η (1 −f) + f/η. (b) Interpretation and further calculation Vneck/V = 1/2 ⇒f/η = 1 −f. η = f/(1 −f) = 1/49 = .0204082. Physical check The same current passes every segment, so voltage-drop fractions equal resistance fractions.

Question 9

A small neck can dominate the resistance of a nearly uniform wire

A wire of total length L and resistivity rho has area A over fraction 1−f1-f of its length and area ηA\eta A over fraction f, with 0<f,η<10<f,\eta<1. Transitions are idealized as short.

Diagram shown in the original SVG worksheet.

  1. Find the total resistance and fraction of the voltage drop across the neck.

  2. For f=.0200f=.0200, find eta that makes the neck account for half the voltage drop.

Question 9 - Solution

A small neck can dominate the resistance of a nearly uniform wire

Diagram shown in the original SVG worksheet.

(a) Method and calculation

R=ρLA[(1−f)+f/η].R=\frac{\rho L}A[(1-f)+f/\eta].VneckV=f/η(1−f)+f/η.\frac{V_{\mathrm{neck}}}V=\frac{f/\eta}{(1-f)+f/\eta}.

(b) Interpretation and further calculation

Vneck/V=1/2⇒f/η=1−f.V_{\mathrm{neck}}/V=1/2\Rightarrow f/\eta=1-f.η=f/(1−f)=1/49=.0204082.\eta=f/(1-f)=1/49=.0204082.

Physical check

The same current passes every segment, so voltage-drop fractions equal resistance fractions.