Question

Figure (32.E1) shows a conductor of length l with a circular cross-section. The radius of the cross-section varies linearly from a to b. The resistivity of the material is ρ. Assuming that b – a << l, find the resistance of the conductor.

Answer 1

Let us consider a small element strip of length dx at a distance x from one end, as shown below.


Let the resistance of the small element strip be dR. Let the radius at that point be c .
Then, the resistance of this small strip,
$$\begin{gathered} \mathrm{d}R=\frac{\rho \mathrm{d}x}{\mathrm{\pi }{c}^2}...\left(\mathrm{i}\right) \\ \mathrm{tan\theta }=\frac{c-\mathrm{a}}{x}=\frac{\mathrm{b}-\mathrm{a}}{L} \\ \Rightarrow \frac{c-\mathrm{a}}{x}=\frac{\mathrm{b}-\mathrm{a}}{L} \\ \Rightarrow L\times \left(c-\mathrm{a}\right)=x\times \left(\mathrm{b}-\mathrm{a}\right) \\ \Rightarrow Lc-\mathrm{La}=xb-xa \end{gathered}$$
Differentiating w.r.t to x, we get:
$$\begin{gathered} \mathrm{L}\frac{\mathrm{d}c}{\mathrm{d}x}-0=\mathrm{b}-\mathrm{a} \\ dx=\frac{Ldc}{\left(b\mathit{-}a\right)}...\left(\mathrm{ii}\right) \end{gathered}$$
Substituting the value of dx in equation (i), we get:
$$\begin{gathered} \mathrm{d}R=\frac{\rho L\mathrm{d}c}{\mathrm{\pi }{c}^2\left(b-a\right)} \\ \mathrm{d}R=\frac{\rho L}{\mathrm{\pi }\left(b-a\right)}­·\frac{\mathrm{d}c}{c^2} \end{gathered}$$
Integrating dR from a to b, we get:
$$\begin{gathered} {\int }_0^{R}dR=\frac{\rho L}{\mathrm{\pi }\left(\mathit{b}\mathit{-}\mathit{a}\right)}{\int }_a^b\frac{\mathrm{d}c}{c^2} \\ \Rightarrow R=\frac{\rho L}{\mathrm{\pi }\left(\mathit{b}\mathit{-}\mathit{a}\right)}{\left[\frac{-1}{c}\right]}_a^b \\ =\frac{\rho L}{\mathrm{\pi }\left(\mathit{b}\mathit{-}\mathit{a}\right)}\left(\frac{-1}{\mathrm{b}}-\frac{-1}{\mathrm{a}}\right) \\ =\frac{\rho L}{\mathrm{\pi ab}} \end{gathered}$$