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 rho. Assuming that b – a << l, find the resistance of the conductor.
Figure 32.E1

Answer 1

Explanation:

Considering the small element length (dx) from distance x

Resistance of the small strip will be

Step 1:

$dR=\frac{\rho dx}{\pi c^2}$        ...(i)

$\begin{aligned}&\tan \theta=\frac{y-a}{x}=\frac{b-a}{L}\\&\Rightarrow \frac{y-a}{x}=\frac{b-a}{L}\\&\Rightarrow L(y-a)=x(b-a)\end{aligned}$

$\begin{aligned}&\Rightarrow L y-L a=x b-x a\\&\left.\Rightarrow L \frac{d y}{d x}-0=b-a \text { (differentiate with respect to } x\right)\\&\Rightarrow L \frac{d y}{d x}=b-a\end{aligned}$

$\Rightarrow \mathrm{dx}=\frac{\mathrm{Ldy}}{\mathrm{b}-\mathrm{a}}$  ...........(ii)

Step 2:

Putting dx in equation (i) we get

$\begin{array}{l}{\mathrm{d} \mathrm{R}=\frac{\mathrm{fLdy}}{\pi \mathrm{y}^{2}(\mathrm{b}-\mathrm{a})}} \\{\Rightarrow \mathrm{d} \mathrm{R}=\frac{\mathrm{fl}}{\pi(\mathrm{b}-\mathrm{a})} \frac{\mathrm{dy}}{\mathrm{y}^{2}}}\end{array}$

$\begin{aligned}&\Rightarrow \int_{0}^{R} d R=\frac{f l}{\pi(b-a)} \int_{a}^{b} \frac{d y}{y^{2}}\\&\Rightarrow R=\frac{f l}{\pi(b-a)} \frac{(b-a)}{a b}=\frac{f l}{\pi a b}\end{aligned}$