Question

Calculate the resistance of this wire which is a combination of a truncated cone and a cylinder, in terms of a, b, c, and o (conductivity). Answers without explicit calculation, especially for the conical section will receive no credit. y(m) с x(m) ylm) b z(m) a 2a

Answer 1

Let's find the resistance of the truncated cone.

See the diagram:

C 16/น dat M

Considering an element of radius 'r' and thickness 'dx' at a distance 'x' from the left end as shown.

The relation between r and x is

$\frac{r-\frac{b}{2}}{x}=\frac{c-\frac{b}{2}}{a}$

$\Rightarrow r=\left (\frac{c-\frac{b}{2}}{a} \right )x+\frac{b}{2}$

Resistance of element is $dR = \frac{dx}{\sigma *\pi r^{2}} = \frac{dx}{\sigma *\pi \left [ \left (\frac{c-\frac{b}{2}}{a} \right )x+\frac{b}{2} \right ]^{2}}$

Total resistance of this cone is

$R_{cone} = \int_{0}^{a}\frac{dx}{\sigma *\pi \left [ \left (\frac{c-\frac{b}{2}}{a} \right )x+\frac{b}{2} \right ]^{2}}$

  $=\frac{a}{\sigma \pi\left ( c-\frac{b}{2} \right )}*\left ( \frac{2}{b}-\frac{1}{c} \right )$

  $=\frac{2a}{\sigma \pi bc}$

Also, the resistance of the cylindrical part is $R_{cyl}=\frac{a}{\sigma \pi(\frac{b}{2})^{2}} = \frac{4a}{\pi \sigma b^{2}}$

As they are connected in series, so the total resistance is $R_{total}=R_{cone}+R_{cyl}=\frac{2a}{\pi \sigma bc}+ \frac{4a}{\pi \sigma b^{2}}$

  $=\frac{2a}{\pi \sigma b}\left ( \frac{1}{c}+\frac{2}{b} \right )$ [answer]