Question

$A$ and $B$ are two points on a uniform ring of resistance $R$. The $\angle ACB=\theta$, where $C$ is the centre of the ring. The equivalent resistance between $A$ and $B$ is:

• A. $\frac{R\theta (2\pi -\theta )}{4{\pi }^2}$
• B. $R(1-\frac{\theta }{2\pi })$
• C. $\frac{R\theta }{2\pi }$
• D. $\frac{R\theta (2\pi -\theta )}{4\pi }$

Answer 1

Answer: (A)

$\frac{R\theta (2\pi -\theta )}{4{\pi }^2}$

Resistance per unit length $=\rho =\frac{R}{2\pi r}$
Lengths of section $APB$ and $AQB$ are $r\theta$ and $r(2\pi -\theta )$
Resistance of sections $APB$ and $AQB$ are
$R_1=\rho r\theta =\frac{R}{2\pi r}r\theta =\frac{R\theta }{2\pi }$
and $R_2=\frac{R}{2\pi r}r(2\pi -\theta )=\frac{R(2\pi -\theta )}{2\pi }$
As $R_1$ and $R_2$ are in parallel between $A$ and $B$, their equivalent resistance is
$R_{eq}=\frac{R_1{R}_2}{R_1+R_2}$
$=\frac{\frac{R\theta }{2\pi }-\frac{R(2\pi -\theta )}{2\pi }}{\frac{R\theta }{2\pi }+\frac{R(2\pi -\theta )}{2\pi }}=\frac{\frac{R^2\theta (2\pi -\theta )}{4{\pi }^2}}{\frac{R}{2\pi }[\theta +2\pi -\theta ]}=\frac{R(2\pi -\theta )\theta }{4{\pi }^2}$.