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

Answer 1

Answer: (A)

Resistance per unit length = $\rho =\frac{R}{2\pi r}$
Lengths of sections $APB$ and $AQB$ are $r\theta$ and $r(2\pi -\theta )$

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