Q. A wire of resistor $R$ is bent into a circular ring of radius $r.$ Equivalent resistance between two points $X$ and $Y$ on its circumference, when angle $XOY$ is $\alpha,$ can be given by
A) $\frac{R\alpha}{4\pi^{2}}(2\pi - \alpha)$
B) $\frac{R}{2\pi}(2\pi - \alpha)$
C) $R(2\pi - \alpha)$
D) $\frac{4\pi}{R\alpha}(2\pi - \alpha)$
Explanation
(a) Here $R_{XWY} = \frac{R}{2\pi r} \times (r\alpha) = \frac{R\alpha}{2\pi}\ \ \left\lbrack \therefore\alpha = \frac{l}{r} \right\rbrack$
and $R_{XZY} = \frac{R}{2\pi r} \times r(2\pi - \alpha) = \frac{R}{2\pi}(2\pi - \alpha)$
$R_{eq} = \frac{R_{XWY}R_{XZY}}{R_{XWY} + R_{XZY}} = \frac{\frac{R\alpha}{2\pi} \times \frac{R}{2\pi}(2\pi - \alpha)}{\frac{R\alpha}{2\pi} + \frac{R(2\pi - \alpha)}{2\pi}} = \frac{R\alpha}{4\pi^{2}}(2\pi - \alpha)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.