Question

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. $\frac{4\pi }{R\alpha }(2\pi -\alpha )$

Answer 1

The correct option is A $\frac{R\alpha }{4{\pi }^2}(2\pi -\alpha )$

Here $R_{XWY}=\frac{R}{2\pi r}\times \left(r\alpha \right)=\frac{R\alpha }{2\pi }(\because \alpha =\frac{1}{r})$ and $R_{XZY}=\frac{R}{2\pi r}\times (2\pi -\alpha )=\frac{R}{2\pi }(2\pi -\alpha )$
$R_{eq}=\frac{R_{XWY}{R}_{XZY}}{R_{XWY}+R_{XZY}}=\frac{\frac{R\alpha }{2\pi }\times \frac{R}{2\pi }(2\pi -\alpha )}{\frac{R\alpha }{2\pi }+\frac{R(2\pi -\alpha )}{2\pi }}=\frac{R\alpha }{4{\pi }^2}(2\pi -\alpha )$