Single Choice

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$$\dfrac{R\alpha}{4\pi^2}( 2\pi -\alpha)$$
Correct Answer
B$$\dfrac{R}{2\pi}( 2\pi -\alpha)$$
C$$R( 2\pi -\alpha)$$
D$$\dfrac{4\pi}{R\alpha}( 2\pi -\alpha)$$

Solution

As we know that Resistance $$\propto$$ length , that's why

Here $$R_{XWY}=\dfrac{R}{2\pi r}\times ( r\alpha )=\dfrac{R\alpha}{2\pi}\ \left( \because \alpha =\dfrac lr \right)$$

and $$R_{XZY}=\dfrac{R}{2\pi r}\times r( 2\pi -\alpha)=\dfrac {R}{2\pi}(2\pi -\alpha)$$

$$R_{eq}=\dfrac{R_{XWY}R_{XZY}}{R_{XWY}+R_{XZY}}=\dfrac{\dfrac{R\alpha}{2\pi}\times \dfrac{R}{2\pi}(2\pi -\alpha)}{\dfrac{R\alpha}{2\pi}+\dfrac{R(2\pi -\alpha )}{2\pi}}=\dfrac{R\alpha }{4\pi^2}(2\pi -\alpha )$$

SIMILAR QUESTIONS

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In the circuit given below, find the total current drawn from the battery:

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