A wire of resistance R is bent into a circle | vedclass

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Question

(A) The total resistance of the wire is $R$. The circumference of the circle is $2\pi r$.The resistance per unit length of the wire is $\lambda = \fra

Vedclass · Accepted Answer

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

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(A) The total resistance of the wire is $R$. The circumference of the circle is $2\pi r$.The resistance per unit length of the wire is $\lambda = \frac{R}{2\pi r}$.The arc length $XWY$ subtends an angle $\alpha$ at the center,so its length is $l_1 = r\alpha$. The resistance of this part is $R_{XWY} = \lambda l_1 = \frac{R}{2\pi r} 	imes r\alpha = \frac{R\alpha}{2\pi}$.The arc length $XZY$ subtends an angle $(2\pi - \alpha)$ at the center,so its length is $l_2 = r(2\pi - \alpha)$. The resistance of this part is $R_{XZY} = \lambda l_2 = \frac{R}{2\pi r} 	imes r(2\pi - \alpha) = \frac{R}{2\pi}(2\pi - \alpha)$.Since these two parts are connected in parallel between points $X$ and $Y$,the equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{R_{XWY} 	imes R_{XZY}}{R_{XWY} + R_{XZY}}$$R_{eq} = \frac{(\frac{R\alpha}{2\pi}) 	imes (\frac{R(2\pi - \alpha)}{2\pi})}{\frac{R\alpha}{2\pi} + \frac{R(2\pi - \alpha)}{2\pi}}$$R_{eq} = \frac{\frac{R^2 \alpha (2\pi - \alpha)}{4\pi^2}}{\frac{R}{2\pi} (\alpha + 2\pi - \alpha)}$$R_{eq} = \frac{R^2 \alpha (2\pi - \alpha)}{4\pi^2} 	imes \frac{2\pi}{R(2\pi)}$$R_{eq} = \frac{R\alpha}{4\pi^2}(2\pi - \alpha)$

$A$ wire of resistance $R$ is bent into a circle of radius $r$. What is the equivalent resistance between points $X$ and $Y$?

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