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

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Question

(A) The total resistance of the wire is $R$. The circumference of the ring is $2\pi r$. The resistance per unit length is $\lambda = \frac{R}{2\pi r}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The total resistance of the wire is $R$. The circumference of the ring is $2\pi r$. The resistance per unit length is $\lambda = \frac{R}{2\pi r}$.The arc length $XWY$ subtending angle $\alpha$ is $l_1 = r\alpha$. Its resistance is $R_1 = \lambda l_1 = \frac{R}{2\pi r} 	imes r\alpha = \frac{R\alpha}{2\pi}$.The remaining arc length $XZY$ subtending angle $(2\pi - \alpha)$ is $l_2 = r(2\pi - \alpha)$. Its resistance is $R_2 = \lambda l_2 = \frac{R}{2\pi r} 	imes r(2\pi - \alpha) = \frac{R}{2\pi}(2\pi - \alpha)$.Since these two segments are connected in parallel between points $X$ and $Y$,the equivalent resistance $R_{eq}$ is:$R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{(\frac{R\alpha}{2\pi}) 	imes (\frac{R}{2\pi}(2\pi - \alpha))}{\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)} = \frac{\frac{R^2 \alpha (2\pi - \alpha)}{4\pi^2}}{\frac{R(2\pi)}{2\pi}} = \frac{R^2 \alpha (2\pi - \alpha)}{4\pi^2 R} = \frac{R\alpha}{4\pi^2}(2\pi - \alpha)$.

$A$ wire of resistance $R$ is bent into a circular ring of radius $r$. The equivalent resistance between two points $X$ and $Y$ on its circumference,where the central angle $XOY$ is $\alpha$ (in radians),is given by:

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