A uniform ring of conducting wire of radius r | vedclass

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(D) The total resistance of the ring is $R$. The ring is divided into two parts by points $A$ and $B$. The length of the arc corresponding to angle $\

Vedclass · Accepted Answer

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

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(D) The total resistance of the ring is $R$. The ring is divided into two parts by points $A$ and $B$. The length of the arc corresponding to angle $	heta$ is $l_1 = r	heta$,and the length of the remaining arc is $l_2 = r(2\pi - 	heta)$. Since the resistance is proportional to the length,the resistances of the two parts are: $R_1 = R \left(\frac{	heta}{2\pi}ight)$ and $R_2 = R \left(\frac{2\pi - 	heta}{2\pi}ight)$. These two parts are connected in parallel between points $A$ and $B$. The equivalent resistance $R_{eq}$ is given by: $R_{eq} = \frac{R_1 R_2}{R_1 + R_2} = \frac{\left(R \frac{	heta}{2\pi}ight) \left(R \frac{2\pi - 	heta}{2\pi}ight)}{R \left(\frac{	heta}{2\pi} + \frac{2\pi - 	heta}{2\pi}ight)}$ $R_{eq} = \frac{R^2 	heta (2\pi - 	heta) / 4\pi^2}{R (2\pi / 2\pi)} = \frac{R 	heta (2\pi - 	heta)}{4\pi^2}$.

$A$ uniform ring of conducting wire of radius $r$ has resistance $R$. $A$ and $B$ are two points on the ring which subtend an angle $\theta$ at the centre of the ring. The equivalent resistance between points $A$ and $B$ is:

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