A and B are two points on a uniform ring of r | vedclass

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(D) Let the total resistance of the ring be $R$. The total length of the ring is $2 \pi r$. The resistance per unit length is $\lambda = \frac{R}{2 \p

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let the total resistance of the ring be $R$. The total length of the ring is $2 \pi r$. The resistance per unit length is $\lambda = \frac{R}{2 \pi r}$.The length of the minor arc $AB$ is $l_1 = r 	heta$. The resistance of this part is $R_1 = \lambda l_1 = \left(\frac{R}{2 \pi r}ight) (r 	heta) = \frac{R 	heta}{2 \pi}$.The length of the major arc $AB$ is $l_2 = r(2 \pi - 	heta)$. The resistance of this part is $R_2 = \lambda l_2 = \left(\frac{R}{2 \pi r}ight) (r(2 \pi - 	heta)) = \frac{R(2 \pi - 	heta)}{2 \pi}$.Since the two arcs are connected in parallel between points $A$ and $B$,the equivalent resistance $R_{AB}$ is given by:$R_{AB} = \frac{R_1 R_2}{R_1 + R_2}$$R_{AB} = \frac{\left(\frac{R 	heta}{2 \pi}ight) \left(\frac{R(2 \pi - 	heta)}{2 \pi}ight)}{\frac{R 	heta}{2 \pi} + \frac{R(2 \pi - 	heta)}{2 \pi}}$$R_{AB} = \frac{\frac{R^2 	heta (2 \pi - 	heta)}{4 \pi^2}}{\frac{R}{2 \pi} (	heta + 2 \pi - 	heta)}$$R_{AB} = \frac{\frac{R^2 	heta (2 \pi - 	heta)}{4 \pi^2}}{\frac{R}{2 \pi} (2 \pi)}$$R_{AB} = \frac{R^2 	heta (2 \pi - 	heta)}{4 \pi^2} \cdot \frac{1}{R} = \frac{R 	heta (2 \pi - 	heta)}{4 \pi^2}$

$A$ and $B$ are two points on a uniform ring of radius $r$. The resistance of the ring is $R$. $\angle AOB = \theta$ as shown in the figure. The equivalent resistance between points $A$ and $B$ is . . . . . . .

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