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(B) To find the equivalent resistance between points $A$ and $B$,we analyze the circuit structure.Looking at the circuit,we can identify that the resi

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$\frac{5 R}{8}$

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(B) To find the equivalent resistance between points $A$ and $B$,we analyze the circuit structure.Looking at the circuit,we can identify that the resistors are arranged in a combination of series and parallel connections.Let the resistance of each resistor be $R$.By simplifying the circuit step-by-step:$1$. The two resistors in the lower branch are in series,giving a resistance of $R + R = 2R$.$2$. This $2R$ is in parallel with the middle resistor $R$,resulting in an equivalent resistance of $\frac{R 	imes 2R}{R + 2R} = \frac{2R}{3}$.$3$. This combination is in series with the resistor in the lower part of the outer loop,but looking at the nodes,the total resistance of the lower section becomes $R + \frac{2R}{3} = \frac{5R}{3}$.$4$. Finally,this $\frac{5R}{3}$ is in parallel with the top resistor $R$,so the equivalent resistance $R_{eq}$ is $\frac{R 	imes \frac{5R}{3}}{R + \frac{5R}{3}} = \frac{\frac{5R^2}{3}}{\frac{8R}{3}} = \frac{5R}{8}$.Thus,the equivalent resistance is $\frac{5R}{8}$.

The equivalent resistance between $A$ and $B$ is

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