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(B) The total resistance of the wire is $R$. Since it is bent into a square of four equal sides,the resistance of each side is $R/4$.Point $E$ is the

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$\frac{7}{64}R$

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(B) The total resistance of the wire is $R$. Since it is bent into a square of four equal sides,the resistance of each side is $R/4$.Point $E$ is the mid-point of side $CD$,so the resistance of segment $DE$ is $R/8$ and the resistance of segment $EC$ is $R/8$.The circuit between $E$ and $C$ consists of two parallel branches:Branch $1$: The segment $EC$ with resistance $R_1 = R/8$.Branch $2$: The path $E-D-A-B-C$ with resistance $R_2 = R_{ED} + R_{DA} + R_{AB} + R_{BC} = R/8 + R/4 + R/4 + R/4 = R/8 + 3R/4 = 7R/8$.The equivalent resistance $R_{eq}$ is given by:$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{R/8} + \frac{1}{7R/8} = \frac{8}{R} + \frac{8}{7R} = \frac{56+8}{7R} = \frac{64}{7R}$.Therefore,$R_{eq} = \frac{7R}{64}$.

$A$ wire of resistance $R$ is bent to form a square $ABCD$ as shown in the figure. The effective resistance between $E$ and $C$ is ( $E$ is the mid-point of arm $CD$ ).

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