Question 153 of 380

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 mid-point of arm $CD$ )

A$R$B$\dfrac{7}{64}R$C$\dfrac{1}{16}R$D$\dfrac{3}{4}R$

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Answer: (B) $\dfrac{7}{64}R$

If the resistance of the wire is R, then each arm of the square has a resistance of $\dfrac{R}{4}$. Accordingly. The resistance of segments DE and EC are ${R}_{DE}={R}_{EC}=\dfrac{R}{8}$.
We can draw the resultant circuit as follows:

Hence,
$\dfrac{1}{{R}_{eq}}=\dfrac{8}{R}+\dfrac{8}{7R}$
$\dfrac{1}{{R}_{eq}}=\dfrac{56+8}{7R}$
${R}_{eq}=\dfrac{7R}{64}$

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