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(A) When a wire of resistance $R$ is bent into a square, each side has a resistance of $R/4$.Let the corners of the square be $A, B, C,$ and $D$ in or

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$8$

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(A) When a wire of resistance $R$ is bent into a square, each side has a resistance of $R/4$.Let the corners of the square be $A, B, C,$ and $D$ in order. The cell is connected between adjacent corners $A$ and $D$.The path $A-B-C-D$ has a total resistance of $R/4 + R/4 + R/4 = 3R/4$.The direct path $A-D$ has a resistance of $R/4$.These two paths are in parallel across the $12 	ext{ V}$ source.The potential difference across any diagonal (e.g., $A$ to $C$) is the potential drop across the path $A-B-C$.The current $I_1$ flowing through the branch $A-B-C$ is given by $I_1 = V / R_{branch} = 12 / (3R/4) = 16/R$.The potential difference across the diagonal $AC$ is the voltage drop across the resistors $AB$ and $BC$ in series:$V_{AC} = I_1 	imes (R/4 + R/4) = (16/R) 	imes (R/2) = 8 	ext{ V}$.

$A$ straight wire of resistance $R$ is bent into the shape of a square. $A$ cell of emf $12 \text{ V}$ is connected between two adjacent corners of the square. The potential difference across any diagonal of the square is (in $\text{ V}$)

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