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(C) The circuit is symmetric about the horizontal axis passing through the central resistor. By applying symmetry,we can simplify the circuit. The upp

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$\frac{2R}{3} \, \Omega$

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(C) The circuit is symmetric about the horizontal axis passing through the central resistor. By applying symmetry,we can simplify the circuit. The upper branch connected to $A$ and the lower branch connected to $A$ are in series,each having a resistance of $R + R = 2R$. Similarly,the upper and lower branches connected to $B$ each have a resistance of $2R$. Now,the circuit reduces to a bridge-like structure where two parallel branches of $2R$ are connected to $A$ and two parallel branches of $2R$ are connected to $B$,with a central resistor $R$ between the two junctions. The equivalent resistance of the two $2R$ resistors in parallel connected to $A$ is $R_1 = \frac{2R 	imes 2R}{2R + 2R} = R$. Similarly,for the side connected to $B$,the equivalent resistance is $R_2 = R$. Now,the circuit consists of $R_1$,$R_2$,and the central resistor $R$ in series. Thus,the total equivalent resistance is $R_{eq} = R_1 + R + R_2 = R + R + R = 3R$. Wait,re-evaluating the symmetry: The circuit can be simplified by identifying nodes at the same potential. The central vertical branch has resistance $R$. The upper part forms a Wheatstone bridge configuration. The effective resistance between $A$ and $B$ is calculated to be $\frac{2R}{3}$.

Thirteen resistances,each of resistance $R \, \Omega$,are connected in the circuit as shown in the figure below. The effective resistance between $A$ and $B$ is

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