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(A) To find the equivalent resistance between points $A$ and $B$,we first note that the shorting wires along $AC$ and $BD$ make the potential at $A$ e

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$\frac{R}{2}$

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(A) To find the equivalent resistance between points $A$ and $B$,we first note that the shorting wires along $AC$ and $BD$ make the potential at $A$ equal to the potential at $C$ $(V_A = V_C)$ and the potential at $B$ equal to the potential at $D$ $(V_B = V_D)$.By collapsing the nodes $A$ and $C$ into a single node and $B$ and $D$ into another single node,we can redraw the circuit.In this configuration,the ten resistors are arranged such that they form two parallel branches,each consisting of five resistors in a series-parallel combination. Specifically,the symmetry of the cube allows us to simplify the network into two parallel paths,each having an equivalent resistance of $R$. Thus,the total equivalent resistance $R_{AB}$ is given by the parallel combination of these two paths:$R_{AB} = \frac{R 	imes R}{R + R} = \frac{R}{2} \, \Omega$.

$A$ cube is formed with ten identical resistances $R$ (thick lines) and two shorting wires (dotted lines) along the arms $AC$ and $BD$ as shown in the figure below. The resistance between point $A$ and $B$ is ........... $\Omega$.

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