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(C) Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. From the circuit diagram,we observe that point $A$ is connected to point $D$ by

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$\frac{R}{2}$

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(C) Let the potential at point $A$ be $V_A$ and at point $B$ be $V_B$. From the circuit diagram,we observe that point $A$ is connected to point $D$ by a wire,so $V_A = V_D$. Similarly,point $C$ is connected to point $B$ by a wire,so $V_C = V_B$. Now,we can redraw the circuit by replacing the nodes: $1$. The resistor of $2R$ is between $A$ and $C$. Since $V_C = V_B$,this resistor is effectively between $A$ and $B$. $2$. The resistor of $2R$ is between $C$ and $D$. Since $V_C = V_B$ and $V_D = V_A$,this resistor is effectively between $B$ and $A$. $3$. The resistor of $R$ is between $D$ and $B$. Since $V_D = V_A$,this resistor is effectively between $A$ and $B$. Thus,all three resistors are connected in parallel between points $A$ and $B$. The equivalent resistance $R_{eq}$ is given by: $\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R} + \frac{1}{R}$ $\frac{1}{R_{eq}} = \frac{1 + 1 + 2}{2R} = \frac{4}{2R} = \frac{2}{R}$ Therefore,$R_{eq} = \frac{R}{2}$.

What is the equivalent resistance between $A$ and $B$?

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