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(C) From the circuit diagram, we can see that the resistors are arranged in a bridge-like structure. By symmetry, the potential at point $F$ and point

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$\frac{V}{2R}$

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(C) From the circuit diagram, we can see that the resistors are arranged in a bridge-like structure. By symmetry, the potential at point $F$ and point $E$ can be analyzed. When a voltage $V$ is applied between $A$ and $B$, the circuit simplifies such that the equivalent resistance between $A$ and $B$ is $R_{eq} = R$. Thus, the total current drawn from the battery is $I = \frac{V}{R}$. Due to the symmetric nature of the circuit, this current splits equally into the two parallel branches connected to $F$. Therefore, the current flowing through the resistor $FC$ is $I_{FC} = \frac{I}{2} = \frac{V}{2R}$.

Five equal resistors, each of resistance $R$, are connected as shown in the figure below. $A$ battery of $V$ volt is connected between $A$ and $B$. The current flowing in $FC$ will be

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