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(B) The circuit can be simplified by observing the symmetry. The points $C$ and $D$ are at the same potential due to the balanced Wheatstone bridge st

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$\frac{V}{2R}$

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(B) The circuit can be simplified by observing the symmetry. The points $C$ and $D$ are at the same potential due to the balanced Wheatstone bridge structure formed by the resistors. Thus,the resistor between $C$ and $D$ carries no current and can be removed. The circuit effectively becomes two parallel branches connected between $A$ and $B$. One branch consists of resistors $R_{FC}$ and $R_{CE}$ in series,with total resistance $R + R = 2R$. The other branch consists of resistors $R_{FD}$ and $R_{DE}$ in series,with total resistance $R + R = 2R$. The equivalent resistance $R_{	ext{eq}}$ of the circuit is $\frac{1}{R_{	ext{eq}}} = \frac{1}{2R} + \frac{1}{2R} = \frac{2}{2R} = \frac{1}{R}$,so $R_{	ext{eq}} = R$. The total current from the battery is $I = \frac{V}{R_{	ext{eq}}} = \frac{V}{R}$. Since the two parallel branches have equal resistance $2R$,the current divides equally. Therefore,the current flowing in the branch $AFCEB$ is $I' = \frac{I}{2} = \frac{V}{2R}$.

Five equal resistances each of resistance $R$ are connected as shown in the figure. $A$ battery of $V$ volts is connected between $A$ and $B$. The current flowing in $AFCEB$ will be

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