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(B) The Wheatstone bridge is in a balanced condition because the ratio of resistances in the arms is equal $(R/R = R/R)$.In a balanced Wheatstone brid

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$R$

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(B) The Wheatstone bridge is in a balanced condition because the ratio of resistances in the arms is equal $(R/R = R/R)$.In a balanced Wheatstone bridge,no current flows through the galvanometer arm $(BD)$.Therefore,the galvanometer resistance $R$ can be ignored for the calculation of equivalent resistance.The circuit simplifies to two parallel branches connected between points $A$ and $C$.The upper branch consists of two resistors $R$ in series: $R_{AB} + R_{BC} = R + R = 2R$.The lower branch consists of two resistors $R$ in series: $R_{AD} + R_{CD} = R + R = 2R$.These two branches ($2R$ and $2R$) are in parallel.The equivalent resistance $R_{eq}$ is given by:$R_{eq} = \frac{(2R 	imes 2R)}{(2R + 2R)} = \frac{4R^2}{4R} = R$.

In a Wheatstone bridge,all the four arms have equal resistance $R$. If the resistance of the galvanometer arm is also $R$,the equivalent resistance of the combination as seen by the battery is

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