In a Wheatstone’s 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
AIPMT 2003, Easy
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(b) As bridge is in balanced condition, no current will flow through $BD.$
$R_{1}=R_{A B}+R_{B C}$
$=R+R=2 R$
$R_{2}=R_{A D}+R_{C D}=R+R=2 R$
Also, $R_{1}$ and $R_{2}$ are in parallel combination
Hence, equivalent resistance between $A$ and $C$ will be
$R_{e q}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=\frac{4 R^{2}}{4 R}=R$
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