In the circuit shown the cells $A$ and $B$ have negligible resistance. For $V _{ A }=12\; V , R _{1}=500\; \Omega$ and $R =100\; \Omega$ the galvanometer $(G)$ shows no deflection. The value of $V_{B}$ is .... $V$
AIPMT 2012, Medium
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In the first loop which contains $V _{ A }$, the current is given by:
$I =\frac{ V _{ A }}{ R _{1}+ R }$
$I =\frac{12}{500+100}$
$I =0.02 A$
Since negative terminal of battery is considered at zero voltage. Now voltage at the intersection point between $R$ and $R_{1}$ is given by:
$V= IR$
$V =0.02 \times 100$\
$V=2 V$
So for battery $V_{B}$ of electric potential $2 V$ there will be no current flowing through the galvanometer as voltage difference between these points is zero.
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