\(I =\frac{ V _{ A }}{ R _{1}+ R }\)
\(I =\frac{12}{500+100}\)
\(I =0.02 A\)
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.