In the circuit shown the cells A and B have negligible resistances. For $V_A=12V,R_1=500\Omega$ and $R=100\Omega$  the galvanometer (G) shows no deflection. The value of $V_B$ is 

In the first loop which contains V_A, The current is given by:

$\begin{array}{l}I=\frac{V_A}{R_1+R}\\ I=\frac{12}{500+100}\\ I=0.02A\end{array}$

Since negative terminal of battery is considered at zero voltage. Now voltage at the intersection point between R and R₁ is given by:

V = IR 

V = 0.02×100

V = 2V

So for battery V_B of electric potential 2V there will be no current flowing through the galvanometer as voltage difference between these points is zero.