The potential difference between points A and B is

From the given figure, current through lower branch of resistances which are joined in series is $i_1=\frac{10}{4+3}=\frac{10}{7} \mathrm{amp}$ Again current through upper branch of resistances which are also connected in series, is $i_2=\frac{10}{8+6}=\frac{10}{14} \mathrm{amp}$ Now according to the Kirchhoff's voltage law $$\begin{aligned} V_B & -V_A=8 \times i_2-4 \times i_1 \\ & =8 \times \frac{10}{14}-4 \times \frac{10}{7}=0\end{aligned}$$