From the given figure, current through lower branch of resistances which are joined in series is
$i_{1}=\frac{10}{4+3}=\frac{10}{7}$ amp
Again current through upper branch of resistances which are also connected in series, is
$i_{2}=\frac{10}{8+6}=\frac{10}{14} \,amp$
Now according to the Kirchhoff's voltage law
$V_{B} -V_{A}=8 \times i_{2}-4 \times i_{1} $
$=8 \times \frac{10}{14}-4 \times \frac{10}{7}=0$
