Let $I_{1}$ is current flowing through first loop
$\therefore I_{1}=\frac{2}{3+2}=\frac{2}{5}A$
and $I_{2}$ is current flowing through the second loop
$\therefore I_{2}=\frac{4}{3+5}=\frac{1}{2}A$
Let $V_{A}$ is potential of point $A$ and $V_{B}$ is potential of point $B$
Now, we apply Kirchhoff’s law between points $A$ and $B$,
$\therefore V_{A}-3I_{1}-4+3I_{2}=V_{B}$
$\Rightarrow V_{A}-V_{B}=3\times\frac{2}{5}+4-3\times\frac{1}{2}$
$=\frac{12+40-15}{10}=\frac{37}{10}$
$\Rightarrow V_{A}-V_{B}=3.7\, V $
