In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then…

Using kirchhoff's law in the circuit $ACDB$, we get

Voltage at $B$= voltage at $A$ + voltage of battery of $1 \mathrm{V}$ + potential difference across $2 \mathrm{\Omega }$ resistance - voltage of battery of $2 \mathrm{V}$

$\Rightarrow {\mathrm{V}}_{\mathrm{A}}+1+2\left(1\right)-2={\mathrm{V}}_{\mathrm{B}}$

Now potential at point $B$ with respect to point $A$ is given by,

$\Rightarrow V_B-V_A=1 \mathrm{V}$