In the circuit, given in the figure currents in different

Let us asssume the potential at $\mathrm{A}=\mathrm{V}_{\mathrm{A}}=0$

Now at junction $\mathrm{C}$, According to $\mathrm{KCL}$

$\mathrm{i}_{1}+\mathrm{i}_{3}=\mathrm{i}_{2}$

$1 \mathrm{~A}+\mathrm{i}_{3}=2 \mathrm{~A}$

$\mathrm{i}_{3}=2 \mathrm{~A}$

Now Analyse potential along $\mathrm{ACDB}$

$\mathrm{v}_{\mathrm{A}}+1+\mathrm{i}_{3}(2)-2=\mathrm{v}_{\mathrm{B}}$

$0+1+2(1)-2=\mathrm{v}_{\mathrm{B}}$

$\mathrm{v}_{\mathrm{B}}=3-2$

$\mathrm{v}_{\mathrm{B}}=1 \mathrm{Amp}$