JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Current Electricity

Take the potential at $A$ as the reference, $V_A=0$. The junction rule at $C$ requires the incoming currents to balance the outgoing one, so $1+i_3=2$, giving $i_3=1$ A. Now trace the potential from $A$ to $B$ along the branch through $C$ and $D$, adding the rise across the source and the drop across the $2\Omega$ resistor: $V_A+1+i_3(2)-2=V_B$. Substituting, $0+1+(1)(2)-2=V_B$, so $V_B=+1$ V. The potential at $B$ with respect to $A$ is therefore $+1$ V.