Current Electricity question for JEE Main (JEE Main 2023 (31 Jan, Shift 2)), with solution

Answer: 1

In the steady state no current flows through any branch containing a capacitor. Between $B$ and $C$ only the $1\ \Omega$ branch conducts.

Branch $ABC$: $2+1=3\ \Omega$, current $2\ \text{A}$, so $V_A-V_B=2\times2=4\ \text{V}$.

Branch $ADC$: $10+2=12\ \Omega$, current $0.5\ \text{A}$, so $V_A-V_D=0.5\times10=5\ \text{V}$.

$|V_B-V_D|=5-4=1\ \text{V}$.

Solution by Sreeraj P, M.Sc Physics