JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Current Electricity

In the steady state no charge flows into the capacitors, so the two $1\mu \text{F}$ branches and the $2\Omega$ arm placed in series with a capacitor between B and C carry no current; the only conducting link between B and C is the $1\Omega$ resistor. The upper path from A is then the $2\Omega$ arm $AB$ in series with the $1\Omega$ arm $BC$, a resistance of $3\Omega$, while the lower path $A-10\Omega -D-2\Omega -C$ has resistance $12\Omega$, and both are driven by the full $6\text{ V}$ of the cell across A and C. The upper branch carries $\frac{6}{3}=2\text{ A}$, so the drop across $AB$ is $2\times 2=4\text{ V}$ and B lies $4\text{ V}$ below A. The lower branch carries $\frac{6}{12}=0.5\text{ A}$, so the drop across $AD$ is $0.5\times 10=5\text{ V}$ and D lies $5\text{ V}$ below A. The magnitude of the difference between B and D is therefore $|(-4)-(-5)|=1\text{ V}$.