For the given circuit, in the steady state, |VB - VD| =......... V.

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Given: In the steady state, the capacitor behaves as an open circuit.

Find: The value of $|V_B - V_D|$ .

So the circuit simplifies to:

Equivalent circuit after removing capacitor branch in steady state, showing top branch 2 ohm and 1 ohm through B, and bottom branch 10 ohm and 2 ohm through D, across 6 V battery.

From the simplified circuit,

i_{AB}=\frac{6}{3}=2\,\text{A}, \qquad i_{AD}=\frac{6}{12}=0.5\,\text{A}

Using the potential relation shown in the solution,

V_B + 2 \times 2 - 10 \times 0.5 = V_D

Hence,

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

Therefore, the required voltage difference is $1$ .

Given: The circuit contains a capacitor and a $6\,\text{V}$ source.

Find: The steady-state voltage difference between points $B$ and $D$ .

In steady state, the capacitor behaves like an open circuit, so no current flows through the capacitor branch. The remaining network has two branches between the source terminals.

Top branch resistance:

2 + 1 = 3\,\Omega

Bottom branch resistance:

10 + 2 = 12\,\Omega

So the branch currents are

I_{top}=\frac{6}{3}=2\,\text{A}

I_{bottom}=\frac{6}{12}=0.5\,\text{A}

Potential drop from $A$ to $B$ is

2 \times 2 = 4\,\text{V}

Potential drop from $A$ to $D$ is

10 \times 0.5 = 5\,\text{V}

Therefore the magnitude of the potential difference is

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

So, the correct answer is $1$ .

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