For the given circuit, in the steady state, $\left|V_B-V_D\right|=$ _____ $\mathrm{V}$.

 

At steady state current will not pass through capacitor, therefore effective resistance between B and C will be only $1 \Omega$. Modified circuit will be as shown below.

$\Rightarrow i_{AB}=\frac{6}{3}=2 \mathrm{A}$  and  $i_{AD}=\frac{6}{12}=0.5 \mathrm{A}$

Applying Kirchoff's law along path BAD, we get

$\Rightarrow V_B+\left(2\times 2\right)-\left(10\times 0.5\right)=V_D$

$\Rightarrow V_B-V_D=1 \mathrm{V}$.