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(C) In the steady state,the capacitor acts as an open circuit. Therefore,no current flows through the branches containing the capacitors.The circuit s

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(C) In the steady state,the capacitor acts as an open circuit. Therefore,no current flows through the branches containing the capacitors.The circuit simplifies to two parallel branches connected across the $6 \ V$ battery.Branch $ABC$ consists of two resistors in series: $2 \ \Omega$ and $1 \ \Omega$. The total resistance of this branch is $R_{ABC} = 2 + 1 = 3 \ \Omega$.The current in this branch is $i_{ABC} = \frac{6 \ V}{3 \ \Omega} = 2 \ A$.The potential at point $B$ relative to $A$ is $V_A - V_B = i_{ABC} 	imes 2 \ \Omega = 2 \ A 	imes 2 \ \Omega = 4 \ V$. Assuming $V_A = 6 \ V$ and $V_C = 0 \ V$,then $V_B = 6 - 4 = 2 \ V$.Branch $ADC$ consists of two resistors in series: $10 \ \Omega$ and $2 \ \Omega$. The total resistance of this branch is $R_{ADC} = 10 + 2 = 12 \ \Omega$.The current in this branch is $i_{ADC} = \frac{6 \ V}{12 \ \Omega} = 0.5 \ A$.The potential at point $D$ relative to $A$ is $V_A - V_D = i_{ADC} 	imes 10 \ \Omega = 0.5 \ A 	imes 10 \ \Omega = 5 \ V$. Thus,$V_D = 6 - 5 = 1 \ V$.The potential difference is $\left| V_{B}-V_{D}ight| = |2 \ V - 1 \ V| = 1 \ V$.

For the given circuit,in the steady state,$\left| V_{B}-V_{D}\right| = .......V.$

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