# In the circuit shown, the steady state potential drop across the capacitor must be

$\begin {array} {1 1} (A)\;V & \quad (B)\;\large\frac{V}{2} \\ (C)\;\large\frac{V}{3} & \quad (D)\;\large\frac{2V}{3} \end {array}$

Let the voltages at the two nodes be 0 and X volts as shown
Writing the KVL equations for the three branches,
$0 + V – I_1R = X$
$0 + V – \Delta V = X$
$0 + 2V – I_2 ^*2R = X$
At steady state, the net current flowing in the branch at potential X is zero, hence
$I_1 + I_2 = 0$
On solving the above equations, $\Delta V = \large\frac{V}{3}$
edited Mar 14, 2014