Find the potential drop across the capacitor in the given circuit.

The current will pass through only resistance part of the circuit. Hence, 

$\mathrm{I}=\frac{9}{1+6+2}=1\mathrm{A}$

Potential drop across $3\mu F$ capacitance is

${\mathrm{V}}_{\mathrm{AC}}={\mathrm{V}}_{\mathrm{AB}}+{\mathrm{V}}_{\mathrm{BC}}=1\times 6+\frac{1}{2}\times 1=6.5\mathrm{V}$

(Apply current division for current in branch BC which is 0.5 A)