• question_answer Find the potential difference across the capacitor in the circuit shown in the figure if the capacitance of the capacitor is $3\mu F.$ A)  5.5 V                          B)  6.0 V  C)    6.5 V               D)  7.0 V

At steady state there will no current through capacitor. The current will only pass through resistance part of the circuit as shown in the figure. Hence, $I=\frac{9}{1+6+\frac{(1+3)\times 4}{(1+3)+4}}=1A$             Pot. Drop across $3\mu \,F$capacitor             $=({{V}_{A}}-{{V}_{B}})+({{V}_{B}}-{{V}_{C}})$             $=6I+1\times I/2=13I/2=13\times 1/2=6.5\,V$ Hence, the correction option is (c).