question_answer

In the circuit shown in the figure, with steady current, the potential drop across the capacitor must be:

A)  V                                

B) \[\frac{1}{2}V\]

C) \[\frac{1}{3}\,V\]                               

D) \[\frac{2}{3}\,V\] 

Correct Answer: C

Solution :

After steady state there will be no current through the capacitor. In steady state \[i=\frac{(2V-V)}{2R+R}=\frac{V}{3R}\] For \[{{V}_{ab}}:{{V}_{a}}+2iR-2V+V={{V}_{b}}\] \[\therefore \]    \[{{V}_{a}}-{{V}_{b}}=V-2iR=V=\frac{V}{3}\] Hence, the correction option is [c].