question_answer

Initially the capacitor was uncharged. Current in the capacitor just after switching on will be?

A) \[\frac{\varepsilon }{R}\]                        

B) \[\frac{\varepsilon }{2R}\] 

C) \[\frac{\varepsilon }{5R}\]                                  

D) \[\frac{\varepsilon }{4R}\]

Correct Answer: C

Solution :

[C]

Just after switching, the capacitor will act like a conducting wire. So effective circuit will be

By applying KVL at point P

\[\frac{V-\varepsilon }{R}+\frac{V-\varepsilon }{R}+\frac{V}{2R}=0\]

\[V=\frac{4\varepsilon }{5}\]

So, current in the branch containing capacitor \[=\frac{\varepsilon }{5R}.\]