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

Options

A. `(epsilon)/(R)`
B. `(epsilon)/(2R)`
C. `(epsilon)/(5R)`
D. `(epsilon)/(4R)`

Answer

C

Explanation

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

By applying KVL at point `P (V - epsilon)/(R)+(V - epsilon)/(R)+(V)/(2R)=0`
`V = (4 epsilon)/(5)`
So, current in the branch containing capacitor `= (epsilon)/(5R)`