At t=0 switch `S_(1)` is closed and remains closed for a long time and `S_(2)` remains open. Now `S_(1)` is opened and `S_(2)` is closed. Find out
(i) The current through the capacitor immediately after that moment
(ii) Charge on the capacitor long after that moment.
(iii) Total charge flown through the cell of emf `2_(epsilon)` after `S_(2)` is closed.

Explanation

(i) Let Potential at point A is zero. Then at point B and C it will be `epsilon` (because current through the circuit is zero).
`V_(B)-V_(A)=(epsilon-0)`
`:.` Charge on capacitor `=C(epsilon-0)=C epsilon `
Now `S_(2)` is closed and `S_(1)` is open. (p.d. across capacitor and charge on it will not change suddenly)
Potential at A is zero so at D it is`-2 epsilon`.

`:.` current through the capacitor `=(epsilon-(-2 epsilon))/(R)=(3 epsilon)/(R) ` (B to D)
(ii) after a long time, i = 0

`V_(B)-V_(A)=V_(D)-V_(A)=-2 epsilon`
`:. Q =C (-2 epsilon-0)=-2 epsilon C`
(iii) The charge on the lower plate (which is connected to the battery) changes from `-epsilon C ` to `2 epsilon C`.
`:. ` this charge will come form the battery,
`:. ` charge flown from that cell is `3 epsilon C` downward.