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

The current through the capacitor immediately after that moment

Explanation

Let Potential at point A is zero. The potential at point B and C will be `epsi` (because current through the circuit is zero.)
`V_(B)-V_(A)=(epsi-0)`
Charge on capacitor `=C(epsi-0)=Cepsi`

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 zeo so at D it is `-2epsi`
`therefore` current through the capacitor `=(e-(-2e))/(R)=(3e)/(R) ("B to D")`