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 . The potential at point B and C will be ` in` (because current through the circuit is zero )
` V_B -V_A =( in -0) `
` therefore " "` Charge on capacitor = C = `(in -0 ) = Cin `
` (##MOT_CON_JEE_PHY_C25_SLV_037_S01.png" width="80%"> 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 `in `
` therefore " " ` Current through the capacitor
` " " = ( in -( 2in ))/( R) =( 3in )/(R) (B"to " D) `
(ii)after a long time ,i =0
` V_a -V_A =V_D -V_A =-2in `
` therefore " "Q =C( -2in -0) =- 2in C `
` (##MOT_CON_JEE_PHY_C25_SLV_037_S02.png" width="80%">
(iii) The charge on the lower plate (which is connected to the battery ) changes from `- in C "to " 2 in C `
` therefore " " ` this charge will come from the battery
` therefore " " ` charge flown from that cell is ` 2in C ` downward
` (##MOT_CON_JEE_PHY_C25_SLV_037_S03.png" width="80%">