At t=0, switch is closed, if initially `C_(1)` is uncharged and `C_(2)` is charged to a potential difference `2epsilon` then find out following (Given `C_(1)=C_(2) =C`)
(a) Charge on `C_(1)` and `C_(2)` as a function of time.
(b) Find out current in the circuit as a function of time.
(c) Also plot the graphs for the relations derived in part (a).

Explanation

Let q charge flow in time 't' from the battery as shown. The charge on various plates of the capacitor is as shown in the figure. Now applying KVL
`epsilon-(q)/(C)-iR-(q-2 epsilon C)/(C)=0`
`epsilon-(q)/(C)-(q)/(C)+2epsilon-iR=0`
` 3epsilon = (2q)/(C)+iR rArr 3 epsilon-iR=(2q)/(C)`
`3 epsilon C -iRC = 2q rArr (dq)/(dt) RC =3 epsilonC-2q`
`int_(0)^(q)(dq)/(3 epsilon C-2q)=int_(0)^(t) (dt)/(RC) rArr -(1)/(2) ln ((3C epsilon - 2q)/(3C epsilon))=(t)/(RC)`
In` ((3 epsilonC-2q)/(3 epsilonC))=-(2t)/(RC) rArr 3 epsilonC-2q=3epsilon C " " e^(-2t//RC)`
`3 epsilon C (1-e ^(-2t//RC))=2q rArr q = (3)/(2) epsilon C (1-e^(-2t//RC))`
(charge on C, as function of time)
`i=(dq)/(dt)=(3 epsilon)/(R) e^(-2t//RC)`
Charge on `C_(2)` as function of time :
`q'=2epsilon C-q`
`=2 epsilon C-(3)/(2) epsilon C + (3)/(2) epsilon C e^(-2t//RC)`
`=(epsilon C)/(2) +(3)/(2) epsilon C e^(-2t//RC)`
`=(epsilon C)/(2) [ 1+3e^(-2t//RC)]`