Initially the capacitor is uncharged find the charge on capacitor as a function of time, if switch is closed at t=0.

Explanation

Applying KVL in loop ABCDA
`epsi-iR -(i-i_(1))R=0`
Applying KVL in loop ABCEFDA `epsi-iR -i_(1)R-q/c=0`

`(2epsi-epsi-i_(1)R-2i_(1)R)/(2)=q/c`
`epsiC-3i_(1)RC=2q`
`epsiC-2q=3(dq)/(dt).RC`
`underset(0)overset(q)int (dq)/(epsiC-2q)=underset(0)overset(t)int (dt)/(3RC) rArr -1/2 ln (epsiC-2q)/(epsiC)=t/(3RC)`
`q-(epsiC)/(2) (1-e^(-2t//3RC)`