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
`epsilon-iR-(i-i_(1))R=0`
`epsilon-2iR-i_(1)R=0 " " ....(i)`
Applying KVL in loop ABCEFDA
`epsilon-iR-i_(1)R-(q)/(C)=0`
by eq (i) `(2 epsilon-epsilon-i_(1)R-2i_(1)R)/(2)=(q)/(C) rArr epsilonC-3i_(1)RC=2q`
`epsilon C-2q=3(dq)/(dt)=3 (Dq)/(dt),RC rArr underset(0) overset(q)int (dq)/(epsilon C-2q)=underset(0)overset(t)int (dt)/(3RC)`
`-(1)/(2) In (epsilon C-2q)/(epsilon C)=(t)/(3RC) rArr q=(epsilon C)/(2) (t-e^(-2t//3RC))`