In the circuit shown in the figure, the switch `S` is initially open and the capacitor is initially uncharged. `I_(1),I_(2)` and `I_(3)` represent the current in the resistance `2Omega, 4Omega` and `8Omega` respectively.

Explanation

(a) Initially the capacitor is uncharged so its behaviour is like a conductor. Let potential at A is zero so at B and C also zero and at F it is `epsilon`. Let potential at E is x so at D also x. Apply Kirchhoff's `I^(st)` law at point E :
`(x-epsilon)/(R) +(x-0)/(R)+(x-0)/(R) =0 rArr (3x)/(R) = (epsilon)/(R)`
`x=(epsilon)/(3) , Q_(c)=0`
`:. I_(1) = (-epsilon//3 +epsilon)/(R)=(2 epsilon)/(3 R) rArr I_(2)=(dQ)/(Dt)=(epsilon)/(3R)` and `I_(3)=(epsilon)/(3 R)`
Alternatively
`i_(1)=(epsilon)/(R_(eq))=(epsilonn)/(R+(R)/(2))=(2 epsilon)/(3R) rArr i_(2)=i_(3)=(i_(1))/(2)=(epsilon)/(3 R)` and `(dQ)/(dt)=i_(2)=(epsilon)/(3 R)`

(b) at `t=oo` (finally)
capacitor completely charged so their will be no current through it.
`I_(2)=0, I_(1)=I_(3)=(epsilon)/(2R)`
`V_(E)-V_(B)=V_(D)-V_(c)=(epsilon//2R)R=epsilon//2`
`rArr Q_(C)=(epsilon C)/(2), (dQ)/(dt)=I_(2)=0`