For the arrangement shown in the figure, the switch is closed at `t = 0`. `C_2` is initially uncharged while `C_1` has a charge of `2muC`

(a) Find the current comingn out of the battery just after the switch is closed.
(b) Find the charge on the capacitors in the steady state condition.

Answer

(a) `(7)/(50)` or `(11)/(50)A`, (b) `Q_(1) = 9muC,Q_(2) = 0`

Explanation

(a) at `t = 0` capacitors act as short circuit applying kirchhoff's loop law
`9 - 30I - 20I - (2)/(I) = 0rArr I = (7)/(50) Amp`

`**` If plates `C_(1)` have charge `-2muC` nad `2muC`.
then `I = (11)/(50)Amp`
(b) For steady state capacitor acts as
open circuit so `I = 0`
`P.D.` across `C_(2) = 0` volt
`q_(2) =0`
`P.D.` across `C_(1) = 9V`
`q_(1) = 1muF xx 9V = 9muC`