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

B

Explanation

At `t=0`, equivalent resistance of an uncharged capacitor is zero and a charged capacitor is like a bttery of emf = potential difference across the capacitor.
a. `V_(C_1)=q_1/C_1=2V`
`:.` Net emf of the circuit `=9-2=7V`
or `92=11V`
Net resistance `=30+(30xx60)/(30+60)=50Omega`
`:.` Current at `t=0` would be `i_0=7/50A`
or `11/50A`
b. In steady state no current will flow through the circuit `C_2` will therefore be short circuited, while `PD` across `C_1` will be 9 V
`:. Q_2=0 and Q_1=9muC`