In the circuit shown in Fig, initially `K_(1)` is closed and `K_(2)` is open . What are the charges on each capacitor.
Then `K_(1)` was opened and `K_(2)` was closed (order is important). What will be the charge on each capacitor now ? `[C = 1muF]`

Explanation

In the circuit when initially `K_(1)` is closed and `K_(2)` is open, the capacitors `C_(1) and C_(2)` acquires potential difference `V_(1) and V_(2)` respectively. So, we have ltBrgt `V_(1)+V_(2)=E`
and `V_(1)+V_(2)=9V`
Also, in series combination, `Vprop1//C`
`V_(1):V_(2)=1//6:1//3`
On solving
`impliesV_(1)=3V and V_(2)=6V`
`thereforeQ_(1)=C_(1)V_(1)=6Cxx3=18muC`
`Q_(2)=9muC and Q_(3)=0`
Then, `K_(1)` was opened and `K_(2)` was clsoed, the parallel combination of `C_(2) and C_(3)` is in series with `C_(1)`
`Q_(2)=Q_(2)'+Q_(3)`
and considering common potential of parallel combination as V, then we have
`C_(2)V+C_(3)V=Q_(2)`
`impliesV=(Q_(2))/(C_(2)+C_(3))=(3//2)V`
On solving, `Q_(2)'=(9//2)muC`
and `Q_(3)=(9//2)muC`