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

Explanation

When `K_(2)` is open and `K_(1)` is closed the capacitor `C_(1)` and `C_(2)` will charge and potential develops across them are `V_(1)` and `V_(2)` respectively.
`E = V_(1)+V_(2)`
`q= V_(1)+V_(2)`
For series connection `V=(q)/(C)` ,q is constant
`:. V prop (1)/( C)`
`:. (V_(1))/(V_(2))=(C_(2))/(C_(1))`
`:. (V_(1))/(V_(2))=(3)/(6)`
`:. V_(2)=2V_(1)`
From equation (1)
`V_(1)+V_(2)=9`
`V_(1)+2V_(1)=9`
`3V_(1)=9 " " V_(1)=3V`
and `V_(2)=2V_(1)=2xx3=6V`
`Q_(1)=C_(1)V_(1)=6Cxx3=18 mu C`
and `Q_(2)=C_(2)V_(2)= 3C xx6 = 18 mu C " " [ because C=1 mu F]`
and ` Q_(3)`=0
Now `K_(1)` is open and `K_(2)` is closed then `C_(2)` and `C_(3)` connected in parallel and `C_(1)` is in series.
`:. Q_(2)= Q_(2)+Q_(3)`
If V is the common potential of parallel connection
`:. V =(Q_(2))/(C_(1)+C_(2))=(18)/(3C+3C)=(18)/(6C) = 3 V `
`:. Q_(2)= C_(2)V = 3C xx3= 9 mu C`
`[ because C = 1 mu C]`
and `Q_(3)= C_(3) V = 3C xx3 = 9 muC`