In the circuit shown in Fig, instially `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 fig, when `K_(1)` is closed and `K_(2)` is open, `C_(3)` is out of contact.
`:. Q_(3) = 0`.
`C_(1), C_(2)` are in series with the battery `E = 9V`,
`(1)/(C_(s)) = (1)/(C_(1)) + (1)/(C_(2)) = (1)/(6) + (1)/(3) = (1+2)/(6) = (3)/(6) = (1)/(2)`
`C_(s) = 2muF`
`:. Q_(1) = E. C_(s) = 9xx2 = 18 muC`,
`Q_(2) = E. C_(s) = 9xx2 = 18 muC`
Later `K_(1)`, is opened and `K_(2)` is closed, charge on `C_(1)` remains uncharged, i.e., `Q'_(1) = Q_(1) = 18 muC`
Charge on `C_(2)` is now shared between `C_(2) and C_(3). As C_(2) = C_(3)`, therefore , `Q'_(2) = Q'_(3) = (Q_(2))/(2) = (18)/(2) = 9 muC`