In the circuit shown , the emf of each battery is `60 V` and `C_(1)=2 muF` and `C_(2)=3 muF`. Find the charges that will flow through the sections `1 , 2` and `3` after the Key is closed.
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Answer

`120 muC`

Explanation

when switch is opened, `2` and `3 muF` capatiors are in series. So,
`C_(eq)=(2xx3)/5=6/5 muF`
Hence, charge on each capacitor is
`q=CV=6/5xx90=108 muC`
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When switch `S` is closed, let `q_(1)` and `q_(2)` be charge on the two capacitors as in (b). So ,
`q_(1)=2xx30=60 muC`
`q_(2)=3xx60=180 muC`
Let charge `q_(B)` goes to the upper plate of `3 muF` capacitor and lower plate of `2muF` capacitor. Initially, both the plates have charge `+q-q=0`. Finally, they have charges `q_(2)-q_(1)`. So,
`q_(2)-q_(1)=q_(B)+0`
or `q_(B)=q_(2)-q_(1)=180-60=120 muC`
Alternatively: After closing the switch, charge floeing out of `2 muF` capacitor is
`Deltaq_(2)=180-108-60=48 muC`
Charge flown into `3 muF` capacitor is
`Deltaq_(2)=180-108=72 muF`
So the net charge flown through the switch is
`Deltaq=Deltaq_(1)+Delta_(2)=48+72=120 muC`
which is same as `q_(B)` obtained earlier.