In the circuit shown in figure `C_(1)= 1mu F` and `C_(2) = 2 mu F`. The capacitor `C_(1)` is charged to `100 V` and the capacitor `C_(2)` is charged to `20 V`. After charging then are connected as shown. When the switches `S_(1),S_(2)` and `S_(3)` are closed, the charged flowing through `S_(1)` is
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Explanation

When `S_(1),S_(2)` and `S_(3)` are closed, both the capacitors are in parallel with unlike charged plates together. So, they attain a common potential.
Before closing the switches,
Charge on `C_(1)` is `q_(1) = 100 xx 1 = 100 mu C`
Charge on `C_(2)` is `q_(2) = 20 xx 2 = 40 mu C`
After closing the switches
Common potential `V=(q_(1)-q_(2))/(C_(1)+C_(2))=(100-40)/(3)=20 V`
Now final charges `q_(1)^(1)=C_(1)V=1 xx 20 = 20 muC`
`q_(2)^(1)=C_(2)V =2 xx 20 = 40 muC`
The charge that flows through `S_(1)` is
`Delta q = 100 - 20 = 80 mu C`.