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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Explanation

When the swich is open (Fig)
`q_(0)=(2xi C_(1)C_(2))/(C_(1)+C_(2))`

and when the swich is closed,
`q_(1) = xi C_(1)` and `q_(2) = xi C_(2)`
Hence, the flow of charge, due to the shortening of swich,
thorugh section `1 = q_(1) - q_(0) xi C_(1) [(C_(1) - C_(2))/(C_(1) + C_(2))] = -24 muC`
through the section `2 = -q_(2) - (q_(0)) = xi C_(2) [(C_(1) - C_(2))/(C_(1) + C_(2))] = -36 muC`
and through the section `3 = q_(2) - (q_(2) - q_(1)) - 0 = xi (C_(2) - C_(1)) = -60 muC`