A capacitor of capacitance `C_(1) = 1.0 muF` charged up to a voltage `V = 110V` is connected in parallel to the teminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitances `C_(s) = 2.0 muF` and `C_(s) = 3.0 muF`. What charge will flow through the connecting wires ?

Explanation

Let, charge `q` flows through the connecting wires, then at the state of equillbrium, charge distribution will be as shows in the fig. In the closed circuit 12341, using `-Delta varphi = 0`
`= (C_(1) V - q)/(C_(1)) + (q)/(C_(2)) + (q)/(C_(3)) = 0`
or, `q = (V)/((1//C_(1) + 1//C_(2) + 1//C_(3))) = 0.06 mC`