Question

A capacitor of capacitance $C_1=1.0\mu F$ charged up to a voltage $V=110V$ is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitances $C_2=2.0\mu F$ and $C_3=3.0\mu F$. What charge will flow through the connecting wires?

Answer 1

$C_{23}=\frac{2\times 3}{2+3}=1.2\mu F$

$q_{total}=C_{1}V=110\mu C$

Common potential in parallel is given by

$V=\frac{\text{Total charge}}{\text{Total capacity}}$

$=\frac{110}{1+1.2}=50V$

$q_{23}=\left(C_{23}\right)V=60\mu C$

So, this much charge flows through the switch.