Three capacitors each having capacitance C = 2$\mathrm{\mu F}$ are connected with a battery of emf = 30V as shown in figure. When the switch S is closed.

When the switch is open,  ${\mathrm{C}}_{\mathrm{eq}}=\left(\frac{4}{3}\right)\mu \mathrm{F}$

Charge flown through the battery = 40 $\mathrm{\mu }$C.

When switch is closed C₁ = 2 $\mathrm{\mu }$F; q₁ = 60 $\mathrm{\mu }$C.

$\mathrm{\Delta q}=20\mu \mathrm{C};{\mathrm{W}}_{\mathrm{b}}=20\times {10}^{-6}\times 30=0.6\mathrm{mJ}$

When switch is closed, it is short circuited. ; Hence all the charge flows through S.