Two capacitors C₁, and C₂, and two resistances are connected between points X and Y as shown in fig. The total loss of electrostatic energy stored in

C₁, and C₂, after switch S is closed is

Initially, the potential difference between points X and Y is (30 -6) = 24 V
Potential energy of C₁$=\frac{1}{2}\left(2\times {10}^{-6}\right)\times (24{)}^2$
Potential energy of C₂ $=\frac{1}{2}\times \left(3\times {10}^{-6}\right)\times (24{)}^2$
Total potential energy of charged capacitors
$\begin{array}{l}\mathrm{U}=\frac{1}{2}\times (24{)}^2\left[\left(2\times {10}^{-6}\right)+\left(3\times {10}^{-6}\right)\right]\\ =1440\times {10}^{-6}\mathrm{J}=1440\mathrm{\mu J}\end{array}$
When switch is closed, then current through resistors i= 24/(5+7)= 2amp
Potential difference across C, and C, becomes 5 x 2 = 10V and 7 x 2 = 14V respectively. Now the energy stored in the capacitors
$\begin{array}{l}{\mathrm{U}}^{\mathrm{\prime }}=\frac{1}{2}\times \left(2\times {10}^{-6}\right)\left({10}^2\right)+\frac{1}{2}\left(3\times {10}^{-6}\right)(14{)}^2\\ =100\mathrm{\mu J}+294\mathrm{\mu J}=394\mathrm{\mu J}\end{array}$
Loss of energy  = U -U ' = 7440 -394 = 1046 $\mathrm{\mu J}$