In the circuit shown in figure  $E=12V,C_1=4\mu F,C_2=2\mu F,C_3=6\mu Fand{C}_4=3\mu F.$  Find the heat produced in the circuit after switch S is shorted. (in  $\mu J$)

 

After closing switch charge on  $C_{1}and{C}_2$  is   $Q_1=\frac{4\times 2}{4+2}\times 12=16\mu C$  
Charge on  $C_3\&C_4$  is   $Q_2=\frac{6\times 3}{6+3}\times 12=24\mu C$  
Total energy stored in capacitors  $=\frac{{\left(16\right)}^2}{2\times \left(\frac{4}{3}\right)}+\frac{{\left(24\right)}^2}{2\times 2}$  $=96+144=240\mu J$
Heat = Work done by cells – U
$=480\mu J-240\mu J=240\mu J$