A capacitor of $2\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position 2 , the percentage of its stored energy dissipated is

 

detailed solution

Correct option is B

Initially, the energy stored in  2μF  capacitor is Ui=12CV2=122×10-6V2=V2×10-6 Jinitially, the charge stored in  2μF capacitor isQi=CV=2×10-6V=2V×10-6coulomb. When switch S is turned to position 2 , the charge flows and both the capacitors share charges till a common potential VCis reached.VC= total charge  total capacitance =2V×10-6(2+8)×10-6=V5 volt  Finally, the energy stored in both the capacitors Uf=12(2+8)×10-6V52=V25×10-6 J% loss of energy, ΔU=Ui-UfUi×100%=V2-V2/5×10-6V2×10-6×100%=80%