A capacitor of 2μF is charged as shown in the diagram. When the switch S is turned to position 2 , the percentage of its stored energy dissipated is
Initially, the energy stored in 2μF capacitor is
initially, the charge stored in 2μF capacitor is
Qi=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
% loss of energy, ΔU=Ui-UfUi×100%