A 2 μF capacitor is charged as shown in figure. The percentage of its stored energy dissipated after the switch S is turned to position 2 is?

U = 1/2 CV²
U = 1/2 * 2V² = V²
qᵢ = 2V
now when switch is turned to position 2 the charge will flow until the two capacitors are at equal potentials i.e V₂ = q/C₁ + C₂ = C₁V/C₁ + C₂ = 2V/10 = V/5
the potential on both the capacitors is given as V/5
Energy stored in the system now is 1/2 (C₁ + C₂) V² = 1/2 (10) V²/25 = V²/5
Thus the energy loss is given as V² − V²/5 / V² × 100 = 80