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:

Initial energy stored in capacitor 2 µF:
Ui = 1/2 * 2(V)² = V²
Final voltage after switch 2 is on:
Vf = (C₁V₁)/(C₁ + C₂) = (2V)/10 = 0.2V
Final energy in both the capacitors,
Uf = 1/2 * (C₁ + C₂) * Vf² = 1/2 * 10 * (2V/10)² = 0.2V²
Therefore, energy dissipated = V² - 0.2V² = 0.8V²
Percentage of energy dissipated = (0.8V²/V²) * 100 = 80