A capacitor of 2 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

Step-by-step solution

Initial energy stored in the capacitor C₁ is: U_i = 1 2 C₁ V^2 When the switch is turned to position 2, the charge is shared between C₁ and C₂ . The common potential V_c is: V_c = C₁ V C₁ + C₂ The final energy of the system is: U_f = 1 2 (C₁ + C₂)V_c^2 = 1 2 (C₁ + C₂) ( C₁ V C₁ + C₂ )^2 = 1 2 C₁^2 V^2 C₁ + C₂ The energy dissipated is: U = U_i - U_f = 1 2 C₁ V^2 - 1 2 C₁^2 V^2 C₁ + C₂ = 1 2 C₁ C₂ C₁ + C₂ V^2 The percentage of stored energy dissipated is: Percentage = U U_i 100 = 1 2 C₁ C₂ C₁ + C₂ V^2 1 2 C₁ V^2 100