COMEDK2024PhysicsCapacitance
A capacitor of 2 ~F is charged as shown in the figure. When the switch S is turned to position 2 , the percentage of its stored energy dissipated is
Options
- A40 %
- B60 %
- C80 %
- D90 %
Correct answer
C. 80 %
Step-by-step solution
Consider the given figure, When the switch S is connected to point 1, then initial energy stored in the capacitor can be given as = 1 2 (2 ~F ) V^2 When the switch S is connected to point 2, energy dissipated on connection across 8 ~F will be aligned & = 1 2 ( C₁ C₂ C₁+C₁ ), V^2= 1 2 2 ~F 8 ~F 10 ~F V^2 & = 1 2 (1.6 ~F ) V^2 aligned Therefore, % loss of energy = 1.6 2 100=80 %