NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B , the energy dissipated in the circuit is
Options
- A1 8 Q 2 C
- B3 8 Q 2 C
- C3 4 Q 2 C
- D5 8 Q 2 C
Correct answer
B. 3 8 Q 2 C
Step-by-step solution
Q 0 = Cε Q 1 C = Q 2 3 C                 Q 1 + Q 2 = Q 0 ⇒ Q 1 = Q 0 4 ;   Q 2 = 3 Q 0 4 Energy dissipated, E = 1 2 Q 0   2 C - 1 2 Q 1   2 C - 1 2 Q 2   2 3 C = 1 2 C Q 0 2 - Q 0 2 16 - 9 Q 0 2 3 × 16 = Q 0 2 32 C 16 - 1 - 3 = 3 8 Q 0 2 C