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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

  1. A1 8 Q 2 C
  2. B3 8 Q 2 C
  3. C3 4 Q 2 C
  4. 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

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