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MHT CET202522 Apr 2025Evening ShiftPhysicsCapacitanceActual

Initially n identical capacitors are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then

Options

  1. Apotential difference and total energy of the combination remain the same.
  2. Bpotential difference remains the same and energy increases n times.
  3. Cpotential difference becomes nV and energy remains the same.
  4. Dpotential difference is nV and energy increases n times.

Correct answer

C. potential difference becomes nV and energy remains the same.

Step-by-step solution

The initial parallel combination of n identical capacitors each with capacitance C gives equivalent capacitance C_p = nC , charged to potential V . The stored energy is U_p = 1 2 nCV^2 , and each capacitor carries charge CV . When reconfigured in series, the equivalent capacitance becomes C_s = C/n . Each capacitor retains its charge CV , so the series combination carries total charge Q_s = CV . The resulting potential difference is V_s = Q_s / C_s = CV / (C/n) = nV . The energy stored is U_s = 1 2 C_s V_s^2 = 1 2

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