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
- Apotential difference and total energy of the combination remain the same.
- Bpotential difference remains the same and energy increases n times.
- Cpotential difference becomes nV and energy remains the same.
- 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