MHT CET202611 April 2026Evening ShiftPhysicsCapacitanceActual
Initially, 'n' identical capacitors are joined in parallel, 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
Initial capacitance of each capacitor is C . In parallel combination, the potential difference across each capacitor is V . Charge on each capacitor, q = CV . Total initial energy, U_i = n 1 2 CV^2 . When the capacitors are separated and connected in series, the charge on each capacitor remains q = CV . The potential difference across the series combination is the sum of individual potential differences. V_f = V + V + n times = nV . Equivalent capacitance in series, C_s = C n . Total final energy, U_f = 1 2 C_s V_f