Question

'n' identical capacitors are joined in parallel and are charged to potential 'V'. Now they are separated and joined in series, then ______

Options

• the potential difference is 'nV' and energy increases 'n' times.

• the potential difference remains the same and energy increases 'n' times.

• the potential difference and the total energy of the combination remain the same.

• the potential difference becomes 'nV' and energy remains the same.

Answer 1

'n' identical capacitors are joined in parallel and are charged to potential 'V'. Now they are separated and joined in series, then the potential difference becomes 'nV' and energy remains the same.

Explanation:

P.D. across each capacitor = V when connected in series the potential difference = nV

Let C be the capacitance of each capacitor then in parallel combination the effective capacitance is C₁ = nC

In series combination `C_2 = C_1/n`

∴ `C_2/C_1 = 1/n^2`

Energy U = `1/2CV^2`

∴ `U_2/U_1 = C_2/C_1 . V_2^2/V_1^2 = 1/n^2 . n^2 = 1`

∴ U₂ = U₁