Initially n identical capacitors are joined in parallel and are charged to potential V. Now they are separated and joined in series. Then (a) potential difference and total energy of the combination remain the same. (b) potential difference remains the same and energy increases n times. (c) potential difference becomes nV and energy remains the same. (d) potential difference is nV and energy increases n times.

Answer: (c). Each capacitor gets q = CV; initial energy Ui = 1/2(nC)V². In series the voltages add: V' = nV. Uf = 1/2/n(nV)² = 1/2nCV² = Ui. So the answer is (c) p.d. becomes nV and energy remains the same.

Initially n identical capacitors are joined in parallel and are… | MHT CET 2025 (22 Apr, Shift 2)

Q20MHT CET 2025 · 22 APR · SHIFT 2Medium

Combination of Capacitors

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

Show answer and solution

Answer: (c)

Each capacitor gets $q = CV$ ; initial energy $U_i = \frac{1}{2}(nC)V^2$ .
In series the voltages add: $V' = nV$ .
$U_f = \frac{1}{2}\cdot\frac{C}{n}(nV)^2 = \frac{1}{2}nCV^2 = U_i$ .
So the answer is (c) p.d. becomes $nV$ and energy remains the same.

Report a mistake

Combination of Capacitors question from MHT CET 2025 (22 Apr, Shift 2)

More questions on Combination of Capacitors