Initially, 'n' identical capacitors are joined in parallel, 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. Cs = C/n, so 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, are… | MHT CET 2026 (11 Apr, Shift 2)

Q7MHT CET 2026 · 11 APR · SHIFT 2Medium

Combination of Capacitors

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

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Answer: (c)

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

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Combination of Capacitors question from MHT CET 2026 (11 Apr, Shift 2)

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