A series combination of n₁ capacitors, each of value C₁ is charged by a source of potential difference 6 V. Another parallel combination of n₂ capacitors, each of value C₂ is charged by a source of potential difference 2 V. Total energy of both the combinations is same. The value of C₂ in terms of C₁ is (a) (3C₁)/(n₁n₂) (b) (9n₂)/(n₁)C₁ (c) (3n₂)/(n₁)C₁ (d) (9C₁)/(n₁n₂)

Answer: (d). Series: 1/2·(C₁)/(n₁)(6)² = (18C₁)/(n₁). Parallel: 1/2(n₂C₂)(2)² = 2n₂C₂. Equal: 2n₂C₂ = (18C₁)/(n₁) ⇒ C₂ = (9C₁)/(n₁n₂). So the answer is (d).

A series combination of n₁ capacitors, each of value C₁ is charged by… | MHT CET 2024 (16 May, Shift 1)

Q32MHT CET 2024 · 16 MAY · SHIFT 1Medium

Combination of Capacitors

A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference 6 V. Another parallel combination of $n_2$ capacitors, each of value $C_2$ is charged by a source of potential difference 2 V. Total energy of both the combinations is same. The value of $C_2$ in terms of $C_1$ is

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Answer: (d)

Series: $\frac{1}{2}\cdot\frac{C_1}{n_1}(6)^2 = \frac{18C_1}{n_1}$ .
Parallel: $\frac{1}{2}(n_2C_2)(2)^2 = 2n_2C_2$ .
Equal: $2n_2C_2 = \frac{18C_1}{n_1} \Rightarrow C_2 = \frac{9C_1}{n_1n_2}$ .
So the answer is (d).

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Combination of Capacitors question from MHT CET 2024 (16 May, Shift 1)

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