Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. When another parallel combination of n₂ capacitors, each of value C₂, is charged by a source of potential difference V, it has the same (total) energy stored in it as the first combination has. The value of C₂ in terms of C₁ is:

2C₁ / (n₁n₂)

16 (n₂/n₁) C₁

2 (n₂/n₁) C₁

16C₁ / (n₁n₂)

Step-by-Step Solution

Series Combination (Case 1):

Number of capacitors = $n_1$ , Capacitance = $C_1$ , Potential = $4V$ .
Equivalent capacitance for series combination ( $C_s$ ) is given by $\frac{1}{C_s} = \frac{n_1}{C_1} \Rightarrow C_s = \frac{C_1}{n_1}$ [1, 2].
Total energy stored ( $U_1$ ) = $\frac{1}{2} C_s (4V)^2 = \frac{1}{2} \left(\frac{C_1}{n_1}\right) 16V^2 = \frac{8 C_1 V^2}{n_1}$ [3, 4].

Parallel Combination (Case 2):

Number of capacitors = $n_2$ , Capacitance = $C_2$ , Potential = $V$ .
Equivalent capacitance for parallel combination ( $C_p$ ) is given by $C_p = n_2 C_2$ [1, 2].
Total energy stored ( $U_2$ ) = $\frac{1}{2} C_p V^2 = \frac{1}{2} (n_2 C_2) V^2$ [3, 4].

Equating Energies:

Given $U_1 = U_2$ .
$\frac{8 C_1 V^2}{n_1} = \frac{n_2 C_2 V^2}{2}$
$\frac{16 C_1}{n_1} = n_2 C_2$
$C_2 = \frac{16 C_1}{n_1 n_2}$

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