A series combination of n₁ capacitors, each of value C₁, is charged by a source of potential difference 4V. Wh

Question

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:

Accepted Answer

1. **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].

2. **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].

3. **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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