A series combination of `n_(1)` capacitors, each of value `C_(1)`, is charged by a source of potential difference `4 V`. When another parallel combination of `n_(2)` capacitors, each of value `C_(2)`, is charged by a source of potential difference `V`, it has same (total) energy stored in it, as the first combination has. the value of `C_(2)`, in terms of `C_(1)`, is then

Options

A. `(2C_(1))/(n_(1)n_(2))`
B. `16(n_(2))/(n_(1))C_(1)`
C. `2(n_(2))/(n_(1))C_(1)`
D. `(16C_(1))/(n_(1)n_(2))`

Explanation

To solve the problem, we need to find the value of \( C_2 \) in terms of \( C_1 \) given the conditions of the series and parallel combinations of capacitors.

### Step-by-Step Solution:

1. **Understanding the Capacitor Combinations**:
- For the series combination of \( n_1 \) capacitors, each of value \( C_1 \), the equivalent capacitance \( C_s \) is given by:
\[ C_s = \frac{C_1}{n_1} \]

2. **Energy Stored in the Series Combination**:
- The energy \( U_s \) stored in the series combination when charged by a potential difference of \( 4V \) is given by the formula:
\[ U_s = \frac{1}{2} C_s (V_1)^2 \]
- Substituting \( C_s \) and \( V_1 = 4V \):
\[ U_s = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (4V)^2 = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (16V^2) = \frac{8C_1 V^2}{n_1} \]

3. **Understanding the Parallel Combination**:
- For the parallel combination of \( n_2 \) capacitors, each of value \( C_2 \), the equivalent capacitance \( C_p \) is given by:
\[ C_p = n_2 C_2 \]

4. **Energy Stored in the Parallel Combination**:
- The energy \( U_p \) stored in the parallel combination when charged by a potential difference of \( V \) is given by:
\[ U_p = \frac{1}{2} C_p (V_2)^2 \]
- Substituting \( C_p \) and \( V_2 = V \):
\[ U_p = \frac{1}{2} (n_2 C_2) V^2 = \frac{1}{2} n_2 C_2 V^2 \]

5. **Setting the Energies Equal**:
- According to the problem, the energies stored in both combinations are equal:
\[ U_s = U_p \]
- Therefore, we have:
\[ \frac{8C_1 V^2}{n_1} = \frac{1}{2} n_2 C_2 V^2 \]

6. **Simplifying the Equation**:
- We can cancel \( V^2 \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{8C_1}{n_1} = \frac{1}{2} n_2 C_2 \]
- Multiplying both sides by 2:
\[ \frac{16C_1}{n_1} = n_2 C_2 \]

7. **Solving for \( C_2 \)**:
- Rearranging the equation gives:
\[ C_2 = \frac{16C_1}{n_1 n_2} \]

### Final Result:
Thus, the value of \( C_2 \) in terms of \( C_1 \) is:
\[ C_2 = \frac{16C_1}{n_1 n_2} \]