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))`

Answer

To solve the problem, we need to find the value of \( C_2 \) in terms of \( C_1 \) given that the total energy stored in a series combination of capacitors is equal to the total energy stored in a parallel combination of capacitors.

### Step-by-step Solution:

1. **Understand the Series Combination of Capacitors:**
For \( n_1 \) capacitors in series, each with capacitance \( C_1 \), the equivalent capacitance \( C_s \) is given by:
\[ C_s = \frac{C_1}{n_1} \]

2. **Calculate the Energy Stored in the Series Combination:**
The energy \( U_s \) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]
For the series combination charged by a potential difference of \( 4V \):
\[ U_s = \frac{1}{2} C_s (4V)^2 = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (4V)^2 \]
Simplifying this gives:
\[ U_s = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (16V^2) = \frac{8C_1 V^2}{n_1} \]

3. **Understand the Parallel Combination of Capacitors:**
For \( n_2 \) capacitors in parallel, each with capacitance \( C_2 \), the equivalent capacitance \( C_p \) is given by:
\[ C_p = n_2 C_2 \]

4. **Calculate the Energy Stored in the Parallel Combination:**
The energy \( U_p \) stored in the parallel combination charged by a potential difference \( V \):
\[ U_p = \frac{1}{2} C_p V^2 = \frac{1}{2} (n_2 C_2) V^2 \]
This simplifies to:
\[ U_p = \frac{1}{2} n_2 C_2 V^2 \]

5. **Set the Energies Equal:**
Since the total energy stored in both combinations is the same, we can set \( U_s = U_p \):
\[ \frac{8C_1 V^2}{n_1} = \frac{1}{2} n_2 C_2 V^2 \]

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

7. **Rearranging to Find \( C_2 \):**
Multiply both sides by \( 2 \):
\[ \frac{16C_1}{n_1} = n_2 C_2 \]
Now, solve for \( C_2 \):
\[ C_2 = \frac{16C_1}{n_1 n_2} \]

### Final Answer:
\[ C_2 = \frac{16C_1}{n_1 n_2} \]