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

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 \( n_1 \) capacitors (each of capacitance \( C_1 \)) charged by a potential difference of \( 4V \) is equal to the total energy stored in a parallel combination of \( n_2 \) capacitors (each of capacitance \( C_2 \)) charged by a potential difference of \( V \).

### Step-by-Step Solution:

1. **Calculate the equivalent capacitance for the series combination:**
For \( n_1 \) capacitors in series, the equivalent capacitance \( C_{eq1} \) is given by:
\[ \frac{1}{C_{eq1}} = \frac{1}{C_1} + \frac{1}{C_1} + \ldots + \frac{1}{C_1} = \frac{n_1}{C_1} \]
Therefore,
\[ C_{eq1} = \frac{C_1}{n_1} \]

2. **Calculate the energy stored in the series combination:**
The energy \( U_1 \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
For the series combination:
\[ U_1 = \frac{1}{2} C_{eq1} (4V)^2 = \frac{1}{2} \left(\frac{C_1}{n_1}\right) (16) = \frac{8C_1}{n_1} \]

3. **Calculate the equivalent capacitance for the parallel combination:**
For \( n_2 \) capacitors in parallel, the equivalent capacitance \( C_{eq2} \) is given by:
\[ C_{eq2} = n_2 C_2 \]

4. **Calculate the energy stored in the parallel combination:**
The energy \( U_2 \) stored in the parallel combination:
\[ U_2 = \frac{1}{2} C_{eq2} V^2 = \frac{1}{2} (n_2 C_2) V^2 \]

5. **Set the energies equal to each other:**
Since the total energy stored in both combinations is the same:
\[ U_1 = U_2 \]
Therefore,
\[ \frac{8C_1}{n_1} = \frac{1}{2} (n_2 C_2) V^2 \]

6. **Solve for \( C_2 \):**
Rearranging the equation gives:
\[ 8C_1 = \frac{1}{2} n_2 C_2 V^2 n_1 \]
\[ C_2 = \frac{16C_1}{n_2} \cdot \frac{1}{V^2} \cdot n_1 \]

7. **Final expression for \( C_2 \):**
Since we want \( C_2 \) in terms of \( C_1 \), we can write:
\[ C_2 = \frac{16C_1 n_1}{n_2} \]

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