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_(1)/n_(2) C_(1)`
C. `2 n_(2)/n_(1) C_(1)`
D. `(16 C_(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 the conditions of the capacitors in series and parallel combinations.

### Step-by-Step Solution:

1. **Determine the Equivalent Capacitance of the Series Combination:**
For \( n_1 \) capacitors in series, each with capacitance \( C_1 \), the equivalent capacitance \( C_{\text{eq1}} \) is given by:
\[ \frac{1}{C_{\text{eq1}}} = \frac{1}{C_1} + \frac{1}{C_1} + \ldots + \frac{1}{C_1} \quad (n_1 \text{ terms}) \]
This simplifies to:
\[ \frac{1}{C_{\text{eq1}}} = \frac{n_1}{C_1} \implies C_{\text{eq1}} = \frac{C_1}{n_1} \]

**Hint:** Remember that in a series combination, the reciprocal of the total capacitance is the sum of the reciprocals of the individual capacitances.

2. **Determine the Equivalent Capacitance of the Parallel Combination:**
For \( n_2 \) capacitors in parallel, each with capacitance \( C_2 \), the equivalent capacitance \( C_{\text{eq2}} \) is given by:
\[ C_{\text{eq2}} = C_2 + C_2 + \ldots + C_2 \quad (n_2 \text{ terms}) = n_2 C_2 \]

**Hint:** In a parallel combination, the total capacitance is simply the sum of the individual capacitances.

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

For the second combination (parallel):
\[ U_2 = \frac{1}{2} C_{\text{eq2}} V^2 = \frac{1}{2} (n_2 C_2) V^2 \]

**Hint:** Make sure to substitute the values of capacitance and voltage correctly into the energy formula.

4. **Set the Energies Equal:**
Since the energies stored in both combinations are equal, we can set \( U_1 = U_2 \):
\[ \frac{8 C_1}{n_1} = \frac{1}{2} (n_2 C_2) V^2 \]

**Hint:** This step involves equating the two expressions for energy, which is crucial for finding the relationship between \( C_2 \) and \( C_1 \).

5. **Simplify the Equation:**
Multiply both sides by 2 to eliminate the fraction:
\[ \frac{16 C_1}{n_1} = n_2 C_2 \]

**Hint:** When manipulating equations, be careful to maintain equality by performing the same operation on both sides.

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

**Hint:** Isolate \( C_2 \) to express it in terms of \( C_1 \) and the other variables.

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