Two capacitors of capacity `4 mu F and 6 mu F` are connected in series and a battery is connected to the combination. The energy stored is `E_1` If they are connected in parallel and if the same battery is connected to this combination the energy stored is `E_2 ` The ratio `E_1 :E_2` is

Options

A. A) `4:9`
B. B) `9:14`
C. C) `6:25`
D. D) `7:12`

Answer

To solve the problem, we need to calculate the energy stored in two capacitors when they are connected in series and in parallel, and then find the ratio of these energies.

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitor 1, \( C_1 = 4 \, \mu F \)
- Capacitor 2, \( C_2 = 6 \, \mu F \)
- Let the voltage of the battery be \( V \).

2. **Calculate the equivalent capacitance when capacitors are connected in series:**
The formula for the equivalent capacitance \( C_s \) of capacitors in series is:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_s} = \frac{1}{4} + \frac{1}{6} \]
Finding a common denominator (which is 12):
\[ \frac{1}{C_s} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \]
Therefore,
\[ C_s = \frac{12}{5} \, \mu F = 2.4 \, \mu F \]

3. **Calculate the energy stored in the series configuration \( E_1 \):**
The energy stored in a capacitor is given by:
\[ E = \frac{1}{2} C V^2 \]
Thus, for the series configuration:
\[ E_1 = \frac{1}{2} C_s V^2 = \frac{1}{2} \times 2.4 \, \mu F \times V^2 \]

4. **Calculate the equivalent capacitance when capacitors are connected in parallel:**
The formula for the equivalent capacitance \( C_p \) of capacitors in parallel is:
\[ C_p = C_1 + C_2 \]
Substituting the values:
\[ C_p = 4 \, \mu F + 6 \, \mu F = 10 \, \mu F \]

5. **Calculate the energy stored in the parallel configuration \( E_2 \):**
Thus, for the parallel configuration:
\[ E_2 = \frac{1}{2} C_p V^2 = \frac{1}{2} \times 10 \, \mu F \times V^2 \]

6. **Find the ratio \( E_1 : E_2 \):**
Now we can find the ratio of the energies:
\[ \frac{E_1}{E_2} = \frac{\frac{1}{2} \times 2.4 \, \mu F \times V^2}{\frac{1}{2} \times 10 \, \mu F \times V^2} \]
The \( \frac{1}{2} \) and \( V^2 \) cancel out:
\[ \frac{E_1}{E_2} = \frac{2.4 \, \mu F}{10 \, \mu F} = \frac{2.4}{10} = \frac{6}{25} \]

7. **Final Answer:**
Thus, the ratio \( E_1 : E_2 \) is:
\[ E_1 : E_2 = 6 : 25 \]