If two capacitors `C_1` & `C_2` are connected in parallel then equivalent capacitance is `10 F`. If both capacitance are connected across 1V battery then energy stored by `C_2` is 4 times of `C_1`. Then the equivalent capacitance if they are connected in series is-

Options

A. `1.6 mu F`
B. `16 mu F`
C. `4 mu F`
D. `1/4 mu F`

Explanation

To solve the problem step by step, let's break down the information given and apply the relevant formulas.

### Step 1: Understand the given information
We have two capacitors, \( C_1 \) and \( C_2 \), connected in parallel with an equivalent capacitance of \( 10 \, \text{F} \). When connected to a \( 1 \, \text{V} \) battery, the energy stored in \( C_2 \) is four times that stored in \( C_1 \).

### Step 2: Write the equation for equivalent capacitance in parallel
For capacitors in parallel, the equivalent capacitance \( C \) is given by:
\[ C = C_1 + C_2 \]
From the problem, we know:
\[ C_1 + C_2 = 10 \quad \text{(Equation 1)} \]

### Step 3: Write the equation for energy stored in capacitors
The energy \( U \) stored in a capacitor is given by:
\[ U = \frac{1}{2} C V^2 \]
For \( C_1 \):
\[ U_1 = \frac{1}{2} C_1 (1^2) = \frac{1}{2} C_1 \]
For \( C_2 \):
\[ U_2 = \frac{1}{2} C_2 (1^2) = \frac{1}{2} C_2 \]
According to the problem, \( U_2 = 4 U_1 \):
\[ \frac{1}{2} C_2 = 4 \left(\frac{1}{2} C_1\right) \]
This simplifies to:
\[ C_2 = 4 C_1 \quad \text{(Equation 2)} \]

### Step 4: Substitute Equation 2 into Equation 1
Now, substitute \( C_2 \) from Equation 2 into Equation 1:
\[ C_1 + 4 C_1 = 10 \]
This simplifies to:
\[ 5 C_1 = 10 \]
Thus, we find:
\[ C_1 = 2 \, \text{F} \]

### Step 5: Calculate \( C_2 \)
Using Equation 2:
\[ C_2 = 4 C_1 = 4 \times 2 = 8 \, \text{F} \]

### Step 6: Find the equivalent capacitance in series
For capacitors in series, the equivalent capacitance \( C_{\text{eq}} \) is given by:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{2} + \frac{1}{8} \]
Finding a common denominator (which is 8):
\[ \frac{1}{C_{\text{eq}}} = \frac{4}{8} + \frac{1}{8} = \frac{5}{8} \]
Thus:
\[ C_{\text{eq}} = \frac{8}{5} = 1.6 \, \text{F} \]

### Final Answer
The equivalent capacitance when the capacitors are connected in series is:
\[ \boxed{1.6 \, \text{F}} \]