Effective capacitance of parallel combination of two capacitors `C_(1) and C_(2)` is `10muF`. When the capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor `C_(2)` is 4 times of `C_(1)`. If these capacitors are connected in series, their effective capacitor will be: lt

Options

A. Option1 `3.2muF`
B. Option2 `4.2muF`
C. Option3 `1.6muF`
D. Option4 `8.4muF`

Answer

To solve the problem step by step, we will follow the information provided in the question and derive the required values.

### Step 1: Understand the effective capacitance in parallel
When two capacitors \( C_1 \) and \( C_2 \) are connected in parallel, the effective capacitance \( C_{eff} \) is given by the formula:
\[ C_{eff} = C_1 + C_2 \]
According to the problem, the effective capacitance is \( 10 \mu F \):
\[ C_1 + C_2 = 10 \mu F \quad \text{(Equation 1)} \]

### Step 2: Use the energy stored in capacitors
The energy stored in a capacitor is given by the formula:
\[ E = \frac{1}{2} C V^2 \]
When each capacitor is connected to a voltage source of \( 1V \):
- The energy stored in capacitor \( C_1 \) is:
\[ E_1 = \frac{1}{2} C_1 (1^2) = \frac{1}{2} C_1 \]
- The energy stored in capacitor \( C_2 \) is:
\[ E_2 = \frac{1}{2} C_2 (1^2) = \frac{1}{2} C_2 \]

According to the problem, the energy stored in capacitor \( C_2 \) is 4 times that in \( C_1 \):
\[ E_2 = 4 E_1 \]
Substituting the expressions for \( E_1 \) and \( E_2 \):
\[ \frac{1}{2} C_2 = 4 \left( \frac{1}{2} C_1 \right) \]
This simplifies to:
\[ C_2 = 4 C_1 \quad \text{(Equation 2)} \]

### Step 3: Substitute Equation 2 into Equation 1
Now we can substitute \( C_2 \) from Equation 2 into Equation 1:
\[ C_1 + 4 C_1 = 10 \mu F \]
This simplifies to:
\[ 5 C_1 = 10 \mu F \]
Thus, we find:
\[ C_1 = 2 \mu F \]

### Step 4: Calculate \( C_2 \)
Using the value of \( C_1 \) in Equation 2:
\[ C_2 = 4 C_1 = 4 \times 2 \mu F = 8 \mu F \]

### Step 5: Calculate the effective capacitance when connected in series
When capacitors are connected in series, the effective capacitance \( C_{eff} \) is given by:
\[ \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values of \( C_1 \) and \( C_2 \):
\[ \frac{1}{C_{eff}} = \frac{1}{2 \mu F} + \frac{1}{8 \mu F} \]
Finding a common denominator (8):
\[ \frac{1}{C_{eff}} = \frac{4}{8 \mu F} + \frac{1}{8 \mu F} = \frac{5}{8 \mu F} \]
Thus, taking the reciprocal gives:
\[ C_{eff} = \frac{8}{5} \mu F = 1.6 \mu F \]

### Final Answer
The effective capacitance when the capacitors are connected in series is:
\[ C_{eff} = 1.6 \mu F \]