When two capacitors are connected in series, the equivalent capacitance is `1.2muF`. When they are connected in parallel, the equivalent capacitance is `5muF`. The capacitances of the capacitors are

Options

A. `3muFand2muF`
B. `4muFand1muF`
C. `2.5muFand2.5muF`
D. `3.5muFand1.5muF`

Explanation

To find the capacitances of the two capacitors \( C_1 \) and \( C_2 \) based on the given conditions, we can follow these steps:

### Step 1: Understand the formulas for capacitors in series and parallel
When capacitors are connected in series, the equivalent capacitance \( C_s \) is given by:
\[ \frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} \]
When capacitors are connected in parallel, the equivalent capacitance \( C_p \) is given by:
\[ C_p = C_1 + C_2 \]

### Step 2: Set up the equations based on the problem statement
From the problem, we know:
- \( C_s = 1.2 \, \mu F \)
- \( C_p = 5 \, \mu F \)

Using the series formula, we can rearrange it to:
\[ C_1 C_2 = C_s (C_1 + C_2) \]
Substituting \( C_s = 1.2 \, \mu F \):
\[ C_1 C_2 = 1.2 (C_1 + C_2) \]

Using the parallel formula:
\[ C_1 + C_2 = 5 \, \mu F \]

### Step 3: Substitute \( C_1 + C_2 \) into the series equation
Let \( C_1 + C_2 = 5 \, \mu F \). We can substitute this into our earlier equation:
\[ C_1 C_2 = 1.2 \times 5 \]
\[ C_1 C_2 = 6 \, \mu F^2 \]

### Step 4: Set up a system of equations
Now we have two equations:
1. \( C_1 + C_2 = 5 \)
2. \( C_1 C_2 = 6 \)

### Step 5: Solve the equations
Let's denote \( C_1 \) and \( C_2 \) as the roots of the quadratic equation:
\[ x^2 - (C_1 + C_2)x + C_1 C_2 = 0 \]
Substituting the known values:
\[ x^2 - 5x + 6 = 0 \]

### Step 6: Factor the quadratic equation
Factoring the equation:
\[ (x - 2)(x - 3) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{or} \quad x = 3 \]

### Step 7: Identify the capacitances
So, we have:
- \( C_1 = 2 \, \mu F \)
- \( C_2 = 3 \, \mu F \)

### Conclusion
The capacitances of the two capacitors are \( C_1 = 2 \, \mu F \) and \( C_2 = 3 \, \mu F \).