Two capacitors have a capacitance of `5 muF` when connected in parallel and `1.2 muF` when connected in series. Calculate their capacitance.

Explanation

To solve the problem of finding the capacitance of two capacitors when they are connected in parallel and in series, we can follow these steps:

### Step 1: Set up the equations based on the given conditions

Let the capacitances of the two capacitors be \( C_1 \) and \( C_2 \).

1. When connected in parallel, the total capacitance is given by:
\[ C_{\text{parallel}} = C_1 + C_2 = 5 \, \mu F \quad \text{(Equation 1)} \]

2. When connected in series, the total capacitance is given by:
\[ \frac{1}{C_{\text{series}}} = \frac{1}{C_1} + \frac{1}{C_2} \implies C_{\text{series}} = \frac{C_1 C_2}{C_1 + C_2} = 1.2 \, \mu F \quad \text{(Equation 2)} \]

### Step 2: Express \( C_2 \) in terms of \( C_1 \)

From Equation 1, we can express \( C_2 \) as:
\[ C_2 = 5 - C_1 \quad \text{(Equation 3)} \]

### Step 3: Substitute \( C_2 \) in Equation 2

Substituting Equation 3 into Equation 2:
\[ C_{\text{series}} = \frac{C_1 (5 - C_1)}{C_1 + (5 - C_1)} = \frac{C_1 (5 - C_1)}{5} = 1.2 \]

### Step 4: Solve for \( C_1 \)

Now, we can rearrange the equation:
\[ C_1 (5 - C_1) = 1.2 \times 5 \]
\[ C_1 (5 - C_1) = 6 \]
\[ 5C_1 - C_1^2 = 6 \]
Rearranging gives us a quadratic equation:
\[ C_1^2 - 5C_1 + 6 = 0 \]

### Step 5: Factor the quadratic equation

Factoring the quadratic equation:
\[ (C_1 - 2)(C_1 - 3) = 0 \]

### Step 6: Find the values of \( C_1 \)

Setting each factor to zero gives:
\[ C_1 - 2 = 0 \quad \Rightarrow \quad C_1 = 2 \, \mu F \]
\[ C_1 - 3 = 0 \quad \Rightarrow \quad C_1 = 3 \, \mu F \]

### Step 7: Find \( C_2 \)

Using Equation 3 to find \( C_2 \):
1. If \( C_1 = 2 \, \mu F \):
\[ C_2 = 5 - 2 = 3 \, \mu F \]

2. If \( C_1 = 3 \, \mu F \):
\[ C_2 = 5 - 3 = 2 \, \mu F \]

Thus, the capacitances of the two capacitors are:
\[ C_1 = 2 \, \mu F \quad \text{and} \quad C_2 = 3 \, \mu F \]

### Final Answer:
The capacitances of the two capacitors are \( 2 \, \mu F \) and \( 3 \, \mu F \).

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