Three capacitors each of capacity `4 muF` are to be connected in such a way that the effective capacitance is `6 muF`. This can be done by

Options

A. connecting them in series
B. connecting them is parallel
C. connecting two in series and one in parallel
D. connecting two in parallel and one is series

Explanation

To solve the problem of connecting three capacitors, each of capacity \(4 \mu F\), to achieve an effective capacitance of \(6 \mu F\), we can analyze the different configurations of connecting capacitors: series and parallel.

### Step 1: Understanding Capacitor Connections

1. **Series Connection**: The formula for the equivalent capacitance \(C_{eq}\) of capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
For three capacitors of equal capacitance \(C\):
\[ C_{eq} = \frac{C}{3} \]
Substituting \(C = 4 \mu F\):
\[ C_{eq} = \frac{4}{3} \mu F \approx 1.33 \mu F \]
This is not equal to \(6 \mu F\).

### Step 2: Parallel Connection

2. **Parallel Connection**: The formula for the equivalent capacitance \(C_{eq}\) of capacitors in parallel is:
\[ C_{eq} = C_1 + C_2 + C_3 \]
For three capacitors of equal capacitance \(C\):
\[ C_{eq} = 3C \]
Substituting \(C = 4 \mu F\):
\[ C_{eq} = 3 \times 4 = 12 \mu F \]
This is also not equal to \(6 \mu F\).

### Step 3: Combination of Series and Parallel

3. **Combination of Two in Series and One in Parallel**: Let's connect two capacitors in series and one capacitor in parallel with the combination.

- **Two Capacitors in Series**:
\[ C_{series} = \frac{C}{2} = \frac{4}{2} = 2 \mu F \]
- **Adding the Third Capacitor in Parallel**:
\[ C_{eq} = C_{series} + C_{parallel} = 2 \mu F + 4 \mu F = 6 \mu F \]
This configuration gives us the desired effective capacitance of \(6 \mu F\).

### Conclusion

The effective capacitance of \(6 \mu F\) can be achieved by connecting two capacitors in series and one capacitor in parallel with the series combination.