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 parallel
B. connecting two in series and one in parallel
C. connecting two in parallel and one in series
D. connecting all of them in series

Answer

To solve the problem of connecting three capacitors of 4 µF each to achieve an effective capacitance of 6 µF, we need to analyze the possible configurations: series and parallel connections.

### Step-by-Step Solution:

1. **Understanding Capacitor Configurations**:
- When capacitors are connected in **parallel**, the total capacitance (C_eq) is the sum of the individual capacitances:
\[ C_{eq} = C_1 + C_2 + C_3 \]
- When capacitors are connected in **series**, the total capacitance is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

2. **Calculating Maximum Capacitance**:
- For three capacitors of 4 µF connected in parallel:
\[ C_{eq} = 4 + 4 + 4 = 12 \, \mu F \]
- This configuration gives the maximum capacitance.

3. **Calculating Minimum Capacitance**:
- For three capacitors of 4 µF connected in series:
\[ \frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4} \]
\[ C_{eq} = \frac{4}{3} \, \mu F \approx 1.33 \, \mu F \]
- This configuration gives the minimum capacitance.

4. **Finding a Combination to Achieve 6 µF**:
- We need to find a combination of series and parallel connections to achieve an effective capacitance of 6 µF.
- One possible way is to connect two capacitors in parallel and one in series with that combination.

5. **Calculating the Effective Capacitance for the Combination**:
- Connect two capacitors in parallel:
\[ C_{parallel} = 4 + 4 = 8 \, \mu F \]
- Now, connect this parallel combination in series with the third capacitor:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{4} \]
- Finding a common denominator (8):
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8} \]
\[ C_{eq} = \frac{8}{3} \, \mu F \approx 2.67 \, \mu F \]
- This does not give us the required capacitance.

6. **Trying Another Configuration**:
- Connect one capacitor in series with two capacitors in parallel:
- Two capacitors in parallel:
\[ C_{parallel} = 4 + 4 = 8 \, \mu F \]
- Now connect this in series with the third capacitor:
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{1}{4} \]
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8} \]
\[ C_{eq} = \frac{8}{3} \, \mu F \approx 2.67 \, \mu F \]
- This is still not yielding 6 µF.

7. **Final Configuration**:
- Connect two capacitors in series:
\[ \frac{1}{C_{series}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \implies C_{series} = 2 \, \mu F \]
- Now connect this series combination in parallel with the third capacitor:
\[ C_{eq} = C_{series} + C_{third} = 2 + 4 = 6 \, \mu F \]

### Conclusion:
To achieve an effective capacitance of 6 µF, connect two capacitors in series (2 µF) and then connect this combination in parallel with the third capacitor (4 µF).