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

Explanation

To solve the problem of finding how to connect 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-by-Step Solution:

1. **Identify the Capacitor Values**:
Each capacitor has a capacitance of \(C = 4 \mu F\).

2. **Understand the Effective Capacitance for Series and Parallel**:
- For capacitors in **series**:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
For three identical capacitors, this simplifies to:
\[ C_{eq} = \frac{C}{3} \]
- For capacitors in **parallel**:
\[ C_{eq} = C_1 + C_2 + C_3 \]
For three identical capacitors, this simplifies to:
\[ C_{eq} = 3C \]

3. **Evaluate Each Option**:
- **Option A**: All capacitors in series.
\[ C_{eq} = \frac{4 \mu F}{3} \approx 1.33 \mu F \quad (\text{Not } 6 \mu F) \]
- **Option B**: All capacitors in parallel.
\[ C_{eq} = 4 \mu F + 4 \mu F + 4 \mu F = 12 \mu F \quad (\text{Not } 6 \mu F) \]
- **Option C**: Two capacitors in series and one in parallel.
- First, calculate the equivalent of the two capacitors in series:
\[ C_{series} = \frac{4 \mu F \cdot 4 \mu F}{4 \mu F + 4 \mu F} = \frac{16}{8} = 2 \mu F \]
- Now add the third capacitor in parallel:
\[ C_{eq} = C_{series} + C_3 = 2 \mu F + 4 \mu F = 6 \mu F \quad (\text{This is correct}) \]
- **Option D**: Two capacitors in parallel and one in series.
- First, calculate the equivalent of the two capacitors in parallel:
\[ C_{parallel} = 4 \mu F + 4 \mu F = 8 \mu F \]
- Now calculate the equivalent with the third capacitor in series:
\[ \frac{1}{C_{eq}} = \frac{1}{C_{parallel}} + \frac{1}{C_3} = \frac{1}{8 \mu F} + \frac{1}{4 \mu F} \]
\[ \frac{1}{C_{eq}} = \frac{1}{8} + \frac{2}{8} = \frac{3}{8} \Rightarrow C_{eq} = \frac{8}{3} \approx 2.67 \mu F \quad (\text{Not } 6 \mu F) \]

4. **Conclusion**:
The only configuration that gives an effective capacitance of \(6 \mu F\) is **Option C**: connecting two capacitors in series and one in parallel.