Three capacitors `2muF,3muF and 6muF` are joined with each other. What is the minimum effective capacitance?

Options

A. `1/2 muF`
B. `1muF`
C. `2muF`
D. `3muF`

Explanation

To find the minimum effective capacitance of the three capacitors (2 µF, 3 µF, and 6 µF), we will consider the configuration that gives the minimum capacitance, which is when the capacitors are connected in series.

### Step-by-Step Solution:

1. **Identify the Capacitors**: We have three capacitors with values:
- C1 = 2 µF
- C2 = 3 µF
- C3 = 6 µF

2. **Capacitors in Series**: When capacitors are connected in series, the formula for the total or equivalent capacitance (C_eq) is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

3. **Substituting the Values**: Plugging in the values of the capacitors:
\[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]

4. **Finding a Common Denominator**: The least common multiple (LCM) of the denominators (2, 3, and 6) is 6. We can rewrite each term:
\[ \frac{1}{C_{eq}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} \]

5. **Adding the Fractions**: Now, add the fractions:
\[ \frac{1}{C_{eq}} = \frac{3 + 2 + 1}{6} = \frac{6}{6} \]

6. **Calculating C_eq**: Therefore, we find:
\[ \frac{1}{C_{eq}} = 1 \implies C_{eq} = 1 \, \mu F \]

7. **Conclusion**: The minimum effective capacitance when the capacitors are connected in series is:
\[ C_{eq} = 1 \, \mu F \]