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 need to connect them in series, as the effective capacitance is minimized in this configuration.

### Step-by-Step Solution:

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

2. **Formula for Series Connection**: The formula for the effective capacitance (C_eff) of capacitors in series is given by:
\[ \frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

3. **Substituting Values**: Substitute the values of the capacitors into the formula:
\[ \frac{1}{C_{eff}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]

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

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

6. **Calculating C_eff**: Now, take the reciprocal to find C_eff:
\[ C_{eff} = \frac{1}{1} = 1 \, \mu F \]

### Final Answer:
The minimum effective capacitance of the three capacitors is **1 µF**.

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