Three capacitors `2 mu f, 3 mu F` and `6muF` are joined in series with each other. The equivalent capacitance is :

Options

A. `1/2 mu F`
B. `1 mu F`
C. `2 mu F`
D. `11mu F`

Answer

To find the equivalent capacitance of three capacitors connected in series, we can use the formula for capacitors in series:

\[ \frac{1}{C_{\text{equiv}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]

### Step 1: Identify the values of the capacitors
Given:
- \( C_1 = 2 \, \mu F \)
- \( C_2 = 3 \, \mu F \)
- \( C_3 = 6 \, \mu F \)

### Step 2: Substitute the values into the formula
Substituting the values into the formula gives:

\[ \frac{1}{C_{\text{equiv}}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} \]

### Step 3: Find 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:

\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6}, \quad \frac{1}{6} = \frac{1}{6} \]

### Step 4: Add the fractions
Now, we can add the fractions:

\[ \frac{1}{C_{\text{equiv}}} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \]

### Step 5: Calculate the equivalent capacitance
Taking the reciprocal gives us:

\[ C_{\text{equiv}} = 1 \, \mu F \]

### Final Answer
The equivalent capacitance of the three capacitors in series is:

\[ C_{\text{equiv}} = 1 \, \mu F \]

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