Three capacitors each of capacitor `2muF` are connected in series. Find resultance capacity in farad.

Answer

To find the resultant capacitance of three capacitors connected in series, we can follow these steps:

### Step 1: Understand the formula for capacitors in series
The formula for the equivalent capacitance (C_eq) of capacitors connected in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
where \(C_1\), \(C_2\), and \(C_3\) are the capacitances of the individual capacitors.

### Step 2: Identify the values of the capacitors
In this case, each capacitor has a capacitance of \(2 \mu F\):
- \(C_1 = 2 \mu F\)
- \(C_2 = 2 \mu F\)
- \(C_3 = 2 \mu F\)

### Step 3: Substitute the values into the formula
Substituting the values into the formula, we get:
\[ \frac{1}{C_{eq}} = \frac{1}{2 \mu F} + \frac{1}{2 \mu F} + \frac{1}{2 \mu F} \]

### Step 4: Simplify the equation
This can be simplified as:
\[ \frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]

### Step 5: Calculate the equivalent capacitance
Now, taking the reciprocal to find \(C_{eq}\):
\[ C_{eq} = \frac{2}{3} \mu F \]

### Step 6: Convert to Farads
To convert microfarads to farads, we use the conversion factor \(1 \mu F = 10^{-6} F\):
\[ C_{eq} = \frac{2}{3} \times 10^{-6} F \]

### Final Result
Thus, the resultant capacitance is:
\[ C_{eq} = \frac{2}{3} \times 10^{-6} F \approx 0.6667 \times 10^{-6} F \]
or approximately \(0.6667 \mu F\).

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