A `5mu F` capacitor is connected to a 200 V, 100 Hz ac source. The capacitive reactance is

Options

A. `212 Omega`
B. `312 Omega`
C. `318 Omega`
D. `412 Omega`

Answer

To find the capacitive reactance (XC) of a capacitor connected to an AC source, we can use the formula:

\[ X_C = \frac{1}{\omega C} \]

Where:
- \( \omega \) (angular frequency) is given by \( \omega = 2\pi f \)
- \( C \) is the capacitance in farads

### Step-by-Step Solution:

1. **Identify the given values:**
- Capacitance, \( C = 5 \mu F = 5 \times 10^{-6} F \)
- Frequency, \( f = 100 Hz \)

2. **Calculate the angular frequency (\( \omega \)):**
\[ \omega = 2\pi f = 2\pi \times 100 \]
\[ \omega = 200\pi \, \text{rad/s} \]

3. **Substitute the values into the capacitive reactance formula:**
\[ X_C = \frac{1}{\omega C} = \frac{1}{200\pi \times 5 \times 10^{-6}} \]

4. **Calculate \( X_C \):**
\[ X_C = \frac{1}{1000\pi \times 10^{-6}} = \frac{10^6}{1000\pi} \]
\[ X_C = \frac{10^3}{\pi} \]

5. **Using the value of \( \pi \approx 3.14 \):**
\[ X_C \approx \frac{1000}{3.14} \approx 318.47 \, \Omega \]

6. **Final Result:**
The capacitive reactance \( X_C \) is approximately \( 318 \, \Omega \).