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 \( X_c \) of a capacitor connected to an AC source, we can use the formula:

\[ X_c = \frac{1}{2 \pi f C} \]

where:
- \( f \) is the frequency in hertz (Hz),
- \( C \) is the capacitance in farads (F).

### Step-by-Step Solution:

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

2. **Substitute the values into the formula**:
\[ X_c = \frac{1}{2 \pi (100) (5 \times 10^{-6})} \]

3. **Calculate the denominator**:
- First, calculate \( 2 \pi \):
\[ 2 \pi \approx 6.2832 \]
- Now calculate \( 2 \pi f C \):
\[ 2 \pi (100) (5 \times 10^{-6}) \approx 6.2832 \times 100 \times 5 \times 10^{-6} = 3.1416 \times 10^{-3} \]

4. **Calculate the capacitive reactance**:
\[ X_c = \frac{1}{3.1416 \times 10^{-3}} \approx 318.31 \, \Omega \]

5. **Round the answer**:
- The capacitive reactance can be rounded to \( 318 \, \Omega \).

### Final Answer:
The capacitive reactance \( X_c \) is approximately \( 318 \, \Omega \).

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