A source of 220 V is connected with a pure inductor of 25 mH. The inductive reactance if the frequency of the source is 50 Hz is given by:

Options

A. `6.5 Omega`
B. `7.85 Omega`
C. `8.75Omega`
D. `785Omega`

Explanation

To find the inductive reactance (X_L) of a pure inductor when connected to an AC source, we can use the formula:

\[ X_L = 2 \pi f L \]

Where:
- \( X_L \) is the inductive reactance in ohms (Ω),
- \( f \) is the frequency in hertz (Hz),
- \( L \) is the inductance in henries (H).

### Step-by-Step Solution:

1. **Identify the given values**:
- Frequency \( f = 50 \, \text{Hz} \)
- Inductance \( L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} \)

2. **Substitute the values into the formula**:
\[ X_L = 2 \pi (50) (25 \times 10^{-3}) \]

3. **Calculate \( 2 \pi \)**:
\[ 2 \pi \approx 6.28 \]

4. **Multiply the values**:
\[ X_L = 6.28 \times 50 \times 25 \times 10^{-3} \]

5. **Calculate \( 6.28 \times 50 \)**:
\[ 6.28 \times 50 = 314 \]

6. **Now multiply by \( 25 \times 10^{-3} \)**:
\[ X_L = 314 \times 25 \times 10^{-3} \]

7. **Calculate \( 314 \times 25 \)**:
\[ 314 \times 25 = 7850 \]

8. **Now apply the \( 10^{-3} \)**:
\[ X_L = 7850 \times 10^{-3} = 7.85 \, \Omega \]

### Final Answer:
The inductive reactance \( X_L \) is \( 7.85 \, \Omega \).