An alternating voltage source `V = 260 sin (628t)` is connected across a pure inductor of 5 mH. Inductive reactance in the circuit is :

Options

A. `3.14Omega`
B. `6.28Omega`
C. `0.318Omega`
D. `0.5Omega`

Answer

To find the inductive reactance in the circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values**:
- The voltage source is given as \( V = 260 \sin(628t) \).
- The inductance \( L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \).

2. **Determine the angular frequency \( \omega \)**:
- From the voltage equation, we can see that \( \omega = 628 \, \text{rad/s} \).

3. **Use the formula for inductive reactance**:
- The formula for inductive reactance \( X_L \) is given by:
\[ X_L = \omega L \]

4. **Substitute the values into the formula**:
- Now substituting the values of \( \omega \) and \( L \):
\[ X_L = 628 \times (5 \times 10^{-3}) \]

5. **Calculate \( X_L \)**:
- Performing the multiplication:
\[ X_L = 628 \times 5 \times 10^{-3} = 3140 \times 10^{-3} \, \Omega \]
- This simplifies to:
\[ X_L = 3.14 \, \Omega \]

6. **Final Answer**:
- Therefore, the inductive reactance in the circuit is:
\[ X_L = 3.14 \, \Omega \]