A 0.01 H inductor and `sqrt(3)pi Omega` resistance are connected in series with a 220 V, 50 Hz AC source. The phase difference between the current and emf is

Options

A. `pi/2rad`
B. `pi/6` rad
C. `pi/3` rad
D. `pi/4` rad

Explanation

To find the phase difference between the current and the emf in a circuit with an inductor and resistance connected in series, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Inductance \( L = 0.01 \, \text{H} \)
- Resistance \( R = \sqrt{3} \pi \, \Omega \)
- Frequency \( f = 50 \, \text{Hz} \)

2. **Calculate Angular Frequency**:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 50 = 100 \pi \, \text{rad/s} \]

3. **Calculate Inductive Reactance**:
\[ X_L = \omega L \]
Substituting the values of \( \omega \) and \( L \):
\[ X_L = 100 \pi \times 0.01 = \pi \, \Omega \]

4. **Use the Formula for Phase Difference**:
The phase difference \( \phi \) between the current and the emf is given by:
\[ \phi = \tan^{-1}\left(\frac{X_L}{R}\right) \]
Substituting the values of \( X_L \) and \( R \):
\[ \phi = \tan^{-1}\left(\frac{\pi}{\sqrt{3} \pi}\right) \]

5. **Simplify the Expression**:
The \( \pi \) in the numerator and denominator cancels out:
\[ \phi = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

6. **Determine the Value of the Phase Difference**:
From trigonometric identities, we know:
\[ \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} \]

7. **Conclusion**:
Therefore, the phase difference between the current and emf is:
\[ \phi = \frac{\pi}{6} \, \text{radians} \]

### Final Answer:
The phase difference between the current and emf is \( \frac{\pi}{6} \) radians.

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