Question

An ideal inductor of \( \frac{1}{2} \, \text{H} \) is connected in series with a \( 300 \, \Omega \) resistor. If a 20 V, 200 Hz alternating source is connected across the combination, the phase difference between the voltage and current is

• A. \( \tan^{-1} \left( \frac{3}{4} \right) \)
• B. \( \tan^{-1} \left( \frac{4}{3} \right) \)
• C. \( \tan^{-1} \left( \frac{1}{4} \right) \)
• D. \( \tan^{-1} \left( \frac{1}{5} \right) \)

Hint:

The phase difference in an AC circuit with both inductive and resistive components depends on the ratio of the inductive reactance to the resistance.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
In an L-R series circuit, there is a phase difference between the total voltage and current due to the inductive reactance.
Step 2: Key Formula or Approach:
1. Inductive reactance \( X_L = 2\pi f L \).
2. Phase difference \( \phi = \tan^{-1} \left( \frac{X_L}{R} \right) \).
Step 3: Detailed Explanation:
Given: \( L = 1/\pi \text{ H} \), \( R = 300 \Omega \), \( f = 200 \text{ Hz} \).
Calculate \( X_L \):
\[ X_L = 2\pi \cdot (200) \cdot \left( \frac{1}{\pi} \right) = 400 \Omega \]
Calculate phase difference \( \phi \):
\[ \tan \phi = \frac{X_L}{R} = \frac{400}{300} = \frac{4}{3} \]
\[ \phi = \tan^{-1} \left( \frac{4}{3} \right) \]
Step 4: Final Answer:
The phase difference is \( \tan^{-1}(4/3) \).