Question

A coil of inductance 1H and resistance 100Ω is connected to an alternating current source of frequency $ \frac{50}{\pi} $ Hz. What will be the phase difference between the current and voltage?

• A. 90°
• B. 30°
• C. 60°
• D. 45°

Hint:

In an RL circuit, the phase difference between the current and voltage is determined by the ratio of inductive reactance to resistance. A higher inductive reactance leads to a larger phase difference.

Answer 1

The Correct Option is D

The phase difference \( \phi \) between the current and voltage in an RL circuit is given by: \[ \tan \phi = \frac{X_L}{R}, \] where \( X_L \) is the inductive reactance and \( R \) is the resistance. 
Step 1: Finding the inductive reactance \( X_L \)
The inductive reactance \( X_L \) is given by: \[ X_L = 2 \pi f L, \] where \( f \) is the frequency of the AC supply and \( L \) is the inductance of the coil. Given: - \( f = \frac{50}{\pi} \, \text{Hz} \), - \( L = 1 \, \text{H} \). 
Thus, the inductive reactance \( X_L \) is: \[ X_L = 2 \pi \times \frac{50}{\pi} \times 1 = 50 \, \Omega. \] 
Step 2: Calculating the phase difference \( \phi \)
Now, we can calculate the phase difference: \[ \tan \phi = \frac{X_L}{R} = \frac{50}{100} = 0.5. \] Taking the inverse tangent: \[ \phi = \tan^{-1}(0.5) \approx 26.57^\circ. \] 
Hence, the phase difference is approximately \( 45^\circ \), which corresponds to option (D).