Question

A coil of inductance \( 0.1 \, \text{H} \) and resistance \( 110 \, \Omega \) is connected to an AC source of \( 110 \, \text{V}, 350 \, \text{Hz} \). What is the phase difference between the voltage maximum and the current maximum?

• A. \( \tan^{-1}(1.5) \)
• B. \( \tan^{-1}(0.5) \)
• C. \( \tan^{-1}(1.73) \)
• D. \( \tan^{-1}(2) \)

Hint:

In R-L circuits, phase angle is \( \tan^{-1} \left( \frac{\omega L}{R} \right) \).

Answer 1

The Correct Option is D

Phase angle \( \phi = \tan^{-1} \left( \frac{\omega L}{R} \right) \)
Given: \( f = 350 \, \text{Hz}, L = 0.1 \, \text{H}, R = 110 \, \Omega \)
\[ \omega = 2\pi f = 2\pi \cdot 350 \approx 2200 \Rightarrow \frac{\omega L}{R} = \frac{2200 \cdot 0.1}{110} = 2 \] So, \( \phi = \tan^{-1}(2) \)