Question

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

• A. $\frac{\pi }{6}$ radians
• B. $\frac{\pi }{3}$ radians
• C. $\frac{\pi }{4}$ radians
• D. $\frac{\pi }{2}$ radians

Answer 1

The correct option is A $\frac{\pi }{6}$ radians

Phase difference between inductor and resistor connected in series = $ta{n}^{-1}\left(\frac{X_L}{R}\right)$

$=ta{n}^{-1}\left(\frac{\omega L}{R}\right)$

$=ta{n}^{-1}\left(\frac{2\pi fL}{R}\right)$

$=ta{n}^{-1}\left(\frac{2\pi 50\times 0.01}{\sqrt{3}\pi }\right)$

$=ta{n}^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$=\frac{\pi }{6}$