Question

A circuit is made up of a resistance $1\, \Omega$ and inductance $0.01\, H$. An alternating emf of $200\, V$ at $50\,Hz$ is connected, then the phase difference between the current and the emf in the circuit is

• A. $\tan^{-1} (\pi)$
• B. $\tan^{-1} (\frac{\pi}{2})$
• C. $\tan^{-1} (\frac{\pi}{4})$
• D. $\tan^{-1} (\frac{\pi}{3})$

Answer 1

Answer: (A)

$\tan^{-1} (\pi)$

Answer 2

$\tan \phi=\left(\frac{X_{L}}{R}\right)$
$X_{L}=\omega L=(2 \pi f L)$
$=(2 \pi)(50)(0.01)=\pi, \Omega$
Also, $R=1 \Omega$
$\therefore \phi=\tan ^{-1}(\pi)$