Question

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

​​​​​​​$[$1 Mark$]$

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

Answer 1

The correct option is A ${\tan }^{-1}\left(\pi \right)$

We know that,
$X_L=\omega L=\left(2\pi fL\right)=\left(2\pi \right)\left(50\right)\left(0.01\right)=\pi \Omega$
Also, $R=1\Omega$
Phase difference is given by formula
$\tan \varphi =\left(\frac{X_L}{R}\right)$
$\therefore \varphi ={\tan }^{-1}\left(\pi \right)$