Question

An $220\text{V}$ $\text{AC}$ voltage at a frequency of $100\text{Hz}$ is applied to a circuit containing a pure inductance of $0.01\text{H}$ and a pure resistance of $6.28\Omega$ in series. Calculate the phase angle between current and voltage.

• A. ${\tan }^{-1}\left(0.1\right)$
• B. ${\tan }^{-1}\left(\sqrt{2}\right)$
• C. ${\tan }^{-1}\left(1\right)$
• D. ${\tan }^{-1}\left(2\right)$

Answer 1

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

In a series $LR$ circuit,

The phase angle between current and voltage is given as,

$\varphi ={\tan }^{-1}\left(\frac{X_L}{R}\right)$

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

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

$={\tan }^{-1}\left(\frac{2\times 3.14\times 100\times 0.01}{6.28}\right)$

$\Rightarrow \varphi ={\text{tan}}^{-1}\left(1\right)$

Hence, option $\left(C\right)$ is correct.