Question

A $12$ ohm resistor and a $0.21$ henry inductor are connected in series to an AC source operating at $20volts$, $50$ cycle/second. The phase angle between the current and the source voltage is

• A. ${30}^o$
• B. ${40}^o$
• C. ${80}^o$
• D. ${90}^o$

Answer 1

The correct option is C ${80}^o$

Phase angle $\varphi ={\tan }^{-1}\frac{X_L}{R}={\tan }^{-1}\frac{\omega L}{R}$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{2\pi \times 50\times 0.21}{12}\right)[\because \omega =50c/s=50\times 2\pi rad/sec]$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{21}{12}\pi \right)={\tan }^{-1}\left(\frac{7}{4}\pi \right)$

$\Rightarrow \varphi ={\tan }^{-1}\left(\frac{11}{2}\right)$

$\Rightarrow \varphi ={79.67}^o={80}^o$