Question

A $0.21H$ inductor and a $12\Omega$ resistance are connected in series to a $220V,50Hz$ ac source. The phase angle between the current and the source voltage is :

• A. ${\tan }^{-1}\left(\frac{7\pi }{4}\right)$
• B. $co{s}^{-1}\left(\frac{7\pi }{4}\right)$
• C. ${\tan }^{-1}\left(\frac{4\pi }{7}\right)$
• D. $co{s}^{-1}\left(\frac{4\pi }{7}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{7\pi }{4}\right)$

Phase angle = $\varphi =ta{n}^{-1}\left(\frac{X_L}{R}\right)=ta{n}^{-1}\left(\frac{0.21\times 2\pi 50}{12}\right)=ta{n}^{-1}\left(\frac{21\pi }{12}\right)=ta{n}^{-1}\left(\frac{7\pi }{4}\right)$