Question

In the shown $AC$ circuit phase different between currents $I_1$ and $I_2$ is

• A. $\frac{\pi }{2}-{\tan }^{-1}\frac{X_L}{R}$
• B. ${\tan }^{-1}\frac{X_L-X_C}{R}$
• C. $\frac{\pi }{2}+{\tan }^{-1}\frac{X_L}{R}$
• D. ${\tan }^{-1}\frac{X_L-X_C}{R}+\frac{\pi }{2}$

Answer 1

Answer: (C)

Let $E=E_0\sin \omega t$, then $I_1=\frac{E_0}{X_C}\sin \left(\omega t+\frac{\pi }{2}\right)$
$I_2=\frac{E_0}{\sqrt{R^2+X^2}}\sin (\omega t+\varphi )$
where $\tan \varphi =-\frac{X_L}{R}\Rightarrow \varphi =-{\tan }^{-1}\left(\frac{X_L}{R}\right)$
Phase difference between $I_1$ and $I_2$
$=\left(\omega t+\frac{\pi }{2}\right)-(\omega t+\varphi )=\frac{\pi }{2}-\varphi =\frac{\pi }{2}+{\tan }^{-1}\left(\frac{X_L}{R}\right)$