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 $I_{1}=\frac{E_{0}}{X_{C}} \sin \left(\omega t+\frac{\pi}{2}\right)$
$I_{2}=\frac{E_{0}}{\sqrt{R^{2}+X^{2}}} \sin (\omega t+\phi)$
where $\tan \phi=-\frac{X_{L}}{R} \Rightarrow \phi=-\tan ^{-1}\left(\frac{X_{L}}{R}\right)$
Phase difference between $I_{1}$ and $I_{2}$
$=\left(\omega t+\frac{\pi}{2}\right)-(\omega t+\phi)=\frac{\pi}{2}-\phi=\frac{\pi}{2}+\tan ^{-1}\left(\frac{X_{L}}{R}\right)$