Question

For the given circuit assumming inductor and source to be ideal, the phase difference between currents $I_1$ and $I_2$ is

• A. $ta{n}^{-1}\left(\frac{X_c}{R}\right)-\frac{\pi }{2}$
• B. $ta{n}^{-1}\left(\frac{X_c}{R}\right)$
• C. $ta{n}^{-1}\left(\frac{X_c}{R}\right)+\frac{\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $ta{n}^{-1}\left(\frac{X_c}{R}\right)+\frac{\pi }{2}$

$V_s=V_c\sin wt$

$I_1=I_{10}\sin (wt-\pi /2)$

$I_2=I_{20}\sin (wt+\theta )$

$\tan \theta =\frac{X_c}{R}$

So, phase difference $=\theta +\frac{\pi }{2},{\tan }^{-1}\left(\frac{X_c}{R}\right)+\frac{z}{2}$

Hence, option C is correct