Q. The initial phase angle for $i = 10\sin{\omega t} + 8\cos{\omega t}$ is
A) $\tan^{- 1}\left( \frac{4}{5} \right)$
B) $\tan^{- 1}\left( \frac{5}{4} \right)$
C) $\sin^{- 1}\left( \frac{4}{5} \right)$
D) $90{^\circ}$
Explanation
(a) Current $i = i_{0}\sin{(\omega t + \phi})$
$i_{p} = i_{0}\sin{\omega t\cos{\phi + i_{0}\cos{\omega t\sin\phi}}}$
Thus, $i_{0}\cos{\phi = 10}$
$i_{0}\sin{\phi = 8}$
Hence, $\tan{\phi = \frac{4}{5}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.