In an a. c. circuit the instantaneous current and e.m.f. are represented as $i=i_o \sin \left(\omega…
In an a. c. circuit the instantaneous current and e.m.f. are represented as $i=i_o \sin \left(\omega t-\frac{\pi}{6}\right)$ and $E=E_o \sin \left(\omega t+\frac{\pi}{3}\right)$ respectively. The voltage leads the current by
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$
Solution
The phase of the current $=-\frac{\pi}{6}$
The phase of voltage $=+\frac{\pi}{3}$
Hence the voltage leads current by phase angle
$\phi=\left(\frac{\pi}{6}\right)-\left(-\frac{\pi}{3}\right)=\frac{\pi}{2}$