Question

The series combination of resistance $R$ and inductance $L$ is connected to an alternating source of e.m.f. $e =311 \sin (100 \pi t )$. If the peak value of watt less current is $0.5 \,A$ and the impedance of the circuit is $311 \,\Omega$, find the power factor.

Answer 1

Amplitude of watt less current is
$I_0 \sin \phi=0.5\, A$
$\bar{Z}=311\, \Omega .$
$E =311 \sin (100\, \pi t )$
$I _{0}=\frac{ E _{0}}{ Z }=\frac{311}{311}=1 A .$
and $ I _{0}=\frac{ E _{0}}{ Z }=\frac{311}{311}=1\, A$
$\therefore \sin \phi=\frac{1}{2} $ or $\phi=30^{\circ}$
Power factor
$\cos \phi=\frac{\sqrt{3}}{2}$