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 $0.5A$ and the impedance of the circuit is $311\Omega$, find the power factor.

Answer 1

Answer: 0.87

Amplitude of watt less current is
$I_0\sin \varphi =0.5A$
$\bar{Z}=311\Omega .$
$E=311\sin (100\pi t)$
$I_0=\frac{E_0}{Z}=\frac{311}{311}=1A.$
and $I_0=\frac{E_0}{Z}=\frac{311}{311}=1A$
$\therefore \sin \varphi =\frac{1}{2}$ or $\varphi =30^{\circ }$
Power factor
$\cos \varphi =\frac{\sqrt{3}}{2}$