Question

The series combination of resistance $R$ and inductance $L$ is connected to an alternating source of e.m.f. $E=311\sin \left(100\pi t\right)$. If the peak value of wattless current is $0.5\text{A}$ and the impedance of the circuit is $311\Omega$, the power factor will be

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\frac{1}{\sqrt{5}}$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

Given,
Amplitude of watt less current is
$I_0\sin \varphi =0.5\text{A}$
Impedance, $Z=311\Omega$
e.m.f = $E=311\sin \left(100\pi t\right)$

Now we can write
and $I_0=\frac{E_0}{Z}=\frac{311}{311}=1\text{A}$

Or,
$\therefore \sin \varphi =\frac{1}{2}$ or $\varphi ={30}^{\circ }$
Power factor
$\cos \varphi =\frac{\sqrt{3}}{2}$

Hence the power factor is $\frac{\sqrt{3}}{2}$