A.

B.

equal to natural frequency of LCR system

The power factor is defined as the cosine of the phase angle between alternating e.m.f. and current in a.c. current. 

Power factor of an a.c. circuit is given by
$\boxed{\cos \mathrm{\varphi } = \frac{\mathrm{R}}{\sqrt{{\mathrm{R}}^2+{\left(\mathrm{\omega L}-{\displaystyle \frac{1}{\mathrm{\omega C}}}\right)}^2}}}$ 

Power factor depends upon the frequency of the a.c. source.

Given,
Instantaneous current, i = 10 sin 314 t A

Instantaneous voltage, V = 50 sin (314 t + $\mathrm{\pi }$/2) V 

Since, 
$$\begin{gathered} \mathrm{I} = {\mathrm{I}}_{\mathrm{O}} \sin \mathrm{\omega t} \mathrm{and}, \\ \mathrm{V} = {\mathrm{V}}_{\mathrm{O} }\sin (\mathrm{\omega t} +\mathrm{\varphi }) \end{gathered}$$
Therefore, from the above equations we get 

Peak current, peak voltage and $\mathrm{\varphi }$ as, 
${\mathrm{i}}_0 = 10, {\mathrm{V}}_0 =50 \mathrm{and} \mathrm{\varphi } = \frac{\mathrm{\pi }}{2}$ 

Power dissipation in the circuit is given by
${\mathrm{E}}_{\mathrm{v}} = \frac{{\mathrm{E}}_{0 }}{\sqrt{2}} \mathrm{and} {\mathrm{I}}_0 = \frac{{\mathrm{I}}_{ \mathrm{o}}}{\sqrt{2}}$ 

Therefore, 

$$\begin{gathered} \mathrm{P} = \left(\frac{50}{\sqrt{2}}\right) \left(\frac{10}{\sqrt{2}}\right) \cos \frac{\mathrm{\pi }}{2} \\ = 0 \end{gathered}$$          
                                                    $\left[\because \cos \frac{\mathrm{\pi }}{2} = 0\right]$