Question 14 Marks
What is power factor? Explain.
Answer
View full question & answer→We know that expression of average power is
$\overline{ P }= E _{ rms } I _{ rms } \cos \phi$
$\cos \phi=\frac{\overline{ P }}{ E _{ rms } I _{ rms }}$
$\cos \phi=\frac{\overline{ P }}{ P _{ app }}$
where $P_{\text {app }}=E_{ rms } I_{ rms }$
Hence the ratio of $\overline{ P }$ and $P _{\text {qpp }}$ is known as power factor.
In other words, cosine of the phase difference $\phi$ between emf and current is known as power factor.
We know that
$\tan \phi=\frac{ X _{ L }- X _{ C }}{ R }$
$\cos \phi=\frac{ AB }{ AC }$
$\cos \phi=\frac{ R }{\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}}$
$\cos \phi=\frac{R}{Z}$
$\because \quad Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$
Power factor is equal to the ratio of resistance and impedance.
If $\phi=0^{\circ}, \cos \phi=+1$,maximum power factor.
If $\phi= \pm 90^{\circ}, \cos \phi=0$, minimum power factor.
$\overline{ P }= E _{ rms } I _{ rms } \cos \phi$
$\cos \phi=\frac{\overline{ P }}{ E _{ rms } I _{ rms }}$
$\cos \phi=\frac{\overline{ P }}{ P _{ app }}$
where $P_{\text {app }}=E_{ rms } I_{ rms }$
Hence the ratio of $\overline{ P }$ and $P _{\text {qpp }}$ is known as power factor.
In other words, cosine of the phase difference $\phi$ between emf and current is known as power factor.
We know that
$\tan \phi=\frac{ X _{ L }- X _{ C }}{ R }$
$\cos \phi=\frac{ AB }{ AC }$
$\cos \phi=\frac{ R }{\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}}$
$\cos \phi=\frac{R}{Z}$
$\because \quad Z =\sqrt{ R ^2+\left( X _{ L }- X _{ C }\right)^2}$
Power factor is equal to the ratio of resistance and impedance.
If $\phi=0^{\circ}, \cos \phi=+1$,maximum power factor.
If $\phi= \pm 90^{\circ}, \cos \phi=0$, minimum power factor.