The input voltage $v(t)$ and current $i(t)$ of a converter are given by,
$v(t) = 300 \sin(\omega t)$ V
$i(t) = 10\sin\left(\omega t - \frac{\pi}{6}\right) + 2\sin\left(3\omega t + \frac{\pi}{6}\right) + \sin\left(5\omega t + \frac{\pi}{2}\right)$ A
where, $\omega = 2\pi \times 50$ rad/s. The input power factor of the converter is closest to
The input power factor (PF) for a converter with non-sinusoidal current is determined by the combined effect of the phase difference between voltage and current fundamentals and the presence of harmonic components.
The power factor can be calculated as:
$ PF = \frac{\text{Real Power (P)}}{\text{Apparent Power (S)}} $
For waveforms containing harmonics, a practical approach is using the product of the fundamental power factor ($PF_{fundamental}$) and the distortion factor (DF):
$ PF = PF_{fundamental} \times DF $
The input voltage is given as:
$ v(t) = 300 \sin(\omega t) \text{ V} $
This is a pure sinusoidal voltage at the fundamental frequency ($\omega$). The phase of the fundamental voltage component is $\phi_v = 0$.
The input current is:
$ i(t) = 10\sin\left(\omega t - \frac{\pi}{6}\right) + 2\sin\left(3\omega t + \frac{\pi}{6}\right) + \sin\left(5\omega t + \frac{\pi}{2}\right) \text{ A} $
The fundamental component of the current is:
$ i_1(t) = 10\sin\left(\omega t - \frac{\pi}{6}\right) \text{ A} $
The phase of the fundamental current component is $\phi_{i1} = -\frac{\pi}{6}$.
The phase difference ($\phi_1$) between the fundamental voltage and current is:
$ \phi_1 = \phi_v - \phi_{i1} = 0 - \left(-\frac{\pi}{6}\right) = \frac{\pi}{6} $
Therefore, the fundamental power factor is:
$ PF_{fundamental} = \cos(\phi_1) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \approx 0.866 $
First, find the RMS (Root Mean Square) values of the current components.
The total RMS current ($I_{rms}$) is calculated as:
$ I_{rms} = \sqrt{I_{1,rms}^2 + I_{3,rms}^2 + I_{5,rms}^2} $
$ I_{rms}^2 = \left(\frac{10}{\sqrt{2}}\right)^2 + \left(\frac{2}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 $
$ I_{rms}^2 = \frac{100}{2} + \frac{4}{2} + \frac{1}{2} = 50 + 2 + 0.5 = 52.5 $
$ I_{rms} = \sqrt{52.5} \text{ A} $
The distortion factor (DF) is the ratio of the fundamental RMS current to the total RMS current:
$ DF = \frac{I_{1,rms}}{I_{rms}} = \frac{10/\sqrt{2}}{\sqrt{52.5}} = \frac{10}{\sqrt{2} \times \sqrt{52.5}} = \frac{10}{\sqrt{105}} $
$ DF \approx \frac{10}{10.247} \approx 0.9759 $
Multiply the fundamental power factor by the distortion factor:
$ PF = PF_{fundamental} \times DF $
$ PF \approx 0.866 \times 0.9759 $
$ PF \approx 0.8451 $
The input power factor of the converter is closest to 0.845.
Which of the following is NOT an advantage of power factor improvement using capacitor?
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The power factor of a circuit is equal to
Which of the following is NOT responsible for poor power factor?