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Question

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 correct answer is
0.845

Understanding Converter Power Factor

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 $

Calculating Fundamental Power Factor ($PF_{fundamental}$)

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 $

Calculating Distortion Factor (DF)

First, find the RMS (Root Mean Square) values of the current components.

  • Fundamental RMS current: $I_{1,rms} = \frac{10}{\sqrt{2}}$ A
  • 3rd Harmonic RMS current: $I_{3,rms} = \frac{2}{\sqrt{2}}$ A
  • 5th Harmonic RMS current: $I_{5,rms} = \frac{1}{\sqrt{2}}$ A

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 $

Calculating Total Power Factor (PF)

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.

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Important Questions from Power Factors

  1. If a circuit load impedance is (25 - j25), find the power factor.
  2. Which of the following is NOT an advantage of power factor improvement using capacitor?

  3. The form factor in reference to alternating current wave form represents the ratio of

  4. The power factor of a circuit is equal to

  5. Which of the following is NOT responsible for poor power factor?

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