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:

• A. 0.845
• B. 0.867
• C. 0.887
• D. 1.0

Hint:

In power electronics, only the fundamental component of current contributes to real power. Harmonics affect apparent power, reducing power factor.

Answer 1

The Correct Option is A

Only the fundamental components contribute to real power. Higher harmonics contribute to distortion but not to real power. Fundamental voltage: \[ v(t) = 300 \sin(\omega t) \] Fundamental current component: \[ i_1(t) = 10 \sin\left(\omega t - \frac{\pi}{6}\right) \] Apparent (RMS) voltage: \[ V_{{rms}} = \frac{300}{\sqrt{2}} \] Fundamental current RMS: \[ I_{1,{rms}} = \frac{10}{\sqrt{2}} \] Real power: \[ P = V_{{rms}} \cdot I_{1,{rms}} \cdot \cos\left(\frac{\pi}{6}\right) = \frac{300}{\sqrt{2}} \cdot \frac{10}{\sqrt{2}} \cdot \cos\left(\frac{\pi}{6}\right) = 1500 \cdot \frac{\sqrt{3}}{2} = 1299 \, {W} \] Now calculate total RMS current including harmonics: \[ I_{{rms}} = \sqrt{ \left( \frac{10}{\sqrt{2}} \right)^2 + \left( \frac{2}{\sqrt{2}} \right)^2 + \left( \frac{1}{\sqrt{2}} \right)^2 } = \sqrt{50 + 2 + 0.5} = \sqrt{52.5} \approx 7.24 \, {A} \] Apparent power: \[ S = V_{{rms}} \cdot I_{{rms}} = \frac{300}{\sqrt{2}} \cdot 7.24 \approx 212.13 \cdot 7.24 \approx 1536 \, {VA} \] Power factor: \[ {PF} = \frac{P}{S} = \frac{1299}{1536} \approx 0.845 \]