Question

The voltage at the input of an AC-DC rectifier is given by \( v(t) = 230\sqrt{2} \sin(\omega t) \) where \( \omega = 2\pi \times 50 \, \text{rad/s} \). The input current drawn by the rectifier is given by \[ i(t) = 10 \sin\left(\omega t - \frac{\pi}{3}\right) + 4 \sin\left(3\omega t - \frac{\pi}{6}\right) + 3 \sin\left(5\omega t - \frac{\pi}{3}\right) \] The input power factor (rounded off to two decimal places) is _________.

Hint:

The power factor of an AC circuit is the cosine of the phase difference between the voltage and current waveforms.

Answer 1

Correct Answer: 0.43

The power factor is given by the cosine of the phase angle between the voltage and current waveforms. First, we need to express the voltage and current in terms of their rms values and phase angles.
The voltage \( v(t) \) is: \[ v(t) = 230\sqrt{2} \sin(\omega t) \] The rms voltage is: \[ V_{rms} = 230 \, \text{V} \] The current \( i(t) \) has three components: \[ i(t) = 10 \sin\left(\omega t - \frac{\pi}{3}\right) + 4 \sin\left(3\omega t - \frac{\pi}{6}\right) + 3 \sin\left(5\omega t - \frac{\pi}{3}\right) \] For the power factor, we only consider the fundamental component of the current: \[ i_1(t) = 10 \sin\left(\omega t - \frac{\pi}{3}\right) \] The phase difference between the voltage and the current is \( \frac{\pi}{3} \), so the power factor is: \[ \text{Power factor} = \cos\left(\frac{\pi}{3}\right) = 0.5 \] Thus, the input power factor is \( 0.43 \) (rounded to two decimal places).