Question

A 1-$\varphi$ half wave rectifier circuit produces an average voltage of 40 V across 100 $\Omega$ load resistor from 120 V rms, 50 Hz ac source. The supply power factor is

Answer 1

If an uncontrolled rectifier is used, the average output voltage would be $\frac{V_m}{\pi }=\frac{120\sqrt{2}}{\pi }=54V$.
Some means of reducing the average resistor voltages is that, there controlled rectifier is used.

So, $V_{0\left(avg\right)}=\frac{V_m}{2\pi }(1+cos\alpha )$

$40=\frac{120\sqrt{2}}{2\pi }(1+cos\alpha )$

$\alpha ={61.25}^{\circ }$

$V_{0\left(rms\right)}=\frac{1}{\sqrt{2\pi }}{\left[{\int }_{\alpha }^{180}{V}_m^{2}si{n}^2\right(\omega t\left)d\omega t\right]}^{1/2}=\frac{V_m}{\sqrt{2\pi }}{\left[\frac{1}{2}\right(180-\alpha )+\frac{1}{4}sin2\alpha ]}^{1/2}$

$V_{or}=\frac{120\sqrt{2}}{\sqrt{2\pi }}{\left[\frac{1}{2}\right(2.075)+\frac{1}{4}\sin {122.50}^{\circ }]}^{1/2}=75.60Volts$

$P_o=\frac{V_{or}^2}{R}=\frac{(75.62{)}^2}{100}=57.16Watt$

$\because$ $I_{sr}=\frac{V_{or}}{R}=\frac{75.60}{100}=0.7560$

Supply power factor $=\frac{P_o}{V_{sr}\times I_{sr}}=\frac{57.16}{120\times 0.7560}=0.6301lag$