Question

A 3-phase; M-3 converter is operated from 3-phase, 230V, 50 Hz supply with load resistance R = 10 $\Omega$ . If average output voltage is 50% of the maximum possible output voltage, then the rectification efficiency in percentage is ___

• A. 54.93

Answer 1

The correct option is A 54.93
Let, $\alpha >{30}^o$

$V_0=\frac{3V_{mp}}{2\pi }[1+cos(\alpha +{30}^o\left)\right]$

Maximum possible output voltage

$V_m=\frac{3V_{ml}}{2\pi }$

$\frac{3V_{ml}}{2\pi \sqrt{3}}(1+cos(\alpha +{30}^o\left)\right)=\left(\frac{3V_{ml}}{2R}\right)\times \frac{1}{2}$

$\alpha ={67.7}^o$

$V_{or}=\frac{V_{ml}}{2\sqrt{\pi }}{[(\frac{5\pi }{6}-\alpha )+\frac{1}{2}sin(2\alpha +\frac{\pi }{3})]}^{1/2}$

$=\frac{\sqrt{2}\times 230}{2\sqrt{\pi }}{[(\frac{5\pi }{6}-\frac{67.7\times }{180})+\frac{1}{2}sin(2\times 67.7+{60}^o\left)\right]}^{1/2}$

= 104.76 V

$I_{or}=\frac{104.765}{10}=10.47A$

Average output current,

$I_o=\frac{V_0}{R}$

$=\frac{3V_{ml}}{2\pi \times 2}\times \frac{1}{10}=\frac{3\times 230\sqrt{2}}{4\pi \times 10}=7.76A$

Rectification efficiency = $\frac{V_0{I}_o}{V_{or}{I}_{or}}$

$=\frac{77.65\times 7.76}{104.76\times 10.47}=54.93\%$