Question

A single phase semiconverter is supplying a resistive load of 100 $\Omega$ from its dc side. The converter is operated from 230 V, 50 Hz ac supply. If the average output is 25% of the maximum possible average output voltage, then value of ratio of rms load current of converter to average load current will be

• A. 1.96
• B. 1.70
• C. 1.11
• D. 1.52

Answer 1

The correct option is A 1.96

$Given,V_s=230V,$
$f=50Hz,$
$R_L=100\Omega$

Average output voltage,
$V_0=25\%ofmaximumpossibleaverageoutputvoltage$
$\frac{V_m}{\pi }(1+cos\alpha )=0.25\times \frac{2V_m}{\pi }$

As maximum possible voltage for $\alpha =0$ for semi converter
$1+cos\alpha =0.5$
$cos\alpha =-0.5$
$or\alpha =\frac{2\pi }{3}$

$Averageloadcurrent,I_0=\frac{V_m}{\pi R}(1+cos\alpha )$
$=\frac{\sqrt{2}\times 230}{\pi \times 100}(1+cos{120}^{\circ })=0.5176A$

$Rmsloadcurrent,I_{rms}=\frac{V_S}{R}{\left[\frac{1}{\pi }\left(\right(\pi -\alpha )+\frac{sin2\alpha }{2})\right]}^{1/2}$

$I_{rms}=\frac{230}{100}{\left[\frac{1}{\pi }((\pi -\frac{2\pi }{3})+\frac{sin\frac{4\pi }{3}}{2})\right]}^{1/2}=2.3{\left[\frac{1}{\pi }(\frac{\pi }{3}+\left(\frac{0.8660}{2}\right))\right]}^{1/2}$

$=2.3\times 0.442=1.0169$

$\frac{Rmsloadcurrent}{Averageloadcurrent}=\frac{1.0169}{0.5176}=1.964$