Question #184245

A single-phase asymmetrical semi-converter feeds an R-L load with R=50 Î© and a large L= 10 mH so that load current is constant. The source voltage is 230 V, 50 Hz. For a firing angle of 600, determine: (i) Average value of output voltage and output current. (ii) Average and rms values of diode, thyristor and source currents.(iii) Input power factor.Â

Expert's answer

**i)Average value of output voltage and output current.**

V_O=\frac{v_m}{\pi}=\frac{230 \sqrt{2}}{\pi}(1+cos 60)=155.31V

I_o=\frac{V_o}{R}=\frac{155.31}{50}=3.11A

**(ii) Average and rms values of diode, thyristor and source currents**.

I_pavg=3.11\times\frac{\pi-\frac{\pi}{3}}{2\pi}=1.04A

I_DAvg=3.11\times\frac{\pi+\frac{\pi}{3}}{2\pi}=2.07A

I_Trms=3.11\times\sqrt{\frac{\pi-\pi /3}{2\pi}}=1.79A

I_Drms=3.11\times\sqrt{\frac{\pi+\pi /3}{2\pi}}=2.54A

Source current I_DAvg=3.11\times\frac{\pi+\frac{\pi}{3}}{2\pi}=2.07A

I_Drms=3.11\times\sqrt{\frac{\pi+\pi /3}{2\pi}}=2.54A

(iii) Input power factor.Â

\frac{I_{SI}}{I_{Sr}}cos\frac{\alpha}{2}=\frac{2.42}{2.54}cos\frac{60}{2}=0.825

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