Answer to Question #184245 in Electrical Engineering for ddhg

Question

A single-phase asymmetrical semi-converter feeds an R-L load with R=50 Ω 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. 

Answer 1

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. 

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