A 60 Hz sinusoidal voltage with an RMS value of 240 V is applied to a single-phase full-wave SCR rectifier. The RMS output voltage of the rectifier is 237 V, the firing angle (in degrees) of SCR is would be: O A. 45 O B. None of the other choices are correct O C. 60 O D. 36 O E. 30

Question (2) A controlled half-wave rectifier has a 240 V rms 60Hz ac source, The load a is R=30 Ohm resistor (a) Determine the delay angle such that the average load current is 2.5A (b) Determine the power absorbed by the load (c) Determine the power factor.

A certain unfiltered center-tapped full wave rectifier is powered by a 120 Vrms, 60 Hz power system. The peak value of the output voltage under loaded conditions is 30 V. The capacitance value is 2000 uF, load current of 2A, determine the following: Peak to peak ripple voltage RMŞ ripple voltage 4. а.

The input to a full-wave rectifier is 20 V AC (RMS). The load resistor is 20 ohms and rectifier losses are 4W. The ripple factor of the rectifier is: O A. 0.96 OB. 0.33 O C. 0.48 O D. None of the other choices are correct O E. 0.75

If a 75V peak sine wave is applied to a half-wave rectifier, the peak inverse voltage across the diode is?

Question (2) A controlled half-wave rectifier has a 240 V rms 60Hz ac source, The load a is R=30 Ohm resistor (a) Determine the delay angle such that the average load current is 2.5A (b) Determine the power absorbed by the load (c) Determine the power factor.

Calculate the average output voltage for a half-wave diode rectifier if the following measurements were made when a sinewave was applied to the input. • Peak output voltage = 18V • Diode conduction time = 7.2ms • Period = 16.66ms O a. 29.45 V O b. 7.78 V O c. 11.00 V O d. 12.73 V O e. 5.50 V

The input to a full-wave rectifier is 10 V AC (RMS). The load resistor is 10 ohms and rectifier losses are 4W. The power efficiency of the rectifier is: O A. 58 % O B. None of the other choices are correct O C. 100 % O D. 80 % O E. 67 %