RL = 1kN. Assuming the cut in voltage of diode=0.7V, solve for the following: (i) total peak output voltage, Vo(m), (ii) average output voltage, Vo(de) and the average load current, lo(dc), (iii) RMS output voltage, Vo(rms) and the RMS load current, L.(rms),and (iv) efficiency of the rectifier. F D + 200V --- n = 2. . n, = 1 - 200V- R D2 Figure Q2

RL = 1kN. Assuming the cut in voltage of diode=0.7V, solve for the following: (i) total peak output voltage, Vo(m), (ii) average output voltage, Vo(de) and the average load current, lo(dc), (iii) RMS output voltage, Vo(rms) and the RMS load current, L.(rms),and (iv) efficiency of the rectifier. F D + 200V --- n = 2. . n, = 1 - 200V- R D2 Figure Q2

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter12: Electronic Control Devices

Section: Chapter Questions

Problem 4RQ: What is the difference between a diode and rectifier?

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