Question

lyue 2 The controlled half-wave rectifier of Figure 3 has an ac input of 120 V rms at 60 Hz, R=20 0, L= 200 mH, and Vdc =80 V. The delay angle a is 30 degrees. Determine (a) an expression for the current, (b) the power absorbed by the resistor and (c) the power absorbed by the de source in the load, R W-0001 th 1.sin(01) Vs Figure 3

please help i do not understand

Answer 1

4 mm Umsim (ut) Vdc analyse the rkt. N There Firstly meed to Thyristor present Vs (ros) 120 V R=20 L = 200 mitt Vdc = sou The thyristor is 1 delay L12 = 30° [ unidirectiond Slari tch Time after which thyristor triggered]. & analyse ve forms Let us draw wave NS Udo 10-30° (2a+33) vo To a 2172

→ Whin thyristor triggered d = 30°, thyristor starts condo conduiting & it behaves like shart- ckt. at 19.2 Vs L also Umsinut) Vda Vdc from the waveform, (66) aug - [14] 1 integration (Vo) ang = the at = 12052 [14 10:30 2n So.425 v Current (i) - RiL Duis + Umsinut (Volary Vdc (2) arg = {To Jang R i. (Iolang = (vo) nung - Van + Vdc [As it current skift type R 아 Loa S0.425 80 -1.478 A 20 & Current back back flows)

6) The by Power the absorbed Pg. 3 venster (Po up) (V. Daug. To lang R (1.478)x 20 93.68 W by the The Power absorbed Source (Load) Basically it delivere, dow mot absorb but it acts a DC Generator Vaco (To) aug (Box 1.478) w 118.24 W X
The Q has been solved keeping in mind the reader knows the working of the thyristor,so it will be easier to understand the waveforms and get the values...Thank you!!!