21. Calculate the fundamental component of source current in 1-Φ Full wave bridge rectifier for the load(Highly inductive) current=78 A.
A. 78 A
B. 45 A
C. 69 A
D. 13 A
Answer: A
The fundamental component of source current in 1-Φ Full wave bridge rectifier is Io. It is the r.m.s value of the source current.
Is(r.m.s)=Io=78 A.
22. Calculate the r.m.s value of source current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=51.2 A and α=15°.
A. 10.53 A
B. 14.52 A
C. 44.92 A
D. 49.02 A
Answer: D
The r.m.s value of source current in 1-Φ Full-wave semi-converter is Io√π-α÷π. It is the r.m.s value of the source current.
I(r.m.s) = Io√π-α÷π
= 51.2(√.916) = 49.02 A.
23. Calculate the r.m.s value of thyristor current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=2.2 A and α=155°. (Asymmetrical configuration)
A. .58 A
B. .57 A
C..51 A
D..52 A
Answer: B
The r.m.s value of source current in 1-Φ Full-wave semi-converter is Io√π-α÷2π. It is the r.m.s value of the thyristor current.
I(r.m.s) = Io√π-α÷2π
=2.2(√.069)=.57 A.
24. Calculate the r.m.s value of diode current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=5.1 A and α=115°. (Asymmetrical configuration)
A. 4.21 A
B. 4.61 A
C. 4.71 A
D. 4.52 A
Answer: B
The r.m.s value of diode current in 1-Φ Full-wave semi-converter is Io√π+α÷2π. It is the r.m.s value of the diode current.
I(r.m.s) = Io√π+α÷π)
=5.1(√.819)=4.61 A.
25. Calculate the average value of thyristor current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=25.65 A and α=18°. (Asymmetrical configuration)
A. 11.54 A
B. 12.15 A
C. 15.48 A
D. 14.52 A
Answer: A
The average value of thyristor current in a 1-Φ Full-wave semi-converter is Io(π-α÷2π). It is the average value of the thyristor current.
Iavg = Io(π-α÷2π)
=25.65(.45)=11.54 A.
26. Calculate the average value of diode current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=75.2 A and α=41°. (Asymmetrical configuration)
A. 46.16 A
B. 42.15 A
C. 41.78 A
D. 41.18 A
Answer: A
The average value of diode current in a 1-Φ Full-wave semi-converter is Io(π+α÷2π). It is the average value of the diode current.
Iavg = Io(π+α÷2π)
=75.2(.61)=46.16 A.
27. Calculate the average value of diode current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=5.2 A and α=11°. (F.D configuration)
A. .32 A
B. .31 A
C. .25 A
D. .27 A
Answer: B
The average value of diode current in a 1-Φ Full-wave semi-converter is Io(α÷π). It is the average value of the diode current.
Iavg= Io(α÷π)
=5.2(.061)=.31 A.
28. Calculate the r.m.s value of diode current in 1-Φ Full-wave semi-converter for the load (Highly inductive) current=.2 A and α=74°. (F.D configuration)
A. .154 A
B. .248 A
C. .128 A
D. .587 A
Answer: C
The r.m.s value of diode current in 1-Φ Full-wave semi-converter is Io√(α÷π). It is the r.m.s value of the diode current.
Ir.m.s = Io√(α÷π)
=.2√(.41)=.128 A.
29. Diodes in 1-Φ Full-wave semi-converter protect the thyristor from short-circuiting.
A. True
B. False
Answer: A
Diodes in 1-Φ Full-wave semi-converter protect the thyristor from short-circuiting. They provide the gap from (α, π+α) to avoid the conduction of one leg thyristors.
30. The problem of short-circuiting in 1-Φ Full-wave semi-converter is very common.
A. True
B. False
Answer: A
The problem of short-circuiting in 1-Φ Full-wave semi-converter is very common. Diodes protect the thyristor from short-circuiting. They provide the gap from (α, π+α) to avoid the conduction of one leg thyristors.
31. Calculate the value of the conduction angle for R-L load if the value of β and α are 19° and 29°.
A. 10°
B. 70°
C. 30°
D. 80°
Answer: A
The conduction angle for the R-L load is
β-α=29°-19°=10°.
R-L load is a current stiff type of load. The current in the circuit only flows from α to β.
