11. 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.
12. 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.
13. 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.
14. 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.
15. 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.
16. 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.
16. Calculate the V0 avg for Single-phase Half Wave rectifier for R-L load using the data: Vm=24 V, ∝=30°, β=60°.
A. 1.39 V
B. 8.45 V
C. 4.55 V
D. 1.48 V
Answer: A
In Half-wave controlled rectifier, the average value of the voltage is
Vm(cos(∝)-cos(β))÷2π
=24(cos(30°)- cos(60°))÷6.28=1.39 V.
The thyristor will conduct from ∝ to β.
17. Calculate the I0 avg for the Single-phase Half Wave rectifier for R-L load using the data: Vm=56 V, ∝=15°, β=30°, R=2 Ω.
A. .56 A
B. .44 A
C. .26 A
D. .89 A
Answer: A
In Half-wave controlled rectifier, the average value of the current is
Vm(cos(∝)-cos(β))÷2πR
=56(cos(15°)- cos(30°))÷12.56=.44 A.
The thyristor will conduct from ∝ to β.
18. Calculate the I0 r.m.s for Single-phase Half Wave rectifier for R-L load (Highly inductive) using the data: Vm=110 V, ∝=16°, β=31°, R=1 Ω.
A. 4.56 A
B. 1.82 A
C. 4.81 A
D. 9.15 A
Answer: B
In Half-wave controlled rectifier, the r.m.s value of the current is
Vm(cos(∝) – cos(β))÷2πR
=110(cos(16°) – cos(31°))÷6.28=1.82 A.
The thyristor will conduct from ∝ to β.
19. Calculate the value of the fundamental displacement factor for 1-φ Full-wave semi-converter if the firing angle value is 0o.
A. 1
B. .8
C. .4
D. .2
Answer: A
The fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. The fundamental displacement factor for 1-φ Full-wave semi-converter is cos(∝÷2)=cos(0o)=1.
20. For highly inductive load current remains continuous.
A. True
B. False
Answer: A
For highly inductive load current remains continuous. The inductor opposes the change in the current. When the current is about to extinct the next conducting cycle starts which currently remains constant.
