RLE load is a current stiff load because of the presence of the inductor. The load current does not suddenly with a change in voltage. The inductor opposes the change of current.
2. In a Half-wave uncontrolled rectifier calculate the average value of the voltage if the supply is 23sin(50t).
A. 7.32 V
B. 8.32 V
C. 9.32 V
D. 7.60 V
Answer: A
In a Half-wave uncontrolled rectifier, the average value of the voltage is
Vm÷π=23÷π=7.32 V.
The diode will conduct only for the positive half cycle. The conduction period of the diode is π.
3. In a Half-wave controlled rectifier calculate the average value of the voltage if the supply is 10sin(50t) and the firing angle is 30°.
A. 2.32 V
B. 2.97 V
C. 4.26 V
D. 5.64 V
Answer: B
In Half-wave controlled rectifier, the average value of the voltage is
Vm(1+cos(∝))÷2π
=10(1+cos(30°)÷6.28=2.97 V.
The thyristor will conduct from ∝ to π.
4. Calculate the extinction angle as a purely inductive load if the firing angle is π÷4.
A. 315°
B. 145°
C. 345°
D. 285°
Answer: A
The extinction angle in the purely inductive load is
2π-∝=360°-45°=315°.
The extinction angle is the angle at which the current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.
5. Calculate the conduction angle as a purely inductive load if the firing angle is π÷2.
A. 205°
B. 175°
C. 180°
D. 195°
Answer: C
The conduction angle in the purely inductive load is
β-α=2(π-∝)=2(180°-90°).
The conduction angle is the angle at which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.
6. RLE load is also known as DC motor load.
A. True
B. False
Answer: A
RLE load is also known as DC motor load because the armature circuit consists of back e.m.f, inductive coils, and armature resistance.
7. In single-phase RLE load, calculate the voltage across the thyristor when current decays to zero using the data: (Vs)r.m.s=220 V, f=40 Hz, R=1 Ω, E=90 V, β=230°.
A. -328.33 V
B. -325.48 V
C. -254.85 V
D. -284.48 V
Answer: A
In single-phase, RLE load the voltage across the thyristor when the current decays to zero
VT=Vmsin(β)-E
=220×√2sin(230°)-90=-328.33 V.
8. Calculate the displacement factor if the fundamental voltage is 24sin(140πt-240°) and the fundamental current is 47sin(140πt-120°).
A. -0.5
B. -0.7
C. 0.9
D. 0.4
Answer: A
The displacement factor is the cosine of the angle difference between the fundamental voltage and fundamental current.
D.F=cos(120°)=-0.5.
9. Calculate the PIV for the Mid-point configuration of the Full-wave rectifier if the peak value of the supply voltage is 311.
A. 622 V
B. 620 V
C. 624 V
D. 626 V
Answer: A
The peak inverse voltage for the Mid-point configuration of the Full-wave rectifier is
2Vm=2×311=622 V.
The peak inverse is the maximum negative voltage across the thyristor.
10. Calculate the average value of the current through the thyristor in the case of a 1-Φ Full wave bridge rectifier if the value of the load current is 42 A.
A. 21 A
B. 12 A
C. 14 A
D. 16 A
Answer: A
The average value of the current through the thyristor in the case of a 1-∅ full-wave bridge rectifier is
