# Solid Sate DC Drives MCQ [Free PDF] – Objective Question Answer for Solid Sate DC Drives Quiz

A. True
B. False

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

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

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°

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°

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.

A. True
B. False

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

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

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

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