# Electric Drives Questions and Answers – Solid State Controlled Drives – DC Motor Systems

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This set of Electric Drives Multiple Choice Questions & Answers (MCQs) focuses on “Solid State Controlled Drives – DC Motor Systems”.

a) True
b) False

Explanation: 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. Inductor opposes the change of current.

2. In 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

Explanation: In 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 Half-wave controlled rectifier calculate the average value of the voltage if the supply is 10sin(50t) and firing angle is 30°.
a) 2.32 V
b) 2.97 V
c) 4.26 V
d) 5.64 V

Explanation: 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 in purely inductive load if the firing angle is π÷4.
a) 315°
b) 145°
c) 345°
d) 285°

Explanation: The extinction angle in the purely inductive load is 2π-∝=360°-45°=315°. The extinction angle is the angle at which current in the circuit becomes zero. The average value of the voltage in a purely inductive load is zero.

5. Calculate the conduction angle in purely inductive load if the firing angle is π÷2.
a) 205°
b) 175°
c) 180°
d) 195°

Explanation: The conduction angle in the purely inductive load is β-α=2(π-∝)=2(180°-90°). The conduction angle is the angle for which the current exists in the circuit. The average value of the voltage in a purely inductive load is zero.

a) True
b) False

Explanation: 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

Explanation: In single phase RLE load the voltage across the thyristor when 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 fundamental current is 47sin(140πt-120°).
a) -0.5
b) -0.7
c) 0.9
d) 0.4

Explanation: 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 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

Explanation: The peak inverse voltage for the Mid-point configuration of 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 case of 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

Explanation: The average value of the current through the thyristor in case of 1-∅ full wave bridge rectifier is Io÷2=21 A. Each thyristor conducts for 180°.

11. Calculate the r.m.s value of the current through the thyristor in case of 1-Φ Full wave bridge rectifier if the value of the load current is 2 A.
a) 1.414 A
b) 1.214 A
c) 1.347 A
d) 1.657 A

Explanation: The r.m.s value of the current through the thyristor in case of 1-∅ full wave bridge rectifier is Io÷√2=√2 A=1.414 A. Each thyristor conducts for 180°.

12. Calculate the value of THD value for 1-Φ Full wave bridge rectifier.
a) 48.43 %
b) 47.25 %
c) 49.26 %
d) 50.48 %

Explanation: The value of the distortion factor is .9. The value of THD value for 1-Φ Full wave bridge rectifier is √(1÷.9)2-1=48.43 %. THD measures the amount of harmonic distortion.

13. Calculate the value of the Input power factor for 1-Φ Full wave bridge rectifier if the firing angle value is 45°.
a) .65
b) .64
c) .61
d) .63

Explanation: The value of the input power factor for 1-Φ Full wave bridge rectifier is .9cos(45°)=.63. The input power factor is a product of distortion factor and displacement factor.

14. Calculate the value of the fundamental displacement factor for 1-Φ Full wave bridge rectifier if the firing angle value is 60°.
a) .5
b) .4
c) .2
d) .8

Explanation: Fundamental displacement factor is the cosine of angle difference between the fundamental voltage and fundamental current. D.F=cos(∝)=cos(60°)=0.5.

15. Calculate the value of the fundamental displacement factor for 1-Φ Full wave semi-converter if the firing angle value is 20°.
a) .82
b) .98
c) .74
d) .26 