DC Motor Application MCQ [Free PDF] – Objective Question Answer for DC Motor Application Quiz

1. Calculate the time period of the waveform y(t)=74cos(81πt+π).

A. .024 sec
B. .027 sec
C. .023 sec
D. .025 sec

Answer: A

The fundamental time period of the cosine wave is 2π. The time period of y(t) is 2π÷81π=.024 sec. The time period is independent of phase shifting and time-shifting. 

 

2. The generated e.m.f from 42-pole armature having 74 turns driven at 64 rev/sec having flux per pole as 21 mWb, with wave winding is ___________

A. 4177.171 V
B. 4177.152 V
C. 4100.189 V
D. 4190.454 V

Answer: B

The generated e.m.f can be calculated using the formula

Eb = Φ×Z×N×P÷60×A

Where

Φ represents flux per pole
Z represents the total number of conductors
P represents the number of poles
A represents the number of parallel paths
N represents speed in rpm.

One turn is equal to two conductors. In wave winding the number of parallel paths is equal to two.

Eb=.021×42×74×2×3840÷60×2=4177.152 V. 

 

3. Calculate the phase angle of the sinusoidal waveform x(t)=20sin(9πt+π÷7).

A. π÷9
B. π÷5
C. π÷7
D. π÷4

Answer: C

The sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, and α represents a phase difference. 

 

4. Calculate the moment of inertia of the solid sphere having a mass of 28 kg and a diameter of 15 cm.

A. 0.01575 kgm2
B. 0.01875 kgm2
C. 0.01787 kgm2
D. 0.01568 kgm2

Answer: A

The moment of inertia of the solid sphere can be calculated using the formula

I=2×miri2÷5.

The mass of the solid sphere and diameter is given.

I =(28)×.4×(.0375)2=.01575 kgm2.

It depends upon the orientation of the rotational axis. 

 

5. R.M.S value of the trapezoidal waveform V=Vmsin(Ωt+α).

A. Vm÷2½
B. Vm÷2¼
C. Vm÷2¾
D. Vm÷3½

Answer: D

R.M.S value of the sinusoidal waveform is Vm÷2½ and r.m.s value of the trapezoidal waveform is Vm÷3½. The peak value of the sinusoidal waveform is Vm. 

 

6. What is the unit of the admittance?

A. ohm
B. ohm-1
C. ohm2
D. ohm.5

Answer: B

The admittance measures how easily the current can flow in the circuit. It is the ratio of current and voltage. It is given in ohm-1. It is reciprocal of impedance. 

 

7. Calculate the value of the frequency if the inductive reactance is 45 Ω and the value of the inductor is 15 H.

A. 0.477 Hz
B. 0.544 Hz
C. 0.465 Hz
D. 0.412 Hz

Answer: A

The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation

XL = 2×3.14×f×L. F

= XL÷2×3.14×L

= 45÷2×3.14×15 = .477 Hz. 

 

8. The slope of the V-I curve is 19°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .3254 Ω
B. .3608 Ω
C. .3543 Ω
D. .3443 Ω

Answer: D

The slope of the V-I curve is resistance. The slope given is 19° so

R=tan(19°)=.3443 Ω.

The slope of the V-I curve is resistance. 

 

9. Calculate the active power in a 41 H inductor.

A. 2 W
B. 1 W
C. 0 W
D. .5 W

Answer: C

The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy.

The voltage and current are 90° in phase in the case of the inductor so the angle between V & I is 90°.

P = VIcos90 = 0 W.

Voltage leads the current in the case of the inductor. 

 

10. Calculate the active power in a 19 F capacitor.

A. 7.8 W
B. 0 W
C. 5.4 W
D. 1.5 W

Answer: B

The capacitor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90° in phase in the case of the capacitor so the angle between V & I is 90°.

P = VIcos90 = 0 W.

Current leads the voltage in the case of the capacitor. 

 

11. Calculate the active power in a 241 H inductor.

A. 21 W
B. 11 W
C. 0 W
D. .51 W

Answer: C

The inductor is a linear element. It only absorbs reactive power and stores it in the form of oscillating energy. The voltage and current are 90o in phase in the case of the inductor so the angle between V & I is 90°.

P = VIcos90 = 0 W. 

 

12. Calculate the active power in a 5 Ω resistor with 5 A current flowing through it.

A. 125 W
B. 110 W
C. 115 W
D. 126 W

Answer: A

The resistor is a linear element. It only absorbs real power and dissipates it in the form of heat. The voltage and current are in the same phase in the case of the resistor so the angle between V & I is 90°.

P=I2R=5×5×5=125 W. 

 

13. Swinburne’s test can be conducted on ___________

A. Series motor
B. Shunt motor
C. Compound motor
D. Shunt and compound motor

Answer: D

The test is practically applicable for machines that have flux constant like the shunt and compound machines as this is a no-load test and the DC series motor should not be run at no-load because of high speed. 

 

14. The generated e.m.f from 20-pole armature having 800 conductors driven at 30 rev/sec having flux per pole as 60 mWb, with 16 parallel paths is ___________

A. 1900 V
B. 1840 V
C. 1700 V
D. 1800 V

Answer: D

The generated e.m.f can be calculated using the formula

Eb = Φ×Z×N×P÷60×A

Where

Φ represents flux per pole
Z represents the total number of conductors
P represents the number of poles
A represents the number of parallel paths
N represents speed in rpm

Eb = .06×20×1800×800÷60×16 = 1800 V. 

 

15. The unit of active power is Watt.

A. True
B. False

Answer: A

The active power in the electrical circuits is useful to power. It determines the power factor of the system. It is expressed in terms of Watt.

P=VIcosΦ. 

 

16. Calculate the mass of the ball having a moment of inertia of 4.5 kgm2 and a radius of 14 cm.

A. 229.59 kg
B. 228.56 kg
C. 228.54 kg
D. 227.52 kg

Answer: A

The moment of inertia of the ball can be calculated using the formula I=∑miri2.

The moment of inertia of the ball and radius is given.

M=(4.5)÷(.14)2 = 229.59 kg.

It depends upon the orientation of the rotational axis. 

 

17. The field control method is suitable for constant torque drives.

A. True
B. False

Answer: B

The field control method is generally used for obtaining speeds greater than the base speed. It is also known as the flux weakening method. It is suitable for constant power drives. 

 

18. What is the unit of the intensity?

A. Watt/m2
B. Watt/m
C. Watt/m4
D. Watt/m3

Answer: A

Intensity is defined as the amount of power incident in a particular area. It is mathematically expressed as

I = Power incident (Watt)÷Area(m2). 

 

19. Calculate the value of the frequency of the signal that completes half of the cycle in 70 sec. Assume the signal is periodic.

A. 0.00714 Hz
B. 0.00456 Hz
C. 0.00845 Hz
D. 0.00145 Hz

Answer: A

The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. It is expressed in Hz. The given signal completes half of the cycle in 70 seconds then it will complete a full cycle in 140 seconds. F = 1÷T=1÷140=.00714 Hz. 

 

20. The slope of the V-I curve is 26°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. .487 Ω
B. .482 Ω
C. .483 Ω
D. .448 Ω

Answer: A

The slope of the V-I curve is resistance. The slope given is 26° so

R=tan(26°) = .487 Ω.

The slope of the V-I curve is resistance. 

 

21. For large DC machines, the yoke is usually made of which material?

A. Cast steel
B. Cast iron
C. Iron
D. Cast steel or cast iron

Answer: A

Yoke in DC machines is made up of cast steel. Yoke provides structural support and mechanical strength to the machine. It helps in carrying the flux from the North pole to the South pole. 

 

22. Calculate the terminal voltage of the Permanent Magnet DC motor having a resistance of 2 Ω and a full load current of 5 A with 20 V back e.m.f.

A. 30 V
B. 25 V
C. 20 V
D. 31 V

Answer: A

A permanent magnet DC motor is a special type of motor in which flux remains constant. The terminal voltage can be calculated using the relation

Vt = Eb+IaRa = 20+5×2 = 30 V. 

 

23. Armature reaction is demagnetizing in nature due to purely lagging load.

A. True
B. False

Answer: A

Due to purely lagging load, armature current is in the opposite phase with the field magneto-motive force. Armature magneto-motive force produced due to this current will be in the opposite phase with the field flux. It will try to reduce the net magnetic field. 

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