DC Series Motors Characteristics MCQ [Free PDF] – Objective Question Answer for DC Series Motors Characteristics Quiz

1. Turn-on and turn-off times of transistor depend on _________

A. Static Characteristic
B. Junction Capacitance
C. Current Gain
D. Voltage Gain

Answer: B

The depletion layer capacitance and diffusion capacitance affect the turn-on and turn-off behavior of transistors. Due to these internal capacitances, transistors do not turn on instantly. 

 

2. The generated e.m.f from 45-pole armature having 400 turns driven at 70 rev/sec having flux per pole as 90 mWb, with 17 parallel paths is ___________
A. 13341.17 V
B. 12370.14 V
C. 14700.89 V
D. 15690.54 V

Answer: A

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.

Eb = .09×45×400×2×4200÷60×17 = 13341.17 V. 

 

3. The unit of Magnetic flux density is Tesla.

A. True
B. False

Answer: A

Magnetic Flux density is defined as the number of magnetic lines passing through a certain point or a surface. It is generally expressed in terms of Tesla. Its C.G.S unit is Gauss. 

 

4. Calculate the moment of inertia of the hollow cylinder having a mass of 78 kg and a radius of 49 cm.

A. 9.363 kgm2
B. 9.265 kgm2
C. 9.787 kgm2
D. 9.568 kgm2

Answer: A

The moment of inertia of the hollow cylinder can be calculated using the formula

I=miri2÷2.

The mass of the hollow cylinder and radius is given.

I=(78)×.5×(.49)2=9.363 kgm2.

It depends upon the orientation of the rotational axis. 

 

5. Calculate the value of the angular acceleration of the motor using the given data: J = 81 kg-m2, load torque = 74 N-m, motor torque = 89 N-m.

A. .195 rad/s2
B. .182 rad/s2
C. .183 rad/s2
D. .185 rad/s2

Answer: D

Using the dynamic equation of motor

J×(angular acceleration) = Motor torque – Load torque:

81×(angular acceleration) = 89-74=15

angular acceleration=.185 rad/s2

 

6. 340 V, 45 A, 1400 rpm DC separately excited motor having a resistance of .7 ohm excited by an external dc voltage source of 90 V. Calculate the torque developed by the motor on full load.

A. 94.73 N-m
B. 94.52 N-m
C. 93.37 N-m
D. 94.42 N-m

Answer: A

Back emf developed in the motor during the full load can be calculated using the equation

Eb = Vt-I×Ra = 308.5 V

machine constant Km = Eb÷Wm which is equal to 2.1053.

Torque can be calculated by using the relation

T = Km × I = 2.1053×45 = 94.73 N-m. 

 

7. Calculate the value of the frequency if the time period of the signal is 99 sec.

A. 0.08 Hz
B. 0.02 Hz
C. 0.01 Hz
D. 0.04 Hz

Answer: C

The frequency is defined as the number of oscillations per second. It is reciprocal to the time period. It is expressed in Hz. F = 1÷T=1÷99=.01 Hz. 

 

8. The slope of the V-I curve is 31°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

A. 0.600 Ω
B. 0.607 Ω
C. 0.543 Ω
D. 0.648 Ω

Answer: A

The slope of the V-I curve is resistance. The slope given is 31° so R=tan(31°)=0.600 Ω. The resistance is the ratio of voltage and current. 

 

9. Calculate the radius of the circular ring having a moment of inertia of 59 kgm2 and a mass of 69 kg.

A. .924 m
B. .928 m
C. .934 m
D. .944 m

Answer: A

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

The moment of inertia of a circular ring and mass is given.

R=((59)÷(69)).5 = .924 m.

It depends upon the orientation of the rotational axis. 

 

10. Calculate the power developed by a motor using the given data: Eb= 48 V and I= 86 A (Assume rotational losses are neglected.)

A. 4128 W
B. 4150 W
C. 4140 W
D. 4170 W

Answer: A

Power developed by the motor can be calculated using the formula

P = Eb×I = 48×86 = 4128 W.

If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor. 

 

11. 780 V, 97 A, 1360 rpm separately excited dc motor with armature resistance (Ra) equal to 9 ohms. Calculate back emf developed in the motor when it operates on one-fourth of the full load. (Assume rotational losses are neglecteD.

A. 564.75 V
B. 561.75 V
C. 562.45 V
D. 565.12 V

Answer: B

Back emf developed in the motor can be calculated using the relation

Eb = Vt-I×Ra.

In question, it is asking for one-fourth load, but the data is given for full load so current becomes one-fourth of the full load current

= 97÷4 = 24.25 A.

250 V is terminal voltage it is fixed so

Eb = 780-24.25×9 = 561.75 V. 

 

12. 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. 

 

13. 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. 

 

14. 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. 

 

15. 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. 

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