Hello Reader, before we see the MCQs on dynamics of electrical drives let us tell you this good news…
We are currently launching complete chapter-wise MCQs on Electrical Drives. And believe us…
…this will completely change your learning on drives subjects!!
The chapters are picked from the book called “A First Course on Electrical Drives By S.K. Pillai and other resources”. All the questions in the books are compiled in content and written separately chapter-wise.
In this article, the question and answers are mostly written from the chapter called “Characteristics of DC & AC Motors”. MCQs of other chapters are also published on our website, to read and learn them, you can simply click on the listed Chapters below:
1. Dynamics of Electrical Drives
2. Characteristics of DC & AC Motors
3. Starting
4. Speed Control of Direct Current Motors & Induction Motors
5. Braking of Electric Motors
6. Braking of DC Motors & Induction Motors
7. Rating and Heating of Motors
8. Introduction to Solid State Controlled Drives
9. Introduction to Solid-State Devices
10. Introduction to Solid-State Switching Circuits
11. Solar and Battery Powered Drives
12. Traction Drives
13. Electrical Drive Systems and Components
With that said… Let’s dive into our main topic MCQs on Characteristics of DC & AC Motors [Electrical Drives Chapterwise Series]
Suppose you want to jump to the specific question directly, expand the table of content below:
1. Swinburne’s test can be conducted on ___________
a) Series motor
b) Shunt motor
c) Compound motor
d) Shunt and compound motor
Answer: d
Explanation: 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 DC series motor should not be run at no-load because of high speed.
2. 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 ___________
b) 1840 V
c) 1700 V
d) 1800 V
Answer: d
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, 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.
3. The unit of active power is Watt.
b) False
Answer: a
Explanation: 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.
4. 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
Explanation: 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.
5. The field control method is suitable for constant torque drives.
a) True
b) False
Answer: b
Explanation: 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.
6. What is the unit of the intensity?
a) Watt/m2
b) Watt/m
c) Watt/m4
d) Watt/m3
Answer: a
Explanation: Intensity is defined as the amount of power incident in a particular area. It is mathematically expressed as I = Power incident (Watt)÷Area(m2).
7. Calculate the value of the frequency if the signal completes half of the cycle in 70 sec. Assume signal is periodic.
b) 0.00456 Hz
c) 0.00845 Hz
d) 0.00145 Hz
Answer: a
Explanation: 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.
8. 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
Answer: a
Explanation: 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.
9. For large DC machines, the yoke is usually made of which material?
b) Cast iron
c) Iron
d) Cast steel or cast iron
Answer: a
Explanation: 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.
10. 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
Explanation: 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.
11. Armature reaction is demagnetizing in nature due to purely lagging load.
b) False
Answer: a
Explanation: 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.
12. The unit of Magento-motive force is Ampere-turns.
b) False
Answer: a
Explanation: The magneto-motive force is defined as the product of current and turns. It is mathematically expressed as F=NI.
13. Calculate the velocity of the wheel if the angular speed is 25 rad/s and radius is 10 m.
a) 250 m/s
b) 260 m/s
c) 270 m/s
d) 240 m/s
Answer: a
Explanation: The velocity of the wheel can be calculated using the relation V=ω×r. The velocity is the vector product of angular speed and radius. V=Ω×r = 25×10 = 250 m/s.
14. Displacement is a ____________ quantity.
b) Vector
c) Scalar and Vector
d) Tensor
Answer: b
Explanation: Displacement is a vector quantity. It depends on the initial and final position of the body. It has both direction and magnitude. Distance is a scalar quantity.
15. When a UJT is used for triggering an SCR, then the wave shape of voltage obtained from the UJT circuit will be ____________
a) Square
b) Pulse
c) Trapezoidal
d) Saw-tooth
Answer: b
Explanation: UJT relaxation oscillator using RC circuit is used for SCR triggering and the wave shape obtained is an exponentially decaying pulse. So, we can say pulse waveform is obtained.
16. Calculate the value of the frequency if the time period of the signal is 0 sec.
b) 0.4 Hz
c) 0.2 Hz
d) 0.78 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷0 = infinity. It signifies signal is aperiodic.
17. The slope of the I-V curve is 87°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
b) 0.0254 Ω
c) 0.0543 Ω
d) 0.0648 Ω
Answer: a
Explanation: The slope of the I-V curve is reciprocal of resistance. The slope given is 87° so R=1÷tan(87°)=.0524 Ω.
18. The most suitable device for high-frequency inversion in switching mode power supply is-
b) BJT
c) MOSFET
d) IGBT
Answer: c
Explanation: MOSFET is the fastest among all the power semiconductor devices and has the highest frequency range. MOSFET stands for metal oxide silicon field-effect transistor.
19. In DC chopper, the waveform for input and output voltages is respectively__________
a) Discontinuous and Continuous
b) Continuous and Discontinuous
c) Both continuous
d) Both discontinuous
Answer: b
Explanation: Chopper has a perfect DC at the input which is chopped into pulses which means the output voltage is discontinuous and by varying the duty cycle we can vary the average output voltage.
20. A chopper behaves as a __________
a) DC equivalent of AC switching device
b) DC equivalent of AC transformer
c) DC equivalent of AC relay
d) AC equivalent of circuit breaker
Answer: b
Explanation: A chopper is used to step up or step down the DC voltage whereas the transformer is used to step up or step down AC voltage so chopper is DC equivalent of AC transformer.
21. A DC chopper feeds an RLE load. If the value of E is increased by 20%, the current ripple ________
b) decreases by 20%
c) increases only 20%
d) remains the same
Answer: d
Explanation: For a buck converter current ripple doesn’t depend on the value of E. ΔIL = Vdc×D×(1-D)×T÷L. The expression is independent of the value of E.
22. The conduction loss versus device current characteristic of a power MOSFET is best approximated by a ________
a) Straight line
b) Rectangular hyperbola
c) Parabola
d) Exponential decaying functions
Answer: c
Explanation: A MOSFET in the ON state behaves as resistance so the conduction power loss is given by, P=I2R. Hence, the power vs current curve will be a parabola.
23. Which of the following device is NOT suitable for parallel operation?
a) MOSFET
b) BJT
c) IGBT
d) TRIAC
Answer: b
Explanation: BJT has a negative temperature coefficient of resistance. If it is operated in parallel operation thermal run-away will take place and the device will damage.
24. SCR is uni-directional in nature.
a) True
b) False
Answer: a
Explanation: SCR is uni-directional in nature. It only allows current to flow from anode to cathode. If the current is greater than the latching current then only it will work in forwarding conduction mode.
25. Which of the following device should be used as a switch in a low power switched-mode power supply (SMPS)?
a) GTO
b) BJT
c) MOSFET
d) TRIAC
Answer: c
Explanation: MOSFET has a low power rating and high-frequency rating and so it can be used in a low power switch mode power supply. MOSFET stands for metal oxide silicon field-effect transistor.
26. The slope of the V-I curve is 78°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) 4.732 Ω
b) 4.608 Ω
c) 4.543 Ω
d) 4.648 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 78° so R=tan(78°)=4.732 Ω. The slope of the I-V curve is reciprocal of resistance.
27. Calculate mark to space ratio if the system is on for 17 sec and the time period is 30 sec.
a) 1.307
b) 1.457
c) 1.478
d) 1.146
Answer: a
Explanation: Mark to space is Ton÷Toff. It is the ratio of time for which the system is active and the time for which is inactive. M = Ton÷Toff=17÷(30-17) = 1.307.
28. Turn-on and turn-off times of transistor depend on _________
a) Static Characteristic
b) Junction Capacitance
c) Current Gain
d) Voltage Gain
Answer: b
Explanation: The depletion layer capacitance and diffusion capacitance affects the turn-on and turn-off behavior of transistors. Due to these internal capacitances, transistors do not turn on instantly.
29. 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
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, 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.
30. The unit of Magnetic flux density is Tesla.
a) True
b) False
Answer: a
Explanation: 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.
31. 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
Explanation: 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.
32. 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
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: 81×(angular acceleration) = 89-74=15, angular acceleration=.185 rad/s2.
33. 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
Explanation: Back emf developed in the motor during the full load can be calculated using equation Eb = Vt-I×Ra = 308.5 V and 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.
34. 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
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. F = 1÷T=1÷99=.01 Hz.
35. 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
Explanation: 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.
36. Calculate the radius of the circular ring having a moment of inertia 59 kgm2 and mass of 69 kg.
a) .924 m
b) .928 m
c) .934 m
d) .944 m
Answer: a
Explanation: 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.
37. 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
Explanation: 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.
38. 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)
b) 561.75 V
c) 562.45 V
d) 565.12 V
Answer: b
Explanation: 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.
39. 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
Explanation: 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.
40. 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
Explanation: The generated e.m.f can be calculated using the formula Eb = Φ×Z×N×P÷60×A, Φ represent 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.
41. 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
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
42. 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
Explanation: 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.
43. 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
Explanation: 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.
44. What is the unit of the admittance?
a) ohm
b) ohm-1
c) ohm2
d) ohm.5
Answer: b
Explanation: The admittance measures how easily 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.
45. 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
Explanation: 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.
46. 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
Explanation: 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.
47. Calculate the active power in a 41 H inductor.
a) 2 W
b) 1 W
c) 0 W
d) .5 W
Answer: c
Explanation: 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 case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W. Voltage leads the current in case of the inductor.
48. 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
Explanation: 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 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.
49. Calculate the active power in a 241 H inductor.
b) 11 W
c) 0 W
d) .51 W
Answer: c
Explanation: 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 case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.
50. 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
Explanation: 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 case of the resistor so the angle between V & I is 90°. P=I2R=5×5×5=125 W.
51. Calculate the frequency of the waveform x(t)=45sin(40πt+5π).
a) 24 Hz
b) 27 Hz
c) 23 Hz
d) 20 Hz
Answer: d
Explanation: The fundamental time period of the sine wave is 2π. The frequency of x(t) is 40π÷2π=20 Hz. The frequency is independent of phase shifting and time-shifting.
52. The generated e.m.f from 16-pole armature having 57 turns driven at 78 rev/sec having flux per pole as 5 mWb, with lap winding is ___________
a) 44.16 V
b) 44.15 V
c) 44.46 V
d) 44.49 V
Answer: c
Explanation: The generated e.m.f can be calculated using the formulaEb = Φ×Z×N×P÷60×A, 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 lap winding, the number of parallel paths is equal to the number of poles. Eb = .005×16×57×2×4680÷60×16=44.46 V.
53. Calculate the phase angle of the sinusoidal waveform x(t)=42sin(4700πt+2π÷3).
a) 2π÷9
b) 2π÷5
c) 2π÷7
d) 2π÷3
Answer: d
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
54. Calculate the moment of inertia of the rod about its end having a mass of 39 kg and length of 88 cm.
a) 9.91 kgm2
b) 9.92 kgm2
c) 9.96 kgm2
d) 9.97 kgm2
Answer: c
Explanation: The moment of inertia of the rod about its end can be calculated using the formula I=ML2÷3. The mass of the rod about its end and length is given.I=(39)×.33×(.88)2=9.96 kgm2. It depends upon the orientation of the rotational axis.
55. Calculate the moment of inertia of the rod about its center having a mass of 11 kg and length of 29 cm.
b) .072 kgm2
c) .076 kgm2
d) .077 kgm2
Answer: d
Explanation: The moment of inertia of the rod about its center can be calculated using the formula I=ML2÷12. The mass of the rod about its center and length is given. I=(11)×.0833×(.29)2=.077 kgm2. It depends upon the orientation of the rotational axis.
56. Calculate the shaft power developed by a motor using the given data: Eb = 404V and I = 25 A. Assume frictional losses are 444 W and windage losses are 777 W.
a) 8879 W
b) 2177 W
c) 8911 W
d) 8897 W
Answer: a
Explanation: Shaft power developed by the motor can be calculated using the formula P = Eb*I-(rotational losses) = 404*25- (444+777) = 8879 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.
57. Calculate the value of the frequency if the capacitive reactance is 13 Ω and the value of the capacitor is 71 F.
a) .0001725 Hz
b) .0001825 Hz
c) .0001975 Hz
d) .0001679 Hz
Answer: a
Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation Xc=1÷2×3.14×f×C. F=1÷Xc×2×3.14×C = 1÷13×2×3.14×71 = .0001725 Hz.
58. The slope of the V-I curve is 27°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
a) .384 Ω
b) .509 Ω
c) .354 Ω
d) .343 Ω
Answer: b
Explanation: The slope of the V-I curve is resistance. The slope given is 27° so R=tan(27°)=.509 ω. The slope of the I-V curve is reciprocal of resistance.
59. Calculate the active power in a 7481 H inductor.
b) 4651 W
c) 0 W
d) 4654 W
Answer: c
Explanation: 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 case of the inductor so the angle between V & I is 90°. P=VIcos90 = 0 W. Voltage leads the current in case of the inductor.
60. Calculate the active power in a 457 F capacitor.
b) 565 W
c) 545 W
d) 0 W
Answer: d
Explanation: 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 case of the capacitor so the angle between V & I is 90°. P = VIcos90 = 0 W. Current leads the voltage in case of the capacitor.
61. Calculate the active power in a 181 H inductor.
a) 2448 W
b) 1789 W
c) 4879 W
d) 0 W
Answer: d
Explanation: 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 case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.
62. Calculate the active power in a 17 ω resistor with 18 A current flowing through it.
b) 5104 W
c) 5554 W
d) 5558 W
Answer: a
Explanation: 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 0°. P=I2R=18×18×17=5508 W.
63. A three-phase slip ring induction motor is fed from the rotor side with the stator winding short-circuited. The frequency of the current flowing in the short-circuited stator is____________
a) Slip frequency
b) Supply frequency
c) The frequency corresponding to rotor speed
d) Zero
Answer: a
Explanation: The relative speed between rotor magnetic field and stator conductors is sip speed and hence the frequency of induced e.m.f is equal to slip frequency.
64. An 8-pole, 3-phase, 50 Hz induction motor is operating at a speed of 720 rpm. The frequency of the rotor current of the motor in Hz is __________
a) 2
b) 4
c) 3
d) 1
Answer: a
Explanation: Given a number of poles = 8. Supply frequency is 50 Hz. Rotor speed is 720 rpm. Ns = 120×f÷P=120×50÷8 = 750 rpm. S=Ns-Nr÷Ns = 750 – 720÷750 = .04. F2=sf=.04×50=2 Hz.
65. Calculate the phase angle of the sinusoidal waveform z(t)=78sin(456πt+2π÷78).
a) π÷39
b) 2π÷5
c) π÷74
d) 2π÷4
Answer: a
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
66. Calculate the moment of inertia of the disc having a mass of 54 kg and diameter of 91 cm.
a) 5.512 kgm2
b) 5.589 kgm2
c) 5.487 kgm2
d) 5.018 kgm2
Answer: b
Explanation: The moment of inertia of the disc can be calculated using the formula I=mr2×.5. The mass of the disc and diameter is given. I=(54)×.5×(.455)2=5.589 kgm2. It depends upon the orientation of the rotational axis.
67. Calculate the moment of inertia of the thin spherical shell having a mass of 73 kg and diameter of 36 cm.
b) 1.47 kgm2
c) 1.38 kgm2
d) 1.48 kgm2
Answer: a
Explanation: The moment of inertia of the thin spherical shell can be calculated using the formula I=mr2×.66. The mass of the thin spherical shell and diameter is given. I=(73)×.66×(.18)2=1.56 kgm2. It depends upon the orientation of the rotational axis.
68. A 50 Hz, 4poles, a single-phase induction motor is rotating in the clockwise direction at a speed of 1425 rpm. The slip of the motor in the direction of rotation & opposite direction of the motor will be respected.
a) 0.05, 0.95
b) 0.04, 1.96
c) 0.05, 1.95
d) 0.05, 0.02
Answer: c
Explanation: Synchronous speed, Ns=120×50÷4=1500 rpm. Given a number of poles = 4. The supply frequency is 50 Hz. Rotor speed is 1425 rpm. S=Ns-Nr÷Ns=1500-1425÷1500=.05. Sb=2-s=1.95.
69. The frame of an induction motor is made of _________
b) Silicon steel
c) Cast iron
d) Stainless steel
Answer: c
Explanation: The frame of an induction motor is made of cast iron. The power factor of an induction motor depends upon the air gap between stator and rotor.
70. The slope of the V-I curve is 5°. 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
Explanation: The slope of the V-I curve is resistance. The slope given is 5° so R=tan(5°)=.3443 ω. The slope of the I-V curve is reciprocal of resistance.
71. In an induction motor, when the number of stator slots is equal to an integral number of rotor slots _________
a) There may be a discontinuity in torque slip characteristics
b) A high starting torque will be available
c) The maximum torque will be high
d) The machine may fail to start
Answer: d
Explanation: When the number of stator slots is an integral multiple of a number of rotor slots the machine fails to start and this phenomenon is called cogging.
72. A 3-phase induction motor runs at almost 1000 rpm at no load and 950 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?
a) 30 revolutions per minute
b) 40 revolutions per minute
c) 60 revolutions per minute
d) 50 revolutions per minute
Answer: d
Explanation: Supply frequency=50 Hz. No-load speed of motor = 1000 rpm. The full load speed of motor=950 rpm. Since the no-load speed of the motor is almost 1000 rpm, hence synchronous speed near to 1000 rpm. Speed of rotor field=1000 rpm. Speed of rotor field with respect to rotor = 1000-950 = 50 rpm.
73. Calculate the active power in a 487 H inductor.
b) 1545 W
c) 4565 W
d) 0 W
Answer: d
Explanation: 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 case of the inductor so the angle between V & I is 90°. P = VIcos90 = 0 W.
74. Calculate the active power in a 788 ω resistor with 178 A current flowing through it.
b) 24.44 MW
c) 24.12 MW
d) 26.18 MW
Answer: a
Explanation: 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 case of the resistor so the angle between V & I is 90°. P=I2R=178×178×788=24.96 MW.
75. The direct axis is taken along ________
a) Inter-polar axis
b) Rotor pole axis
c) In between interpolar and rotor axis
d) Parallel to the interpolar axis
Answer: b
Explanation: The direct axis is oriented along the rotor pole axis and the quadrature axis is 90° electrical to the rotor pole axis. The direct axis is not oriented along the inter-polar axis.
76. A 4-pole, 3-phase, 60 Hz induction motor is operating at a speed of 1500 rpm. The frequency of the rotor current of the motor in Hz is __________
b) 4
c) 2
d) 7
Answer: b
Explanation: Given a number of poles = 4. Supply frequency is 60 Hz. Rotor speed is 1500 rpm. Ns = 120×f÷P = 120×60÷4 = 1800 rpm. S=Ns-Nr÷Ns = 1800-1500÷1800 = .166. F2=sf=.166×60=4 Hz.
77. Calculate the phase angle of the sinusoidal waveform z(t)=8cos(45t+2π÷15).
b) 2π÷15
c) π÷4
d) 2π÷44
Answer: b
Explanation: Sinusoidal waveform is generally expressed in the form of V=Vmsin(ωt+α) where Vm represents peak value, ω represents angular frequency, α represents a phase difference.
78. The leakage flux paths are ________ on the angular position of the rotor.
a) Dependent
b) Proportional
c) Independent
d) Dependent and independent
Answer: c
Explanation: Leakage flux and leakage reactance are constant irrespective of rotor angular position. The armature reaction though is dependent on the angular position of the rotor in the salient pole machine but in the cylindrical rotor machine, both the quantities are independent of the rotor position.
79. The hunting phenomenon in a synchronous motor is also referred to as_________
a) Surging
b) Phase swinging
c) Cogging
d) Surging and phase swinging
Answer: d
Explanation: During hunting, power oscillates so power surges are observed and hence it can be called as surging. Also, the rotor phase angle oscillates and it is called as phase swinging.
80. Which of the following are used in preventing the hunting phenomenon in synchronous generators?
a) Damper bars
b) Short pitch chords
c) Distributed winding
d) Damper bars and short pitch chords
Answer: a
Explanation: Damper bars try to maintain synchronism between the rotating magnetic field and the rotor so they help in preventing hunting. It produces surges in the machine.
81. In a synchronous machine, the phase sequence can be reversed by reversing the_________
b) Field polarities
c) Armature terminal
d) Rotor direction and armature terminal
Answer: a
Explanation: In synchronous generator, the phase sequence is governed by the direction of rotation of the rotor and in a synchronous motor, the phase sequence governs the direction of rotation of the rotor.
82. The slope of the V-I curve is 7°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.
b) .360 Ω
c) .377 Ω
d) .578 Ω
Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 7° so R=tan(7°)=.122 Ω. The slope of the I-V curve is reciprocal of resistance.
83. In an induction motor, when the number of stator slots is not equal to an integral number of rotor slots _________
a) There may be a discontinuity in torque slip characteristics
b) A high starting torque will be available
c) The machine performs better
d) The machine may fail to start
Answer: c
Explanation: When the number of stator slots is not an integral multiple of a number of rotor slots the machine will not fail to start. It does not cause the cogging phenomenon.
84. A 3-phase induction motor runs at almost 1500 rpm at no load and 900 rpm at full load when supplied with power from a 50 Hz, 3-phase supply. What is the corresponding speed of the rotor field with respect to the rotor?
a) 300 revolutions per minute
b) 400 revolutions per minute
c) 600 revolutions per minute
d) 500 revolutions per minute
Answer: c
Explanation: Supply frequency=50 Hz. No-load speed of motor = 1500 rpm. The full load speed of motor=900 rpm. Since the no-load speed of the motor is almost 1500 rpm, hence synchronous speed near to 1500 rpm. Speed of rotor field=1500 rpm. Speed of rotor field with respect to rotor = 1500-900 = 600 rpm.
85. For a practical synchronous motor, the pull-out torque will occur when the torque angle is nearly equal to ________
a) 0°
b) 30°
c) 45°
d) 75°
Answer: d
Explanation: In a practical synchronous motor, the armature resistance cannot be neglected and hence the pull-out occurs at delta=beta which is the impedance angle and is practically 75°.
86. Calculate the active power in an 8 Ω resistor with 8 A current flowing through it.
a) 512 W
b) 514 W
c) 512 W
d) 518 W
Answer: a
Explanation: 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 case of the resistor so the angle between V & I is 90°. P = I2R = 8×8×8 = 512 W.
