60+ MCQs On Starting of Electrical Drives

July 4, 2022

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 “Starting of Electrical Drives”. 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 Starting of Electrical Drives [Electrical Drives Chapterwise Series]

Suppose you want to jump to the specific question directly, expand the table of content below:

Table of Contents

1. The ferrite cores are used for ____________ transformers.

a) Small transformers
b) Medium transformers
c) Large transformers
d) Medium and small transformers

Answer: a
Explanation: Ferrite cores are used for cores of small transformers used in communication circuits at high frequencies and low energy levels. Because ferrites have high resistivity they will have lower eddy current losses.

2. A 2-pole, 3-phase, 50 Hz induction motor is operating at a speed of 400 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 43.33
b) 42.54
c) 43.11
d) 41.47

Answer: a
Explanation: Given a number of poles = 2. Supply frequency is 50 Hz. Rotor speed is 400 rpm. N_s = 120×f÷P=120×50÷2 = 3000 rpm.

S=N_s-N_r÷N_s = 3000-400÷3000= .8666. F₂=sf=.8666×50=43.33 Hz.

3. Calculate the phase angle of the sinusoidal waveform w(t)=.45sin(87πt+8π÷787).

a) 2π÷39
b) 8π÷787
c) 5π÷74
d) 42π÷4

Answer: b
Explanation: Sinusoidal waveform is generally expressed in the form of V=V_msin(ωt+α) where V_m represents peak value, ω represents angular frequency, α represents a phase difference.

4. Calculate the moment of inertia of the disc having a mass of 1 kg and radius of 1 m.

a) 1 kgm²
b) .5 kgm²
c) 2 kgm²
d) 3 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=(1)×.5×(1)²=.5 kgm². It depends upon the orientation of the rotational axis.

5. Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and a diameter of 6 cm.

a) .0178 kgm²
b) .0147 kgm²
c) .0398 kgm²
d) .0144 kgm²
View Answer

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=(3)×.66×(.03)²=.0178 kgm². It depends upon the orientation of the rotational axis.

6. The power factor of a synchronous motor __________

a) Improves with an increase in excitation and may even become leading at high excitations
b) Decreases with increase in excitation
c) Is independent of its excitation
d) Increase with loading for a given excitation

Answer: a
Explanation: From the inverted V-curve we can see when the power factor is leading power factor decreases when the excitation increases and it is over-excited conditions and when the power factor is lagging if the motor power factor is increasing if excitation increases.

7. The frame of a synchronous motor is made of _________

a) Aluminum
b) Silicon steel
c) Cast iron
d) Stainless steel

Answer: c
Explanation: The frame of a synchronous motor is made of cast iron. The power factor of a synchronous motor depends upon maximum power transfer capability.

8. The slope of the V-I curve is 45°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) 2 Ω
b) 3 Ω
c) 4 Ω
d) 1 Ω

Answer: d
Explanation: The slope of the V-I curve is resistance. The slope given is 45° so R=tan(45°)=1 Ω. The slope of the I-V curve is reciprocal to resistance.

9. Which one of the following methods would give a higher than the actual value of regulation of the alternator.

a) ZPF method
b) MMF method
c) EMF method
d) ASA method

Answer: c
Explanation: EMF method is a pessimistic method of voltage regulation as it gives higher than the actual value of voltage regulation. EMF method will the values that are greater than the actual value.

10. A 3-phase induction motor runs at almost 1100 rpm at no load and 640 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) 430 revolution per minute
b) 440 revolution per minute
c) 460 revolution per minute
d) 450 revolution per minute

Answer: c
Explanation: Supply frequency=50 Hz. No-load speed of motor = 1100 rpm. The full load speed of motor=640 rpm. Since the no-load speed of the motor is almost 1100 rpm, hence synchronous speed near to 1100 rpm. Speed of rotor field=1100 rpm. Speed of rotor field with respect to rotor = 1100-640 = 460 rpm.

11. Which of the following core has linear characteristics?

a) Air core
b) Steel core
c) CRGO core
d) Iron core

Answer: a
Explanation: Air core coils have linear magnetization characteristics that they do not saturate. The open circuit characteristics graph is linear in the case of a synchronous machine.

12. Calculate the active power in a 1 Ω resistor with 2 A current flowing through it.

a) 2 W
b) 4 W
c) 7 W
d) 1 W

Answer: b
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 90°. P = I²R = 2×2×1=4 W.

13. The slope of the V-I curve is 13°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .2544 Ω
b) .7771 Ω
c) .2308 Ω
d) .5788 Ω

Answer: c
Explanation: The slope of the V-I curve is resistance. The slope given is 13° so R=tan(13°)=.2308 Ω. The slope of the I-V curve is reciprocal of resistance.

14. A 2-pole lap wound generator with 44 armature conductors and a flux per pole of .07 Wb has an armature current of 80 A. The developed torque is _________

a) 39.7 N-m
b) 39.2 N-m
c) 38.4 N-m
d) 37.2 N-m

Answer: b
Explanation: The developed torque in the motor is K_m×I. The value of machine constant(K_m) is E_b÷ωm = Φ×Z×P÷2×3.14×A = .07×44×2÷2×3.14×2 = .49 Vs/rad. The developed torque is .49×80 = 39.2 N-m.

15. Calculate the active power in a .45 H inductor.

a) 0.11 W
b) 0.14 W
c) 0.15 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.

16. A 10-pole, 3-phase, 60 Hz induction motor is operating at a speed of 100 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 52.4
b) 54.8
c) 51.66
d) 51.77

Answer: c
Explanation: Given a number of poles = 10. Supply frequency is 60 Hz. Rotor speed is 100 rpm. N_s = 120×f÷P=120×60÷10 = 720 rpm. S=N_s-N_r÷N_s = 720-100÷720=.86. F₂=sf=.86×60=51.66 Hz.

17. Calculate the phase angle of the sinusoidal waveform z(t)=.99sin(4578πt+78π÷78).

a) π÷3
b) 2π
c) π÷7
d) π

Answer: d
Explanation: Sinusoidal waveform is generally expressed in the form of V=V_msin(ωt+α) where V_m represents peak value, Ω represents angular frequency, α represents a phase difference.

18. Calculate the moment of inertia of the disc having a mass of 4 kg and diameter of 1458 cm.
a) 106.288 kgm²
b) 104.589 kgm²
c) 105.487 kgm²
d) 107.018 kgm²

Answer: a
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=(4)×.5×(7.29)²=106.288 kgm². It depends upon the orientation of the rotational axis.

19. Calculate the moment of inertia of the thin spherical shell having a mass of 703 kg and a diameter of 376 cm.

a) 1639.89 kgm²
b) 1628.47 kgm²
c) 1678.12 kgm²
d) 1978.19 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=(703)×.66×(1.88)²=1639.89 kgm². It depends upon the orientation of the rotational axis.

20. Calculate the value of the torque when 89 N force is applied perpendicular to a 78 m length of stick fixed at the center.

a) 6942 N-m
b) 3000 N-m
c) 1000 N-m
d) 4470 N-m

Answer: a
Explanation: Torque can be calculated using the relation T = (length of stick) × (Force applied) = r×F×sin90. F is given as 89 N and r is 78 m then torque is 89×78 = 6942 N-m. (the angle between F and r is 90 degrees).

21. 100 V, 2 A, 90 rpm separately excited dc motor with armature resistance (R_a) equal to 8 ohms. Calculate back emf developed in the motor when it operates on 3^th/4 of the full load. (Assume rotational losses are neglected)

a) 100 V
b) 87 V
c) 88 V
d) 90 V

Answer: c
Explanation: Back emf developed in the motor can be calculated using the relation E_b = V_t-I×R_a. In question, it is asking for 3^th/4 load, but the data is given for full load so current becomes 3^th/4 of the full load current = 2÷1.33 = 1.5 A. 100 V is terminal voltage it is fixed so E_b = 100-1.5×8 = 88 V.

22. The slope of the V-I curve is 16.8°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .324 Ω
b) .301 Ω
c) .343 Ω
d) .398 Ω

Answer: b
Explanation: The slope of the V-I curve is resistance. The slope given is 16.8° so R=tan(16.8°)=.301 Ω. The slope of the I-V curve is reciprocal to resistance.

23. Calculate the value of the torque when 1 N force is applied perpendicular to a 1 m length of chain fixed at the center.

a) 1 N-m
b) 3 N-m
c) 2 N-m
d) 4 N-m

Answer: a
Explanation: Torque can be calculated using the relation T = (length of chain) × (Force applied) = r×F×sin90. F is given as 1 N and r is 1 m then torque is 1×1 = 6942 N-m. (the angle between F and r is 90 degrees).

24. A 3-phase induction motor runs at almost 140 rpm at no load and 50 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) 20 revolutions per minute
b) 80 revolutions per minute
c) 90 revolutions per minute
d) 70 revolution per minute

Answer: c
Explanation: Supply frequency=50 Hz. No-load speed of motor= 140 rpm. The full load speed of the motor=50 rpm. Since the no-load speed of the motor is almost 140 rpm, hence synchronous speed is near 140 rpm. Speed of rotor field=140 rpm. Speed of rotor field with respect to rotor = 140-50 = 90 rpm.

25. Calculate the active power in a .7889 H inductor.

a) .123 W
b) .155 W
c) 0 W
d) .487 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.

26. Calculate the active power in a .8 Ω resistor with 2 A current flowing through it.

a) 2.4 W
b) 3.4 W
c) 2.2 W
d) 3.2 W

Answer: d
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=2×2×.8=3.2 MW.

27. The unit of angular frequency is Hz.

a) True
b) False

Answer: a
Explanation: Angular frequency is defined as the rate of change of angular displacement with respect to time. Angular displacement is generally expressed in terms of a radian. The unit of angular frequency is Hz. ω=2×3.14×f.

28. Calculate the active power in a .56 F capacitor.

a) 37.8 W
b) 0 W
c) 15.4 W
d) 124.5 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.

29. Calculate the value of the angular acceleration of the motor using the given data: J= .01 kg-m², load torque= 790 N-m, motor torque= 169 N-m.

a) -62100 rad/s²
b) -62456 rad/s²
c) -34056 rad/s²
d) -44780 rad/s²

Answer: a
Explanation: Using the dynamic equation of motor J*(angular acceleration) = Motor torque – Load torque: .01*(angular acceleration) = 169-790=-621, angular acceleration=-62100 rad/s².The motor will decelerate and will fail to start.

30. A 14-pole, 3-phase, 50 Hz induction motor is operating at a speed of 99 rpm. The frequency of the rotor current of the motor in Hz is __________

a) 39.5
b) 40
c) 38.45
d) 39.9

Answer: c
Explanation: Given a number of poles = 14. Supply frequency is 50 Hz. Rotor speed is 699 rpm. N_s=120×f÷P=120×50÷14 = 428.57 rpm. S=N_s-N_r÷N_s=428.57-99÷428.57=.769. F₂=sf=.769×50=38.45 Hz.

31. Calculate the phase angle of the sinusoidal waveform z(t)=18cos (1546πt + 1900π/76).

a) 25π÷39
b) 25π÷5
c) 25π÷1
d) π÷4

Answer: c
Explanation: Sinusoidal waveform is generally expressed in the form of V=V_msin(ωt+α) where V_m represents peak value, Ω represents angular frequency, α represents a phase difference.

32. Calculate the mass of the solid sphere having a moment of inertia of 17 kgm² and a radius of 4 cm.

a) 10624 kg
b) 10625 kg
c) 10628 kg
d) 10626 kg

Answer: b
Explanation: The moment of inertia of the ball can be calculated using the formula I=Σm_ir_i². The moment of inertia of the ball and radius is given. M=(17)÷(.04)² = 10625 kg. It depends upon the orientation of the rotational axis.

33. Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and a diameter of 66 cm.

a) .2156 kgm²
b) .2147 kgm²
c) .2138 kgm²
d) .2148 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=(3)×.66×(.33)²=.2156 kgm². It depends upon the orientation of the rotational axis.

34. Calculate the value of the time period if the frequency of the signal is 1 sec.

a) 1 sec
b) 2 sec
c) .5 sec
d) 1.5 sec

Answer: a
Explanation: The time period is defined as the time after the signal repeats itself. It is expressed in second. T = 1÷F=1÷1=1 sec.

35. The slope of the V-I curve is 270°. Calculate the value of resistance.

a) 112 Ω
b) 178 Ω
c) infinite Ω
d) 187 Ω

Answer: c
Explanation: The slope of the V-I curve is resistance. The slope given is 270° so R=tan(270°)=infinite Ω. It behaves as an open circuit.

36. The slope of the V-I curve is 23.56°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .464 Ω
b) .436 Ω
c) .443 Ω
d) .463 Ω

Answer: b
Explanation: The slope of the V-I curve is resistance. The slope given is 23.56° so R=tan(23.56°)=.436 Ω. The slope of the I-V curve is reciprocal of resistance.

37. Calculate the reactive power in a 23 Ω resistor.

a) 45 VAR
b) 10 VAR
c) 245 VAR
d) 0 VAR

Answer: d
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°. Q = VIsin0° = 0 VAR.

38. A 3-phase induction motor runs at almost 50 rpm at no load and 25 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) 25 revolutions per minute
b) 20 revolutions per minute
c) 10 revolutions per minute
d) 30 revolutions per minute

Answer: a
Explanation: Supply frequency=50 Hz. No-load speed of motor= 50 rpm. The full load speed of motor=25 rpm. Since the no-load speed of the motor is almost 50 rpm, hence synchronous speed near to 50 rpm. Speed of rotor field=50 rpm. Speed of rotor field with respect to rotor = 50-25 = 25 rpm.

39. Calculate the value of the frequency of the 220 V DC supply.

a) 10 Hz
b) 0 Hz
c) 20 Hz
d) 90 Hz

Answer: b
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. DC supply magnitude is constant. It does not change with time so the frequency of DC supply is 0 Hz.

40. Calculate the value of the duty cycle if the system is on for 48 sec and off for 1 sec.

a) .979
b) .444
c) .145
d) .578

Answer: a
Explanation: The duty cycle is Ton÷T_total. It is the ratio of time for which the system is active and the time taken by the signal to complete one cycle. D = Ton÷T_total = 48÷49 = .979.

41. Calculate the value of the frequency if the signal completes half of the cycle in 30 sec. Assume the signal is periodic.

a) 0.028 Hz
b) 0.016 Hz
c) 0.054 Hz
d) 0.045 Hz

Answer: b
Explanation: The frequency is defined as the number of oscillations per second. It is reciprocal of the time period. It is expressed in Hz. The given signal completes half of the cycle in 30 seconds then it will complete a full cycle in 60 seconds. F = 1÷T = 1÷60 = .016 Hz.

42. Calculate the velocity of the disc if the angular speed is 5 rad/s and radius is 2 m.

a) 25 m/s
b) 20 m/s
c) 25 m/s
d) 10 m/s

Answer: d
Explanation: The velocity of the disc can be calculated using the relation V=ω×r. The velocity is the vector product of angular speed and radius. V = ω×r = 5×2 = 10 m/s.

43. 440 V, 77 A, 700 rpm DC separately excited motor having a resistance of 0.11 ohms excited by an external dc voltage source of 24 V. Calculate the torque developed by the motor on full load.

a) 453.51 N-m
b) 451.24 N-m
c) 440.45 N-m
d) 452.64 N-m

Answer: a
Explanation: Back emf developed in the motor during the full load can be calculated using equation E_b = V_t-I×R_a = 431.53 V and machine constant K_m = E_b÷W_m which is equal to 5.88. Torque can be calculated by using the relation T = K_m × I = 5.88×77 = 453.51 N-m.

44. Calculate the power developed by a motor using the given data: E_b = 55 V and I = 6 A.

a) 440 W
b) 220 W
c) 330 W
d) 550 W

Answer: c
Explanation: Power developed by the motor can be calculated using the formula P = E_b×I = 55×6 = 330 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

45. Calculate the value of the angular acceleration of the motor using the given data: J = 36 kg-m², load torque = 66 N-m, motor torque = 26 N-m.

a) 1.11 rad/s²
b) 2.22 rad/s²
c) 3.33 rad/s²
d) 4.44 rad/s²

Answer: a
Explanation: Using the dynamic equation of motor J×(angular acceleration) = Motor torque – Load torque: 36×(angular acceleration) = 66-26 = 40, angular acceleration = 1.11 rad/s².

46. Calculate the moment of inertia of the apple having a mass of .4 kg and diameter of 12 cm.

a) .0008 kgm²
b) .0007 kgm²
c) .0009 kgm²
d) .0001 kgm²

Answer: b
Explanation: The moment of inertia of the apple can be calculated using the formula I=mr²×.5. The mass of the apple and diameter is given. I=(.4)×.5×(.06)² = .0007 kgm². It depends upon the orientation of the rotational axis.

47. Calculate the moment of inertia of the thin spherical shell having a mass of 3.3 kg and diameter of .6 cm.

a) .00125 kgm²
b) .00196 kgm²
c) .00145 kgm²
d) .00178 kgm²

Answer: b
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=(3.3)×.66×(.03)²=.00196 kgm². It depends upon the orientation of the rotational axis.

48. Calculate the time period of the waveform y(t)=7cos(54πt+2π÷4).

a) .055 sec
b) .037 sec
c) .023 sec
d) .017 sec

Answer: b
Explanation: The fundamental time period of the cosine wave is 2π. The time period of y(t) is 2π÷54π=.037 sec. The time period is independent of phase shifting and time-shifting.

49. Calculate the useful power developed by a motor using the given data: P_in= 1500 W, I_a= 6 A, R_a=.2 Ω. Assume frictional losses are 50 W and windage losses are 25 W.

a) 1400 W
b) 1660.5 W
c) 1417.8 W
d) 1416.7 W

Answer: c
Explanation: Useful power is basically the shaft power developed by the motor that can be calculated using the formula P_sh = P_dev-(rotational losses). P_dev = P_in-I_a²R_a = 1500-6²(.2)=1492.8 W. The useful power developed by the motor is P_sh = P_dev-(rotational losses) = 1492.8 –(50+25) = 1417.8 W.

50. The slope of the V-I curve is 15.5°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .277 Ω
b) .488 Ω
c) .443 Ω
d) .457 Ω

Answer: a
Explanation: The slope of the V-I curve is resistance. The slope given is 15.5° so R=tan(15.5°)=.277 Ω. The slope of the I-V curve is reciprocal of resistance.

51. The generated e.m.f from 2-pole armature having 2 conductors driven at 3000 rpm having flux per pole as 4000 mWb, with 91 parallel paths is ___________

a) 8.64 V
b) 8.56 V
c) 8.12 V
d) 8.79 V

Answer: d
Explanation: The generated e.m.f can be calculated using the formula E_b = 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. E_b = 4×2×3000×2÷60×91 = 8.79 V.

52. A 3-phase induction motor runs at almost 888 rpm at no load and 500 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) 388 revolutions per minute
b) 400 revolutions per minute
c) 644 revolutions per minute
d) 534 revolutions per minute

Answer: a
Explanation: Supply frequency=50 Hz. No-load speed of motor = 888 rpm. The full load speed of motor=500 rpm. Since the no-load speed of the motor is almost 888 rpm, hence synchronous speed near to 888 rpm. Speed of rotor field=888 rpm. Speed of rotor field with respect to rotor = 888-500 = 388 rpm.

53. Calculate the active power in a 157.1545 H inductor.

a) 4577 W
b) 4567 W
c) 4897 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.

54. Calculate the active power in a 1.2 Ω resistor with 1.8 A current flowing through it.

a) 3.88 W
b) 3.44 W
c) 3.12 W
d) 2.18 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 90^o. P=I²R=1.8×1.8×1.2=3.88 W.

55. Calculate the total heat dissipated in a resistor of 12 Ω when 9.2 A current flows through it.

a) 2.01 KW
b) 3.44 KW
c) 1.01 KW
d) 2.48 KW

Answer: c
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 90°. P=I²R=9.2×9.2×12=1.01 kW.

56. Calculate mark to space ratio if the system is on for 9 sec and the time period is 11 sec.

a) 4.6
b) 4.8
c) 4.5
d) 4.9

Answer: c
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=9÷(11-9) = 4.5.

1. Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).

a) 40π Hz
b) 60π Hz
c) 70π Hz
d) 20π Hz

Answer: a
Explanation: The fundamental time period of the sine wave is 2π. The sinusoidal waveform is generally expressed in the form of V=V_msin(Ωt+α) where V_m represents peak value, Ω represents angular frequency, and α represents a phase difference. Ω can be directly calculated by comparing the equations. Ω = 40π Hz.

57. The generated e.m.f from 22-pole armature having 75 turns driven at 78 rpm having flux per pole as 400 mWb, with lap winding is ___________

a) 76 V
b) 77 V
c) 78 V
d) 79 V

Answer: c
Explanation: The generated e.m.f can be calculated using the formula E_b=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.

E_b = .4×22×75×2×78÷60×22 = 78 V.

58. Calculate the phase angle of the sinusoidal waveform u(t)=154sin(9.85πt-π÷89).

a) -78π÷9
b) -12π÷5
c) -π÷89
d) -2π÷888

Answer: c
Explanation: Sinusoidal waveform is generally expressed in the form of V=V_msin(Ωt+α) where V_m represents peak value, Ω represents angular frequency, α represents a phase difference.

59. Calculate the moment of inertia of the stick about its end having a mass of 22 kg and a length of 22 cm.

a) .088 kgm²
b) .087 kgm²
c) .089 kgm²
d) .086 kgm²

Answer: b
Explanation: The moment of inertia of the stick about its end can be calculated using the formula I=ML²÷3. The mass of the stick about its end and length is given. I = (22)×.33×(.11)² =.087 kgm². It depends upon the orientation of the rotational axis.

60. Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and a length of 2.9 m.

a) .66 kgm²
b) .77 kgm²
c) .88 kgm²
d) .47 kgm²
View Answer

Answer: b
Explanation: The moment of inertia of the stick about its center can be calculated using the formula I=ML²÷12. The mass of the stick about its center and length is given. I=(1.1)×.0833×(2.9)²=.77 kgm². It depends upon the orientation of the rotational axis.

61. Calculate the useful power developed by a motor using the given data: E_b = 4V and I = 52 A. Assume frictional losses are 3 W and windage losses are 2 W.

a) 203 W
b) 247 W
c) 211 W
d) 202 W

Answer: a
Explanation: Useful power developed by the motor can be calculated using the formula P = E_b*I -(rotational losses) = 4*52 – (5) = 203 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

62. Calculate the value of the frequency if the capacitive reactance is .1 Ω and the value of the capacitor is .02 F.

a) 71.25 Hz
b) 81.75 Hz
c) 79.61 Hz
d) 79.54 H

Answer: c
Explanation: The frequency is defined as the number of oscillations per second. The frequency can be calculated using the relation X_c = 1÷2×3.14×f×C. F = 1÷X_c×2×3.14×C = 1÷.1×2×3.14×.02 = 79.61 Hz.

63. The slope of the V-I curve is 6.9°. Calculate the value of resistance. Assume the relationship between voltage and current is a straight line.

a) .38 Ω
b) .59 Ω
c) .34 Ω
d) .12 Ω

Answer: d
Explanation: The slope of the V-I curve is resistance. The slope given is 6.9° so R=tan(6.9°)=.12 Ω. The slope of the I-V curve is reciprocal to resistance.

64. Calculate the active power in an 8764 H inductor.

a) 8645 W
b) 6485 W
c) 0 W
d) 4879 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 the case of the inductor.

65. Calculate the active power in a 543 F capacitor.

a) 581 W
b) 897 W
c) 0 W
d) 892 W

Answer: c
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.

66. Calculate the active power in a 32 H inductor.

a) 28 W
b) 189 W
c) 4 W
d) 0 W