Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

51.	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-mI have been asked this question in my homework.The doubt is from Methods of Starting Electric Motors in section Starting of Electric Drives
Answer» The correct option is (a) 6942 N-m To EXPLAIN I would SAY: 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).

52.	100 V, 2 A, 90 rpm separately excited dc motor with armature resistance (Ra) 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 VThe question was asked by my school principal while I was bunking the class.My question comes from Methods of Starting Electric Motors topic in chapter Starting of Electric Drives
Answer»

53.	Calculate the moment of inertia of the thin spherical shell having a mass of 703 kg and diameter of 376 cm.(a) 1639.89 kgm^2(b) 1628.47 kgm^2(c) 1678.12 kgm^2(d) 1978.19 kgm^2I have been asked this question in exam.Question is taken from Methods of Starting Electric Motors in section Starting of Electric Drives
Answer» Correct answer is (a) 1639.89 kgm^2 The explanation is: The MOMENT of inertia of the thin spherical shell can be calculated using the formula I=mr^2×.66. The mass of the thin spherical shell and diameter is given. I=(703)×.66×(1.88)^2=1639.89 kgm^2. It depends UPON the orientation of the ROTATIONAL axis.

54.	Calculate the moment of inertia of the disc having a mass of 4 kg and diameter of 1458 cm.(a) 106.288 kgm^2(b) 104.589 kgm^2(c) 105.487 kgm^2(d) 107.018 kgm^2I had been asked this question in an interview.I'm obligated to ask this question of Methods of Starting Electric Motors topic in division Starting of Electric Drives
Answer» The correct OPTION is (a) 106.288 kgm^2 The BEST explanation: The moment of inertia of the disc can be calculated using the formula I=mr^2×.5. The mass of the disc and diameter is GIVEN. I=(4)×.5×(7.29)^2=106.288 kgm^2. It depends upon the ORIENTATION of the rotational AXIS.

55.	Calculate the phase angle of the sinusoidal waveform z(t)=.99sin(4578πt+78π÷78).(a) π÷3(b) 2π(c) π÷7(d) πI have been asked this question during a job interview.This is a very interesting question from Methods of Starting Electric Motors topic in section Starting of Electric Drives
Answer» The correct option is (d) π To explain: 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.

56.	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) 51.77I had been asked this question in semester exam.The above asked question is from Methods of Starting Electric Motors topic in division Starting of Electric Drives
Answer» Correct answer is (c) 51.66 The explanation: GIVEN a NUMBER of poles = 10. Supply frequency is 60 HZ. Rotor speed is 100 rpm. Ns = 120×f÷P=120×60÷10 = 720 rpm. S=Ns-Nr÷Ns = 720-100÷720=.86. F2=sf=.86×60=51.66 Hz.

57.	Calculate the active power in a .45 H inductor.(a) 0.11 W(b) 0.14 W(c) 0.15 W(d) 0 WI got this question in class test.The origin of the question is Methods of Starting Electric Motors in portion Starting of Electric Drives
Answer» The correct ANSWER is (d) 0 W Easiest 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.

58.	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-mThis question was addressed to me during an online exam.I'd like to ask this question from Effect of Starting on Power Supply, Motor and Load topic in section Starting of Electric Drives
Answer» Right answer is (b) 39.2 N-m Easy explanation: The developed torque in the MOTOR is KM×I. The VALUE of machine CONSTANT(Km) is Eb÷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.

59.	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 ΩI have been asked this question in a national level competition.The query is from Effect of Starting on Power Supply, Motor and Load topic in division Starting of Electric Drives
Answer» The CORRECT ANSWER is (c) .2308 Ω For 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.

60.	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 WI have been asked this question during an interview.My doubt stems from Effect of Starting on Power Supply, Motor and Load in chapter Starting of Electric Drives
Answer» The correct answer is (b) 4 W To EXPLAIN: 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 = I^2R = 2×2×1=4 W.

61.	Which of the following core has linear characteristics?(a) Air core(b) Steel core(c) CRGO core(d) Iron coreThe question was asked in an international level competition.I'm obligated to ask this question of Effect of Starting on Power Supply, Motor and Load topic in portion Starting of Electric Drives
Answer» Correct choice is (a) AIR CORE The explanation is: Air core coils have LINEAR magnetization CHARACTERISTICS that are they do not SATURATE. Open circuit characteristics graph is linear in case of synchronous machine.

62.	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 ΩI got this question during an online interview.I need to ask this question from Effect of Starting on Power Supply, Motor and Load in chapter Starting of Electric Drives
Answer» The CORRECT CHOICE is (d) 1 Ω Easiest 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 of resistance.

63.	The frame of a synchronous motor is made of _________(a) Aluminum(b) Silicon steel(c) Cast iron(d) Stainless steelThis question was posed to me in semester exam.My question is taken from Effect of Starting on Power Supply, Motor and Load topic in portion Starting of Electric Drives
Answer» The correct OPTION is (c) Cast iron Easiest 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.

64.	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 excitationI had been asked this question during an online interview.This key question is from Effect of Starting on Power Supply, Motor and Load in portion Starting of Electric Drives
Answer» Right option is (a) Improves with an increase in excitation and may even become leading at high excitations Easiest explanation: From 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 power factor is LAGGING if the MOTOR power factor is increasing if excitation increases.

65.	Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and diameter of 6 cm.(a) .0178 kgm^2(b) .0147 kgm^2(c) .0398 kgm^2(d) .0144 kgm^2I had been asked this question in examination.I'd like to ask this question from Effect of Starting on Power Supply, Motor and Load topic in section Starting of Electric Drives
Answer» Correct choice is (a) .0178 kgm^2 The best I can explain: The moment of inertia of the THIN spherical shell can be calculated USING the FORMULA I=mr^2×.66. The mass of the thin spherical shell and diameter is given. I=(3)×.66×(.03)^2=.0178 kgm^2. It DEPENDS upon the ORIENTATION of the rotational axis.

66.	Calculate the moment of inertia of the disc having a mass of 1 kg and radius of 1 m.(a) 1 kgm^2(b) .5 kgm^2(c) 2 kgm^2(d) 3 kgm^2The question was posed to me in quiz.This intriguing question comes from Effect of Starting on Power Supply, Motor and Load in chapter Starting of Electric Drives
Answer» CORRECT OPTION is (b) .5 kgm^2 The explanation is: The moment of inertia of the disc can be calculated using the formula I=mr^2×.5. The mass of the disc and DIAMETER is given. I=(1)×.5×(1)^2=.5 kgm^2. It depends upon the orientation of the ROTATIONAL axis.

67.	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÷4The question was posed to me in an interview for job.My question is based upon Effect of Starting on Power Supply, Motor and Load topic in portion Starting of Electric Drives
Answer» Right CHOICE is (b) 8π÷787 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.

68.	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) 41.47This question was posed to me in final exam.This is a very interesting question from Effect of Starting on Power Supply, Motor and Load in portion Starting of Electric Drives
Answer» CORRECT ANSWER is (a) 43.33 For explanation: Given a NUMBER of poles = 2. Supply frequency is 50 HZ. Rotor speed is 400 rpm. Ns = 120×f÷P=120×50÷2 = 3000 rpm. S=Ns-Nr÷Ns = 3000-400÷3000 = .8666. F2=sf=.8666×50=43.33 Hz.

69.	The ferritecores are used for ____________ transformers.(a) Small transformers(b) Medium transformers(c) Large transformers(d) Medium and small transformersThis question was addressed to me in class test.My question comes from Effect of Starting on Power Supply, Motor and Load topic in division Starting of Electric Drives
Answer» Correct option is (a) Small transformers To EXPLAIN I would say: 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.