69 + Interview Questions in Starting in Electric Drives Page 1 InterviewSolution

1.	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 revolution per minute(b) 20 revolution per minute(c) 10 revolution per minute(d) 30 revolution per minuteThis question was addressed to me at a job interview.The origin of the question is Starting topic in portion Starting of Electric Drives
Answer» Correct OPTION is (a) 25 revolution per minute To ELABORATE: 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.

Discussion

2.	Calculate the value of the frequency of the 220 V DC supply.(a) 10 Hz(b) 0 Hz(c) 20 Hz(d) 90 HzI have been asked this question in an interview for internship.Question is taken from Starting in section Starting of Electric Drives
Answer» Correct answer is (B) 0 HZ Easiest 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.

Discussion

3.	Calculate the active power in an 8965 Ω resistor with .23 A current flowing through it.(a) 547.12 W(b) 474.24 W(c) 554.78 W(d) 123.88 WThis question was posed to me in a job interview.The doubt is from Methods to Reduce the Energy Loss During Starting topic in section Starting of Electric Drives
Answer» Right option is (b) 474.24 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 0°. P=I^2R=.23×.23×8965=474.24 W.

Discussion

4.	Which one of the following methods would give a lower than the actual value of regulation of the alternator?(a) ZPF method(b) MMF method(c) EMF method(d) ASA methodI got this question in an interview for internship.My query is from Methods to Reduce the Energy Loss During Starting topic in division Starting of Electric Drives
Answer» The correct answer is (b) MMF method For explanation I would say: MMF method is an optimistic method of voltage regulation as it gives lower than the ACTUAL VALUE of voltage regulation. MMF method will give the values that are lesser than the actual value.

Discussion

5.	Calculate the active power in a 32 H inductor.(a) 28 W(b) 189 W(c) 4 W(d) 0 WThis question was posed to me in unit test.My query is from Methods to Reduce the Energy Loss During Starting in portion Starting of Electric Drives
Answer» RIGHT option is (d) 0 W Easy 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.

Discussion

6.	Calculate the active power in a 543 F capacitor.(a) 581 W(b) 897 W(c) 0 W(d) 892 WThis question was addressed to me during a job interview.This key question is from Methods to Reduce the Energy Loss During Starting in section Starting of Electric Drives
Answer» Correct option is (C) 0 W The BEST I can explain: 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.

Discussion

7.	Calculate the active power in an 8764 H inductor.(a) 8645 W(b) 6485 W(c) 0 W(d) 4879 WThis question was posed to me in an interview.This intriguing question comes from Methods to Reduce the Energy Loss During Starting topic in chapter Starting of Electric Drives
Answer» The correct answer is (c) 0 W The 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.

Discussion

8.	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 ΩThe question was posed to me in exam.My question is from Methods to Reduce the Energy Loss During Starting topic in section Starting of Electric Drives
Answer» RIGHT choice is (d) .12 Ω To explain: 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 of resistance.

Discussion

9.	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 HzI have been asked this question in semester exam.The origin of the question is Methods to Reduce the Energy Loss During Starting in portion Starting of Electric Drives
Answer» The correct CHOICE is (c) 79.61 Hz Easy 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÷.1×2×3.14×.02 = 79.61 Hz.

Discussion

10.	Calculate the useful power developed by a motor using the given data: Eb = 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 WThis question was posed to me in quiz.This key question is from Methods to Reduce the Energy Loss During Starting topic in section Starting of Electric Drives
Answer» Correct option is (a) 203 W The EXPLANATION is: Useful power DEVELOPED by the motor can be calculated using the formula P = EbI -(rotational LOSSES) = 452 – (5) = 203 W. If rotational losses are neglected, the power developed becomes equal to the shaft power of the motor.

Discussion

11.	Calculate the moment of inertia of the stick about its center having a mass of 1.1 kg and length of 2.9 m.(a) .66 kgm^2(b) .77 kgm^2(c) .88 kgm^2(d) .47 kgm^2The question was asked during an interview for a job.This interesting question is from Methods to Reduce the Energy Loss During Starting in portion Starting of Electric Drives
Answer» Correct option is (B) .77 kgm^2 The best I can explain: The moment of inertia of the STICK about its CENTER can be calculated using the formula I=ML^2÷12. The mass of the stick about its center and LENGTH is given. I=(1.1)×.0833×(2.9)^2=.77 kgm^2. It depends upon the orientation of the rotational axis.

Discussion

12.	Calculate the moment of inertia of the stick about its end having a mass of 22 kg and length of 22 cm.(a) .088 kgm^2(b) .087 kgm^2(c) .089 kgm^2(d) .086 kgm^2This question was posed to me in final exam.Question is taken from Methods to Reduce the Energy Loss During Starting topic in portion Starting of Electric Drives
Answer» Correct answer is (b) .087 kgm^2 The explanation: The moment of inertia of the stick about its end can be CALCULATED USING the formula I=ML^2÷3. The mass of the stick about its end and length is given. I = (22)×.33×(.11)^2 =.087 kgm^2. It depends upon the ORIENTATION of the rotational axis.

Discussion

13.	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÷888I got this question by my college director while I was bunking the class.My doubt stems from Methods to Reduce the Energy Loss During Starting in portion Starting of Electric Drives
Answer» RIGHT answer is (C) -π÷89 To ELABORATE: 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.

Discussion

14.	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 VI had been asked this question in homework.The query is from Methods to Reduce the Energy Loss During Starting topic in chapter Starting of Electric Drives
Answer» Correct option is (c) 78 V Best 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 lap winding, the number of parallel paths is equal to a number of poles. Eb = .4×22×75×2×78÷60×22 = 78 V.

Discussion

15.	Calculate the angular frequency of the waveform y(t)=69sin(40πt+4π).(a) 40π Hz(b) 60π Hz(c) 70π Hz(d) 20π HzI got this question in final exam.I would like to ask this question from Methods to Reduce the Energy Loss During Starting in chapter Starting of Electric Drives
Answer» The correct option is (a) 40π Hz Easy explanation: The fundamental time period of the SINE wave is 2π. The 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. Ω can be directly calculated by COMPARING the equations. Ω = 40π Hz.

Discussion

16.	Calculate mark to space ratio if the system is on for 9 sec and time period is 11 sec.(a) 4.6(b) 4.8(c) 4.5(d) 4.9This question was posed to me during an online interview.Question is taken from Energy Relations During Starting topic in section Starting of Electric Drives
Answer» The correct answer is (c) 4.5 Easiest 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.

Discussion

17.	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 KWI got this question in unit test.The origin of the question is Energy Relations During Starting topic in division Starting of Electric Drives
Answer» Correct answer is (c) 1.01 KW 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=I^2R=9.2×9.2×12=1.01 kW.

Discussion

18.	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 WI got this question during an interview for a job.This question is from Energy Relations During Starting in portion Starting of Electric Drives
Answer» Right choice is (a) 3.88 W 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^o. P=I^2R=1.8×1.8×1.2=3.88 W.

Discussion

19.	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 revolution per minute(b) 400 revolution per minute(c) 644 revolution per minute(d) 534 revolution per minuteThis question was posed to me during an internship interview.The above asked question is from Energy Relations During Starting in chapter Starting of Electric Drives
Answer» Correct choice is (a) 388 revolution per minute Best 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.

Discussion

20.	Calculate the active power in a 157.1545 H inductor.(a) 4577 W(b) 4567 W(c) 4897 W(d) 0 WThe question was asked by my school principal while I was bunking the class.Query is from Energy Relations During Starting in division Starting of Electric Drives
Answer» The correct choice is (d) 0 W The 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.

Discussion

21.	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 VI had been asked this question in a job interview.I would like to ask this question from Energy Relations During Starting topic in chapter Starting of Electric Drives
Answer» The correct answer is (d) 8.79 V To elaborate: 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. Eb = 4×2×3000×2÷60×91 = 8.79 V.

Discussion

22.	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 ΩThe question was posed to me in final exam.My doubt is from Energy Relations During Starting topic in division Starting of Electric Drives
Answer» The correct ANSWER is (a) .277 Ω To ELABORATE: 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.

Discussion

23.	Calculate the time period of the waveform y(t)=7cos(54πt+2π÷4).(a) .055 sec(b) .037 sec(c) .023 sec(d) .017 secI had been asked this question in an online quiz.The above asked question is from Energy Relations During Starting topic in chapter Starting of Electric Drives
Answer» CORRECT answer is (B) .037 sec The explanation is: 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.

Discussion

24.	Calculate the useful power developed by a motor using the given data: Pin= 1500 W, Ia= 6 A, Ra=.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 WThe question was asked by my college professor while I was bunking the class.This key question is from Energy Relations During Starting topic in division Starting of Electric Drives
Answer» The correct choice is (C) 1417.8 W Easiest explanation: USEFUL power is basically the shaft power DEVELOPED by the motor that can be calculated USING the formula Psh = Pdev-(rotational LOSSES). Pdev = Pin-Ia^2Ra = 1500-6^2(.2)=1492.8 W. The useful power developed by the motor is Psh = Pdev-(rotational losses) = 1492.8 –(50+25) = 1417.8 W.

Discussion

25.	Calculate the moment of inertia of the thin spherical shell having a mass of 3.3 kg and diameter of .6 cm.(a) .00125 kgm^2(b) .00196 kgm^2(c) .00145 kgm^2(d) .00178 kgm^2I got this question by my school principal while I was bunking the class.I need to ask this question from Energy Relations During Starting in section Starting of Electric Drives
Answer» Correct choice is (b) .00196 kgm^2 To 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=(3.3)×.66×(.03)^2=.00196 kgm^2. It depends UPON the orientation of the ROTATIONAL axis.

Discussion

26.	Calculate the moment of inertia of the apple having a mass of .4 kg and diameter of 12 cm.(a) .0008 kgm^2(b) .0007 kgm^2(c) .0009 kgm^2(d) .0001 kgm^2I have been asked this question at a job interview.The doubt is from Energy Relations During Starting in portion Starting of Electric Drives
Answer» The correct option is (b) .0007 kgm^2 For explanation I would SAY: The moment of inertia of the apple can be calculated using the formula I=mr^2×.5. The MASS of the apple and diameter is given. I=(.4)×.5×(.06)^2 = .0007 kgm^2. It depends upon the orientation of the rotational axis.

Discussion

27.	Calculate the value of the angular acceleration of the motor using the given data: J = 36 kg-m^2, load torque = 66 N-m, motor torque = 26 N-m.(a) 1.11 rad/s^2(b) 2.22 rad/s^2(c) 3.33 rad/s^2(d) 4.44 rad/s^2This question was posed to me in an online interview.The query is from Energy Relations During Starting topic in portion Starting of Electric Drives
Answer» Correct choice is (a) 1.11 rad/s^2 The best I can EXPLAIN: Using the dynamic equation of motor J×(ANGULAR ACCELERATION) = Motor torque – Load torque: 36×(angular acceleration) = 66-26 = 40, angular acceleration = 1.11 rad/s^2.

Discussion

28.	Calculate the power developed by a motor using the given data: Eb = 55 V and I = 6 A.(a) 440 W(b) 220 W(c) 330 W(d) 550 WThe question was asked in an international level competition.I'm obligated to ask this question of Energy Relations During Starting topic in section Starting of Electric Drives
Answer» Right choice is (c) 330 W The explanation is: Power developed by the motor can be CALCULATED using the formula P = Eb×I = 55×6 = 330 W. If rotational LOSSES are neglected, the power developed BECOMES equal to the shaft power of the motor.

Discussion

29.	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-mI have been asked this question in a job interview.My question is based upon Energy Relations During Starting topic in portion Starting of Electric Drives
Answer» The correct CHOICE is (a) 453.51 N-m To explain: Back emf developed in the motor during the FULL load can be calculated using EQUATION EB = Vt-I×Ra = 431.53V and machine constant Km = Eb÷Wm which is equal to 5.88. Torque can be calculated by using the relation T = Km × I = 5.88×77 = 453.51 N-m.

Discussion

30.	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/sI had been asked this question in a national level competition.This interesting question is from Starting in section Starting of Electric Drives
Answer» Right choice is (d) 10 m/s 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.

Discussion

31.	Calculate the value of the frequency if the signal completes half of the cycle in 30 sec. Assume signal is periodic.(a) 0.028 Hz(b) 0.016 Hz(c) 0.054 Hz(d) 0.045 HzI had been asked this question in quiz.My doubt is from Starting in division Starting of Electric Drives
Answer» Right option is (B) 0.016 Hz To explain: 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.

Discussion

32.	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) .578This question was addressed to me in an interview for job.This interesting question is from Starting in portion Starting of Electric Drives
Answer» The correct ANSWER is (a) .979 Easy EXPLANATION: DUTY cycle is Ton÷Ttotal. 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÷Ttotal = 48÷49 = .979.

Discussion

33.	Calculate the reactive power in a 23 Ω resistor.(a) 45 VAR(b) 10 VAR(c) 245 VAR(d) 0 VARThis question was addressed to me during a job interview.This interesting question is from Starting topic in section Starting of Electric Drives
Answer» The correct answer is (d) 0 VAR Easy 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.

Discussion

34.	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 ΩThe question was asked in an online quiz.This question is from Starting in division Starting of Electric Drives
Answer» Correct ANSWER is (b) .436 Ω To EXPLAIN: 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.

Discussion

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 ΩThis question was addressed to me during an interview.My query is from Starting in section Starting of Electric Drives
Answer» The correct choice is (C) INFINITE Ω To explain I would say: 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.

Discussion

36.	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 secThis question was posed to me in class test.This intriguing question originated from Starting topic in portion Starting of Electric Drives
Answer» RIGHT answer is (a) 1 sec To EXPLAIN: 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.

Discussion

37.	Calculate the moment of inertia of the thin spherical shell having a mass of 3 kg and diameter of 66 cm.(a) .2156 kgm^2(b) .2147 kgm^2(c) .2138 kgm^2(d) .2148 kgm^2This question was posed to me during a job interview.I would like to ask this question from Starting in chapter Starting of Electric Drives
Answer» The correct answer is (a) .2156 kgm^2 For explanation: 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×(.33)^2=.2156 kgm^2. It depends UPON the ORIENTATION of the rotational axis.

Discussion

38.	Calculate the mass of the solid sphere having a moment of inertia 17 kgm^2 and radius of 4 cm.(a) 10624 kg(b) 10625 kg(c) 10628 kg(d) 10626 kgThe question was posed to me during an internship interview.Asked question is from Starting in division Starting of Electric Drives
Answer» Correct option is (b) 10625 KG To explain: The moment of inertia of the ball can be calculated using the formula I=Σmiri^2. The moment of inertia of ball and radius is given. M=(17)÷(.04)^2 = 10625 kg. It DEPENDS upon the orientation of the rotational axis.

Discussion

39.	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) 39.9The question was asked in quiz.Enquiry is from Starting topic in portion Starting of Electric Drives
Answer» The correct option is (c) 38.45 Explanation: Given a number of poles = 14. SUPPLY FREQUENCY is 50 Hz. ROTOR speed is 699 RPM. Ns=120×f÷P=120×50÷14 = 428.57 rpm. S=Ns-Nr÷Ns=428.57-99÷428.57=.769. F2=sf=.769×50=38.45 Hz.

Discussion

40.	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÷4This question was posed to me in my homework.The origin of the question is Starting topic in portion Starting of Electric Drives
Answer» Right OPTION is (c) 25π÷1 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.

Discussion

41.	Calculate the value of the angular acceleration of the motor using the given data: J = .01 kg-m^2, load torque= 790 N-m, motor torque= 169 N-m.(a) -62100 rad/s^2(b) -62456 rad/s^2(c) -34056 rad/s^2(d) -44780 rad/s^2The question was asked in an interview for internship.This is a very interesting question from Starting topic in section Starting of Electric Drives
Answer»

Discussion

42.	Calculate the active power in a .56 F capacitor.(a) 37.8 W(b) 0 W(c) 15.4 W(d) 124.5 WThe question was asked by my school teacher while I was bunking the class.Question is from Methods of Starting Electric Motors topic in division Starting of Electric Drives
Answer» Correct option is (b) 0 W To explain: 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.

Discussion

43.	The unit of angular frequency is Hz.(a) True(b) FalseThe question was asked in an international level competition.The origin of the question is Methods of Starting Electric Motors in portion Starting of Electric Drives
Answer» Correct CHOICE is (a) True The best I can explain: 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.

Discussion

44.	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 minuteThe question was posed to me in an interview for internship.I'm obligated to ask this question of Effect of Starting on Power Supply, Motor and Load in division Starting of Electric Drives
Answer» Correct choice is (c) 460 REVOLUTION per minute 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.

Discussion

45.	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 methodThe question was posed to me in unit test.Origin of the question is Effect of Starting on Power Supply, Motor and Load topic in chapter Starting of Electric Drives
Answer» Correct choice is (c) EMF method 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.

Discussion

46.	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 WI had been asked this question in semester exam.The question is from Methods of Starting Electric Motors topic in chapter Starting of Electric Drives
Answer» RIGHT option is (d) 3.2 W The BEST I can 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×.8=3.2 MW.

Discussion

47.	Calculate the active power in a .7889 H inductor.(a) .123 W(b) .155 W(c) 0 W(d) .487 WI got this question in homework.This question is from Methods of Starting Electric Motors topic in chapter Starting of Electric Drives
Answer»

Discussion

48.	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 revolution per minute(b) 80 revolution per minute(c) 90 revolution per minute(d) 70 revolution per minuteThe question was asked during an interview for a job.My question is taken from Methods of Starting Electric Motors topic in section Starting of Electric Drives
Answer» The CORRECT answer is (c) 90 revolution per minute Easiest explanation: Supply frequency=50 HZ. No-load SPEED of motor= 140 rpm. The full load speed of motor=50 rpm. SINCE the no-load speed of the motor is almost 140 rpm, hence synchronous speed near to 140 rpm. Speed of rotor FIELD=140 rpm. Speed of rotor field with respect to rotor = 140-50 = 90 rpm.

Discussion

49.	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-mThis question was addressed to me during a job interview.The origin of the question is Methods of Starting Electric Motors topic in chapter Starting of Electric Drives
Answer» The correct answer is (a) 1 N-m To EXPLAIN I WOULD say: 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).

Discussion

50.	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 ΩThe question was asked in quiz.Origin of the question is Methods of Starting Electric Motors topic in division Starting of Electric Drives
Answer» The CORRECT OPTION is (b) .301 Ω Easy 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 of resistance.

Discussion

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