31 + Interview Questions in Circuit Model in Dc Machines Page 1 InterviewSolution

1.	What is the armature current for a DC motor if terminal voltage is 255V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?(a) 10A(b) 100A(c) 1KA(d) 1AThis question was posed to me in my homework.Query is from Electromagnetic Power and Circuit Models in section Circuit Model, Emf and Torque of DC Machines
Answer»

Discussion

2.	When torque of the electromagnetic origin is in the direction of rotation of armature, machine is said to be operating in which of the following mode?(a) Depends on other parameters(b) Generating(c) Motoring(d) InducingI had been asked this question by my college director while I was bunking the class.This question is from Electromagnetic Power and Circuit Models in chapter Circuit Model, Emf and Torque of DC Machines
Answer» The correct choice is (c) Motoring Explanation: In a motoring MODE, torque of the electromagnetic origin is in the DIRECTION of rotation of armature, implying electrical power is absorbed and a prime mover is not NEEDED to run the machine. Ea and Ia are in the OPPOSITE directions.

Discussion

3.	What is the armature current for a DC generator if terminal voltage is observed to be 245V and open circuit voltage is equal to 250V, where armature resistance is 0.05Ω?(a) 10A(b) 100A(c) 1KA(d) 1AThis question was posed to me by my college professor while I was bunking the class.Question is taken from Electromagnetic Power and Circuit Models topic in division Circuit Model, Emf and Torque of DC Machines
Answer» Right choice is (B) 100A Explanation: OPEN circuit voltage means the voltage at armature current equal to 0, i.e. Vt=Ea= 250V. Actual Vt at loaded condition is 245V. For motoring mode, Ea – Vt= Ia*Ra. By substituting all the GIVEN values, we get Ia= 100A.

Discussion

4.	For a DC generator feeding 100kW power into 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____(a) 436.8 A(b) 434.8 a(c) 432.8 A(d) Data insufficientThe question was asked in an online interview.This is a very interesting question from Electromagnetic Power and Circuit Models topic in section Circuit Model, Emf and Torque of DC Machines
Answer» RIGHT choice is (a) 436.8 A Best EXPLANATION: Field current is equal to 230/115 = 2A, by OHM’s law. When running as a generator line current IL= 100k/230= 434.8A. SINCE power is supplied to 230V mains, Ia=Il+If = 434.8+2 =436.8A. For motoring MODE Ia= IL- If.

Discussion

5.	When torque of the electromagnetic origin is in the opposite direction of rotation of armature, machine is said to be operating in which of the following mode?(a) Depends on other parameters(b) Generating(c) Motoring(d) InducingThe question was posed to me by my school teacher while I was bunking the class.This key question is from Electromagnetic Power and Circuit Models topic in portion Circuit Model, Emf and Torque of DC Machines
Answer» RIGHT option is (b) Generating The explanation is: In a generating mode, torque of the electromagnetic origin is in the opposite direction of rotation of armature, implying mechanical power is ABSORBED and a prime MOVER is needed to run the MACHINE. Ea and Ia are in the same directions.

Discussion

6.	Simple equation of DC machine operating in generating mode, with non-zero Ra value is________(a) Vt = Ea – Ia/Ra(b) Vt = Ea + IaRa(c) Vt = – Ea + IaRa(d) Vt = Ea – Ia*RaI have been asked this question in an internship interview.This intriguing question originated from Electromagnetic Power and Circuit Models in division Circuit Model, Emf and Torque of DC Machines
Answer» The correct ANSWER is (d) Vt = Ea – IaRa Easy explanation: The machine operates in GENERATING mode (puts out electrical power) when Ia is in the direction of INDUCED emf Ea, also Ea>Vt. For the GIVEN armature CIRCUIT, with non-zero value of Ra, Vt= armature terminal voltage= Ea- IaRa.

Discussion

7.	When Ea
Answer» CORRECT choice is (c) Motoring Explanation: The machine operates in motoring mode (puts in electrical power) when Ia is in the DIRECTION opposite of induced EMF Ea. For the given armature circuit Vt = armature TERMINAL voltage= Ea + Ia * Ra; which IMPLIES Vt> Ea.

Discussion

8.	Product of torque and mechanical angular velocity ω is_____(a) Ea/ω(b) Ea*Ia(c) ω/Ea(d) Can’t tellI had been asked this question at a job interview.My doubt stems from Electromagnetic Power and Circuit Models topic in portion Circuit Model, Emf and Torque of DC Machines
Answer» Right choice is (b) EaIa To ELABORATE: According to statement of energy conversion, electrical and MECHANICAL power of the machine must BALANCE in a machine. EaIa is referred to as electromagnetic power. Thus, TORQUE = (Electromagnetic power)/ ω.

Discussion

9.	For an ideal DC machine, which phenomenon will reduce the terminal voltage?(a) Armature reaction(b) Commutation(c) Armature ohmic losses(d) All will contribute in reducing the terminal voltageI have been asked this question in quiz.Origin of the question is EMF and Torque Production topic in division Circuit Model, Emf and Torque of DC Machines
Answer» Right answer is (B) Commutation For explanation I would say: In an ideal case, Commutation does not REDUCE the TERMINAL voltage of a dc machine. In a non-ideal case, commutation takes place improperly at desired timings, thus losses contribute to the terminal voltage reduction. Armature reaction, OHMIC losses DUE to winding resistance contribute to the losses in the terminal voltage.

Discussion

10.	In a DC machine, what is the torque induced beyond the pole shoes?(a) 0(b) 2/π ∅i(c) π ∅i/2(d) Can’t be calculatedThe question was posed to me during an internship interview.I want to ask this question from EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines
Answer» Right option is (a) 0 Explanation: In a construction of a DC machine POLES are LOCATED in magnetically neutral REGION. The magnetic field at the pole terminals in a DC machine will be equal to 0. Thus, cross product with the current flowing through the armature yields zero.

Discussion

11.	If the average coil emf of a DC motor is doubled and flux is halved (keeping other parameters constant) then its shaft speed will become ___________(a) a. Twice of the original speed(b) b. Square of the original speed(c) c. Four times of the original speed(d) d. Half of the original speedThe question was asked during an online interview.My question is based upon EMF and Torque Production in section Circuit Model, Emf and Torque of DC Machines
Answer» The correct option is (c) c. Four times of the original speed The BEST explanation: Induced emf in a DC MACHINE is equal to, From the emf equation we GET speed of the shaft i.e. n α E/Z, when all other parameters are kept constant. So, when E is doubled n becomes twice the original, HALVING flux on reduced emf will QUADRUPLED the speed of a DC machine.

Discussion

12.	If the no load speed of DC motor is 1300 rpm and full load speed is 1100 rpm, then its voltage regulation is ____________(a) a. 12.56%(b) b. 18.18 %(c) c. 17.39%(d) d. 18.39%I have been asked this question in my homework.My enquiry is from EMF and Torque Production topic in section Circuit Model, Emf and Torque of DC Machines
Answer» Correct answer is (b) b. 18.18 % For explanation I would SAY: For A DC machine when all other parameters are fixed average coil emf generated is proportional directly to the speed of the dc motor. Voltage regulation is defined as ratio of difference of no load voltage and FULL load voltage to the full load voltage. VR= (1300-1100/1100)*100%.

Discussion

13.	A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of induced emf?(a) 188 V(b) 276 V(c) 94 V(d) 188 mVI have been asked this question in homework.The question is from EMF and Torque Production in chapter Circuit Model, Emf and Torque of DC Machines
Answer» The CORRECT OPTION is (a) 188 V The BEST explanation: Induced EMF in a DC machine is equal to,

Discussion

14.	A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of flux/pole?(a) 0.188 Wb(b) 18.88 Wb(c) 0.0188 Wb(d) 1.888 WbThis question was addressed to me during an interview.I want to ask this question from EMF and Torque Production topic in portion Circuit Model, Emf and Torque of DC Machines
Answer» Right option is (c) 0.0188 Wb To explain I would say: Flux is DEFINED as flux density for a GIVEN surface AREA. Here, Surface area can be calculated and multiplied with B to GIVE the value of flux. Flux= 2πrlB. now, for calculating flux per pole, DIVIDE it by P=4. So, Flux per pole after substituting all values is equal to 0.0188 Wb.

Discussion

15.	What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?(a) BavIclZr(b) BavIcl/Zr(c) BavIc*Zr/l(d) Can’t be expressedThe question was asked during an internship interview.Enquiry is from EMF and Torque Production topic in section Circuit Model, Emf and Torque of DC Machines
Answer» CORRECT choice is (a) BavIcLZr To explain I WOULD say: Avg. conductor force f= BavlIc. Here, Bav= Average flux density over pole, l= acyive conductor length. Thus, torque T= Zf = BavlIAZ. This torque is CONSTANT because both the flux density wave and current distribution is fixed in space at all times. T developed= BavIclZr(Here, r= mean air gap radius).

Discussion

16.	What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?(a) 2(b) 3(c) 2.5(d) 4The question was posed to me during a job interview.I want to ask this question from EMF and Torque Production in section Circuit Model, Emf and Torque of DC Machines
Answer» CORRECT CHOICE is (C) 2.5 The best I can explain: In an emf equation Nc= CP * Np. Here, Cp= coils/ parallel path. Np is defined as number of turns PER parallel paths which is also called as ratio of total armature conductors to the twice of number of parallel paths. Np= 10/(2*2)= 10/4= 2.5.

Discussion

17.	Emf and torque produced in a DC machine are proportional to ________ and _________ respectively.(a) Armature speed and armature emf(b) Armature emf and armature speed(c) Armature current and armature emf(d) Armature speed and armature currentI had been asked this question during an interview.This interesting question is from EMF and Torque Production topic in portion Circuit Model, Emf and Torque of DC Machines
Answer» Right answer is (d) Armature SPEED and armature CURRENT Explanation: Average coil emf GENERATED= ∅ωNP/π. Machine torque = KA∅Ia. Thus, average coil emf generated can also be represented as ka∅ω. So, average coil emf is directly PROPORTIONAL to ω (armature speed) and average torque is directly proportional to Ia (armature current).

Discussion

18.	In which mode machine is operating, given that conductor current is in the same direction of conductor emf?(a) Motoring(b) Generating(c) Can’t be determined using directions(d) In both modes for different cyclesThis question was addressed to me during a job interview.My enquiry is from EMF and Torque Production topic in chapter Circuit Model, Emf and Torque of DC Machines
Answer» Correct choice is (b) Generating The explanation is: If the conductor current is in the same direction of conductor emf then machine OUTPUTS electrical POWER and absorbs mechanical power. So, when mechanical power is absorbed machine is said to be in a generating mode. When conductor emf and conductor current are in OPPOSITE DIRECTIONS then machine is said to be in a motoring mode.

Discussion

19.	In a DC machine, average energy stored in the magnetic field remains constant independent of the armature rotation.(a) True(b) FalseThis question was addressed to me in an interview.I would like to ask this question from EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines
Answer» RIGHT CHOICE is (a) True The best I can explain: In a DC MACHINE, barring the irrecoverable losses of both electric and magnetic origin, there is balance between electrical and mechanical powers of the machine; the average energy STORED in the magnetic field remains CONSTANT irrespective of armature rotation.

Discussion

20.	Nature of the flux density wave in the air gap is__________ (for armature current equal to 0)(a) Flat topped with quarter wave symmetry(b) Point topped with quarter wave symmetry(c) Flat topped with half wave symmetry(d) Point topped with half wave symmetryThis question was addressed to me in class test.The question is from EMF and Torque Production topic in section Circuit Model, Emf and Torque of DC Machines
Answer» The CORRECT CHOICE is (a) Flat TOPPED with quarter wave symmetry To elaborate: In a DC machine magnetic structure is such that the flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is DISTURBED because of armature reaction.

Discussion

21.	Simple equation of DC machine operating in motoring mode, with non-zero Ra value is________(a) Vt = Ea – Ia/Ra(b) Vt = Ea + IaRa(c) Vt = – Ea + IaRa(d) Vt = Ea – Ia*RaThis question was addressed to me during an interview.This is a very interesting question from Electromagnetic Power and Circuit Models topic in chapter Circuit Model, Emf and Torque of DC Machines
Answer» The correct answer is (B) Vt = EA + Ia*Ra To explain I would say: The machine operates in MOTORING mode (PUTS in electrical power) when Ia is in the direction opposite of induced emf Ea, also Ea

Discussion

22.	What is the average coil emf generated in a 4-pole DC machine having flux/pole equal to 0.1 wb rotating at 1500 rpm? (No. of coil sides = 100)(a) 19 kV(b) 1.9 kV(c) 190 V(d) 19 VThis question was posed to me in an online interview.Enquiry is from EMF and Torque Production topic in chapter Circuit Model, Emf and Torque of DC Machines
Answer» The correct option is (a) 19 KV To explain: Average coil emf generated= ∅ωNP/π. E= 0.11500100*4/3.14 E= 60000/3.14 E≅ 19 Kv.

Discussion

23.	Average coil emf for 20 coil turns (E1) and 40 coil turns (E2), will have ratio E1/E2=____ (assuming all other parameters same for both machines).(a) 1/2(b) 2/1(c) 1/4(d) 4/1I had been asked this question in a national level competition.My enquiry is from EMF and Torque Production in chapter Circuit Model, Emf and Torque of DC Machines
Answer» Right OPTION is (a) 1/2 The best explanation: Emf generated in a DC machine is directly proportional to NUMBER of coil turns, Flux per POLE, number of poles and armature speed in rad/s. Thus, ratio E1/E2= 20/40 (assuming all other parameters same for both MACHINES).

Discussion

24.	When Ea>Vt machine is said to be operating in which of the following mode?(a) Depends on the Shaft speed(b) Generating(c) Motoring(d) InducingThis question was posed to me in exam.I want to ask this question from Electromagnetic Power and Circuit Models in section Circuit Model, Emf and Torque of DC Machines
Answer» Right choice is (b) Generating To explain I would SAY: The machine operates in generating mode (puts out electrical POWER) when Ia is in the direction of induced emf Ea. For the GIVEN armature CIRCUIT VT = armature terminal voltage= Ea – Ia * Ra; which implies Vt < Ea.

Discussion

25.	For a constant emf, if field current is reduced then the speed of the DC motor will_____(a) Remains same(b) Increases(c) Decreases(d) Can’t sayThe question was asked in an interview for internship.My question comes from EMF and Torque Production topic in chapter Circuit Model, Emf and Torque of DC Machines
Answer» Correct OPTION is (b) Increases To explain I WOULD say: When the FIELD current is reduced, the field produced by the field winding also reduces. Thus, the term Φ from the emf equation also decreases. For all other parameters kept CONSTANT speed of the DC machine is inversely proportional to field. Hence, speed of DC motor will increase.

Discussion

26.	Condition for linear magnetization is______(a) φ α If(b) φ α Ia(c) φ α 1/If(d) φ α 1/IaThe question was posed to me in an international level competition.I need to ask this question from Electromagnetic Power and Circuit Models in portion Circuit Model, Emf and Torque of DC Machines
Answer» Right choice is (a) φ α If Best explanation: For a LINEAR magnestization to take place, FLUX produced by the current flowing through particaluar flux producing coil must VAY in direct PROPORTION. Here, φ is produced DUE to field winding which carried current If.

Discussion

27.	Coil torque for 20 kA armature current (T1) and 40 mA armature current (T2), will have ratio T1/T2=____ (assuming all other parameters same for both machines).(a) 1/2(b) 2/1(c) 1/4(d) 4/1I had been asked this question in an interview.The above asked question is from EMF and Torque Production in section Circuit Model, Emf and Torque of DC Machines
Answer» Correct CHOICE is (a) 1/2 The BEST explanation: Torque produced in a DC machine is directly proportional to number of coil turns, Flux PER pole, number of POLES and armature current. Thus, ratio T1/T2= 20/40 (ASSUMING all other parameters same for both machines).

Discussion

28.	A 4-pole wave wound DC motor drawing an armature current of 20 A has provided with 360 armature conductors. If the flux per pole is 0.015 Wb then the torque developed by the armature of motor is _______(a) a. 10.23 N-m(b) b. 34.37 N-m(c) c. 17.17 N-m(d) d. 19.08 N-mThis question was posed to me by my college professor while I was bunking the class.My question is taken from EMF and Torque Production topic in division Circuit Model, Emf and Torque of DC Machines
Answer» The correct option is (b) b. 34.37 N-m For EXPLANATION I would say: DC Machine torque equation: T = ka∅Ia. Here, ka= ZP/(2πA), Z= total armature conductors, P= No. of poles, A= No. of parallel paths. For a wave winding A=2. So, SUBSTITUTING all the values in the torque equation we GET torque equal to 34.37 N-m.

Discussion

29.	For a DC motor taking 10kW power from 230V mains, having armature resistance and field resistance equal to 0.08Ω and 115Ω resp. The value of armature current is_____(a) 45.68 A(b) 43.48 a(c) 41.48 A(d) Data insufficientI had been asked this question by my college professor while I was bunking the class.The query is from Electromagnetic Power and Circuit Models in division Circuit Model, Emf and Torque of DC Machines
Answer» CORRECT OPTION is (c) 41.48 A Best explanation: Field current is equal to 230/115 = 2A, by Ohm’s LAW. When running as a motor line current IL= 10k/230= 43.48A. Since power is taken from 230V MAINS, Ia=IL- If = 43.48-2 =41.48A. For generating mode Ia= IL+ If.

Discussion

30.	What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?(a) 0.785* 10^-6(b) 0.785* 10^-3(c) 0.785* 10^-2(d) 0.785* 10^-4This question was posed to me in unit test.Query is from EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines
Answer» Correct option is (B) 0.785* 10^-3 The best explanation: POLE pitch is called as center to center distance between two adjacent poles. When measured in ELECTRICAL degrees one pole itch is equal to 1800. Pole pitch can be calculated as RATIO of 2πr/P. Pole pitch= 23.14 0.5* 10^-3 / 4= 0.785* 10^-3 m.

Discussion

31.	Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as__________(a) E1 = E2(b) E1 > E2(c) E1 < E2(d) Can’t be determinedI got this question in semester exam.Origin of the question is EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines
Answer» Right option is (a) E1 = E2 Best explanation: In a DC machine FLUX density wave in the air gap is flat TOPPED with quarter wave symmetry as long as armature current is equal to 0. For non-zero VALUE of armature current, this quarter wave symmetry is disturbed because of armature reaction. Emf produced is independent of B-wave shape, thus we will get same value for both cases.

Discussion

Explore topic-wise InterviewSolutions in .

When Ea

Product of torque and mechanical angular velocity ω is_____(a) Ea/ω(b) Ea*Ia(c) ω/Ea(d) Can’t tellI had been asked this question at a job interview.My doubt stems from Electromagnetic Power and Circuit Models topic in portion Circuit Model, Emf and Torque of DC Machines

In a DC machine, what is the torque induced beyond the pole shoes?(a) 0(b) 2/π ∅i(c) π ∅i/2(d) Can’t be calculatedThe question was posed to me during an internship interview.I want to ask this question from EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines

What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?(a) BavIcl*Zr(b) BavIcl/Zr(c) BavIcZr/l(d) Can’t be expressedThe question was asked during an internship interview.Enquiry is from EMF and Torque Production topic in section Circuit Model, Emf and Torque of DC Machines

What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?(a) 2(b) 3(c) 2.5(d) 4The question was posed to me during a job interview.I want to ask this question from EMF and Torque Production in section Circuit Model, Emf and Torque of DC Machines

Condition for linear magnetization is______(a) φ α If(b) φ α Ia(c) φ α 1/If(d) φ α 1/IaThe question was posed to me in an international level competition.I need to ask this question from Electromagnetic Power and Circuit Models in portion Circuit Model, Emf and Torque of DC Machines

What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?(a) 0.785* 10^-6(b) 0.785* 10^-3(c) 0.785* 10^-2(d) 0.785* 10^-4This question was posed to me in unit test.Query is from EMF and Torque Production in division Circuit Model, Emf and Torque of DC Machines