Electrical Machines Questions

7/25/2019 Electrical Machines Questions

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SITAMS

DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

GATE-2014

Electrical Machines

1. The single phase, 50 Hz iron core transformer in thecircuit has both the vertical arms of cross sectionalarea 20 cm2 and both the horizontal arms of crosssectional area 10 cm2. If the two windings shown werewound instead on opposite horizontal arms, the mutualinductance will

(A) double (B) remain same(C) be halved (D) become one quarter

SOl: Given single-phase iron core transformer has both the vertical arms

of cross section area 20 cm2, and both the horizontal arms of crosssection are 10 cm2So, Inductance NBA/ 1 (proportional to cross section area)When cross section became half, inductance became half.Hence (C) is correct option.

2. Figure shows the extended view of a 2-pole dc machine with 10

armature conductors. Normal brush positions are shown by A and B,placed at the interpolar axis. If the brushes are now shifted, in thedirection of rotation, to A and B as shown, the voltage waveformVA'B' will resemble


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The figure above shows coils-1 and 2, with dot markings as shown,


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having 4000 and 6000 turns respectively. Both the coils have a ratedcurrent of 25 A. Coil-1 is excited with single phase, 400 V, 50 Hzsupply. [ a ]

3.The coils are to be connected to obtain a single-phase,400/1000V, auto-transformer

to drive a load of 10 kVA. Which of the option given should be exercised to realize therequired auto-transformer ?

(A) Connect A and D; Common B(B) Connect B and D; Common C(C) Connect A and C; Common B(D) Connect A and C; Common D

Sol: fraction and windage losses = 1050 WCore losses = 1200W = 1.2 kW

So,Input power in stator = 3 400500.8 = 27.71 kW

Air gap power = 27.71 1.5 1.2= 25.01 kW

Hence (C) is correct option.

Common Data for Questions 4 and 5.

A 3-phase, 440 V, 50 Hz, 4-pole slip ring induction motor is feedfrom the rotor side through an auto-transformer and the stator isconnected to a variable resistance as shown in the figure.

The motor is coupled to a 220 V, separately excited d.c generatorfeeding power to fixed resistance of 10 . Two-wattmeter method isused to measure the input power to induction motor. The variableresistance is adjusted such the motor runs at 1410 rpm and thefollowing readings were recorded W1 = 1800 W, W2 = 200 W.


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4.The speed of rotation of stator magnetic field with respect to rotorstructure will be

(A) 90 rpm in the direction of rotation(B) 90 rpm in the opposite direction of rotation

(C) 1500 rpm in the direction of rotation(D) 1500 rpm in the opposite direction of rotation.

Sol:Given 3-, 440 V, 50 Hz, 4-Pole slip ring motorMotor is coupled to 220 V

N = 1410 rpm,W1 = 1800 W,W2 = 200 WSo,Ns = 120f / p

= (12050)/4 =1500 rpmRelative speed = 1500 1410

= 90 rpm in the direction of rotation.Hence (A) is correct option.

5.Neglecting all losses of both the machines, the dc generator poweroutput and the current through resistance (Rex) will respectively be

(A) 96 W, 3.10 A (B) 120 W, 3.46 A(C) 1504 W, 12.26 A (D) 1880 W, 13.71 A

Sol: Neglecting losses of both machines

Slip(S ) =

=

=0.06

total power input to induction motor isPin= 1800 200 = 1600 W

Output power of induction motorPout = (1 S)Pin

= (1 0.06)1600= 1504 W

Losses are neglected so dc generator input power = output power= 1504W

So,I2R = 1504

I = = 12.26 AHence (C) is correct option.

6. The fifth harmonic component of phase emf(in volts), for a threephase star connection is

(A) 0 (B) 269 (C) 281 (D) 808


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Sol:Fifth harmonic component of phase emf

SoAngle = 180/5 =36othe phase emf of fifth harmonic is zero.

Hence (A) is correct option.

7.In a transformer, zero voltage regulation at full load is [ c ]

(A) not possible(B) possible at unity power factor load(C) possible at leading power factor load(D) possible at lagging power factor load

8.The dc motor, which can provide zero speed regulation at full load without anycontroller is [ b ]

(A) series (B) shunt (C) cumulative compound (D) differential compound

9.A three-phase, three-stack, variable reluctance step motor has 20 poles on each rotorand stator stack. The step angle of this step motor is [ b ]

(A)30 (B) 60 (C) 90 (D) 180

Sol: Given 3-, 3-stackVariable reluctance step motor has 20-poles

Step angle = 360/(320) = 60Hence (B) is correct option.

10.For a single phase capacitor start induction motor which of the following statementsis valid?

[ b ](A) The capacitor is used for power factor improvement(B) The direction of rotation can be changed by reversing the main winding terminals(C) The direction of rotation cannot be changed(D) The direction of rotation can be changed by interchanging the supply terminals.

Ans: A single-phase capacitor starts induction motor. It has cage rotor and its statorhas two windings.The two windings are displaced 90oin space. The direction of rotationcan be changed by reversing the main winding terminals.Hence (B) is correct option.


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11.In a DC machine, which of the following statements is true ? [ b ](A) Compensating winding is used for neutralizing armature reaction while interpole

winding is used for producing residual flux.(B) Compensating winding is used for neutralizing armature reaction while interpole

winding is used for improving commutation.

(C) Compensating winding is used for improving commutation while interpole winding isused for neutralizing armature reaction.(D) Compensation winding is used for improving commutation while interpole winding is

used for producing residual flux.

Sol: In DC motor, compensating winding is used for neutralizing armature reactancewhile interpole winding is used for improving commutation.Interpoles generate voltage necessary to neutralize the e.m.f of self induction inthe armature coils undergoing commutation. Interpoles have a polarity opposite tothat of main pole in the direction of rotation of armature.Hence (B) is correct option.

12.A 220 V DC machine supplies 20 A at 200 V as a generator. The armatureresistance is 0.2 ohm. If the machine is now operated as a motor at same terminalvoltage and current but with the flux increased by 10%, the ratio of motor speed togenerator speed is

(A) 0.87 (B) 0.95 (C) 0.96 (D) 1.06

Sol: Given: A 230 V, DC machine, 20 A at 200 V as a generator.Ra = 0.2

The machine operated as a motor at same terminal voltage andcurrent, flux increased by 10%So for generator

Eg = V + IaRa= 200 + 200.2

Eg = 204 volt

for motor Em = V IaRa= 200 200.2

Em = 196 volt

So

=

= . ; = . =0.87



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13.A synchronous generator is feeding a zero power factor (lagging) load at ratedcurrent. The armature reaction is [ b ]

(A) magnetizing (B) demagnetizing (C) cross-magnetizing (D) ineffective

Ans:A synchronous generator is feeding a zero power factor(lagging) loadat rated current then the armature reaction is demagnetizing.Hence (B) is correct option.

14.Two transformers are to be operated in parallel such that they share load inproportion to their kVA ratings. The rating of the first transformer is 500 kVA ratings.The rating of the first transformer is 500 kVA and its pu leakage impedance is 0.05pu. If the rating of second transformer is 250 kVA, its pu leakage impedance is

(A) 0.20 (B) 0.10 (C) 0.05 (D) 0.025 [ b ]

Sol:Given the rating of first transformer is 500 kVAPer unit leakage impedance is 0.05 p.u.Rating of second transformer is 250 kVASo,Per unit impedance =actual impedance/base impedanceand,Per unit leakage impedance 1/ kVAThen

500 kVA 0.05 = 250 kVAx

x = =0.1. .

Hence (B) is correct option.

15. The speed of a 4-pole induction motor is controlled by varying the supply frequencywhile maintaining the ratio of supply voltage to supply frequency (V/f ) constant. At ratedfrequency of 50 Hz and rated voltage of 400 V its speed is 1440 rpm. Find the speed at30 Hz, if the load torque is constant [ b ](A) 882 rpm (B) 864 rpm (C) 840 rpm (D) 828 rpm

Sol:Given speed of a 4-pole induction motor is controlled by varying thesupply frequency when the ratio of supply voltage and frequency isconstant.f = 50 Hz, V = 400 V, N = 1440 rpm

So Vf = V2= 400 (30/50) = 240VIn induction motor speed is directly proportional to voltage.N2= N1 (V2/ V1) = 1440 (240/400) = 864 rpm


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Statement for Linked Answer Questions 16 and 17.A 300 kVA transformer has 95% efficiency at full load 0.8 p.f. laggingand 96% efficiency at half load, unity p.f.

16.The iron loss (Pi) and copper loss (Pc) in kW, under full load operationare(A) Pc = 4.12,Pi = 8.51 (B) Pc = 6.59,Pi = 9.21(C) Pc = 8.51,Pi = 4.12 (D) Pc = 12.72,Pi = 3.07

Sol: Given that:A 300 kVA transformerEfficiency at full load is 95% and 0.8 p.f. lagging96% efficiency at half load and unity power factorSo

For Ist condition for full load95% = ..++Second unity power factor half load

96% =.

.++So

Wcu +Wi = 12.630.25Wcu + 0.96Wi = 6.25

ThenWcu = 8.51, Wi = 4.118

Hence (C) is correct option.

17. What is the maximum efficiency (in %) at unity p.f. load ?(A) 95.1 (B) 96.2 (C) 96.4 (D) 98.1

Efficiency () =.

++So

X = .. =0.6956

% = ..+.+.. = 96.20%Hence (B) is correct option.

18 . The core flux of a practical transformer with a resistive load [ a ](A) is strictly constant with load changes(B) increases linearly with load


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(C) increases as the square root of the load(D) decreases with increased load

Sol: We know that in case of practical transformer with resistive load, thecore flux is strictly constant with variation of load.


19.If a 400 V, 50 Hz, star connected, 3-phase squirrel cage inductionmotor is operated from a 400 V, 75 Hz supply, the torque that themotor can now provide while drawing rated current from the supply [ b ](A) reduces(B) increases(C) remains the same(D) increases or reduces depending upon the rotor resistance

Sol: Motor is overloaded, and magnetic circuit is saturated. Than Torquespeed characteristics become linear at saturated region.as shown in figure

actual torque-speed characteristics is given by curve B.Hence (B) is correct option.

20.In relation to DC machines, match the following and choose thecorrect combination

List-I List-II

Performance Variables Proportional toP. Armature emf (E) 1. Flux(), speed () andarmature current (Ia)Q. Developed torque (T) 2. and onlyR. Developed power (P) 3. and Ia only

4. Ia and only5. Ia only

Codes:P Q R

(A) 3 3 1


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(B) 2 5 4(C) 3 5 4(D) 2 3 1Sol: In DC motor

E=PN (Z/A)

or E = KnSo

Armature emf E depends upon and only.and torque developed depends uponT = PZIa/2A

So, torque(T ) is depends of and Ia and developed power(P) isdepend of flux , speed and armature current Ia .Hence (D) is correct option.

21.Under no load condition, if the applied voltage to an induction motoris reduced from the rated voltage to half the rated value, [ b ]

(A) the speed decreases and the stator current increases(B) both the speed and the stator current decreases(C) the speed and the stator current remain practically constant(D) there is negligible change in the speed but the stator currentDecreases

Sol:Given a three-phase cage induction motor is started by direct on lineswitching at rated voltage. The starting current drawn is 6 time the

full load current.Full load slip = 4%)So

=

= (6)20.04 = 1.44Hence (B) is correct option.

22 .Xd,Xd andXd are steady state d -axis synchronous reactance,

transient d -axis reactance and sub-transient d -axis reactance of asynchronous machine respectively. Which of the following statementsis true ? [ a ]

(A)Xd >Xd >Xd (B)Xd >Xd >Xd

(C)Xd >Xd >Xd (D)Xd >Xd >Xd

Sol:In synchronous machine it is known that

Xd >Xd >Xd


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whereXd = steady state d -axis reactance

Xd = transient d -axis reactance

Xd = sub-transient d -axis reactance


23.A dc series motor driving and electric train faces a constant powerload. It is running at rated speed and rated voltage. If the speedhas to be brought down to 0.25 p.u. the supply voltage has to beapproximately brought down to [ A ](A) 0.75 p.u (B) 0.5 p.u(C) 0.25 p.u (D) 0.125 p.u

Sol:Given that:

A dc series motor driving a constant power load running at ratedspeed and rated voltage. Its speed brought down 0.25 pu. Then

Emf equation of dc series motorE = V (Ra + Rse)Ra + Rse = R

so, E = V IR= KN

Then N= E/K

In series motor Iso, N = (VIR)/ KI

At constant power load

EI = TW = Const ...(1)T = KI = KI 2 ...(2)

If W is decreased then torque increases to maintain power constant.T I 2

W = then T = 4

So current is increased 2 time and voltage brought down to 0.5 pu.Hence (B) is correct option.

24. If a 400 V, 50 Hz, star connected, 3-phase squirrel cage inductionmotor is operated from a 400 V, 75 Hz supply, the torque that themotor can now provide while drawing rated current from the supply [ a ]

(A) reduces(B) increases(C) remains the same


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(D) increases or reduces depending upon the rotor

Sol:Given 400 V, 50 Hz, Y-connected, 3- squirrel cage induction motoroperated from 400 V, 75 Hz supply. Than Torque is decreased.

Machine is rated at 400 V, 50 Hz

and it is operated from 400 V, 75 Hzso, speed of Motor will increase as N =120 f / PN f

and we know Torque in induction motor

= => If speed increases, torque decreases.Hence (A) is correct option.

25.A single-phase induction motor with only the main winding excitedwould exhibit the following response at synchronous speed [ d ](A) Rotor current is zero(B) Rotor current is non-zero and is at slip frequency(C) Forward and backward rotaling fields are equal(D) Forward rotating field is more than the backward rotating field

Sol: Given that:1- induction motor main winding excited then the rotating field ofmotor changes, the forward rotating field of motor is greater then theback ward rotating field.Hence (D) is correct option.

26.A 4-pole, 3-phase, double-layer winding is housed in a 36-slot statorfor an ac machine with 60c phase spread. Coil span is 7 short pitches.Number of slots in which top and bottom layers belong to differentphases is

(A) 24 (B) 18 (C) 12 (D) 0Sol: Given that:

A 4-pole, 3-, double layer winding has 36 slots stator with 60phase

spread, coil span is 7 short pitchedso,

Pole pitch =

= = 9

Slot/pole/phase = 3

so,3-slots in one phase, if it is chorded by 2 slots then


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Out of 3 2 have different phaseOut of 36 24 have different phase.

Hence (A) is correct option

27.The following table gives four set of statement as regards poles andtorque. Select the correct set corresponding to the mmf axes as shownin figure.

StatorSurface

ABC forms

StatorSurfaceCDA forms

RotorSurfaceabc forms

Rotorsurfacecda forms

Torqueis

(A) North Pole(B) South Pole

(C) North Pole

(D) South Pole

South PoleNorth Pole

South Pole

North Pole

North PoleNorth Pole

South Pole

South Pole

South PoleSouth Pole

North Pole

North Pole

ClockwiseCounterClockwise

CounterClockwiseClockwise

Sol:

Given thatFs is the peak value of stator mmf axis. Fr is the peak value ofrotor mmf axis. The rotor mmf lags stator mmf by space angle. The direction of torque acting on the rotor is clockwise or counterclockwise.When the opposite pole is produced in same half portion of statorand rotor then the rotor moves. So portion of stator is north-pole in

ABC and rotor abc is produced south pole as well as portion surfaceCDA is produced south pole and the rotor cda is produced Northpole.The torque direction of the rotor is clock wise and torque at surfaceis in counter clockwise direction.Hence (C) is correct option.

28.Group-I lists different applications and Group-II lists the motorsfor these applications. Match the application with the most suitable

motor and choose the right combination among the choices giventhereafter [ c ]

Group-I Group-IIP. Food mixer 1. Permanent magnet dc motorQ. Cassette tape recorder 2. Single-phase induction motorR. Domestic water pump 3. Universal motorS. Escalator 4. Three-phase induction motor


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5. DC series motor6. Stepper motor

Codes:

P Q R S(A) 3 6 4 5

(B) 1 3 2 4(C) 3 1 2 4(D) 3 2 1 4

Sol:In food mixer the universal motor is used and in cassette tap recorderpermanent magnet DC motor is used. The Domestic water pumpused the single and three phase induction motor and escalator usedthe three phase induction motor.Hence (C) is correct option.

29. A stand alone engine driven synchronous generator is feeding a partly

inductive load. A capacitor is now connected across the load tocompletely nullify the inductive current. For this operating condition.(A) the field current and fuel input have to be reduced(B) the field current and fuel input have to be increased(C) the field current has to be increased and fuel input leftunaltered(D) the field current has to be reduced and fuel input left unaltered

Sol:Given a engine drive synchronous generator is feeding a partlyinductive load. A capacitor is connected across the load to completelynullify the inductive current. Then the motor field current has to bereduced and fuel input left unaltered.Hence (D) is correct option.


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30.A 500 kVA, 3-phase transformer has iron losses of 300 W and full loadcopper losses of 600 W. The percentage load at which the transformeris expected to have maximum efficiency is [ b ]

(A) 50.0% (B) 70.7% (C) 141.4% (D) 200.0%

31.


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Answer:- D

32.


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33.

34.

[34] A field excitation of 20A in a certain alternator results in an armature current of 400A in

short circuit and a terminal voltage of 2000V on open circuit. The magnitude of the internalvoltage drop within the machine at a load current of 200A is [GATE2009] [ d ]

A.1V

B.10V

C.100VD.1000V

[35] Figure shows the extended view of a 2 pole dc machine with 10 armature

conductors.Normal brush positions are shown by A and B, placed at the interpolar axis. If thebrushes are now shifted, in the direction of rotation, to A' and B' as shown, the voltage waveform

VA'B'will resemble. [GATE] [ a ]


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[36]A 220V,50Hz,single-phase induction motor has the following connection diagram and winding

orientations shown.MM' is the axis of the main stator winding(M1M2) and AA' is that of the auxiliary

winding (A1A2). Directions of the winding axes indicate direction of flux when currents in the windings

are in the directions shown. Parameters of each winding are indicated. When switch S is closed, the

motor [GATE] [ c ]

A. rotates clockwise

B. rotates anticlockwise

C. does not rotate

D. rotates momentarily and comes to a halt

[37] The electromagnetic torque Te of a drive, and its connected load torque Tlare as shown below. Outof the operating points A,B, C and D, the stable ones are [GATE 2007] [ c ]


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[38] In a transformer, zero voltage regulation at full load is [GATE 2007] [ c ]

A. Not possible

B. Possible at unity Power factor load

C. Possible at leading Power factor load

D. Possible at lagging Power factor load

[39] The DC motor, which can provide zero speed regulation at full load without any controller is [GATE

2007] [ b ]

A. Series

B. Shunt

C. Cumulative Compound

D. Differential Compound

[40] A single phase 10kVA, 50 Hz transformer with 1kV primary winding draws 0.5A and 55W, at rated

voltage and frequency, on no load. A second transformer has a core with all its linear 2 times the

corresponding dimensions of the first transformer. The core material and lamination thickness are the

same in both transformers. The primary windings of both the transformers have the same number of

turns. If a rated voltage of 2kV at 50Hz is applied to the primary of the second transformer, then the noload current and power, respectively, are [GATE2012] [ b ]

a. 0.7 A,77.8W

b. 0.7A,155.6 W

c. 1A,110W


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d. 1A,220W

[41] The locked rotor current in a 3-phase, star connected 15kW, 4-pole, 230V, 50Hz induction motor at

rated conditions is 50A. Neglecting losses and magnetizing current, the approximate locked rotor line

current drawn when the motor is connected to a 236V, 57Hz supply is [GATE2012] [b ]

a. 58.5A

b. 45.0A

c. 45.7A

d. 55.6A

[42] A220V, 15kW,1000rpm shunt motor with armature resistance of 0.25, has a rated line current of

68A and a rated field current of 2.2A. The change in field flux required to obtain a speed of 1600 rpm

while drawing a line current of 52.8A and a field current of 1.8A is [GATE2012] [ d]

a. 18.18% increase

b. 18.18% decrease

c. 36.36% increase

d. 36.36% decrease

[43] In 8 - pole wave connected motor armature, the number of parallel paths are [ c ]

A. 8

B. 4

C. 2

D. 1

[44] The crawling in an induction motor is caused by

A. improper design of the machine

B. low voltage supply

C. high loads

D. harmonics developed in the motor


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Ans = D

[45] The speed of an induction motor

A. decreases too much with the increase of load

B. increase with the increase of load

C. decreases slightly with the increase of load

D. remains constant with the increase of load

Ans = C

[46] The effect of increasing the length of the air gap in an induction motor will increase

A. power factor

B. speed

C. magnetising current

D. air-gap flux

Ans = C

[47] the difference between the synchronous speed and the actual speed of an induction motor is

known as

A. Regulation

B. back lash

C. slip

D. lag

Ans = C

[48] Rotating magnetic field is produced in a....A. single - phase induction motor

B. three phase induction motor

C. dc series motor

D. ac series motor

Ans= B

[49] The stator core of the induction motor is made of

A. Laminated cast iron

B. Mild steel

C. Silicon steel stampingsD. Soft wood

Ans = C

[50] Star- delta starter of an induction motor

A. Inserts resistance in rotor circuit

B. Inserts resistance in stator circuit


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C. Applies reduced voltage to rotor

D. Applies reduced voltage to stator

Ans = D

[51] The starting torque of a 1-phase induction motor is

A. High

B. Medium

C. Low

D. Zero

Ans = D

[52] The thrust developed by a linear induction motor depends on

A. Synchronous speed

B. Rotor input

C. Number of poles

D. both A and B

Ans = D

[53] The slip in actual induction motor is generally [ b ]

(a) 1% (b) 3% to 5% (c) 10% to 12% (d) 15% to 20%

[54] In terms of slip s, the ratio of rotor copper loss to rotor output is equal to [ b ]

(a) s-1 (b) 1/(1-s) (c) (1-s2) (d) 1/(1-s)2

[55] Which motor has the highest power to the weight ratio [ a ]

(a)Universal motor (b) Induction motor (c) Capacitor motor (d)Synchronous motor

[56] Slip rings are made of [ c ]

(a) carbon (b) cast iron (c) Phosphor bronze (d) aluminium

[57] The number of slip rings on a squirrel cage induction motor is [ d ]

(a) 1 (b) 3 (c) 6 (d) none


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[58] The colour of fresh dielectric oil for a transformer is [ a ]

(a) Pale yellow (b) dark brown (c) White to gray (d) colour less

[59] Buchholz relay is generally not provided on transformers below [ d ]

(a) 1KVA (b) 5KVA (c) 50KVA (d) 500KVA

[60] A transformer working under the conditions of maximum efficiency has a total losses of 1600W.

While operating at half load, the copper loss will be [ c ]

(a)800W (b)400W (c) 200W (d)100W

[61] A properly shunted a centre zero galvanometer is connected in a rotor circuit 6 poles 50Hzinduction motor, if the galvanometer makes 90 complete oscillations in 1min. Calculate the rotor speed?

[ b ]

(a) 980rpm (b)970rpm (c) 1000rpm (d) 960rpm

Sol: Given

Fr= no. of oscillations per second; F = supply frequency = 50Hz

Fr= 90/60 =1.5Hz

Slip = Fr/ F = 1.5 / 50 = 0.03

Rotor speed (Nr) = Ns(1-s) = 1000 ( 1 - 0.03) = 970 rpm


A 3- phase, 50Hz , induction motor runs at a speed 576 Rpm at full load.

[62] How many poles does the motor have [ a ]

(a)10 (b)12 (c) 8 (d) 6

Sol: Given

Nr= 576 rpm, Ns= 120 f / p;

The Synchronous speed nearer to 576 is 600rpm, for 50Hz supply and Ns= 600 rpm, the no. of poles

should be

P = 120* f /Ns= 120* 50 / 600 = 10 poles


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[63] What is the slip and the frequency of the rotor currents at full load [ b ]

(a)0.02 , 2Hz (b)0.04, 2Hz (c) 0.06,4Hz (d)none

Sol: Slip = (Ns- Nr)/Ns

= (600-570) / 600 =0.04

Fr= S F = 0.04 * 50 =2Hz

[64] find the rotor speed w.r.t rotating field [ c ]

(a)10rpm (b)12 rpm (c) 24 rpm (d) 20 rpm

Sol: The speed of the rotating field is = 600 rpm

The speed of the rotor w.r.t stator structure is = 576 rpm

The rotor speed w.r.t rotating field is = 600 -576 = 24 rpm.

[65] What is the motor speed at twice the full load silp is [ b ]

(a)550rpm (b)552rpm (c) 524 rpm (d)5 20 rpm

Sol: Given

Rotor speed at full load Nfl1=576 rpm

Slip at full load Sfl1= 0.04

New slip is twice the full load slip Snew= 2* Sfl1 = 2 * 0.04 = 0.08

New Rotor speed is Nr new = Ns(1- Snew) = 600 (1- 0.08) = 552 rpm

[66] By what factor should the rotor resistance to increase to run the rotor at 528 rpm at full load

torque? [ b ]

(a)4 times (b)3 times (c)2 times (d) 6 times

Sol: Given Sfl = 0.04 ; Nfl = 576 rpm ;

Ns = 600 rpm;

Rext is inserted to decrease the rotor speed to N2= 528 rpm

Snew= (Ns - N2) /Ns = (600528)/ 600 = 0.12

T = SV2/ R2 => S R2


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Sfl / Snew = R2old /R2 new

R2new =( Snew/ Sfl ) R2old = 3 R2 old

[67] A Centre 0 Ammeter connected in the rotor circuit of a 6 pole, 50Hz induction motor makes 30oscillations in one minute, the rotor speed is . [ ]

(a)990rpm (b)982rpm (c) 970 rpm (d)10 20 rpm

Sol : Given P= 6; f=50Hz; Ns= 120*f /P = 120* 50 / 6 =1000rpm

In rotor circuit Ammeter is connected, it makes 30 oscillations in one minute,

Fr = 30 / 60 = = 0.5

S = Fr /F = 0.5 / 50 = 0.01

Nr = Ns(1- S) = 1000(1- 0.01) =990rpm.


A 3- phase, 4 pole 50Hz has a full load speed is 1440 rpm, for this calculate the following

[68] The speed of the Stator field w.r.t Stator Structure, . [ a ]

(a)1500rpm (b)12 00rpm (c) 2400 rpm (d) 2000 rpm

Sol:

Stator Structure is stationary, so its speed is zero.

Stator field is rotating at synchronous speed i.e., Ns= 120 * f / P = 120 * 50 /4 =1500 rpm

So, The speed of the Stator field w.r.t Stator Structure = 1500 -0 = 1500 rpm

[69].The speed of the Stator field w. r.t Rotor Structure is [ a ]

a)60rpm (b)12 0rpm (c) 240 rpm (d) 200 rpm

Sol: The speed of the rotor structure is Nr= 1440rpm

The speed of the Stator field w.r.t Rotor Structure is = 15001440 =60 rpm


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[70]. The speed of the rotor field w.r.t Stator Structure is [ a ]

(a)1500rpm (b)12 00rpm (c) 2400 rpm (d) 2000 rpm

Sol: The speed of the rotor field w.r.t Stator Structure is equal to the speed of the stator field.

So, The speed of the rotor field w.r.t Stator Structure is =1500 rpm.

[71]. The speed of the rotor field w.r.t rotor Structure is [ a ]


Sol: Given

The speed of the rotor structure is Nr= 1440rpm.

The speed of the rotor field w.r.t Stator Structure is =1500 rpm.

The speed of the rotor field w.r.t rotor Structure is =1500 -1440 rpm = 60 rpm.

[72]. The speed of the Stator field w.r.t Rotor field is .. [ d ]


Sol:



So, The speed of the rotor field w.r.t Stator field is =1500 - 1500 rpm = 0 rpm.

[73]. The magnetizing current component of a no-load current of an induction motor is much larger

than a corresponding transformer because of [ d ]

(a) additional frictional and windage losses in motor (b) different in winding, configuration on rotor

& stator. (c) increased flux requirement (d) an air gap in magnetic circuit

[74]. In an Induction motor, if the air gap is increased [ c ]

(a) Its speed will reduce (b) its efficiency will improve (c) its power factor reduces (d)its

break down torque will reduce


27/34

[75]. A 3- phase, 4-pole Squirrel cage induction motor has 36 stator slots and 28 rotor slots. The no. ofphases in rotor is [ b ]

(a)10 (b)7 (c) 8 (d) 6

Sol: No. of phases in rotor is = no. of rotor slots / no of poles

= 28 / 4 =7

[76]. The rotor of a 3-, 5KW, 400V, 50Hz slip ring induction motor is wound for 6- poles, while its stator

is wound for 4 poles. The approximate average no load steady state speed, when this motor is

connected to 400V, 50Hz supply is .. [ d ]

a)1460rpm (b)1500rpm (c) 500 rpm (d) 0 rpm

Sol: Given that slip ring IM, whose Stator poles are = 4;

Rotor poles are = 6;

Induction motor rotates only when Stator poles must be equal to rotor poles

So, in this problem Stator and Rotor poles are not equal , so speed of induction motor if 0rpm

[77]. The direction of rotation of a 3-IM is clockwise when it is supplied with 3-sinusoidal voltage

having phase sequence ABC. For the counter clockwise rotation of motor, the phase sequence of the

power supply should be [ c ]

(a)BCA (b)CAB (c) ACB (d)Either (a) or (b)

[78]. The mmf produced by the currents of a 3-induction motor .. [ b ]

(a) rotates at the speed of the rotor in air gap (b) is stand still w.r.t rotor mmf

(b) rotate at slip speeds w.r.t stator mmf (c) Rotate at synchronous speed w.r.t rotor

Sol: The mmf produced by the currents of a 3-induction motor either in Stator or Rotor are equal.

And also its speeds are equal and both rotate at synchronous speeds.

So, Stator mmf is stand still w.r.t rotor mmf.

[79] If a 400V, 50Hz, Y- connected, 3-Squirrel cage IM is operated from a 400V, 75Hz supply, the

torque that the motor can provide while drawing a rated current from the supply. [ a ]

(a) reduces (b) increases (c) remains same (d) increases or reduces depending on rotorresistance.


28/34

Sol: Given initially, Y- connected, 3-Squirrel cage IM is operating at 400V, 50Hz.

Now it is operating at 400V, 75Hz

Then Torque (T) SV2/f

T 1/f

As f - increases , T -reduces

[80]. When the applied voltage per phase is to halved, the starting torque of a 3-cage motor becomes

a .. [ b ]

(a) half of initial value (b) 1/4thof initial value (c) Twice of initial value (d)4 times of initial

value.

Sol: We Know that starting torque is directly proportional to square of the supply voltage.

Tst V2

=

; =

(.) ; =

[81] For a slip ring induction motor if the rotor resistance is increased, then [ b ]

(a)Tst & increases (b) Tst increases and decreases (c) Tst decreases and increases

(d) Tst decreases and decreases

Sol: We Know that starting torque is directly proportional to rotor resistance

Tst R2

As R2increases Tst increases, but if R2increases => cu losses increases => so efficiency decreases

[82] A 4-pole, 50Hz, 3-IM has blocked reactance per phase which is 4 times the rotor resistance per

phase. The speed at which maximum torque developed is .. [ a ]

(a)1125 rpm (b) 1500 rpm (c) 1050 rpm (d) 1210 rpm

Sol: Given that Ns= 120 * f / P = 120 * 50 /4 =1500 rpm

blocked reactance per phase which is 4 times the rotor resistance per phase

X20 = 4 R20

Slip at maximum torque STmax= R20/ X20


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=R20/ 4R20 = 0.25

The speed at which maximum torque developed is

NrTmax = Ns (1STmax)

= 1500 (10.25)

= 1125 rpm.

[83]. A 3Squirrel cage IM has a starting torque of 150% and a torque of 300% with respect to rated

torque at a rated voltage and rated frequency. Neglect the stator resistance and rotational losses. The

value of slip for maximum torque : [ d ]

(a)13.48% (b) 16.42% (c) 18.92% (d) 26.79%

Sol: Given Tst=150%, Tmax= 300% both w.r.t Tfl,E,& f

Tst = 1.5 TFL ; Tmax= 3.0 TFL

= +

. = +

1 4 = 0STmax= 26.79%

[84] In an IM what is the ratio of rotor cu losses and the rotor input [ b ]

(a) 1/s (b) s (c) 1s (d) s / (1 -s)

Sol: we know that Pri : Pcl : Pgm = 1 : s : 1-s

Where Pri = rotor input ; Pcl= rotor cu loss ; Pgm= gross mechanical power

the ratio of rotor cu losses and the rotor input ( Pcl / Pri ) = s

[85]. An IM has a rotor resistance of 0.02/phase. If the resistance is increased to 0.04/phase then

the maximum torque will [ d ]

(a) reduced to half (b) increased by 100% (c) increased by 200% (d)remains unaltered


30/34

Sol: We know

= that is Tmax is independent of rotor resistance per phase[86]. If the rotor p.f of a 3-IM is 0.866. Then special displacement between the stator

magnetic field & rotor magnetic field is . [ c ]

(a) 300 (b) 900 (c) 1200 (d) 1500

Sol:

The Spacial displacement between the stator magnetic field & rotor magnetic field is = 900+

= 900+ 300

=1200

[87]. A 400V, 50Hz, 30h.p , 3 IM is drawing 50A current at 0.8 p.f lagging. The Stator and

rotor cu losses are 1.5kW and 900W respectively. The frictional and windage losses are 1050W

and core losses are 1200W. The air gap power of the motor will be .. [ c ]

(a) 23.06kW (b) 24.11kW (c) 25.01kW (d) 26.21kW

Sol: we know Pag = Psi - PsL

Where Pag is the air gap power ; Psi = stator input power ; PsL= stator losses

= = 3 cos = = 3 400500.8=27.7

PsL= Core loss + copper loss

= 1200 + 1500 =2.7kW

Pag = Psi - PsL

Pag = 27.7kW -2.7kW = 27.0kW

[88]. No load test on a 3-IM was conducted on different supply voltages and a plot of input

power verses voltage was drawn. The curve was extrapolated to intersect the Y- axis. This

intersection point yields [ d ]


31/34

(a) core losses (b) Stator cu losses (c) Stray load losses (d) windage &

frictional losses

[89]. what is the shunt resistance component in equivalent circuit obtained by no-load test of

an induction motor representative [ a ]

(a) Core losses only (b) cu losses (c) windage & frictional losses (d) core losses, windage

& frictional losses.

[90]. An IM having full load torque of 60 N-m when delta connected develops a starting torque

of 120 N-m. For the same supply if the motor is changed to Y- connection, the starting torque

developed will be .. [ a ]

(a) 40N-m (b) 60N-m (c) 90N-m (d) 120 N-m

Sol: = 1 2 0 = 4 0 [91]. In an induction motor rotor slots are usually not quite parallel to the shaft but given aslight skew [ c ]

(a)To reduce the magnetic hum (b) To reduce the locking tendency (c)a &b (d) none

[92].The rotor power output of a three phase induction motor is 15KW and corresponding slip

is 4%, the rotor copper loss will be [ b ]

(a)600W (b) 620W (c)650W (d)700W

Sol: Pcu= Pgm* (s/1 - s) = 15 1000 (0.04/(1-0.04)) = 620W

[93]. A permanent magnet DC commutator motor has no-load speed of 6000rpm, when

connected to a 120V DC supply. The armature resistance is 2.5, and other losses may beneglected. The speed of the motor with supply voltage of 60V developing a torque 0.5N-m is

.. [ b ]

(a)3000rpm (b)2673rpm (c)2836rpm (d)5346rpm

Sol:

Given

CASE 1: No load speed = 6000 rpm

DC supply = 120 vArmature resistance = 2.5 ohms

CASE 2: DC supply = 60 v

Armature resistance = 2.5 ohmsTorque = 0.5 N-m

Speed =?

For constant flux KT = KB = K


32/34

KT = torque constant

KB= Back emf constant

EB speed wEB= KB* wT = EB * Ia/ w = KB* w *Ia/ w

= KB* Ia= KT* IaCASE 1:

No load speed = 6000 rpm ; DC supply = 120 v; Armature resistance = 2.5 ohms

K= EB/w= 120/ (2* 3.14*6000/ 60)

= 0.19

CASE 2:

DC supply = 60 v; Armature resistance = 2.5 ohms; Torque = 0.5 N-mIa= T/K = 0.5/0.19 = 2.61A

EB = VIaRa

= 602.16 * 2.5 = 53.47V = [ since 1= 2=permanent magnet ] = 53.47120 6000 = 2673

[94]. A 4-point starter is used to start and control the speed of a [ a ]

(a) DC Shunt Motor with armature resistance control

(b) DC Shunt Motor with field weakening control(c) DC Series Motor

(d) DC Compound Motor

[95]. A 3-, Salient Pole Synchronous motor is connected to an infinite bus. It is operated at no-load at normal excitation. The field excitation of the motor is first reduced to zero and then

increased in the reverse direction gradually. Then the armature current [ b ]

(a) Increases continuously (b) First increases and then decreases steeply

(c) First decreases and then increases steeply (d) remains constant

[96]. Consider the following statements:

(i) The compensating coil of a low power factor wattmeter compensates the effect of theimpedance of the current coil.

(ii) The compensating coil of a low power factor wattmeter compensates the effect of the

impedance of the voltage coil circuit. [ b ]

(a) (i) is true but (ii) is false (b) (i) is false but (ii) is true


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(c) Both (i) and (ii) are true (d) Both (i) and (ii) are false

[97]. A 220 V, DC Shunt motor is operating at a speed of 1440 rpm. The armature resistance is

1.0 and armature current is 10 A. If the excitation of the machine is reduced by 10%, the extraresistance to be put in armature circuit to maintain the same speed and torque will be [ a ]

(a) 1.79 (b) 2.1 (c) 3.1 (d) 18.9

Sol: Ia1= 10

Now flux is decreased by 10 %, so 2= 0.9 1

Torque is constant, Therefore Ia11= Ia22

Ia2=(10 / 0.9) = 11.11 A

N Eb/

N1/ N2= (Eb1/ Eb2) (2/ 1)

= (+) . = ..(+)1+ R =2.79

R = 1.79

[98]. A 3- 440 V, 6-pole, 50Hz Squirrel cage Induction motor is running at a slip of 5 %. Thespeed of stator magnetic field with respect to rotor magnetic field and speed of the rotor with

respect to stator magnetic field are.. [ c ]

(a) zero, -5 rpm (b) zero, 955 rpm

(c) 1000 rpm, -50 rpm (d)1000 rpm, 955 rpm

Sol: Ns=

= =1000 rpmRotor speed = NsSNs=1000(0.05 1000) = 950 rpm.

Stator magnetic field speed Ns= 1000 rpm.

Rotor magnetic field speed, Ns= 1000 rpm.

Relative speed between stator and rotor magnetic fields is zero.


34/34

Rotor speed with respect to Stator magnetic field = 950 -1000 =-50 rpm.

[99]. A field excitation of 20 A in a certain alternator results in an armaturecurrent of 400 A in

short circuit and a terminal voltage of 2000 V onopen circuit. The magnitude of the internal

voltage drop within the machine at a load current of 200 A is

(A) 1 V (B) 10 V (C) 100 V (D) 1000 V

Sol: Given field excitation of = 20 AArmature current = 400 AShort circuit and terminal voltage = 200 V

On open circuit, load current = 200 A

So, Internal resistance =2000 /400 = 5

Internal vol. drop = 5 200= 1000 V

Hence (D) is correct option.

[100]. For an induction motor, operation at a slips , the ration of grosspower output to air gap

power is equal to [ b ]

(A) (1 s)2 (B) (1 s) (C) (1 s)1/2 (D) (1s1/2)

Sol: copper loss = sPg

Gross power output = Pg(1-s)

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