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# Multiple choice question for engineering

## Set 1

1. The circuit depicting the equation V=Ea + I*Ra.
a)
b)
c)
d)

Answer: a [Reason:] This is a motor performance equation.

2. The circuit depicting the equation V=Ea – I*Ra.
a)
b)
c)
d) None of the mentioned

Answer: a [Reason:] This is a generator performance equation.

3. If the terminal voltage of 220-V dc generator having armature resistance of 1 ohms. The induced emf produced is 200-V. The armature current for the above machine is
a) 20 A
b) -20 A
c) -10 A
d) 10 A

Answer: a [Reason:] I = V-E/r = 220-200/1 =20 A.

4. The voltage drop at brush-commutator contact is variable (1-2V) and dependent of armature current.
a) True
b) False

Answer: b [Reason:] The voltage drop at brush-commutator contact is variable (1-2V) and independent of armature current.

5. The shaft power at the DC Generator is
a) sum of mechanical power and rotational losses
b) only mechanical losses
c) no-load losses and mechanical power
d) any of the mentioned

Answer: a [Reason:] The shaft power is sum of mechanical power and rotational losses.

6. If the electromagnetic torque in a DC shunt-generator is opposite, what can be further concluded?
a) Mechanical power is absorbed by the machine
b) Mechanical power is delivered by the machine
c) Electromagnetic torque is in same direction of prime mover
d) None of the mentioned

Answer: a [Reason:] When the electromagnetic torque is in opposite direction, it is of motoring nature.

7. The conductor EMF and current are in _____ direction and developed torque is in _____ for generating mode.
a) same, opposite
b) same,same
c) opposite, same
d) opposite, opposite

Answer: a [Reason:] The conductor emf and current will be in same direction and the developed torque is in opposite direction for a generator.

8. If the armature terminal voltage is more that its induced EMF, the DC machine given is
a) motoring mode
b) generating mode
c) regenerative mode
d) none of the mentioned

answer: a [Reason:] As the terminal voltage is lesser than armature voltage, the supply is fed to the machine and so it will be acting like a motor.

9. Consider a 200V, 25kW, 30A DC machine lap connected with armature resistance of 0.4 ohms. If the machine is later wave wound, then the developed power is
a) 25 kW
b) 12.5 kW
c) 20 kW
d) 50 kW

Answer: a [Reason:] The power of the machine remains unaltered by the type of connections.

10. If the DC machine is held constant at 3000 rpm. The DC voltage is 250V. If the field is held constant with 250V. Is this machine generator or motor?
a) Motor
b) Generator
c) None of the mentioned
d) Any of the mentioned

Answer: a [Reason:] From the speed and emf relation, E = 250*2950/3000 = 245.8 V This is less than the terminal voltage. Hence it is a motor.

11. A shunt generator has an induced voltage on open circuit of 127 V. When the machine is on load the terminal voltage is 120 V. The load current if the field resistance be 15 ohm and armature resistance be 15 ohm
a) 342 A
b) 350 A
c) 358 A
d) 8 A

Answer: a [Reason:] Ia*Ra = E-V = 127 – 120 = 7 V Ia = 350 A Ish = 120/15 = 8 A I = 350 – 8 = 342 A

## Set 2

1. The armature mmf affects
a) commutation
b) generated voltage
c) torque
d) any of the mentioned

Answer: a [Reason:] Armature reaction thus armature mmf affects all of the mentioned above.

2. The space distribution of flux produced in a DC machine is
a)
b)
c)
d) None of the mentioned

Answer: d [Reason:] Distorted triangular waveform shape is that of flux distribution under poles, which is due to the armature reaction.

3. Armature reaction in a dc machine is
a) cross magnetizing in nature
b) magnetizing in nature
c) demagnetizing in nature
d) none of the mentioned

Answer: a [Reason:] It is cross magnetizing in nature which happens due to interaction between stator and field mmf.

4. The air-gap flux density waveform, has decreased flux under one pole tip and measured under the other is due to
a) cross magnetization
b) magnetization
c) demagnetization
d) saturation of iron magnetic circuit

Answer: a [Reason:] The flux density increases under one pole tip while it decreases under the other due to the saturation of iron magnetic circuit.

5. The flux distortion caused by cross magnetizing armature reaction is more pronounced in a
a) dc shunt motor
b) dc series motor
c) any of the mentioned
d) differential Compound Motors

Answer: a [Reason:] Since field excitation remains substantially constant, while aarmature mmf reaches higher values at loading.

6. The method needed to limit cross-magnetizing effect is
a) by increasing reluctance of cross flux path
b) by chamfering the pole faces
c) by compensating winding
d) any of the mentioned

Answer: a [Reason:] By ultimately increasing the reluctance of the magnetic circuit at the pole tips so that the armature reaction effect is neutralized.

7. The brushes of a dc motor, are shifted by 5′, from the main field axis, then
a) demagnetizing or magnetizing effect may occur depending on machine
b) cross-magnetization will be pronounced
c) both cross-magnetization as well as magnetization and demagnetization will occur
d) none of the mentioned

Answer: a [Reason:] When the brushes are shifted from GNA, there will either be magnetizing or demagnetizing effect will be observed.

8. A 100KW, 250 V, 400 A, a long shunt compound generator has an armature resistance of 0.025 ohms. There are 1000 shunt fields turns per pole and 3 series field turns per pole. The series field is connected in a such a fashion that positive armature current produces direct-axis MMF which adds to that of the shunt field. Compute the gross MMF at the rated terminal current when shunt field current is 4.7A and speed is 1150 rpm.
a) 5.9AT
b) 3.5AT
c) 4.7AT
d) 1.2AT

Answer: a [Reason:] Series field current = Is = I(l)+I(f) = 400+4.7 = 405 A(approx) Main field mmf = I(f)+(Ns/Nf)*I(s) = 4.7+(3/1000)*405 = 5.9 AT

9. Compensating windings are embedded in pole faces and having the same polarity as that of adjoining armature winding.
a) True
b) False

Answer: b [Reason:] Compensating windings are embedded in pole faces and having an opposite polarity as that of adjoining armature winding.

10. Compensating winding has advantage of
a) eliminating flux distortion
b) importing speed of response
d) any of the mentioned

Answer: a [Reason:] Any of the corrective measures will be applicable.

11. The correct positioning of the windings in the below diagrams is
a)
b)
c)
d) Any of the mentioned

Answer: a [Reason:] Commutating and compensating fields act along armature axis. Shunt and series fields act along main field axis.

12. A dc shunt motor is connected to the source through 3-point starter. If the field id kept open and starter handle is moved from off to on position, then
a) motor will not start
b) armature will draw large current from source
c) no sparking would occur
d) all of the mentioned

Answer: d [Reason:] Leaving the field winding open, the machine will not start as the exciting flux is zero.

13. A dc shunt motor is running at 1000 rpm at the rated load torque. If we reverse the filed winding terminals, then
a) direction of rotation reverses and commutation will be unaffected
b) direction of rotation will be same and commutation will be unaffected
c) direction of ration reverses and commutation will be affected
d) direction of ration will be same and commutation will be affected

Answer: a [Reason:] Reversing the field will change the direction of rotation but will not affect the commutation.

14. A dc shunt motor is running at 1000 rpm at the rated load torque. If we reverse the supply terminals, then
a) direction of rotation reverses and commutation will be unaffected
b) direction of rotation will be same and commutation will be unaffected
c) direction of ration reverses and commutation will be affected
d) direction of ration will be same and commutation will be affected

Answer: b [Reason:] Reversing the supply in dc machine will not change the direction of rotation and will not affect the commutation too.

15. A dc shunt motor is running at 1000 rpm at the rated load torque. If few of the field windings get shot circuited then
a) motor speed will increase and more armature current is drawn
b) motor speed will decrease and more armature current is drawn
c) motor speed will increase and lesser armature current is drawn
d) motor speed will decrease and lesser armature current is drawn

Answer: a [Reason:] If field winding get short circuited, field excitation reduces causing rise in speed and more armature current.

## Set 3

1. Choose the correct representation, which depict a non-zero torque in the armature.
a)
b)
c)
d) any of the mentioned

Answer: a [Reason:] Torque is produced when the net force is zero and there is a couple generated. The directions must obey Lorentz’s law and vector product.

2. What is the principle of torque production in a dc machine?
a) Lorentz’s law
b) Lenz’s law
d) Self inductance

Answer: a [Reason:] Though all the mentioned laws contribute towards the torque generation, but primarily its the Lorentz’s forces which initiate and define the torque direction.

3. The torque induced beyond the pole shoes in the DC machine is
a) 0
b) 2/pi *phi *i
c) 4/pi *phi *i
d) none of the mentioned

Answer: a [Reason:] The magnetic field vanishes at the pole terminals and thus the cross product with the magnetic field yields zero.

4. The simple rotating loop between pole faces connected to a battery and resistor through a switch, the specifications of this machine are radius = 0.5m, length 1m, resistance = 0.3 ohms and magnitude strength = 0.25T is supplied with 120V. Suddenly the switch is closed at t=0, what is observed in the circuit?
a) Current will flow but zero induced EMF
b) Current will not flow and zero induced EMF
c) Current will not flow but EMF is induced
d) Current will flow and EMF will also be induced

Answer: a [Reason:] Initially the induced emf in the armature is zero. But the current will flow as there is terminal source voltage.

5. The simple rotating loop between pole faces connected to a battery and resistor through a switch, the specifications of this machine are radius = 0.5m, length 1m, resistance = 0.3 ohms and magnitude strength = 0.25T is supplied with 120V. What will be the magnitude of the following current at t=0+?
a) 400A
b) 200A
c) 0
d) any of the mentioned

Answer: a [Reason:] I=(V-E)/r =(120-0)/0.3 =400 A.

6. If the torque induced is zero in the dc machine, it can be said that
a) current is zero
b) flux can be zero
c) current or flux=0
d) any of the mentioned

Answer: a [Reason:] The torque at steady state of operation is zero as the emf induced is equal to the source voltage. Also the the very much starting the current will be zero which will not let the emf to get induced.

7. The simple rotating loop between pole faces connected to a battery and resistor through a switch, the specifications of this machine are radius = 0.5m, length 1m, resistance = 0.3 ohms and magnitude strength = 0.25T is supplied with 120V. What is the steady state angular velocity at no-load?

8. If the torque is in the direction of rotation, the DC machine acts as
a) generator
b) motor
c) amplidyne
d) any of the mentioned

Answer: a [Reason:] The electromagnetic torque generated in the dc generator is in the direction in which the prime mover is rotated.It is produced due to the action of armature current and stator field.

9. The cogging torque is absent in the permanent magnet dc machine is due to
a) non-magnetic nature of rotor
b) magnetic nature of rotor
c) absence of ac supply
d) any of the mentioned

Answer: a [Reason:] It is due to the fact that no field poles are induced in PMDC to create magnetic flux leading to harmonics.

10. DC Motor torque depends on
a) geometry
b) magnetic properties
c) any of the mentioned
d) both geometry and magnetic properties of the structure

Answer: d [Reason:] The torque will indeed depend on the geometry as well the magnetic materials used in the construction of the dc machine. Though this remains constant with the machine.

11. For a universal motor, the rated supply current is 10A giving a rated torque of 100 N-m. If a current of 5A is applied, what will be generated torque vary to?
a) 250 N-m
b) 400 N-m
c) 500 N-m
d) 1000 N-m

Answer: b [Reason:] The torque is proportional to the square of the current. So the new torque will be 100*(2^2)= 400 N-m.

12. Pulsating torque frequency in universal motor
a) 2f
b) f
c) f/2
d) 4f

Answer: a [Reason:] The torque reverses its direction in the negative cycle of the ac supply and thus it always unidirectional.

13. The variation of current in the armature can be best expressed by which of the below curves?
a)
b)
c)
d) An impulse signal

Answer: a [Reason:] The emf induced is initially zero which causes a high current in the begining and slowly as the emf is build up, the current reduces to zero at steady state.

14. The velocity of the dc machine at the final state is
a) E/Bl
b) Bl/E
c) EBl
d) None of the mentioned

## Set 4

1. A starting resistance is inserted at the starting in an induction motor as well as dc motor.
a) Induction motor has to control starting torque whereas in dc motor, it is done to avoid large current
b) To limit starting current in both the machines
c) To limit starting speed
d) All of the mentioned

Answer: a [Reason:] For both the machines the purpose is different.

2. Considering a human handed control system for the dc motor speed control, if the resistance wire cut out too slowly, then
a) starting resistance would burn
b) field winding would burn
c) speed will rise steeply
d) any of the mentioned

Answer: a [Reason:] Due to the slow cut off the resistance the current will be high for a large amount of time and it will cause the machine to burn.

3. A 100 hp, 250 V, 350 A shunt dc motor with an armature resistance of 0.05 ohms. To limit maximum starting current to twice the rated of its value, what will be the number of stages of starting resistances?
a) 3
b) 2
c) 4
d) 5

Answer: a [Reason:] n= log(Ra/Rt)/log(Imin/Imax) ; Rt = Vt/Imax= 250/700 = 0.357 ohms = log(0.05/0.357)/log(350/700) = 2.84 = 3 (approx).

4. Which of the following express the starting current nature of the dc motor.
a)
b)
c)
d)

Answer: a [Reason:] This is due to the resistance steps used in the speed control mechanism.

5. Following are the observations due to large starting current in dc motor.
(i) Sparking at brushes
(ii) Mechanical shock at shaft
(iii) Source current fluctuations
(iv) Burning of the field winding
a) (i), (ii), (iii), (iv)
b) (i), (iii)
c) (ii), (iii), (iv)
d) (i), (ii)

Answer: a [Reason:] All the observations can be seen due to large starting current in dc motor.

6. The shunt motor starters that can be used is/are
a) 3-point and 4-point starter
b) 5-point starter
c) 4-point starter
d) 5-point and 3-point starter

Answer: a [Reason:] Both 3-point and 4-point starters can be used.

7. How does the speed build up takes place in a dc motor with time?
a)
b)
c)
d)

Answer: a [Reason:] Due to subsequent addition of the step resistances in the dc motor, the speed build also occurs like steps.

8. The direct-on-line starter is used to start a small dc motor because it limits initial current drawn by armature circuit.
a) True
b) False

Answer: b [Reason:] The direct-on-line starter can not limit the speed by limiting the current.

9. Thyristor controlled starter is preferred over DOL starter due to
a) lesser losses
b) controlled direction
c) least resistance offered
d) all of the mentioned

Answer: a [Reason:] A thyristor controlled method is more efficient as it has lesser resistance and losses while operating.

10. For a 7.46 kW, 200 V dc shunt motor with full load efficiency of 85% has armature resistance of 0.25 ohms. Calculate the value of starting resistance in ohms for a current 1.5 times of the full load current.
a) 2.788
b) 3.038
c) 2.688
d) 2.588

Answer: a [Reason:] Full load current = 7460/(200*0.85) = 43.88 A Starting current = 1.5*43.88 = 65.883 A R = V/I = 200/65.883 = 3.038 ohms Starting resistance = 3.038 – 0.25 = 2.788 ohms.

11. The effect of fringing increases as we
a) increase air gap
b) decrease air gap
c) increase in flux density
d) introduce more ferric core material

Answer: a [Reason:] Fringing is the effect in which the magnetic flux lines bulge out of the the flux. and it is introduced at the air gaps.

12. The post effects of the armature reaction is
a) main field distortion
b) shift in MNA
c) reduction in main field
d) none of the mentioned

Answer: d [Reason:] There is no such post effect of armature reaction . The armature reaction phenomena affects the working condition only.

13. If the students give a forward shift of 10° to the dc generator, then it
a) reduces flux per pole
b) improves flux per pole
c) increases the flux density in core
d) none of the mentioned

Answer: a [Reason:] A generator given a forward brush shift will get its flux density reduced at the pole as the flux will be in the opposite direction.

14. A dc machine is run at rated speed in forward direction and then in backward direction. It is observed that, speeds of the rotation are different, then
it leads to the conclusion of
a) incorrect brush placement
b) incorrect pole and core alignment
c) incorrect field supply
d) all of the mentioned

Answer: a [Reason:] Given at the same speed of operation, if the speeds are different then the brushes are placed in not aligned manner.

15. For a dc machine , its commutator has a diameter of 50 c rotating at 1000 rpm. For a brush width of 1 cm, the time commutation taken by the machine will be
a) 0.382 ms
b) 0.456 ms
c) 0.573 ms
d) 0.312 ms

Answer: a [Reason:] Time of commutation = brush width/peripheral velocity = 1*60/(pi*50*1000) = 0.382 ms.

## Set 5

1. The short circuit load losses is/are
d) field current loss

Answer: a [Reason:] Short circuit losses comprise of direct load loss and stray losses too.

2. The open circuit voltage and open circuit core loss variation is
a)
b)
c)
d)

Answer: a [Reason:] Core losses vary as a square of the voltage.

3. Rotational losses vary as following with respect to field current.
a)
b)
c)
d)

Answer: a [Reason:] The y-intercept depicts the friction and windage losses.

4. A 100 KVA, 400V, 3-phase, star connected alternator due to following data
Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra=0.02ohms.
The voltage applied to the field winding is 220V. The short circuit load loss at half full load is
a) 258W
b) 268W
c) 480w
d) 340W

Answer: a [Reason:] SC losses at half load = 3*((Ia/2)2)*ra = 3*((131.22/2)2)*0.02 = 258.3 A.

5. A 100 KVA, 400V, 3-phase, star connected alternator due to following data
Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra=0.02ohms.
The voltage applied to the field winding is 220V. The field current loss will be
a) 270W
b) 258W
c) 480W
d) 250W

Answer: a [Reason:] Field circuit loss = 220^2/180 = 169W

6. A 100 KVA, 400V, 3-phase, star connected alternator due to following data
Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra=0.02ohms.
The voltage applied to the field winding is 220V. The efficiency of the machine at half load is
a) 96.7%
b) 95%
c) 94.2%
d) 97

Answer: a [Reason:] Total losses at half load = 340+480+258+268.9 = 1347W Efficiency = 1-(1347/40000+1347) = 96.47%.

7. A 100 KVA, 400V, 3-phase, star connected alternator due to following data
Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra=0.02ohms.
The voltage applied to the field winding is 220V. The load losses at full load is
a) 1033W
b) 500W
c) 540W
d) 940W

Answer: a [Reason:] Short circuit losses at full load = 3*(131.22)2*0.02 = 1033W.

8. A 100 KVA, 400V, 3-phase, star connected alternator due to following data
Friction and windage losses = 340W
Open circuit core loss = 480W
Rf = 180ohms, ra=0.02ohms.
The voltage applied to the field winding is 220V. Efficiency at full load is
a) 97%
b) 96.5%
c) 92%
d) 95%

Answer: a [Reason:] Total losses at full load = 340+480+1033+268.9 = 2122W Efficiency = 1 – (2122/2122+80000) = 97.4%.

9. The given variable losses are 5kW fr a 500 KVA, 11 kV, 3-phase star connected alternator having armature resistance of 4 ohms. Calculate the current at which maximum efficiency occurs
a) 20.4A
b) 10.2A
c) 40A
d) 26A

Answer: a [Reason:] I2=5000/3*4 I = 20.4 A.

10. The given variable losses are 5kW fr a 500 KVA, 11 kV, 3-phase star connected alternator having armature resistance of 4 ohms. Calculate the
full load armature current per phase
a) 26.24 A
b) 20.41 A
c) 79 A
d) 40 A

Answer: a [Reason:] I = P/1.73*V = 5000/1.73*11000 = 26.24 A.

11. A synchronous machine with its field winding on stator and polyphase armature winding on rotor. At steady state, its air gap field is
I. stationary w.r.t. stator
II. rotating at double the speed Ns w.r.t. rotor
III. rotating in direction opposite to rotor
a) I,II,III
b) I
c) II,III
d) II,I

Answer: a [Reason:] ALl the statements are correct.

12. A synchronous machine with its field winding on rotor and polyphase armature winding on stator. At steady state running condition, its air gap field is
a) rotating at synchronous speed w.r.t. stator
b) stationary w.r.t. rotor
c) rotating in the direction of the rotor rotor rotation
d) all of the mentioned

Answer: a [Reason:] As the field is on the rotor, the field will be set up w.r.t. rotor.

13. Consider a 3-phase cylindrical-rotor alternator
A. E.m.f. generated by armature reaction lags armature current by 90°.
B. Air gap voltage leads the field flux by 90°
C. Air gap voltage lags the field flux by 90°
D. Armature reaction mmf lags the field flux by (90°+ internal p.f. angle)
a) A,B,D
b) A,C,D
C) B,D
D) C,D

Answer: a [Reason:] Air gap voltage will lag the field flux by 90° to generate the emf.

14. In a 3-phase cylindrical-rotor alternator, synchronous reactance is sum of
a) mutual and leakage reactance
b) magnetizing and leakage reactance
c) magnetizing and mutual reactance
d) mutual, magnetizing and leakage reactance

Answer: a [Reason:] Xs = magnetizing reactance + leakage reactance.

15. The reactive power output of a synchronous generator is limited by
a) armature current and field current
b) field current and load angle