Multiple choice question for engineering

Set 1

1. The speed torque characteristics of the dc motor is best described by following the equation
w=(Vt/k*flux)-Ra*T/((R*flux)^2)
a)
b)
c)
d)

Answer

Answer: a [Reason:] The speed reduced slightly due to armature drops practically.

2. Mark the correct option which affects the terminal voltage of a dc shunt motor.
a) Armature reaction
b) Source voltage variations
c) Compensating winding
d) Any of the mentioned

Answer

Answer: c [Reason:] If the dc motor has compensating winding, flux will remain constant regardless of load.

3. For a dc shunt motor of 5 kW, running at 1000 rpm, the induced torque will be
a) 47.76 N
b) 57.76 N
c) 35.76 N
d) 37.76 N

Answer

Answer: a [Reason:] Torque = Power/(speed in rev/s)
= 5000/(2*pi*1000/60)
= 47.76 N.

4. The flux and the internally generated voltage of a dc machine is a ________ function of its magneto-motive force.
a) non-linear
b) linear
c) constant
d) inverse

Answer

Answer: a [Reason:] The flux and induced emf are non linear function of its mmf.

5. It is advised not to run dc series motor with no load. Why?
a) Because zero torque at no load will make speed infinite
b) Because zero torque as no load will not let machine start
c) Because infinite torque will be produced
d) None of the mentioned

Answer

Answer: a [Reason:] If the torque is zero in dc series motor then the speed will be infinite causing the damage of the machine permanently.

6. Identify the speed torque characteristics of a dc series motor.
a)
b)
c)
d)

Answer

Answer: a [Reason:] It is inverse in nature. At start it is infinite due to zero flux.

7. Choose the best option which identifies about the below characteristics.

a) (1)series motor (2) Cumulative compound motor (3) Shunt motor
b) (1)Cumulative compound motor (2) series motor (3) Shunt motor
c) (1)Cumulative compound motor (2) Shunt motor (3) series motor
d) (1)Shunt motor (2) series motor (3) Cumulative compound motor

Answer

Answer: a [Reason:] Shunt machine has almost constant speed while the series varies as inverse relation with torque.

8. How will the speed torque characteristic of a dc shunt motor and cumulatively compound dc motor perform at no load?
a)
b)
c)
d)

Answer

Answer: a [Reason:] The shunt motor has speed characteristic of linearly falling characteristics as the load increases the speed also falls.

9. The speed torque of the differential compound dc motor is shown below. What conclusions can be made?

a) This is an unstable machine
b) There is regenerative increment in the speed
c) This is impractical to be used
d) All of the mentioned

Answer

Answer: d [Reason:] This is basically an unstable scenario which makes the machine unable to be used.

10. It is impossible to start a differential compounded dc motor.
a) True
b) False

Answer

Answer: a [Reason:] It is so because of the unstable speed and torque characteristics. At starting, armature current and series field current are high. Since the series flux subtracts from shunt flux, series flux actually reverses the magnetic polarity.This motor will typically be still or turn slowly in wrong direction while burning up because of excessive current.

11. A student is given a differential compound motor and he has been asked to make it start. How will he try?
a) By shorting series field at start
b) To run as shunt motor at start
c) By making rated current at start
d) All of the mentioned

Answer

Answer: d [Reason:] At starting , the flux should be rated so that there is no abnormal situation arise.

12. For a 100 hp 250 V, compound dc motor with compensating winding has a field current of 5 A to produce a voltage of 250 V at 1200 rpm. What will be the shunt field current of this machine at no load?
a) 5 A
b) 5.6 A
c) 4 A
d) 0 A

Answer

Answer: a [Reason:] At no load, armature current is zero, so the internal generated voltage is 250 V. So 5 A will be the shunt field current at no load.

13. The torque characteristic is best described for a dc series motor with respect to armature current is
a)
b)
c)
d) None of the mentioned

Answer

Answer: a [Reason:] After saturation, flux becomes constant. After that we observe a linear region.

14. The torque vs armature current of a differential compound motor in the strong field is
a)
b)
c)
d)

Answer

Answer: a [Reason:] Initially it will act as dc shunt motor but as the speed increases it starts behaving like dc series motor.

15. A dc shunt motor is connected to the source through 3-point starter. Suddenly if we starter handle is moved fastly from off to on position, then
a) motor will draw large current
b) motor will not start
c) motor will burn
d) all of the mentioned

Answer

Answer: d [Reason:] As the starter is placed very rapidly from off to on the resistance seen by the machine will be very less and it will draw a large starting current.

Set 2

1. A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted.
The maximum torque equal to twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit.
The slip at the full load torque will be
a) 2.7%
b) 5%
c) 3.7%
d) 10%

Answer

Answer: a [Reason:] Tfl/Tem = 2/((sm/sf)+ (sf/sm))
1/2 = 2/((0.1/sf)+(sf/0.1))
s^2-0.4s+0.01=0
s=0.0268.

2. A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted.
The maximum torque is twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit.
The speed at the full load torque will be
a) 1460 rpm
b) 1400 rpm
c) 1360 rpm
d) 1470 rpm

Answer

Answer: a [Reason:] 1/2 = 2/((0.1/sf)+(sf/0.1))
s=2.68%
rotor speed = (1-0.0268)=1460rpm.

3. A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted.
The maximum torque is twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit.
An external resistance is now tripled in the rotor circuit. Then the slip at which maximum torque occur for the same load torque is
a) 0.3
b) 0.268
c) 0.03
d) 3

Answer

Answer: a [Reason:] The load torque is maintained constant.
So, r2/smT = X;
r2/x = 0.1
new value of slip will be, 3*r2/x=0.1
smT = 0.3.

4. A 10 kW, 400 V, 3-phase, 4-pole, 50 Hz induction motor develops the rated output at rated voltage with its slip rings shorted. The maximum torque is twice the full load torque occurs at the slip of 10% with zero external resistance in the rotor circuit. An external resistance is now tripled in the rotor circuit. Then the stator current will
a) remain constant
b) increase
c) decrease
d) any of the mentioned

Answer

Answer: a [Reason:] As the ratio of r2/s is constant, the changes will not be reflected to the stator.

5. Introducing the external resistance increases
(i) speed
(ii) output power
(iii) starting torque
a) (iii)
b) (ii),(iii)
c) (ii)
d) (i), (ii), (iii)

Answer

Answer: a [Reason:] External resistance only increases the starting torque.

6. Introducing the external resistance decreases
(i) speed
(ii) output power
(iii) starting torque
a) (iii)
b) (ii),(iii)
c) (i),(ii)
d) (i), (ii), (iii)

Answer

Answer: c [Reason:] Speed and the output power decrease with insertion of the external resistance in the circuit.

7. For a three phase induction motor, maximum torque is double the full load torque and starting torque is 1.6 times the full load torque.
To get a full load slip of 5%, percentage reduction in the rotor resistance should be
a) 62.7%
b) 75%
c) 60%
d) 35%

Answer

Answer: a [Reason:] Slip at which maxim torque occurs is calculated as smT=0.5.
r2/x2=0.5
for full load slip of 0.05, 1/2 = 2/((0.05/sf)+(sf/0.05))
smT2 = 0.186
reduction in resistance = (0.5×2-0.186×2)/0.5×2 = 63%.

8. The speed-power output characteristic of a 3-phase induction motor is
a)
b)
c)
d)

Answer

Answer: a [Reason:] At no load, rotor speed is near to synchronous speed and the operating slip is not very large, so there is small difference in operating speed and the synchronous speed.

9. The power factor variation of the 3-phase induction motor is given by
a)
b)
c)
d)

Answer

Answer: a [Reason:]

10. Four students had been conducting conducting experiments on the same induction motor for its efficiency characteristics. Which student has taken the best results?
a)
b)
c)
d)

Answer

Answer: a [Reason:] At no load the shaft output is zero so the efficiency is also zero. At lower loads, fixed losses are more comparable to variable losses. As the load is increased, efficiency also rises.

11. The stator current of the induction motor is given by which of the below?
a)
b)
c)
d)

Answer

Answer: a [Reason:] The no-load current is about 30-50% of rated current.With increase in load current rises correspondingly. It also follows a semi-circle.

12. Slip vs the torque developed in an induction motor is given as
a)
b)
c)
d)

Answer

Answer: a [Reason:] Try to check the axes.

13. The stable point in the operation of the induction motor is

a) C
b) D
c) B
d) A

Answer

Answer: a [Reason:] If the load torque shift to the Tl2, the motor speed will increase while load torque decreases.

14. At which point shall the induction machine be operated to attain a stable operation?

a) The machine will not start
b) A
c) B
d) Both A as well as B.

Answer

Answer: a [Reason:]

15. The stable point to use the induction motor possesing the below characteristic.

a) C
b) D
c) C and D
d) Machine will not start

Answer

Answer: a [Reason:] For the stable operation, (dTl/dn)>(dTe/dn).

16. Mark the stable operating point when the load torque shifts from Tl1 to Tl2.

a) A
b) A,B
c) B,C
d) A,C

Answer

Answer: a [Reason:] For the stable operation, (dTl/dn)>(dTe/dn).

17. A 3-phase, 50 Hz, 6-pole induction motor has a shaft output of 10kW at 930 rpm. Friction and windage losses is 1% of the output.Total stator losses is 600W.
The rotor input is
a) 10860 W
b) 10100 W
c) 11460 W
d) 11000 W

Answer

Answer: a [Reason:] Full load slip = 1000-930/1000 = 0.07.
Friction losses = 0.01*10000 = 100 W
Pm = 10000+100 = 10100 W
Rotor input = 10100/0.93 = 108060 W

18. A 3-phase, 50 Hz, 6-pole induction motor has a shaft output of 10kW at 930 rpm. Friction and windage losses is 1% of the output.Total stator losses is 600W.
The rotor input is
a) 10860 W
b) 10100 W
c) 11460 W
d) 11000 W

Answer

Answer: c [Reason:] Full load slip = 1000-930/1000 = 0.07.
Friction losses = 0.01*10000 = 100 W
Pm = 10000+100 = 10100 W
Rotor input = 10100/0.93 = 108060 W
Stator input = Pg+600 = 11460 W

Set 3

1. The qualities aspired to obtain a good permanent magnet is/are
a) high residual flux
b) low coercivity
c) high coercivity
d) high residual flux and high coercivity

Answer

Answer: d [Reason:] Its both high residual flux as well as coercivity is desired for a good permanent magnet.

2. The type of field winding required in PMMDC is
a) series winding
b) shunt winding
c) cumulative winding
d) None of the mentioned

Answer

Answer: d [Reason:] There is no field winding used in PMMDC to create the flux.

3. PMMDC are smaller in size due to
a) ansence of field winding
b) presence of smaller field winding
c) present of magnets
d) Any of the mentioned

Answer

Answer: a [Reason:] There is no field winding used in PMMDC to create the flux. So it is lighter.

4. PMMDC offers ______ characteristics.
a) shunt
b) series
c) armature
d) cumulative

Answer

Answer: a [Reason:] As the field remains constant due to the constant field, it has shunt characteristic.

5. Select the possible dc magnetization characteristics of permanent magnetic materials.
a)
b)
c)
d)

Answer

Answer: a [Reason:] The magnetic field has a positive value when the magnetic field intensity, H has negative value due to coercivity property. We use second quadrant of hysteresis loop.

6. Mark the speed torque characteristic of the PMDC.
a)
b)
c)
d) None of the mentioned

Answer

Answer: a [Reason:] It is linear in nature and also analogous to shunt machine.

7. PMDC had no ________ excited machine.
a) electrically
b) magnetically
c) none of the mentioned
d) least

Answer

Answer: a [Reason:] It has no electrical excitation.

8. How to manage run-away issue for dc shunt motor in industry?
a) By using PMDC motor
b) By using constant field motors
c) This can not be avoided
d) None of the mentioned

Answer

Answer: a [Reason:] A PMDC motor can be used to replace the dc shunt motor in industries as they have quite identical characteristics.

9. A PMMDC motor has as armature resistance of 4.2 ohms. When 6 V supply is applied to motor it runs at 1215 rpm drawing 14.5 mA on no-load. The rotational losses is
a) 86.1 mW
b) 86.1 W
c) 8.6 W
d) 8.6 mW

Answer

Answer: a [Reason:] Rotational losses = E*Ia = 5.939* 14.5/1000 = 86.1 mW.

10. PMMDC can only be armature controlled.
a) True
b) False

Answer

Answer: a [Reason:] Due to absence of the field winding there is no option of field control.

Set 4

1. Floating of synchronous machine on infinite bus means
a) Fa = 0 and machine neither generator nor motor
b) Fa is not zero and machine neither generator nor motor
c) Fa = 0 and machine is not excited
d) frequency is zero

Answer

Answer: a [Reason:] Floating means the armature flux is zero and it neither motoring nor generating.

2. If the field poles are ahead of resultant mmf, then the 3-phase synchronous machine is operating as
a) generator
b) motor
c) reluctance motor
d) any of the mentioned

Answer

Answer: a [Reason:] Field mmf leads the resultant mmf so it is generator.

3. At steady state operation in a 3-phase synchronous generator
a) Tem = Tpm
b) Tem > Tpm
c) Tem < Tpm
d) can not be coupled
where Tem is developed electromagnetic torque developed in the machine and Tpm is the prime mover torque.

Answer

Answer: a [Reason:] At steady state the net torque is zero in the machine.

4. Active power in synchronous machine is proportional to
a) torque
b) excitation
c) sinϕ
d) all of the mentioned

Answer

Answer: a [Reason:] Active power is proportional to the torque of the synchronous machine.

5. The reactive power in synchronous machine is proportional to
a) torque
b) excitation
c) sinϕ
d) all of the mentioned

Answer

Answer: b [Reason:] Reactive power depends on the excitation of the synchronous machine.

6. If the synchronous machine is loaded from floating conditions, the alternator begins to operate at
a) leading p.f
b) lagging p.f
c) unity p.f
d) zero p.f

Answer

Answer: a [Reason:] The resultant flux will be leading in nature, so after loading it from loading it from floating, alternator will operate at leading p.f.

7. If the synchronous machine is loaded from floating conditions, the synchronous motor begins to operate at
a) leading p.f
b) lagging p.f
c) unity p.f
d) zero p.f

Answer

Answer: b [Reason:] The resultant flux will be lagging in nature, so after loading it from loading it from floating, alternator will operate at lagging p.f.

8. Reactive power flow in synchronous machine can be controlled by
a) varying field excitation
b) varying field current
c) varying field flux
d) any of the mentioned

Answer

Answer: d [Reason:] All the alternatives are same, varying the field flux matters at the end to change the reactive power in a synchronous machine.

9. If the 3-phase alternator has 4 poles and has synchronous speed of 120 rad/s.
Then the mechanical speed in rad/sec is
a) 60
b) 240
c) 120
d) 30

Answer

Answer: a [Reason:] Mechanical speed = electrical speed*2/Number of Poles.

10. The power load angle characteristic as maximum for a load angle of
a) 90°
b) 45°
c) more than 90°
d) 180°

Answer

Answer: a [Reason:] As per the parabolic graph, the maxima is obtained at 90°.

11. For running a 750 kW compressor, a synchronous motor is preferred because of
(i) lower noise
(ii) faster build up of pressure
(iii) better power factor
(iv) low starting torque
a) (ii),(iii)
b) (i),(ii),(iv)
c) (i),(ii),(iii),(iv)
d) (iii),(iv)

Answer

Answer: a [Reason:] It is preferred due to better power factor and low starting torque.

12. A poly phase synchronous motor will be used for the load of ______ over poly phase induction motor.
a) 600 kW 500 rpm
b) 600 kW 1500 rpm
c) 600 kW 3500 rpm
d) 600 kW 1200 rpm

Answer

Answer: a [Reason:] A synchronous motor is used at low speed.

13. During the starting of the 3-phase synchronous motor by damper bars, the field winding is usually short circuited so that starting torque is equal to
a) induction motor torque plus an additional torque produced by short circuited field winding
b) induction motor torque
c) electromagnetic torque
d) reluctance torque due to rotor saliency

Answer

Answer: a [Reason:] Using dampers to start the synchronous motor, it will produce induction motor torque initially along with the torque due to shorted field winding.

14. To start the synchronous motor, its field winding should be
a) short circuited
b) kept open
c) connected to dc excitation source
d) any of the mentioned

Answer

Answer: a [Reason:] To start the synchronous motor, it is first run as field excitation as zero, so we short circuit them.

15. Squirrel cage bars placed in the rotor pole faces of a 3-phase alternator help in reducing hunting
I. above synchronous speed
II. below synchronous speed
III. at synchronous speed
a) I,II
b) III
c) I,II,III
d) none of the mentioned

Answer

Answer: a [Reason:] This method is based on the operation of the induction motor, hence it will not operate at synchronous speed.

Set 5

1. Choose the correct power angle characteristics for the synchronous motor and generator
a)
b)
c)
d) None of the mentioned

Answer

Answer: a [Reason:] In the generator mode, the power is positive and it is taken out of the terminals.

2. For a 3 phase 400V alternator having 0.5pu of synchronous reactance and its excitation voltage of 1.2pu and Vt of 1pu. Then the power delivered to the infinite bus is
a) 2.4pu
b) 4.8pu
c) 1.44pu
d) 1pu

Answer

Answer: a [Reason:] P = EV/X = 1.2*1/0.5 = 2.4 pu.

3. For a 3 phase 400V alternator supplying a load of 0.8 pf lagging at an armature current of 1.1pu. The power delivered to load is
a) 0.88pu
b) 0.8pu
c) 0.66pu
d) 1.375pu

Answer

Answer: c [Reason:] P = EVsinδ/X = 1.1*1*sin36.67 = 6.6 pu.

4. The power-angle characteristic for a three phase salient pole alternator is most correctly shown by
a)
b)
c)
d)

Answer

Answer: a [Reason:] The power angle curve is not completely sinusoidal due to the salient nature of the poles.

5. A 3 phase 400V salient pole alternator is running at no-load. Suddenly the excitation is removed then the net power in the machine is proportional to
a) sin 2d
b) sin d
c) cos 2d
d) 0

Answer

Answer: a [Reason:] The reluctance power generated will be proportional to sin(2δ).

6. A 3 phase 400V cylindrical rotor is running at 1500rpm. Suddenly the field excitation zero. Then the net reluctance power is
a) 0
b) sin 2d
c) cos d
d) cos 2d

Answer

Answer: a [Reason:] The net reluctance power for a cylindrical rotor type alternator will be zero if the field excitation is made zero.

7. A cylindrical rotor alternator can also run as reluctance motor
a) True
b) False

Answer

Answer: b [Reason:] The net reluctance power for a cylindrical rotor type alternator will be zero if the field excitation is made zero. There will not be any torque generated.

8. If the synchronous machine is connected to an infinite bus of constant voltage Vt, through a transformed, a transmission line of reactance ‘X’. then power for a cylindrical machine is
a) P = (Ef*Vt sinδ)/(Xs + X)
b) P = (Ef*Ef sin2δ)/(Xs + X)
c) P = (Ef*Vt sinδ)/(Xs – X)
d) P = (Vt*Vt sinδ)/(Xs + X)

Answer

Answer: a [Reason:] The total reactance will be the reactance offered to the synchronous motor in the machine.

9. Maximum power in cylinder rotor alternator occurs at load angle of
a) 90 degree
b) 45 degree
c) less than 90 degree
d) 180 degree

Answer

Answer: a [Reason:] The maximum power will occur at exact 90° for a cylindrical rotor alternator.

10. A synchronous generator is operating with a Ef=1.40pu. This machine having Xs of 1.2pu, is delivering a synchronous power of 0.5pu to the bus. If the prime-mover torque is increased by 1%, by how much will the synchronous power P change?
a) 1%
b) 10%
c) -1%
d) 4%

Answer

Answer:a [Reason:] dP/dδ = EVcosδ/X
So the change in P will also be 1%.

11. A synchronous generator is running over excited with a Ef=1.40pu. This machine, with a synchronous reactance of 1.2pu, is delivering a synchronous power of 0.5pu to the bus. If the prime-mover torque is increased by 1%, by how much will the reactive power, Q change?
a) -0.475 %
b) 0.475%
c) 4.75%
d) -4.75%

Answer

Answer: a [Reason:] dQ/dδ = -EVsinδ/X
dP/dδ = EVcosδ/X
dQ/dP = -tanδ = -tan25.4 = -0.475
dQ = -0.475%.

12. A synchronous machine which is synchronized with an infinite bus. If it is desired to obtain a condition when the machine delivers the real power to the IBB without changing the field excitation, then
a) reactive power will be consumed by the machine
b) reactive power will be delivered by the machine
c) no reactive power flow will take place
d) none of the mentioned

Answer

Answer: a [Reason:] Due to the condition here that it is providing the real power to the infinite bus bar here, the machine will need more flux to produce the real power in the machine.

13. The phasor addition of stator and rotor mmfs in a cylindrical rotor synchronous machine, is possible because
a) mmfs are rotating in opposite direction
b) mmfs are rotating in same direction at different speeds
c) mmfs are stationary with respect to each other
d) one mmf is stationary and the other mmf is rotating

Answer

Answer: c [Reason:] The phasor addition is possible due to the fact that the two mmfs are stationary with respect to each other.

14. For the effective electromechanical energy conversion in the device, the developed torque depends upon
a) stator field and torque angle
b) stator field and rotor field
c) stator field and rotor field and the torque angle
d) stator field only

Answer

Answer: c [Reason:] T α E*V*sinδ

15. In a synchronous machine, hunting is predominantly damped by
a) mechanical losses in the rotor
b) iron losses in rotor
c) copper losses of stator
d) copper losses of rotor

Answer

Answer: d [Reason:] The hunting is damped by the copper losses in the rotor.