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Multiple choice question for engineering

Set 1

1. Characteristics of separately excited DC generator are drawn by keeping _____
a) Field current and speed both constant
b) Field current and speed both variable
c) Field current constant and speed variable
d) Field current variable and speed constant

View Answer

Answer: b [Reason:] The operation considered here assumes that the armature is driven at constant speed (by means of prime mover) and the field excitation (If) is adjusted to give rated voltage at no-load and is then held constant at this value throughout the operation considered.

2. What is the reason behind dropping down of Ea with load?
a) Field resistance
b) Load resistance
c) Internal factors
d) Armature reaction

View Answer

Answer: d [Reason:] In spite of fixed excitation, Ea drops off with load owing to the demagnetizing effect of the armature reaction. As the voltage drop is caused by magnetic saturation effect, it increases with load non-linearity.

3. External characteristic differ from an internal characteristic in separately excited DC generator by _________
a) Ia *Ra
b) If *Ra
c) IL *RL
d) If *Rf

View Answer

Answer: a [Reason:] Internal characteristic is drawn across armature generated voltage, it doesn’t account the presence of armature resistance. External characteristic takes into account the presence of armature resistance as it is drawn across terminal voltage.

4. The variation in terminal voltage of DC shunt generator with respect to variation in separately excited DC generator is ___________
a) Much rapid
b) Much slower
c) Remains constant
d) Can’t say

View Answer

Answer: a [Reason:] The terminal voltage drops off much more rapidly with load in a shunt generator than in a separately-excited generator because of fall in field current with terminal voltage. The external characteristic is a double-valued curve with a certain IL (max).

5. Which of the following characteristic lies above of all others?
a) Differential compound
b) Under compound
c) Level compound
d) Over compound

View Answer

Answer: d [Reason:] All the graphs when drawn across voltage and load current, start from the same point but with increase in values of load current all machines show different elevation according to the values of series field resisitor.

6. Why differential compound generator is not used in practice?
a) High cost
b) High maintenance
c) High drop down in voltage
d) Difficult construction

View Answer

Answer: c [Reason:] At a given value of load current, differential compound machine gives lowest voltage output. As the load current increases drop down in terminal voltage of DC differential compound generator is maximum.

7.How compounding level in a compound machine is adjusted?
a) By adding variable resistance in series with series field resistance
b) By adding variable resistance in parallel with series field resistance
c) By adding fixed resistance in parallel with series field resistance
d) By adding fixed resistance in series with series field resistance

View Answer

Answer: b [Reason:] When fixed resistance is added in parallel with series field resistance we get only one other compounding level. So, by adding variable resistance in parallel with series field we can get various other compounding levels.

8. Which of the following have different external characteristic than other?
a) Self excited DC shunt generator
b) Separately excited generator
c) Compound DC generator
d) Series DC generator

View Answer

Answer: d [Reason:] Unlike Self excited DC shunt generator, separately excited generator, compound DC generator the voltage at zero load current for DC series motor doesn’t start from some positive non-zero value, instead it starts from origin.

9. External characteristic differ from internal characteristic in DC series motor by ______
a) Ia *(Ra + RSE)
b) Ia *Ra
c) Ia *RSE
d) If *RSE

View Answer

Answer: a [Reason:] In series DC generator series resistor RSE is added with armature circuit, thus total drop in terminal voltage from generated armature voltage is equal to Ia *(Ra + RSE). While drawing internal characteristics presence of armature resistance and series resistance is not taken into account.

10. For series DC generator, internal/external characteristic start from ____________
a) Positive non-zero voltage
b) Zero voltage
c) Negative non-zero voltage
d) Can start from anywhere

View Answer

Answer: b [Reason:] For all other DC generators other than series DC generators, the external and internal characteristic as well, start from non-zero positive value of voltage. While in series DC generator both internal and external characteristic start from origin.

11. For a DC series generator what is the condition for self-excitation?
a) (Ra+Rse) > RC
b) (RSE+Rse+RL) < RC
c) (RSE+Rse+RL) > RC
d) (Rse+RL) > RC

View Answer

Answer: c [Reason:] Summation of armature resistance, series field resistance, load resistance must be less than the critical resistance. As seen in self-excited DC generator characteristics if this value is greater than critical resistance, voltage build-up will not be possible.

12. For a given DC series generator with critical resistance equal to 100 Ω, armature resistance is equal to 50 Ω, and series field resistance is equal to 20 Ω, is connected across load of 50 Ω. What will be the load voltage?
a) 20 kV
b) 0 V
c) 2 kV
d) Data insufficient

View Answer

Answer: b [Reason:] Here, DC series motor fails to excite as addition of armature resistance, load resistance and field resistance is greater than the critical resistance of the machine by 20 Ω. Thus, machine fails to self-excite, as a result we’ll get zero terminal voltage.

Set 2

1. Ward Leonard method is ___________
a) Armature control method
b) Field control method
c) Combination of armature control method and field control method
d) Totally different from armature and field control method

View Answer

Answer: c [Reason:] Ward Leonard method is the combination of armature control method and field control method, which can also be called as voltage control method. This is the most efficient method of speed control over wide range.

2. Which of the following component is not used in Ward Leonard method?
a) AC motor
b) DC generator
c) DC motor
d) AC generator

View Answer

Answer: d [Reason:] Whole unit of Ward Leonard speed control unit consists of various units like DC generator, DC motor, AC motor, exciter circuit and various pots which are used for carrying out smooth operation.

3. In Ward Leonard speed control method for lowering the speed of the motor ______________
a) Reduce armature voltage
b) Increase armature voltage
c) Increase field current
d) Decrease field current

View Answer

Answer: a [Reason:] In Ward Leonard speed control method, speed can be reduced under base value by reducing armature voltage. By increasing field current speed can be reduced but this is not employed in Ward Leonard method.

4. Reducing the armature voltage will give us _______________
a) Variable torque speed control
b) Constant torque speed control
c) Variable and constant both can be achieved
d) Cannot comment on torque

View Answer

Answer: b [Reason:] As seen from speed torque characteristics, reducing armature voltage will reduce the speed of the motor below base value but torque will remain same. Thus, reducing armature voltage will give constant torque speed control.

5. In Ward Leonard speed control method for increasing the speed of the motor ______________
a) Reduce armature voltage
b) Increase armature voltage
c) Increase field current
d) Decrease field current

View Answer

Answer: d [Reason:] In Ward Leonard speed control method, speed can be increased above base value by weakening of the field, which can be done by lowering field current value. By increasing armature voltage speed can be increased but this is not employed in Ward Leonard method.

6. Reducing the field current will give us _______________
a) Constant torque and variable power speed control
b) Constant torque speed control
c) Variable power speed control
d) Constant power speed control

View Answer

Answer: b [Reason:] As seen from speed torque characteristics, reducing field current will increase the speed of the motor above base value but power will remain same. Thus, reducing armature voltage will give constant power speed control, with variable torque.

7. Speed-power characteristic for Ward Leonard speed control method _________________
a) Will start from origin
b) Will start from some positive value on power axis
c) Will start from some positive value on speed axis
d) Depends on other parameters

View Answer

Answer: a [Reason:] Speed power characteristic of DC motor is plotted when, Ward Leonard speed control method is employed. For speed equal to zero, which is less than base speed, we get constant torque but variable power operation. Thus, power will start increasing from origin.

8. Efficiency of Ward Leonard method is ____________
a) Higher than rheostatic control method but lower than shunted field control method
b) Lower than rheostatic control method
c) Higher than rheostatic control method and shunted field control method
d) Depends on load

View Answer

Answer: c [Reason:] Unlike all other methods, external resistance is not added in the circuit of control system. Thus, efficiency of Ward Leonard control method is always highest at various different speeds.

9. Ward Leonard method is an ideal choice for motor which undergoes frequent starting, stopping, speed reversal.
a) True
b) False

View Answer

Answer: a [Reason:] Absence of external resistance increases efficiency. Also, when the generator emf becomes less than the back emf of the motor, electrical power flows back from motor to generator, is converted to mechanical form and is returned to the mains via the driving ac motor. This aspect makes Ward Leonard method perfect for given application.

10. Starting gear used in Ward Leonard method___________
a) Is of small size
b) Is of large size
c) Size depends on application
d) Is absent

View Answer

Answer: d [Reason:] No special starting gear is required in Ward Leonard method of speed control. As the induced voltage by generator is gradually raised from zero, the motor starts up smoothly. Speed reversal is smoothly carried out.

11. To get the speed of DC motor below the normal speed without wastage of electrical energy we use __________________
a) Ward Leonard control
b) Rheostatic control
c) Any of the Ward Leonard or rheostatic method can be used
d) Not possible

View Answer

Answer: a [Reason:] Ward Leonard method of speed control is most efficient method of speed control in all aspects. We can get constant torque operation and constant power operations as well, with this method.

12. Speed control by Ward Leonard method, can give uniform speed variation _______________
a) In both directions
b) In one direction
c) Below normal speed only
d) Above normal speed only.

View Answer

Answer: a [Reason:] Speed control by Ward Leonard method, gives uniform speed variation in both the directions and above and below of normal speed as well. Speed reversal is carried out smoothly by this control method.

13. Ward Leonard control is basically a _____________
a) Voltage control method
b) Field diverter method
c) Field control method
d) Armature resistance control method

View Answer

Answer: a [Reason:] Ward Leonard speed control method is combination of rheostatic series control method and shunted armature control method with field control as well. Thus, it can be called as voltage control method also.

14. In Ward Leonard control of DC motor, the lower limit of speed is imposed by ____________
a) Residual magnetism of the generator
b) Core losses of motor
c) Mechanical losses of motor and generator together
d) Cannot be determined

View Answer

Answer: a [Reason:] We get the speed below the base value by reducing armature voltage, which is the simple method of reducing back emf which is proportional to the residual magnetism. Thus, lower limit of speed is imposed by residual magnetism of the generator.

15. The disadvantage of the Ward Leonard control method is ___________
a) High initial cost
b) High maintenance cost
c) Low efficiency at high loads
d) High cost, high maintenance and low efficiency

View Answer

Answer: d [Reason:] Ward Leonard speed control method requires large number of building blocks like generators, motors. Thus, installing cost and maintenance cost of the whole unit is very high. Apart from cost, it gives low efficiency at very high loads.

Set 3

1. What will be the value of “Yf + Yb” for a wave winding?
a) Equal to Yc
b) Half of the Yc value
c) Double of the Yc value
d) Four times Yc value

View Answer

Answer: c [Reason:] In the wave winding, as the number of coil-sides is double the number of segments, the top coil-side of the second coil will be numbered as (1+2*Yc). After numbering other coil sides, 1 + 2*Yc – Yf = 1+ Yb So Yf + Yb = 2Yc.

2. For a progressive wave winding Yc = ______
a) 2C/P
b) 2(C+1)/P
c) 2(C-1)/P
d) 2C/(P+1)

View Answer

Answer: b [Reason:] Starting at segment 1 and after going through P/2 coils or Yc (P/2) segments, the winding should end in segment 2 for progressive winding or segment (C) for retrogressive winding. That is mathematically, Yc (P/2) = (C+1) Yc = 2(C+1)/P

3. Number of parallel paths in wave winding are ______
a) Equal to P
b) Equal to P/2
c) 2
d) Depends on other parameters

View Answer

Answer: c [Reason:] In wave winding all coils are divided into 2 groups- all coils carrying clockwise current are series connected and so are all coils with counter-clockwise current- and these 2 groups are in parallel because the winding is closed. Thus, a wave winding has 2 parallel paths irrespective of number of poles.

4. What is the spacing between the brushes for a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coils.
a) 4 segments
b) 8 segments
c) 16 segments
d) 12 segments

View Answer

Answer: b [Reason:] Only 2 brushes are required in this case as the number of poles in wave winding is equal to 2. So, spacing between the brushes is equal to total number of segments i.e. total slots divided by 2. Spacing between brushes = C/A = 16/2 = 8 segments.

5. What is the relation between conductor current and armature current in wave winding?
a) Ic = Ia
b) Ic = 2Ia
c) Ic = 4Ia
d) Ic = Ia/2

View Answer

Answer: d [Reason:] the number of parallel paths in the in a wave winding is equal to 2. So, armature current will get divided equally into total number of conductors/paths. Conductor current in a wave wounded machine is half of the Ia.

6. For a conductor current equal to 4mA, Current carried by a particular brush in a 2-pole machine will be _____
a) 16mA
b) 8mA
c) 2mA
d) 10mA

View Answer

Answer: b [Reason:] Conductor current in a wave wounded machine is half of the Ia. So, Ia= 8mA. All positive and all negative brushes are respectively connected in parallel to feed the external circuit. Thus, IBRUSH = Ia /(P/2). Solving we get Brush current = 8mA.

7. Equalizer rings are needed in the wave winding.
a) True
b) False

View Answer

Answer: b [Reason:] The armature coil forms 2 parallel paths under the influence of all pole-pairs so that the effect of the magnetic circuit asymmetry is equally present in both the parallel paths resulting in equal parallel-path voltages. Thus, equalizer rings are not needed in wave winding.

8. For a wave winding when a machine is 6-pole DC armature with 16 slots having 2-coil sides per slot and single-turn coil, Yf value is ____
a) 5
b) 3
c) 2
d) 7

View Answer

Answer: a [Reason:] Ycs = 16/6 = 2 slots (nearest lower integral value) Yb= 2*2+1 = 5 Yc= 2(16-1)/6 = 5 segments Yf = 2Yc – Yb = 5.

9. Wave winding machines are used in ______ currents applications.
a) High
b) Moderate
c) Low
d) Can be used anywhere

View Answer

Answer: c [Reason:] Lap winding machine has the advantage of large number of parallel paths and lower conductor current and is therefore used in low voltage and high current applications. Wave winding has fixed number of parallel paths so, wave wounded machine is used in low currents application.

10. For a wave wounded machine number of brushes for small, large machines respectively is ________ _________
a) 2, 2
b) 4, 2
c) 2, P
d) Both values depend on the given conditions

View Answer

Answer: c [Reason:] For a small wave wounded machine number of parallel paths are 2, thus 2 brushes are used. For a large machine total number of brushes is equal to the total number of poles. The spacing between adjacent brushes is C/P commutator segments.

Set 4

1. The armature reaction in a DC machine causes distortion in the main field flux. Effect of armature reaction can be reduced by_________
a) Increasing the length of air gap
b) Decreasing the length of air gap
c) Increasing the number of poles
d) Decreasing the number of poles

View Answer

Answer: b [Reason:] Decreasing the air gap is simple to say but hard to achieve, due to various other limitations. But it is one of the way to reduce dropping down of armature mmf flux density distribution, which further reduces the distortion of resultant flux density.

2. In order to neutralize armature mmf perfectly under the pole shoe, the ampere-conductors of compensating winding must be (pole arc/pole pitch=1) _______
a) Not equal to the total armature ampere conductors under the pole shoe
b) Equal to the total armature ampere conductors under the pole shoe
c) Half of the total armature ampere conductors under the pole shoe
d) Twice of the total armature ampere conductors under the pole shoe

View Answer

Answer: b [Reason:] The number of ampere-turns required for compensating winding is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)]*(pole arc/pole pitch). Here, pole arc= pole pitch, is given. Thus ampere-conductors required are equal to the total armature ampere conductors under the pole shoe.

3. What is the ampere turns per pole for compensating winding in DC machines?
a) (pole arc / pole pitch) * armature ampere turns per pole
b) (pole pitch / pole arc) * armature ampere turns per pole
c) (pole arc / pole pitch) * total ampere conductors per pole
d) Cannot be determined

View Answer

Answer: a [Reason:] The number of ampere-turns required for this purpose is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)]*(pole arc/pole pitch). The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization.

4. Where are no equalizer rings connected?
a) Only wave winding
b) Only lap winding
c) Both wave winding and lap winding
d) Cannot be determined

View Answer

Answer: a [Reason:] The armature coils forming each of the two parallel paths are under the influence of all pole-pairs so that the effect of the magnetic circuit asymmetry is equally present in both the parallel paths resulting in equal parallel-path voltages. Thus, equalizer rings are not needed in a wave winding.

5. Inequality in brush arm currents caused due to different emf induced in different parallel paths may give rise to copper losses. These effects can be avoided by using _________
a) Compensating windings
b) Interpoles
c) Equalizer rings
d) Cannot be determined

View Answer

Answer: c [Reason:] The armature coil in lap windings have equal to or more than 2 parallel paths. Under the influence of pole pairs there may arise asymmetry in various paths resulting in different parallel path voltages. So, equalizer rings are connected in order to achieve symmetry again.

6. Which of the following method will completely neutralize the armature reaction in a DC machine?
a) Only compensating winding
b) Only interpoles
c) Compensating winding and interpoles
d) Cannot be determined

View Answer

Answer: c [Reason:] To neutralize the cross magnetizing effect of armature reaction, a compensating winding is used. The compensating windings consist of a series of coils embedded in slots of the pole faces. Interpoles are designed to overcome the effects of the armature reactance and the self-induction of the machine.

7. What is the exact location of compensating winding?
a) Across armature
b) In series with armature
c) Across armature or in series
d) Cannot be determined

View Answer

Answer: b [Reason:] By adding a compensating winding in the pole face plate which carries armature current in the opposite direction of current in the adjacent armature windings, the position of the flux at the pole face plate can be restored to the position it would have with zero armature current.

8. Exact purpose of compensating winding in a DC generator is _________
a) Mainly to reduce the eddy currents by providing local short-circuits
b) To provide path for the circulation of cooling air
c) To neutralise the cross-magnetising effect of the armature reaction
d) Cannot be determined

View Answer

Answer: c [Reason:] Armature reaction can be reduced with the help of compensating winding. To neutralize the cross-magnetizing effect of armature reaction in a DC machine, a compensating winding is used. The compensating windings consist of a series of coils embedded in slots of the pole faces.

9. Function of interpole flux is_________
a) Neutralise the commutating self-induced emf
b) Neutralise the armature reaction flux
c) Neutralise both the armature reaction flux as well as commutating emf induced in the coil
d) Perform none of the above functions

View Answer

Answer: c [Reason:] Interpoles are designed to overcome the effects of the armature reactance and the self-induction of the machine i.e. to neutralise both the armature reaction flux as well as commutating emf induced in the coil.

10. Why dummy coils are connected in DC machine?
a) To reduce eddy current losses
b) To enhance flux density
c) To amplify voltage
d) To provide mechanical balance for the rotor

View Answer

Answer: d [Reason:] Dummy coils are connected in armature winding of type wave winding. In a lap winding Yc= +/-1 irrespective of the number of armature coils so that coils can always be chosen to completely fill all the slots (C =1/2 US). In a wave winding the number of coils must fulfil the condition C =P/2 Yc +- 1 while at the same time C must also be governed by C =1/2 US.

11. The most likely cause(s) of sparking at the brushes in a DC machine is /are ____________
a) Open coil in the armature
b) Defective interpoles
c) Incorrect brush spring pressure
d) Open coil in armature, defective interpoles and incorrect brush spring pressure

View Answer

Answer: d [Reason:] Brushes are the point of contacts between rotating and non-rotating parts. So, if point of contact is rough then brushes will face damage and will lead to sparking in them. Defective interpoles will contribute in irregular commutation ultimately it will lead to sparking.

12. Each of the following statements regarding interpoles is true except______________________
a) They are small yoke-fixed poles spaced in between the main poles
b) They are connected in parallel with the armature so that they carry part of the armature current
c) Their polarity, in the case of generators is the same as that of the main pole ahead
d) They automatically neutralize not only reactance voltage but cross-magnetisation as well

View Answer

Answer: b [Reason:] Interpoles are located in interpolar regions hence, they are called as interpoles. They are small narrow poles which speed up the commutation process (Also called commutating poles/ compoles).

Set 5

1. Regenerative braking is used when duty cycle ____________
a) Requires braking of machine
b) Requires accelerating of machine
c) Requires constancy of machine
d) Cannot comment on duty cycle

View Answer

Answer: a [Reason:] Regenerative braking is used specially where the duty cycle requires the braking or slowing of the machine more frequently and is most useful in holding a descending load of high potential energy at a constant speed.

2. Regeneration is not easily possible for ____________
a) DC shunt motor
b) Separately excited motor
c) Compounding motor with weak series compounding
d) DC series motor

View Answer

Answer: d [Reason:] Regeneration is possible with a shunt and separately excited motors and with compound motors with weak series compounding. Series motors need a reversal of either the field or the armature connections.

3. Which of the following method is not used for regeneration?
a) Increasing field current
b) Increasing armature speed
c) Increasing supply voltage
d) Reducing supply voltage

View Answer

Answer: c [Reason:] Regeneration is achieved by either increasing field current, increasing armature speed, or by reducing supply voltage. Increasing supply voltage is not the method which is employed in regeneration process.

4. If the terminals of armature of DC motor are interchanged, this action will offer following kind of electrical braking ______________
a) Regenerative
b) Plugging
c) Dynamic braking
d) Depends on other parameters

View Answer

Answer: b [Reason:] Plugging is electrical braking method, where field or armature connections are reversed technically. But field reversal is not employed as results obtained from field reversal are not good compare to armature reversal.

5. The plugging braking gives the _____________
a) Zero torque braking
b) Smallest torque braking
c) Highest torque braking
d) Variable torque braking

View Answer

Answer: c [Reason:] In electrical braking called plugging direct reversal of connections is done, which causes maximum torque to act on shaft but in opposite direction. As the speed decreases this torque also starts decreasing.

6. Regenerative method of braking is based on ___________
a) Back emf is less than the applied voltage
b) Back emf is equal to the applied voltage
c) Back emf of rotor is more than the applied voltage
d) Cannot be determined

View Answer

Answer: c [Reason:] The condition for regeneration is that the rotational emf is more than the applied voltage so that the current is reversed and the mode of operation changes from motoring to generating.

7. During regenerative braking of DC motors ____________
a) Motor will run as a generator
b) Motor will reverse in direction
c) Motor will run at reduced speed
d) Motor will run as free rotating shaft

View Answer

Answer: a [Reason:] In regenerative method of electrical braking, motor is suddenly forced to act as a generator, all the energy then obtained is pushed back into the supply unlike in dynamic braking this energy is wasted.

8. Where dynamic braking is used?
a) Shunt motors
b) Series motors
c) Compound motors
d) All DC motors

View Answer

Answer: d [Reason:] Dynamic braking is used in all DC motors though its implantation in series DC motor requires one more additional step of reversal of connections. Only care taken is, addition of braking resistance, armature resistance and series field resistance is lower than the critical resistance at that speed.

9. Which method of braking is generally used in elevators?
a) Plugging
b) Regenerative braking
c) Rheostatic braking
d) Mechanical braking

View Answer

Answer: a [Reason:] Plugging braking provides maximum torque in opposite direction at the instant of braking, this characteristic of braking suits perfectly with the application that is in elevators. If switch is kept ON, we get reverse rotation also.

10. For which of the following motor dynamic braking is very effective?
a) Shunt motors
b) Separately excited motors
c) Series motors
d) Differential compound motors

View Answer

Answer: b [Reason:] Dynamic braking is very effective for separately excited DC motors. As in separately excited motors the direction of field can be very easily altered by altering the terminals of the field, which is the condition in dynamic braking.

11. When is the dynamic braking is employed?
a) Non-reversing drive
b) Reversing drive
c) Both Reversing and Non-reversing
d) Cannot tell

View Answer

Answer: c [Reason:] Dynamic braking is employed to brake both reversing drives and non-reversing drives. In dynamic braking, the electrical energy generated during stopping action is released as heat through a voltage regulated transistor and resistor.