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# Multiple choice question for engineering

## Set 1

1. In which mode machine is operating, given that conductor current is in the same direction of conductor emf?
a) Motoring
b) Generating
c) Can’t be determined using directions
d) In both modes for different cycles

Answer: b [Reason:] If the conductor current is in the same direction of conductor emf then machine outputs electrical power and absorbs mechanical power. So, when mechanical power is absorbed machine is said to be in a generating mode. When conductor emf and conductor current are in opposite directions then machine is said to be in a motoring mode.

2. Nature of the flux density wave in the air gap is__________ (for armature current equal to 0)
a) Flat topped with quarter wave symmetry
b) Point topped with quarter wave symmetry
c) Flat topped with half wave symmetry
d) Point topped with half wave symmetry

Answer: a [Reason:] In a DC machine magnetic structure is such that the flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction.

3. In a DC machine, average energy stored in the magnetic field remains constant independent of the armature rotation.
a) True
b) False

Answer: True [Reason:] In a DC machine, barring the irrecoverable losses of both electric and magnetic origin, there is balance between electrical and mechanical powers of the machine; the average energy stored in the magnetic field remains constant irrespective of armature rotation.

4. Emf produced by DC machine, for zero armature current (E1) and non-zero armature current (E2) can be related as__________
a) E1 = E2
b) E1 > E2
c) E1 < E2
d) Can’t be determined

Answer: a [Reason:] In a DC machine flux density wave in the air gap is flat topped with quarter wave symmetry as long as armature current is equal to 0. For non-zero value of armature current, this quarter wave symmetry is disturbed because of armature reaction. Emf produced is independent of B-wave shape, thus we will get same value for both cases.

5. Average coil emf for 20 coil turns (E1) and 40 coil turns (E2), will have ratio E1/E2=____ (assuming all other parameters same for both machines).
a) 1/2
b) 2/1
c) 1/4
d) 4/1

Answer: a [Reason:] Emf generated in a DC machine is directly proportional to number of coil turns, Flux per pole, number of poles and armature speed in rad/s. Thus, ratio E1/E2= 20/40 (assuming all other parameters same for both machines).

6. What is the average coil emf generated in a 4-pole DC machine having flux/pole equal to 0.1 wb rotating at 1500 rpm? (No. of coil sides = 100)
a) 19 kV
b) 1.9 kV
c) 190 V
d) 19 V

Answer: a [Reason:] Average coil emf generated= ∅ωNP/π. E= 0.1*1500*100*4/3.14 E= 60000/3.14 E≅ 19 Kv.

7. Emf and torque produced in a DC machine are proportional to ________ and _________ respectively.
a) Armature speed and armature emf
b) Armature emf and armature speed
c) Armature current and armature emf
d) Armature speed and armature current

Answer: d [Reason:] Average coil emf generated= ∅ωNP/π. Machine torque = ka*∅*Ia. Thus, average coil emf generated can also be represented as ka*∅*ω. So, average coil emf is directly proportional to ω (armature speed) and average torque is directly proportional to Ia (armature current).

8. What is the value of Np in an average coil emf equation, for 10 armature conductors with 2 parallel paths?
a) 2
b) 3
c) 2.5
d) 4

Answer: c [Reason:] In an emf equation Nc= Cp * Np. Here, Cp= coils/ parallel path. Np is defined as number of turns per parallel paths which is also called as ratio of total armature conductors to the twice of number of parallel paths. Np= 10/(2*2)= 10/4= 2.5.

9.What is the torque equation in terms of B, Ic, l, Zr (r= mean air gap radius)?
a) Bav*Ic*l*Zr
b) Bav*Ic*l/Zr
c) Bav*Ic*Zr/l
d) Can’t be expressed

Answer: a [Reason:] Avg. conductor force f= Bav*l*Ic. Here, Bav= Average flux density over pole, l= acyive conductor length. Thus, torque T= Z*f = Bav*l*Ia*Z. This torque is constant because both the flux density wave and current distribution is fixed in space at all times. T developed= Bav*Ic*l*Zr (Here, r= mean air gap radius).

10. What is the value of pole pitch (in SI unit) for mean air gap radius= 0.5mm and P=4?
a) 0.785* 10-6
b) 0.785* 10-3
c) 0.785* 10-2
d) 0.785* 10-4

Answer: b [Reason:] Pole pitch is called as center to center distance between two adjacent poles. When measured in electrical degrees one pole itch is equal to 1800. Pole pitch can be calculated as ratio of 2πr/P. Pole pitch= 2*3.14* 0.5* 10-3 / 4= 0.785* 10-3 m.

## Set 2

1. The condition for maximum efficiency for a DC generator is __________
a) Eddy current losses = stray losses
b) Hysteresis losses = eddy current losses
c) Copper losses = 0
d) Variable losses = constant losses

Answer: d [Reason:] All losses in a given DC machine can be categorized into variable losses and constant losses. Variable losses are proportional to the square of armature current while constant losses are almost constant for a given DC machine throughout its application.

2. DC generators are normally designed for maximum efficiency at or near ____________
c) Rated voltage
d) At all loads

Answer: c [Reason:] The efficiency of a machine is different at different values of power output. All electrical machines are generally designed to give maximum efficiency at or near the rated output of the machine. Thus, maximum efficiency occurs at rated voltage.

3. In a DC generator, the iron losses mainly take place in ____________
a) Yoke
b) Commutator
c) Armature conductors
d) Armature rotor

Answer: d [Reason:] Iron losses take place in the form of hysteresis loss and eddy current loss. These losses are maximum, where field is maximum. Thus, when armature is rotated in presence of flux we get maximum iron loss.

4. If DC generators are located near load centres, which losses can be minimised?
a) Iron losses
b) Eddy current losses
c) Line losses
d) Corona losses

Answer: c [Reason:] Line losses occur in long transmission lines while sending output power to the loading stations. Thus, by locating the generators near loading stations losses occurring due transmission line can be eliminated.

5. Nature of efficiency curve of DC machine is _________________
a) First decreases then increases
b) First constant then decreases
c) First constant then increases
d) First increases then decreases

Answer: d [Reason:] The efficiency of a DC machine is different at different values of power output. As the output increases, the efficiency increases till it reaches to a maximum value. As the output is further increased, the efficiency starts decreasing slowly.

6. Why retardation test is carried on DC machine?
a) To find stray losses
b) To find eddy current losses
c) To find field copper losses
d) To find windage losses

Answer: a [Reason:] This test is generally employed to shunt generators and shunt motors. From this method we can get stray losses of a machine. Thus, if armature and shunt copper losses at any given load current are known then efficiency of a DC machine can be easily estimated.

7. In the DC motor the iron losses occur in __________
a) Field
b) Rotor
c) Brushes
d) Commutator

Answer: b [Reason:] Armature winding in a DC machine is located on rotor. Iron losses occur in an armature, hence in rotor. Iron losses are hysteresis loss and eddy current loss, which are seen prominently in armature rotor.

8. Which of the following losses is not under constant losses?
a) Friction and windage losses
b) No load core losses
c) Shunt field losses
d) Hysteresis losses

Answer: d [Reason:] All the losses that is friction and windage losses, no load core losses, shunt field core losses in shunt field and compound motors come under the category of constant losses, while iron losses come under category of variable losses.

9. Variable losses are proportional to ________
a) Armature current
b) Square of armature current
c) Inverse of armature current
d) Inverse of square of armature current

Answer: b [Reason:] Variable losses include losses in armature resistance and losses in series resistance, which are directly proportional to the square of armature current. It also includes stray load loss (iron plus copper) which is proportional to square of armature current.

## Set 3

1. Axis undergo shifting as a result of armature reaction, can be balanced by ______
a) Increase in armature current
b) Decrease in armature current
c) Introducing interpoles
d) Removing interpoles

Answer: c [Reason:] Apart from distortion of the resultant flux density wave, its MNA also gets shifted from its GNA by a small angle α so that the brushes placed in GNA are no longer in MNA as is the case in the absence of armature current, due to armature reaction. This effect is countered by the interpoles placed in GNA.

2. The choice of average coil voltage determines the minimum number of commutator segments for its design.
a) True
b) False

Answer: a [Reason:] The maximum allowable voltage between adjacent segments is 30–40 V, limiting the average voltage between them to much less than this figure. The choice of the average coil voltage determines the minimum number of commutator segments for its design, to avoid any flashover and ultimately short circuit.

3. Compensating winding will provide incomplete neutralization ____________
a) Under pole region
b) In interpolar region
c) Everywhere
d) Complete neutralization

Answer: b [Reason:] The compensating winding neutralizes the armature mmf directly under the pole while in the interpolar region, there is incomplete neutralization. Further, the effect of the resultant armature mmf in interpolar region is rendered insignificant because of large interpolar gap.

4. Cross-magnetizing effect of armature reaction can be reduced by __________
a) Removing saturation in teeth and pole-shoe
b) Making smooth pole shoes
c) Introducing saturation in teeth and pole-shoe
d) Cannot be determined

Answer: c [Reason:] The cross-magnetizing effect of the armature reaction can be reduced by making the main field ampere-turns larger compared to the armature ampere-turns such that the main field mmf exerts predominant control over the air-gap flux. This is achieved by introducing saturation in the teeth and pole-shoe.

5. Cross-magnetizing effect of armature reaction can be reduced by __________
a) Removing saturation in teeth and pole-shoe
b) Making smooth pole shoes
c) Chamfering the pole shoes
d) Cannot be determined

Answer: c [Reason:] By chamfering the pole-shoes which increases the air-gap at the pole tips. This method increases the reluctance to the path of main flux in a DC machine but its influence on the cross-flux is much greater. This is because the cross flux has to cross the air-gap twice.

6. To counter the effect of shift in MNA due to armature reaction, which of the following component can be shifted?
a) Poles
b) Commutator
c) Brushes
d) Cannot be determined

Answer: c [Reason:] To counter the effect of shift in MNA due to armature reaction, the brushes could be shifted. A small brush shift in appropriate direction, in the direction of rotation for generator and in opposite direction for motor, also helps in commutation.

7. Calculate the number of conductors on each pole piece required in a compensating winding for a 6-pole lap-wound dc armature containing 286 conductors. The compensating winding carries full armature current. Assume ratio of pole arc/ pole pitch = 0.7.
a) 6
b) 8
c) 9
d) 7

Answer: a [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). Ncw/pole = (Z/2AP) * (pole arc/pole pitch) = [286 / (2*6*6)] 0.7 = 2.78. Compensating conductors/pole = 2 * 2.78 = 6 (nearest integer).

8. A compensating winding with ampere-turns greater than peak ampere turns is required in order to neutralize the effect of armature reaction because _____________
a) Pole arc = Pole pitch
b) Pole arc > Pole pitch
c) Pole arc < Pole pitch
d) Can’t be determined using pole arc, pole pitch

Answer: b [Reason:] The number of ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)] * (pole arc/pole pitch). Thus, if compensating winding ampere turns are more then, pole arc is definitely greater than pole pitch.

9. If pole arc is less than pole pitch, a compensating winding will have ampere-turns _________ (compare to peak ampere turns).
a) Less
b) Equal
c) More
d) Can’t be specified

Answer: a [Reason:] The ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch) = [IaZ/(AP2)] * (pole arc/pole pitch). Thus, if compensating winding ampere turns are less then, pole arc is smaller than pole pitch and vice-versa.

10. What is the pole arc/pitch ratio, if 360 AT compensating winding is used where 1960AT is peak value?
a) 0.7
b) 0.8
c) 0.9
d) 0.6

Answer: d [Reason:] The ampere-turns required for compensating winding in a DC machine is ATcw /pole = ATa (peak) *(pole arc/pole pitch). If compensating winding of 360AT is used the, 360/1960 will give ratio of pole pitch /pole arc, equal to 0.6.

11. Only drawback of compensating winding is _______
a) Cost
b) Unavailability of material
c) Construction
d) Not a single drawback

Answer: a [Reason:] Compensating winding is the best method in order to prevent the effect of armature reaction and its consequences. Only problem is compensating winding is expensive, but it is must to use them in machines with heavy overloads occur.

## Set 4

1. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of flux/pole?
a) 0.188 Wb
b) 18.88 Wb
c) 0.0188 Wb
d) 1.888 Wb

Answer: c [Reason:] Flux is defined as flux density for a given surface area. Here, Surface area can be calculated and multiplied with B to give the value of flux. Flux= 2πr*l*B. now, for calculating flux per pole, divide it by P=4. So, Flux per pole after substituting all values is equal to 0.0188 Wb.

2. A 4-pole Dc wound machine is lap wound with 400 conductors. The pole shore is 20 cm long and average flux density over one-pole pitch is 0.4 T, the armature diameter being 30 cm. What is the value of induced emf?
a) 188 V
b) 276 V
c) 94 V
d) 188 mV

Answer: a [Reason:] Induced emf in a DC machine is equal to,

3. Coil torque for 20 kA armature current (T1) and 40 mA armature current (T2), will have ratio T1/T2=____ (assuming all other parameters same for both machines).
a) 1/2
b) 2/1
c) 1/4
d) 4/1

Answer: a [Reason:] Torque produced in a DC machine is directly proportional to number of coil turns, Flux per pole, number of poles and armature current. Thus, ratio T1/T2= 20/40 (assuming all other parameters same for both machines).

4. If the no load speed of DC motor is 1300 rpm and full load speed is 1100 rpm, then its voltage regulation is ____________
a. 12.56%
b. 18.18 %
c. 17.39%
d. 18.39%

Answer: b [Reason:] For A DC machine when all other parameters are fixed average coil emf generated is proportional directly to the speed of the dc motor. Voltage regulation is defined as ratio of difference of no load voltage and full load voltage to the full load voltage. VR= (1300-1100/1100)*100%.

5. If the average coil emf of a DC motor is doubled and flux is halved (keeping other parameters constant) then its shaft speed will become ___________
a. Twice of the original speed
b. Square of the original speed
c. Four times of the original speed
d. Half of the original speed

Answer: c [Reason:] Induced emf in a DC machine is equal to, From the emf equation we get speed of the shaft i.e. n α E/Z, when all other parameters are kept constant. So, when E is doubled n becomes twice the original, halving flux on reduced emf will quadrupled the speed of a DC machine.

6. A 4-pole wave wound DC motor drawing an armature current of 20 A has provided with 360 armature conductors. If the flux per pole is 0.015 Wb then the torque developed by the armature of motor is _______
a. 10.23 N-m
b. 34.37 N-m
c. 17.17 N-m
d. 19.08 N-m

Answer: b [Reason:] DC Machine torque equation: T = ka*∅*Ia. Here, ka= ZP/(2πA), Z= total armature conductors, P= No. of poles, A= No. of parallel paths. For a wave winding A=2. So, substituting all the values in the torque equation we get torque equal to 34.37 N-m.

7. In a DC machine, what is the torque induced beyond the pole shoes?
a) 0
b) 2/π *∅*i
c) π *∅*i/2
d) Can’t be calculated

Answer: a [Reason:] In a construction of a DC machine poles are located in magnetically neutral region. The magnetic field at the pole terminals in a DC machine will be equal to 0. Thus, cross product with the current flowing through the armature yields zero.

8. For a constant emf, if field current is reduced then the speed of the DC motor will_____
a) Remains same
b) Increases
c) Decreases
d) Can’t say

Answer: b [Reason:] When the field current is reduced, the field produced by the field winding also reduces. Thus, the term Φ from the emf equation also decreases. For all other parameters kept constant speed of the DC machine is inversely proportional to field. Hence, speed of DC motor will increase.

9. For an ideal DC machine, which phenomenon will reduce the terminal voltage?
a) Armature reaction
b) Commutation
c) Armature ohmic losses
d) All will contribute in reducing the terminal voltage

Answer: b [Reason:] In an ideal case, Commutation does not reduce the terminal voltage of a dc machine. In a non-ideal case, commutation takes place improperly at desired timings, thus losses contribute to the terminal voltage reduction. Armature reaction, ohmic losses due to winding resistance contribute to the losses in the terminal voltage.

## Set 5

1. No load point of DC generator is __________
a) Intersection of OCC and Rf line
b) Point on the Y axis at rated field current
c) Point on the X axis at rated terminal voltage
d) Can’t find through graphical interpretation

Answer: a [Reason:] Intersection of OCC with field resistance line gives the no-load point. All the value so obtained from x and y axis respectively gives the terminal voltage and field current of a DC generator at no-load.

2. Rf line intersects with OCC in ________

Answer: a [Reason:] Rf line is a straight line passing through the origin and having constant slope, so rising always in a positive direction. OCC starts from some positive value on Y axis and increases till maximum point, afterwards it starts becoming constant, where generally it intersects with Rf line.

3. How armature resistance effect is shown graphically?
a) By adding IaRa product horizontally with Rf line
b) By subtracting IaRa product vertically with Rf line
c) By adding IaRa product vertically with Rf line
d) By subtracting IaRa product horizontally with Rf line

Answer: c [Reason:] For representing the voltage drop in armature resistance, we add product IaRa vertically with Rf line at minimum 2 points and draw line parallel to Rf to get line with V + IaRa = constant.

4. For determining IaRa maximum _____________
a) Distance between Rf line and v+ IaRa line is taken
b) Distance between OCC and v+ IaRa line is taken
c) Maximum Distance between OCC and v+ IaRa line is taken
d) Can’t calculated graphically

Answer: c [Reason:] OCC when starts from some residual voltage on y axis, goes on increasing till some maximum value and then starts reducing slightly. The bulk is formed where maximum distance between OCC and v+ IaRa line is taken, to get effective frop in armature.

5. How demagnetization effect of armature reaction is shown graphically?
a) By shifting origin towards +ve y axis
b) By shifting origin towards +ve x axis
c) By shifting origin towards -ve y axis
d) By shifting origin towards -ve x axis

Answer: d [Reason:] The demagnetization caused by armature reaction can be quantified by equivalent field current Ifd which can be taken as proportional to the armature current Ia. So, by shifting the origin towards -ve x axis by Ifd we can show armature reaction graphically.

6. External characteristics of DC shunt motor lies in ______________
c) 1st and 2nd quadrant

Answer: c [Reason:] When external characteristics are plotted on graphs, and if effect of armature reaction is not considered graph lies in 1st quadrant only. When effect of armature reaction is considered we shift the origin thus, characteristics lies in 2nd quadrant as well.

7. From magnetization characteristic at If = 7.1 A, Ea = 225 V at 1000 rpm. What will be the terminal voltage at speed 950 rpm?
a) 225 V
b) 235 V
c) 214 V
d) 220 V

Answer: c [Reason:] Speed is directly proportional to the back emf of a machine. So, Ea = 225 V at 1000 rpm Ea (950 rpm) = 225 x 950 / 1000, this will give speed of given DC machine at 950 rpm. Upon calculations we get, Ea = 213.7 V= 214 V approx.

8. For a given compound DC machine, Net field current obtained from characteristic is equal to 7.5 A, where shunt field current is equal to 5 A, armature current is 505 A, demagnetizing current equal to 0.95 A and shunt field winding of 1000 turns at rated speed of 1000 rpm. What will be the series field turns?
a) 7
b) 8
c) 5
d) 9

Answer: a [Reason:] From the excitation balance equation, If + [Nse/Nf] Ia – Ifd =If (net). 5 + 505(Nse /1000) – 0.95 = If (net). Calculating for Nse by substituting If (net), we get Nse = 6.8 that is 7 turns.

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