Design Electrical Machines Pune University MCQs

 This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Electrical Conducting Materials”.

1. What is the basic property of electrical conducting materials?

a) allows the passage of current through the materials

b) blocks the passage of current through the materials

c) leaks the current through the materials

d) reverses the direction of current in the materials

Answer: a

Explanation: The basic property of conducting materials is to allow the flow of charges, and align them in a particular direction. The process is nothing but the flow of current in the materials.

2. What is the correct classification of the conducting materials?

a) low resistivity, low conductivity

b) low resistivity, high conductivity

c) high resistivity, high conductivity

d) medium resistivity, medium conductivity

Answer: b

Explanation: Resistivity is inversely proportional to the conductivity. So, if the material consists of high resistivity then it will obviously have low conductivity and vice versa.

3. Example of low resistivity material is _____________

a) silver

b) manganese

c) magnesium

d) tungsten

Answer: a

Explanation: Silver is the low resistivity material of all given materials. Tungsten is a part of high resistivity materials. The other two materials do not have a fixed resistivity and they vary with temperature.

4. Example of high resistivity material is ________________

a) copper

b) gold

c) aluminum

d) carbon

Answer: d

Explanation: Carbon is the highly resistivity material of all the materials. Whereas the other 3 materials are associated with low resistivity property in nature.

5. High resistivity material used in making the filaments of incandescent lamps.

a) true

b) false

Answer: a

Explanation: It is because to protect the lamps from getting over-heated. If the filaments get over-heated, it can lead to the bursting of the lamps.

6. What materials are used as conductors in the Transmission and Distribution sector?

a) copper

b) silver

c) tungsten

d) carbon

Answer: a

Explanation: Conductors in power system require less resistivity, highly malleable, highly ductile and less cost. Silver has all the above properties, but it is highly costly. So that makes copper highly suitable.

7. What are the properties of Conducting Materials with respect to temperature coefficient of resistance and tensile strength?

a) low temperature coefficient, low tensile strength

b) low temperature coefficient, high tensile strength

c) high temperature coefficient, low tensile strength

d) high temperature coefficient, high tensile strength

Answer: b

Explanation: The resistance of the material should not increase with temperature rise. This can lead to the loss of conduction property. High tensile strength allows in withstanding external disturbances, for smooth functioning.

8. What are the conditions of the conducting materials with respect to melting point and resistance to corrosion?

a) high melting point, low resistance to corrosion

b) low melting point, low resistance to corrosion

c) high melting point, high resistance to corrosion

d) low melting point, high resistance to corrosion

Answer: c

Explanation: High melting point, allows the materials to withstand low temperatures. High resistance to corrosion allows the material to avoid corrosion, to conduct effectively.

9. How should the conducting materials be in terms of malleability and ductility?

a) highly malleable, less ductile

b) less malleable, less ductile

c) highly malleable, highly ductile

d) less malleable, highly ductile

Answer: c

Explanation: The materials, having high malleability allow smooth conduction in transmission and distribution. The materials having high ductility help in producing wires flexibly for conduction.

10. Aluminum has high conductivity than Copper.

a) true

b) false

Answer: b

Explanation: Copper has high conductivity than Aluminum. The conductivity of Copper is 58.14*10 6 s/m and the conductivity of aluminum is 37.2*10 6 s/m.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “High Conductivity Materials”.

1. Which material has the highest conductivity of all materials?

a) Silver

b) Copper

c) Gold

d) Tungsten

Answer: a

Explanation: On a scale of 100, silver has 100 percent on high conductivity, copper has 97. When compared to silver and copper gold has only 76 percent. Tungsten is not a material of this group.

2. High conductivity materials are used in electrical machines.

a) True

b) False

Answer: a

Explanation: These materials have low resistivity. Hence they allow for the good flow of current, which in turn allows the proper operation of the machine.

3. What are the characteristics of high conductivity materials based on cost and flexibility?

a) Low cost, low flexibility

b) Low cost, high flexibility

c) High cost, low flexibility

d) High cost, high flexibility

Answer: b

Explanation: Cost should be always less, in order to help in purchase of many quantities of the material for more applications. It should also be highly flexible, in order to mould according to people’s choice.

4. What is the temperature coefficient of silver?

a) 0.00386 per 0 C

b) 0.0034 per 0 C

c) 0.00429 per 0 C

d) 0.0038 per 0 C

Answer: d

Explanation: 0.0034 per 0 C relates to the temperature coefficient of Gold, whereas 0.00429 per 0 C is the temperature coefficient of Aluminum. 0.00386 per 0 C corresponds to temperature coefficient of Copper.

5. Silver is not used in practical electrical machines.

a) True

b) False

Answer: a

Explanation: Silver has lots of properties which can make it suitable to be used in practical use. But the high cost factor which occurs to Silver makes it used only for important instruments.

6. What is the conductivity of Copper?

a) 0.6329*10 6 mho/cm

b) 0.5952*10 6 mho/cm

c) 0.4529*10 6 mho/cm

d) 0.3773*10 6 mho/cm

Answer: b

Explanation: 0.6329*10 6 mho/cm relates to the conductivity value of Silver. 0.4529*10 6 mho/cm relates to the conductivity value of Gold and 0.4529*10 6 mho/cm relates to the conductivity value of Aluminum.

7. What is the melting point of aluminum?

a) 660 0 C

b) 1085 0 C

c) 962 0 C

d) 1064 0 C

Answer: b

Explanation: 660 0 C is the melting point of Aluminum. 962 0 C relates to the melting point of Silver and 1085 0 C is the melting point of Copper.

8. What is the specific gravity of aluminum?

a) 8.96 gm/cm 3

b) 19.30 gm/cm 3

c) 2.70 gm/cm 3

d) 10.49 gm/cm 3

Answer: c

Explanation: 8.96 gm/cm 3 is the specific gravity of Copper. 19.30 gm/cm 3 relates to the specific gravity of Gold and 10.49 gm/cm 3 is the specific gravity of Silver.

9. Which two elements are used in precious instruments?

a) Copper, Silver

b) Gold, Silver

c) Copper, Aluminum

d) Gold, Aluminum

Answer: b

Explanation: Silver is used only in precious instruments because of its high cost. Gold, on the other hand, is not only costly but also not suitable for many practical applications and can lose its properties easily.

10. Which property of aluminum it the most preferred element?

a) good conductivity

b) highly malleable, highly ductile

c) most abundant element

d) good corrosion resistant

Answer: c

Explanation: All the other elements among Silver, Copper, Gold have the other 3 properties along with Aluminum. But all the above mentioned materials are not highly abundant, which is also an important factor.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Magnetic Materials”.

1. What is the property of magnetic materials?

a) Resistivity

b) Conductivity

c) Permeability

d) Ductility

Answer: c

Explanation: There are many properties of magnetic materials, and permeability is one among them. The other 3 properties are related to other materials like conducting and insulating materials.

2. What is the property of permeability in magnetic materials?

a) how easily the magnetic flux is broken/clear

b) how easily the magnetic flux is set up

c) how long the magnetic flux takes to form

d) how long the magnetic flux takes to clear

Answer: b

Explanation: The basic operation of magnetic material is to form magnetic flux. Permeability is the ability of the material to determine how easily the magnetic flux is set up.

3. What is the representation of permeability?

a) coercivity/retentivity

b) flux/flux density

c) magnetic force/magnetic flux density

d) magnetic flux density/magnetic force

Answer: d

Explanation: Permeability is the property which deals, with the relationship with magnetic flux density and magnetic force. Magnetic force/Magnetic flux density deals with the reciprocal of permeability. Coercivity/Retentivity deals with the terms of B-H curve.

4. How should the permeability and number of ampere turns for good magnetic materials be?

a) high permeability, high ampere turns

b) high permeability, low ampere turns

c) low permeability, low ampere turns

d) low permeability, high ampere turns

Answer: b

Explanation: High permeability is always required in magnetic materials for its good operation. At the same time high permeability leads to less ampere turns in the materials.

5. Is retentivity associated with B-H curve?

a) Yes

b) No

Answer: a

Explanation: B-H curve deals with the concepts of retentivity and coercivity. The property of retentivity can be shown in the B-H curve by an increasing curve in the curve.

6. What is the property of retentivity in magnetic materials?

a) After removal of external magnetic fields, magnetization exists

b) After removal of external magnetic fields, magnetization doesn’t exist

c) After removal of internal magnetic fields, magnetization exists

d) After removal of internal magnetic fields, magnetization doesn’t exist

Answer: a

Explanation: Magnetic materials have the property of retentivity in which the magnetic flux produced acts according to the external magnetic field. When the external field is removed, the magnetization in the materials doesn’t deform immediately.

7. What is coercivity force in magnetic materials?

a) The force required to add upon the existing magnetization

b) The force required to remove the existing magnetization

c) The force required to produce magnetic flux

d) The force required to break magnetic flux

Answer: b

Explanation: Magnetic materials generally have the property of retaining magnetization, even if the external magnetic field is removed. So, coercive force is the force that is required to reduce the magnetization.

8. What are magnetic hard materials?

a) High retentivity, low coercivity

b) High retentivity, high coercivity

c) Low retentivity, low coercivity

d) Low retentivity, high coercivity

Answer: b

Explanation: High retentivity is required for protecting the magnetic materials from losing its magnetic property. High coercivity is required to reduce the effect of retentivity to protect the material.

9. What is reluctance in magnetic materials?

a) Allows the buildup of magnetic flux

b) Reduces the buildup of magnetic flux

c) Resists the buildup of magnetic flux

d) Increases the buildup of magnetic flux

Answer: c

Explanation: Reluctance, as the name suggests, is something which is reluctant or hesitant to do. As per the magnetic terms it resists the building up of magnetic flux in the materials.

10. High Reluctance affects the performance of magnetic materials.

a) True

b) False

Answer: a

Explanation: High reluctance means the materials resist in building up the magnetic flux to a higher extent. So, for the proper functioning the reluctance values should be as low as possible.

11. What is the unit of reluctance in magnetic materials?

a) Henry/m

b) Weber/m 2

c) Ampere-turns/Weber

d) Ampere-turns/m

Answer: c

Explanation: Henry/m deals with the unit of permeability. Weber/m 2 deals with the unit of magnetic field. Reluctance is the opposite of permeance.

12. How many classifications of magnetic materials are present?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: There are basically 4 properties in magnetic materials and 5 classifications. They are diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, ferrimagnetic.

13. What is the property of ferromagnetic materials?

a) Negative magnetization

b) Magnetization slightly less than 1

c) Magnetization slightly greater than 1

d) Magnetization very much higher than 1

Answer: d

Explanation: Negative magnetization denotes the property of Diamagnetic materials. Magnetization slightly greater than 1 denotes the property of Paramagnetic materials. Ferromagnetic materials have magnetization in the range of 1000+.

14. What is the example of diamagnetic materials?

a) Quartz

b) Pyrite

c) Montmorillonite

d) Biotite

Answer: a

Explanation: The other 3 materials are paramagnetic in nature, which means magnetization is slightly above 1. Quartz is a diamagnetic material in which the magnetization is negative.

15. What is the example of ferromagnetic materials is?

a) Magnetite

b) Hematite

c) Nickel

d) Biotite

Answer: a

Explanation: Hematite denotes the example of antiferromagnetic materials. Nickel denotes an example of ferromagnetic materials. Biotite denotes the example of paramagnetic materials.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Insulating Materials”.

1. What is the property of insulating materials?

a) Prevents the unwanted flow of current

b) Allows the unwanted flow of current

c) Increases the unwanted flow of current

d) Decreases the unwanted flow of current

Answer: a

Explanation: Conductors, allow the flow of current through the material. Insulators are the opposite of conductors. The material doesn’t allow the flow of current through them.

2. In the Transmission and Distribution sector, where should the insulators be placed?

a) Between towers and poles

b) Between towers and ground

c) Between towers and conductors

d) Between conductors and ground

Answer: c

Explanation: The insulators are used to block the flow of unwanted current. In power system, already the tower and the conductors are grounded. Thus the insulators are connected between towers and conductors.

3. What is the main cause for the failure of overhead line insulators?

a) Surges

b) Flashover

c) Arching

d) Grounding

Answer: b

Explanation: In overhead lines, there occurs a flow of abnormal over voltages. This abnormal over voltages, causes flashover. This flashover causes damage to overhead line insulators.

4. What happens when some serious phenomenon occurs in the insulators?

a) Puncher is produced in the insulator body

b) Insulator body bulges

c) Insulator body bursts

d) Insulator body tears apart

Answer: a

Explanation: The serious phenomenon is the abnormal over voltage, produced in the insulators. Due to that, flashover occurs in the insulators. This causes puncher of the insulator body.

5. Insulation Resistance should be high in insulators.

a) True

b) False

Answer: a

Explanation: Insulation Resistance is very important in the performance of insulating materials. If the insulation resistance becomes low, high flow of current occurs and can damage the material.

6. How should the properties of strength and dielectric strength in insulating materials?

a) High strength, low Dielectric strength

b) Low strength, low Dielectric strength

c) High strength, high Dielectric strength

d) Low strength, high Dielectric strength

Answer: c

Explanation: The insulator should have high strength in order to prevent the insulating materials. The insulator should have high dielectric strength, in order to hold the electric field without breaking down.

7. What is property of porosity and temperature change in insulating materials?

a) Less, less affected

b) Less, highly affected

c) High, highly affected

d) High, less affected

Answer: a

Explanation: The insulating materials should have less porosity as it should not lose the internal properties due to holes. The material should also be less affected by temperature change in order to preserve its properties.

8. What is the dielectric strength of porcelain insulators?

a) 60 kV/cm

b) 140 kV/cm

c) 50 kV/cm

d) 40 kV/cm

Answer: a

Explanation: Porcelain has a dielectric strength of 60kV/cm. 140 kV/cm denotes the dielectric strength of glass insulator.

9. What is the dielectric strength, coefficient of thermal expansion of glass with respect to porcelain insulators?

a) High, high

b) High, low

c) Low, low

d) Low, high

Answer: b

Explanation: Glass has a higher dielectric strength  when compared to porcelain  and glass has a lower coefficient of thermal expansion when compared to porcelain.

10. Glass has lower tensile strength compared to porcelain insulators.

a) True

b) False

Answer: b

Explanation: Glass insulators have all properties better than that of porcelain. Glass has high dielectric strength, low coefficient of thermal expansion and then High tensile strength than that of porcelain.

11. What is the other name of Polymer Insulator?

a) Moisture insulator

b) Core insulator

c) Composite insulator

d) Mixed insulator

Answer: c

Explanation: It is also known as composite insulator. It is known as composite insulator because it consists of both core and the weather sheds in them.

12. How many classifications of overhead line insulators are there?

a) 3

b) 4

c) 5

d) 6

Answer: a

Explanation: There are basically 3 types of overhead line insulators. They are Pin type, Suspension type and Stray Insulator type.

13. How many types of electrical insulators are present on the basis of voltage application?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are two types of insulators based on voltage application. They are Stay Insulators and Shackle Insulators.

14. How many discs are used in suspension insulators for 220kV?

a) 3

b) 4

c) 8

d) 14

Answer: d

Explanation: 3 discs are used when voltage is 33kV. 4 discs are used when voltage is 66kV. 8 discs are used when voltage application is 132kV.

15. What is the other name of the shackle insulator?

a) String

b) Hanging

c) Spool

d) Post

Answer: c

Explanation: String is the other name of strain insulators, whereas, hanging is the other name of suspension insulators. Post insulator is otherwise Pin insulators.

This set of Design of Electrical Machines Interview Questions and Answers focuses on “Temperature Rise and Insulating Materials”.

1. How many number of insulation classes are present with respect to electrical equipment?

a) 5

b) 6

c) 7

d) 8

Answer: c

Explanation: There are 7 classes of insulation with respect to electrical equipment. They are Class Y, Class A, Class E, Class B, Class F, Class H, Class C.

2. How many classes have their temperatures above 100°C?

a) 5

b) 6

c) 7

d) 8

Answer: b

Explanation: There are 7 insulation classes present in relation with temperature. Of the 7 classes, there are 6 classes whose temperatures are greater than 100°C. Class Y is the only class having temperature less than 100.

3. How many classes have their temperatures above 150°C?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: Of 7 classes of insulation, there are 3 classes whose temperatures are above 150°C. They are classes F, H, C of insulation.

4. Which class has the lowest and the highest temperature?

a) Class Y, Class C

b) Class Y, Class H

c) Class H, Class C

d) Class B, Class H

Answer: a

Explanation: Class Y belongs to the lowest insulation class of having temperature of about 90°C. Class C is the highest insulation class of having temperature above 180°C.

5. Class A has higher temperature than Class E.

a) True

b) False

Answer: b

Explanation: Class E has a higher temperature than that of Class A. The temperature of Class A is 105°C and the temperature of Class E is 120°C.

6. What is the temperature of Class B?

a) 120°C

b) 130°C

c) 155°C

d) 180°C

Answer: b

Explanation: 120°C refers to the temperature of Class E. 155°C refers to the temperature of Class F. 180°C refers to the temperature of Class H.

7. Which among the following is the example of Class Y?

a) Varnish

b) Insulation oil

c) Paper

d) Resins

Answer: c

Explanation: Varnish is an example of Class A. Even insulation oil is an example of Class A. Resins is an example of Class E.

8. Which among the following is the example of Class B?

a) Inorganic material with adhesives

b) Hard fiber

c) Wood

d) Impregnated oil

Answer: a

Explanation: Impregnated oil is the example of Class H. Hard Fiber is the example of Class Y and Wood is the example of Class A.

9. Which among the following is an example of Class F?

a) Paper lamination

b) Nitrile rubber

c) Asbestos

d) Silicone

Answer: d

Explanation: Paper lamination is the example of Class E. Nitrile Rubber is the example of Class A. Asbestos is the example of Class B.

10. Silicone rubber is an example of Class H.

a) True

b) False

Answer: a

Explanation: Class H is one of the insulation classes having temperature about 180°C. Silicone rubber is one of the examples of Class H.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Rating of Machines”.

1. What is the concept of power rating of machines with respect to voltage?

a) the required supply voltage for smooth running of the machine

b) the required supply voltage for stopping the machine

c) the required supply voltage for speeding the machine

d) the required supply voltage for slowing up the machine

Answer: a

Explanation: Power rating is nothing but the standard value at which the machine is said to be safe in operation. Rating determines the voltage which allows the smooth running of the machine.

2. What is the concept of power rating of machines with respect to current?

a) Maximum permissible amount of current that can easily flow

b) Minimum permissible amount of current that can easily flow

c) Maximum permissible amount of current that can stop the machine

d) Maximum permissible amount of current that can stop the machine

Answer: a

Explanation: Power rating always deals with two variables, one is current and the other is voltage. When it comes to the current, it is the maximum permitted current that can be allowed into the machine.

3. What happens if there is insufficient rating of the machine?

a) The efficiency of the machine increases

b) The efficiency of the machine improves

c) Damage and shutdown occurs

d) Loading problems occur

Answer: c

Explanation: When there is insufficient rating, it can lead to the damage of the windings of the machine. It indirectly, leads to the shutdown of the machine to avoid more hazards.

4. What happens if the power ratings of the machine are decided liberally?

a) Damage occurs to the machine

b) Efficiency of the machine improves

c) Long life of the machine

d) Uneconomical usage of the machine

Answer: d

Explanation: If the power rating becomes very liberal, then it causes a high initial cost. Along with the high initial cost, loss of energy also occurs and leads to uneconomical usage.

5. If the power ratings are crossed, machine breakdown occurs.

a) True

b) False

Answer: a

Explanation: Every machine has a permissible limit for both voltage and current, for its efficient operation. If the limit is crossed, it will lead to the breakdown of the machine.

6. Which are the important criteria related to the power ratings of the machine?

a) Heat should be prevented from generation

b) Heat should be dissipated through power ventilation, irrespective of the time

c) Heat should be prevented through power ventilation within a short time period

d) Heat should be converted to some useful form

Answer: c

Explanation: Due to the components present in the machine, I 2 R losses occur in the machine. Due to this, heat is produced, and proper thermal ventilation should be provided to prevent the machine from breakdown.

7. What is the concept of thermal loading?

a) Output power is indirectly proportional to the temperature rise

b) Output power is indirectly proportional to the square of temperature rise

c) Output power is directly proportional to the temperature rise

d) Output power is directly proportional to the square of temperature rise

Answer: c

Explanation: Thermal loading is nothing, but the increase of the output power with respect to the temperature rise. It can lead to the power rating levels being crossed.

8. What is the ideal condition for thermal dissipation?

a) Heat generated > Heat Dissipated

b) Heat generated < Heat Dissipated

c) Heat generated = Heat Dissipated

d) Heat generated = 0

Answer: c

Explanation: For ideal thermal dissipation, the heat dissipated should be equal to the heat generated. In that case, there will be very less power loss and high efficiency.

9. What is the main objective of power ratings of machines?

a) helps in building a suitable thermal model of machines

b) helps in building a suitable physical model of machines

c) helps in classifying the machines into different types

d) helps to improve the machine efficiency

Answer: a

Explanation: The power ratings, mainly help in building the thermal model. It helps in reducing the heat losses and helps in bringing out a smooth and efficient operation of the machines.

10. Power ratings help in classifying machines to different classes of duties.

a) true

b) false

Answer: a

Explanation: The classification of the machines into different classes of duties depends on the power ratings. The ratings also helps in identifying the different types of machines available under different classes.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Armature”.

1. How many type of armature windings are present in the DC machine armature?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are basically two types of armature windings present. They are known as lap winding and wave winding.

2. Which factor determines the difference between the types of armature windings?

a) brush connection

b) slip ring connection

c) commutator connection

d) pole connection

Answer: c

Explanation: The armature windings are all connected to the commutator in the case of DC machines. Commutator is nothing but device which converts dc voltage to ac voltage.

3. What is the meaning of pole pitch?

a) type of armature slot

b) number of armature slots/pole

c) number of poles/armature slot

d) number of poles/number of armature slots

Answer: b

Explanation: Pole pitch is the concept which is used in the armature slot design. It is nothing but the ratio between the number of armature slots present to the number of poles used in the machine design.

4. A 4 pole DC machine has 36 number of armature conductors. What is the pole pitch?

a) 9

b) 1/9

c) 9 2

d) 1/9 2

Answer: a

Explanation: Pole pitch = Number of armature slots/pole

Pole Pitch = 36/4 = 9.

5. The choosing of the slot values for armature depends on the type of windings.

a) true

b) false

Answer: a

Explanation: The slot values are chosen based on the type of windings. If wave winding is used, then multiples of pole pair is chosen as slot values and if lap windings are used the slot values are continuous and not multiples of pole pair.

6. What is the condition for choosing the armature current/parallel path in armature design?

a) > 200 ampere for both lap and wave windings

b) >200 ampere for lap winding, <200 ampere for wave winding

c) < 200 ampere for both lap and wave windings

d) <200 ampere for lap winding, >200 ampere for wave winding

Answer: c

Explanation: While designing the armature slots, the armature current should be less than 200 ampere. The armature current should also never cross 200 ampere for the lap winding.

7. What is the range of armature slot pitch in the armature slot design?

a) 20mm – 25mm

b) 25mm – 35mm

c) 30mm – 35mm

d) > 35mm

Answer: b

Explanation: For the armature design, the lowest value of slot pitch is generally chosen as 25mm. The highest value of slot pitch is generally chosen as 35mm.

8. For a DC machine, the armature slot pitch is 35mm and the diameter is 0.2m. What is the number of armature slots for the machine?

a) 17

b) 18

c) 19

d) 20

Answer: b

Explanation: Number of armature slots = /Number of armature slot pitch

Number of armature slots = /35*10 -3

Number of armature slots = 17.94 = 18.

9. What is the formula to reduce flux pulsations?

a) armature slot/number of poles = integer

b) armature slot/number of poles = integer + 0.5

c) armature slot/number of poles = integer – 0.5

d) armature slot/number of poles = integer +-0.5

Answer: d

Explanation: First the number of armature slots is calculated. Then the ratio of the number of armature slots to the number of poles is approximately equal to integer+-0.5.

10. The number of coils chosen should be minimum in number.

a) true

b) false

Answer: a

Explanation: The number of coils chosen for the armature slot design should always be minimum in number. This is because the machine cost reduces and it becomes cost efficient.

11. What should be the range of the integer value while calculating the formula to reduce the flux pulsations?

a) 8-14

b) 9-15

c) 9-16

d) 9-17

Answer: c

Explanation: While calculating the formula to reduce the flux pulsations, the lowest value is generally chosen as 9. The highest value of integer chosen is generally 16.

12. What is ‘Z/Sa’ calculation during design of armatures?

a) armature poles/slots

b) armature conductors/slots

c) armature slots/poles

d) armature slots/conductors

Answer: b

Explanation: This is the second step in the armature design. Firstly, the number of armature conductors is found out. Then the number of slots used is taken and the ratio of both gives the value of ‘Z/Sa’.

13. What is the formula for calculating the minimum number of coils?

a) Cmin = Voltage

b) Cmin = Voltage/5

c) Cmin = Voltage/10

d) Cmin = Voltage/15

Answer: d

Explanation: The voltage through the armature is first calculated from the data given. Then the value of voltage is divided by 15, to get the value of minimum number of coils.

14. Given Z=228 conductors, number of coils C=38 for a DC machine. What is the turns/coils ratio?

a) 4

b) 3

c) 7

d) 3

Answer: d

Explanation: Turns/coils = Number of armature conductors/

Turns/coils = 228/38*2 = 3.

15. For a DC generator, given the load current is 510 ampere, field current is 1.4 ampere. What is the value of armature current?

a) 508.6 ampere

b) 714 ampere

c) 511.4 ampere

d) 364.28 ampere

Answer: c

Explanation: For DC generator, Armature Current = Load Current + Field Current

Armature Current = 510 + 1.4 = 511.4 ampere.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Commutator and Brushes”.

1. What is a commutator in DC machines?

a) electrical device which reverses the current direction between the rotor and external circuit

b) mechanical device which reverses the current direction between the rotor and external circuit

c) electrical device which allows the current flow between the rotor and external circuit

d) mechanical device which allows the current flow between the rotor and external circuit

Answer: a

Explanation: Commutator is an electrical device found only in DC machines. It is used to reverse the current direction between rotor and external circuit. It also changes dc voltage to alternating voltage.

2. What is the use of brushes in DC machines?

a) to connect the parts of the machine to the external circuit

b) to conduct current between moving parts

c) to conduct current between stationary wires and moving parts

d) used for smooth conduction of current

Answer: c

Explanation: Brushes are another important part in the construction of DC machine. They are connected in the lower end of the machine to allow the current flow between the moving parts and stationary wires.

3. Which material is commonly used in brushes?

a) copper

b) carbon

c) silicon

d) steel

Answer: b

Explanation: Carbon is the most commonly used material in the manufacture of brushes. It is because carbon has high melting point, and is also less prone to high temperatures.

4. What is the total number of design steps available for the commutators in DC machines?

a) 4

b) 3

c) 2

d) 5

Answer: a

Explanation: There are basically 4 design steps available for the commutators. They include the finding of the number of commutator segments, voltage across the commutator, width of the commutator and length of the commutator.

5. What is the total number of design steps for the brushes in DC machines?

a) 4

b) 3

c) 5

d) 6

Answer: c

Explanation: There are basically 5 steps in the brush design. They include the calculation of the brush current, number of brushes, thickness of the brush, width of the brush, total commutator losses.

6. What factor does the diameter of the commutator depend on?

a) length of the commutator

b) speed of the commutator

c) peripheral speed of the commutator

d) opening of the commutator

Answer: c

Explanation: Peripheral speed is a term which is used in the design of the commutators of DC machines. If the peripheral speed of the machine gets crossed, then the diameter should be reduced.

7. For a DC machine, given Diameter of the commutator= 0.48 m, Speed = 600 rpm. What is the voltage across the commutator?

a) 15 V

b) 15.5 V

c) 15.2 V

d) 15.1 V

Answer: d

Explanation: Voltage across the commutator = 3.14**Speed

Voltage = 3.14*0.48*600 = 15.1 V.

8. What is the other name for commutator segment pitch of DC machines?

a) width of the commutator

b) length of the commutator

c) breadth of the commutator

d) height of the commutator

Answer: a

Explanation: Commutator segment pitch is otherwise known as the width of the commutator. It is one of the commutator design steps.

9. What is the formula for finding out the width of the commutator of the DC machines?

a) width of the commutator = 3.14*

b) width of the commutator = 3.14**

c) width of the commutator = 3.14* / 

d) width of the commutator = Number of armature coils / 3.14*

Answer: c

Explanation: First by the design of the armature, the number of armature coils is found out. Then the diameter of the commutator is measured.

10. The diameter of the commutator should be 0.2-0.4 times the main diameter for a good design.

a) true

b) false

Answer: b

Explanation: For a good design of the commutator, the diameter of the commutator should be in the range of 0.6-0.8 times of the main diameter. If the value goes above or below this range, we should not choose that value.

11. What should be the range for the width of the commutator in a good design?

a) < 4 mm

b) > 4 mm

c) 3-4 mm

d) < 3 mm

Answer: b

Explanation: For a good design of the commutator, the width of the commutator should be greater than 4 mm. If the value goes below 4 mm, we should not choose that value.

12. The design of commutator and the brushes of DC machines are interconnected.

a) true

b) false

Answer: a

Explanation: The design of brushes and commutators are interconnected, because in the calculation of the length of the commutator, the width of the brush, number of brushes are used. In the same way, brush values are also used in calculation of commutator loss.

13. What is the formula for calculation of brush current of DC machine for wave winding?

a) brush Current = 2* / P

b) brush Current =  / P

c) brush Current = Armature Current

d) brush Current = P/2*

Answer: c

Explanation: Brush Current = 2* / P is the formula for the calculation of brush current of DC machine for lap winding.

14. What is the formula for brush contact loss of DC machines?

a) brush contact loss = armature current + brush contact voltage

b) brush contact loss = armature current – brush contact voltage

c) brush contact loss = armature current * brush contact voltage

d) brush contact loss =  / 2

Answer: c

Explanation: For calculation of the brush contact loss, first the armature current is obtained. Then the voltage through the brush contacts is calculated and the product gives the brush contact loss.

15. What is the formula for total commutator loss for DC machine?

a) brush contact loss + brush friction loss

b) brush contact loss – brush friction loss

c) brush contact loss * brush friction loss

d) brush contact loss / brush friction loss

Answer: a

Explanation: Brush contact loss = brush contact loss = armature current * brush contact voltage

Brush friction loss =  * Brush Pressure * Area of the brush*Commutator voltage.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Output Equations”.

1. With what component is the output equation of DC machines related to?

a) power

b) voltage

c) current

d) losses

Answer: a

Explanation: The output equation generally deals with the power generated in the machine. Power of the machine relates the voltage and current flowing though the machine.

2. What can be found out using the output equation of the DC machine?

a) main dimensions

b) angle of rotation

c) losses

d) efficiency

Answer: a

Explanation: The output equation of DC machine is mainly used to obtain the main dimensions of the machine. Main dimensions are very important in calculation of various factors related to the machine.

3. What are the components of the main dimensions of output equation of DC machine?

a) diameter

b) length

c) diameter and length

d) voltage

Answer: c

Explanation: Main dimensions generally deal with the diameter of the machine. It also deals with the length of the machine.

4. What is the starting equation for deriving the output equation of DC Machines?

a) P = Generated Emf + Armature Current

b) P = Generated Emf – Armature Current

c) P = Generated Emf * Armature Current

d) P = Generated Emf / Armature Current

Answer: c

Explanation: The first equation for deriving the output equation of DC machines starts from this equation. The generated emf is multiplied along with the armature current.

5. The output equation of the DC machines can be used to calculate the speed of the machine.

a) true

b) false

Answer: a

Explanation: The output equation can be used to calculate the speed of the machine. The output coefficient of the DC machine, diameter and length of the conductors must be provided.

6. What is the output equation of DC machine?

a) Output power = Output Coefficient of the machine* Diameter 2 * Length * Speed in rpm

b) Output power = Output Coefficient of the machine* Diameter 2 * Length * Speed in rps

c) Output power = Output Coefficient of the machine* Diameter 2 * Length / Speed in rps

d) Output power = Output Coefficient of the machine* Diameter 2 * Length / Speed in rpm

Answer: b

Explanation: While calculating the output power of the DC machine, the coefficient of output equation, the diameter of the conductor, the length of the conductor and the speed is required. Speed should be in rotations per second only.

7. What are the terms related to deriving the output equation of the DC machine?

a) specific electric loading

b) specific magnetic loading

c) thermal coefficient of machine

d) specific electric and magnetic loading

Answer: d

Explanation: For deriving the output equation, both the specific magnetic and electric loading formulas are made use of. By substituting the 2 formulas, the output equation is derived.

8. For a DC generator, what is the output power equation?

a) Output power = Generated Power * efficiency

b) Output power = Generated Power / efficiency

c) Output power = Generated Power – efficiency

d) Output power = efficiency / generated power

Answer: b

Explanation: For DC generator, the efficiency also taken into account, while calculating the final output power value. Whereas, the same is not considered while calculating for motor.

9. For a DC motor, what is the output power equation?

a) Output power = Generated Power / efficiency

b) Output power = Generated Power * efficiency

c) Output power = Generated Power

d) Output power = Generated Power + efficiency

Answer: c

Explanation: Output power = Generated Power / efficiency – For DC Generator

Output power = Generated Power – For DC Motor.

10. For a DC generator, given D = 0.35 m, L = 0.21 m, Coefficient of output = 108.5, efficiency = 0.9, speed = 1400 rpm. What is the output power of the DC generator?

a) 65.12 W

b) 72.35 KW

c) 72.35 W

d) 65.12 KW

Answer: b

Explanation: Generated power = 108.5 * 0.35 * 0.35 * 0.21 *  = 65.12 KW

Output Power = Generated Power / Efficiency = 65.12 / 0.9 = 72.35 KW.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Selection of Number of Poles”.

1. How is the selection of number of poles made in AC and DC machines?

a) any number of poles can be used for both AC and DC machines

b) fixed number of poles in both AC and DC machines

c) fixed number of poles in DC machines and any number of poles in AC machines

d) fixed number of poles in AC machines and any number of poles in DC machines

Answer: d

Explanation: When it comes to DC machines, any number of poles can be made use of, but it should be within a range. Whereas in AC machines the number of poles is fixed by supply frequency and speed.

2. How many considerations are present in the selection of number of poles?

a) 4

b) 5

c) 6

d) 7

Answer: d

Explanation: There are 7 consideration in selection of number of poles. They are frequency, weight of iron parts, weight of copper, length of commutator, labor charges, flash over, distortion of field form.

3. What is the formula for frequency of flux reversals?

a) f = p*n

b) f = p/n

c) f = n/p

d) f = /2

Answer: d

Explanation: Frequency is directly proportional to the number of poles . It is also proportional to the speed of the machine as well.

4. What is the range of frequency during the selection of number of poles?

a) 20-50 Hz

b) 25-40 Hz

c) 25-50 Hz

d) >50 Hz

Answer: c

Explanation: While selecting the number of poles, the lowest value of frequency should be minimum 25 Hz. The highest value of frequency should be limited to 50 Hz.

5. What is the relation of hysteresis loss and weight of iron parts with respect to increase of number of poles?

a) decrease in hysteresis loss, increase in weight

b) decrease in hysteresis loss, decrease in weight

c) increase in hysteresis loss, increase in weight

d) increase in hysteresis loss, decrease in weight

Answer: b

Explanation: With larger number of poles, the area of cross section can be reduced, henceforth decreasing the hysteresis loss. Also by increasing pole number, weight of iron parts is reduced.

6. What happens to the weight of copper in both armature and field windings when the number of poles increase?

a) weight of copper in armature winding decreases and weight of copper in field winding increases

b) weight of copper in armature winding increases and weight of copper in field winding decreases

c) weight of copper in armature winding and field winding decreases

d) weight of copper in armature winding and field winding increases

Answer: c

Explanation: The weight of copper is indirectly proportional to the number of poles. As the number of poles increases, the weight of the copper decreases.

7. What happens to the length of the commutators with the increase in number of poles?

a) The length of commutators are increased

b) The length of commutators are decreased

c) The length of commutators are stable

d) The length of commutators are higher

Answer: b

Explanation: The area of the brushes decreases if the number of poles are being increased. As the area of the brushes are decreased, the length of the commutators also decrease.

8. What happens to the labor charges when there is an increase in number of poles?

a) labor charges are reduced

b) labor charges are increased

c) labor charges are fixed always

d) labor charges vary

Answer: b

Explanation: With increase in the number of poles, the armature windings increase, and more work increases to insulate. The commutator segments also increase, and the work increases.

9. What is the effect of the distortion of field form with respect to the small number of poles?

a) small number of poles cause no distortions

b) small number of poles clears all distortions

c) small number of poles reduces distortions

d) small number of poles increases distortions

Answer: d

Explanation: When there is small number of poles, that time the armature mmf per pole increases. As the armature mmf increases, it results in increase of distortion.

10. Large number of poles lead to large flashover between brushes.

a) true

b) false

Answer: a

Explanation: The number of brushes is equal to number of poles. For the same diameter of the commutator, the distance between the adjacent brush arms decreases and this increases the possibility of flashover.

11. What is the dependency of the cost of the armature and field windings with respect to large number of poles?

a) high cost for armature windings, low cost for field windings

b) high cost for armature windings, high cost for field windings

c) low cost for armature windings, high cost for field windings

d) low cost for armature windings, low cost for field windings

Answer: d

Explanation: With large number of poles the armature and the field windings reduce in number. Thus the cost of the field and armature windings also decrease.

12. Lower values of frequency are used for small machines.

a) true

b) false

Answer: b

Explanation: Lower values of frequency are actually used for the large machines. Also, higher values of frequency are actually used for small machines.

13. What is the range of the current per parallel path for the choice of number of poles?

a) limited to 100 A

b) limited to 150 A

c) limited to 200 A

d) limited to 250 A

Answer: c

Explanation: The current per parallel path should be limited to maximum of 200 A. If the limit gets exceeded then there occurs damage to the machine.

14. What should be the range of the current per brush arm?

a) limited to 400 A

b) limited to 200 A

c) limited to 100 A

d) limited to 300 A

Answer: a

Explanation: The current per brush arm should be limited to maximum of 400 A. If the limit gets exceeded then there occurs damage to the machine.

15. What should be the armature mmf per pole for output over 1500 kW?

a) 5000 A

b) 5000-7500 A

c) 7500-10000 A

d) upto 12500

Answer: d

Explanation: 5000 A is for output of about 100 kW. 5000 to 7500 A for output voltage of 100 to 500 kW. 7500 to 10000 A is for the output of 500 to 1500 kW.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Pole Design”.

1. What are the factors the design of poles of the DC machine depends on?

a) length, breadth, height of the conductors

b) area of cross section of poles

c) area of cross section of poles and height of the poles

d) area of cross section of poles and height of the poles and the design of field windings

Answer: d

Explanation: For designing the poles, first the area of cross section of the poles and the height of the poles should be obtained. Then the field winding design details are also required during the pole design.

2. What is the relationship between flux in the pole body and the useful flux per pole?

a) flux in the pole body is directly proportional to useful flux per pole

b) flux in the pole body is indirectly proportional to the useful flux per pole

c) flux in the pole body is directly proportional to the square of useful flux per pole

d) flux in the pole body is indirectly proportional to the square of useful flux per pole

Answer: a

Explanation: According to the flux in the pole body formula the flux in the pole body is directly proportional to the useful flux per pole. It is also proportional to the leakage coefficient.

3. What is the flux in the pole body, given leakage coefficient = 1.2 and the useful flux per pole is 10 weber?

a) 12 weber

b) 11.2 weber

c) 8.2 weber

d) 20 weber

Answer: a

Explanation: Flux in the pole body = leakage coefficient * useful flux per pole

Flux in the pole body = 1.2 * 10 = 12 weber.

4. What is the meaning of useful flux?

a) the flux which is being created in the machine

b) the flux which can be used

c) the flux which can produce the output

d) the flux that is wasted

Answer: c

Explanation: Total flux is the maximum amount of flux that is being generated by the current flowing in the circuit. Useful flux is nothing but the flux which can produce the output in the machine.

5. What is the range of leakage coefficient for output of 100kW?

a) 1.12-1.25

b) 1.11-1.22

c) 1.10-1.20

d) 1.11-1.15

Answer: b

Explanation: 1.12-1.25 is the leakage coefficient when the output is 50kW. 1.10-1.20 is the leakage coefficient when the output is 200kW.

6. What is the range of leakage coefficient for output of 1000kW?

a) 1.12-1.25

b) 1.11-1.22

c) 1.09-1.18

d) 1.08-1.16

Answer: d

Explanation: 1.12-1.25 is the leakage coefficient when the output is 50kW. 1.11-1.22 is the leakage coefficient when the output is 100kW. 1.09-1.18 is the leakage coefficient when the output is 500kW.

7. What is the range of the flux density in the pole shrank for laminated poles?

a) 1.1-1.7 Wb per m 2

b) 1.2-1.6 Wb per m 2

c) 1.3-1.7 Wb per m 2

d) 1.2-1.7 Wb per m 2

Answer: d

Explanation: The flux density in the pole shrank of laminated poles should have a minimum value of 1.2. The flux density in the pole shrank of the laminated poles should not exceed 1.7 at the same time.

8. What is the formula for the area of the poles shrank of the laminated poles?

a) area of the pole shrank = flux in the pole body * magnetic field

b) area of the pole shrank = flux in the pole body + magnetic field

c) area of the pole shrank = flux in the pole body – magnetic field

d) area of the pole shrank = flux in the pole body / magnetic field

Answer: a

Explanation: For finding out the area of the pole shrank first the flux in the pole body is found out using the product of the leakage coefficient and the useful flux in the pole. Next, the magnetic field is measured and the product gives the area.

9. What should be the length of pole with respect to the length of the armature and what should be the range of the length of pole?

a) length of pole < length of armature, 10-15 m

b) length of pole > length of armature, 10-15 mm

c) length of pole > length of armature, 10-15 cm

d) length of pole < length of armature, 10-15 mm

Answer: d

Explanation: The length of the pole should be very much less than the length of the armature in order to permit the end play and t avoid magnetic centering. It should be in the range of 10-15 mm.

10. The formula for length of pole is L = Total length of armature – .

a) true

b) false

Answer: a

Explanation: First the total length of armature is calculated. Then the value is subtracted by 10-15 mm in order to obtain the length of the pole.

11. What is the formula for the width of pole of DC machines?

a) width of pole body = area of the pole * length of the pole

b) width of pole body = area of the pole + length of the pole

c) width of pole body = area of the pole – length of the pole

d) width of pole body = area of the pole / length of the pole

Answer: d

Explanation: For obtaining the width of the pole, the area of the pole is first obtained. Then the length of the pole is calculated. The ratio of area of the pole to the length of the pole gives the width.

12. Height of the pole depends on the mmf to be provided on the pole at full load.

a) true

b) false

Answer: a

Explanation: The height of the pole totally depends on the mmf provided to the poles. The mmf provided at full load is only taken into consideration for the height measurement.

13. How is the mmf required at full load obtained for the calculation of height of poles?

a) using closed circuit characteristics

b) using open circuit characteristics

c) using formula

d) using equivalent circuit

Answer: b

Explanation: The mmf at full load is calculated using the magnetization curve. The open circuit characteristics are obtained which help in finding out the mmf at full load.

14. How should the field mmf be with respect to armature mmf to reduce the armature reaction?

a) armature mmf > field mmf

b) armature mmf >= field mmf

c) armature mmf < field mmf

d) armature mmf = field mmf

Answer: c

Explanation: To reduce the armature reaction, the field system should be designed such that the field mmf should be dominant over the armature mmf. If the armature mmf becomes low, the armature reaction reduces.

15. What should be the range of the field mmf to armature mmf ratio at full load?

a) 1.0-1.2

b) 1.1-1.3

c) 1.3-1.5

d) 1.1-1.25

Answer: d

Explanation: The minimum value of the ratio should be atleast 1.1. The maximum value of the ratio should be not greater than 1.25.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Window Space Factor”.

1. What is window space factor?

a) window space factor = copper area in the window – total window area

b) window space factor = copper area in the window + total window area

c) window space factor = copper area in the window * total window area

d) window space factor = copper area in the window / total window area

Answer: d

Explanation: Window space factor is the ratio of the copper area in the window to the total window area. It is a constant used in the output equation of transformers.

2. What does the window space factor depend on?

a) it depends on the core

b) it depends on the armature

c) it depends on the insulation

d) it depends on the insulation and copper

Answer: d

Explanation: Window space factor totally depends on the insulation of the machine. It also depends on the copper windings provided.

3. What does the insulation and copper of the transformer depend on?

a) current rating

b) voltage rating

c) output power

d) voltage rating and output power

Answer: d

Explanation: The insulation and copper of the transformer depend on the voltage rating of the transformer. It also depends on the output power produced by the transformer.

4. What is the empirical formula for calculating the value of window space factor?

a) window space factor = 10 / 

b) window space factor = 5 * 

c) window space factor = 10 * 

d) window space factor = 5 / 

Answer: a

Explanation: The output voltage in kV is calculated first. Then the value is substituted in the formula to obtain the empirical value of the window space factor.

5. What is the empirical value of window space factor, given the output is 1000kV?

a) 0.09

b) 0.9

c) 0.009

d) 0.0009

Answer: c

Explanation: Empirical value of window space factor = 10 / 

kV = 1000 kV, empirical value = 10/1030 = 0.009.

6. What is the ratings of the transformers for using the empirical value of window space factor?

a) 50-100 kVA

b) 50-150 kVA

c) 50-200 kVA

d) 100-200 kVA

Answer: c

Explanation: The empirical formula of the window space factor is used for the transformers of the rating 50-200 kVA. If the transformer rating is higher than 200 kVA then the empirical formula isn’t used.

7. What is the relationship of the space factor value with the large and small outputs?

a) small value for both large and small outputs

b) large values for both large and small outputs

c) large value for small output and small value for large outputs

d) small value for small output and large value for large outputs

Answer: d

Explanation: The space factor is directly proportional to the output. If large output is obtained, space factor is high and vice versa.

8. What is the formula for the window space factor, when the output is 1000 kVA?

a) 12 / 

b) 10 / 

c) 9 / 

d) 11 / 

Answer: a

Explanation: 10 /  denotes the empirical value of window space factor for rating between 50-200 kilo-volt-amp. When the output is near 1000 kilo-volt-amp then the formula used is 12/.

9. What is the formula of window space factor, when the transformer rating is 20 kVA?

a) 10 / 

b) 12 / 

c) 8 / 

d) 19 / 

Answer: c

Explanation: 10 /  denotes the empirical value of window space factor for rating between 50-200 kilo-volt-amp. 12 /  is the window space factor for transformers having rating around 1000 kVA.

10. The area of the window depends on the window space factor.

a) true

b) false

Answer: a

Explanation: Area of the window = Total conductor area/window space factor.

The area of the window is indirectly proportional to the window space factor.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Core and Winding – 1”.

1. What is stacking factor?

a) the allowance made for the power loss

b) the allowance made for the space loss between laminations

c) the allowance made for the heat loss between laminations

d) the allowance made for the power loss between laminations

Answer: b

Explanation: The net cross sectional area is obtained from the dimensions of various packets and an allowance is made for the space lost between laminations. This allowance is called stacking factor.

2. What is utilization factor?

a) utilization factor= cross sectional area + gross area of the core

b) utilization factor= cross sectional area – gross area of the core

c) utilization factor= cross sectional area * gross area of the core

d) utilization factor= cross sectional area / gross area of the core

Answer: d

Explanation: The utilization factor is equal to the ratio of the cross sectional area to the gross area of the core. The cross sectional area and the gross area of the core are first found out, and the ratio gives utilization factor.

3. What is the relationship between utilization factor and the number of core steps?

a) utilization factor is directly proportional to the number of core steps

b) utilization factor is indirectly proportional to the number of core steps

c) utilization factor is indirectly proportional to the square of number of core steps

d) utilization factor is directly proportional to the square of number of core steps

Answer: a

Explanation: The utilization factor increases with the increase in the number of core steps used. This eventually increases the manufacturing cost.

4. What is the optimum number of steps for small and large transformers?

a) 5, 10

b) 10, 15

c) 6, 15

d) 1, 10

Answer: c

Explanation: The optimum number of steps used for the large transformers is maximum of 15. The optimum number of steps for the small transformers is maximum of 6.

5. What happens if the utilization factor gets improved?

a) core area increases and the volt/turns decreases

b) core area increases and the volt/turns increases

c) core area decreases and the volt/turn decreases

d) core area decreases and the volt/turn increases

Answer: b

Explanation: When the utilization factor increases the core area gets increased. This leads to the increase in the volt/turn for any particular core diameter and specified flux density.

6. How many types of cores are available for core type of transformer?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are basically 3 types of core section available for core type of transformer. They are rectangular, square or stepped type of core sections.

7. What type of core section is used for shell type transformer?

a) rectangular

b) square

c) stepped

d) cruciform

Answer: a

Explanation: Shell type transformers prefer only rectangular core section. Shell type transformer are moderate and low voltage transformer which use only rectangular core section.

8. What is the range of the ratio of depth to width of core in rectangular core?

a) 1-2

b) 1.5-2.5

c) 1.4-2

d) 1.5-2

Answer: c

Explanation: In rectangular core, the ratio of the depth to core should be minimum 1.4. The maximum value of ratio of depth to core is 2.

9. When is square and stepped cores used?

a) when circular coils are required for low voltage distribution

b) when rectangular coils are required for low voltage distribution

c) when circular coils are required for high voltage distribution

d) when rectangular coils are required for high voltage distribution

Answer: c

Explanation: Circular coils are required for high voltage distribution and power transformer. When circular coils are required square and stepped cores are used.

10. Circular coils are preferred because of their electrical characteristics.

a) true

b) false

Answer: b

Explanation: Circular coils are preferred because of their high mechanical strength. Their high mechanical strength allows them to be used in high voltage distribution and power transformer.

11. What is the ratio of the net core area to the area of the circumscribing circle in square cores?

a) 0.58

b) 0.64

c) 0.70

d) 0.80

Answer: a

Explanation: 0.64 is the ratio of the gross core area to the area of the circumscribing circle. Net core area is the product of stacking factor and gross iron area.

12. The laminations are manufactured in standard size to minimize the wastage of steel during punching of laminations.

a) true

b) false

Answer: a

Explanation: The laminations are manufactured in the standard size of width, 0.75m to 1 m. This is used to avoid excessively wide assortment of laminations and to minimize wastage of steel during punching of laminations.

13. What is the value of ratio of gross core area to the area of circumscribing circle in stepped cores?

a) 0.71

b) 0.79

c) 0.89

d) 0.91

Answer: b

Explanation: 0.71 is the ratio of net core area to the area of circumscribing circle in stepped cores. The gross core area for stepped cores is 0.618 * d 2 .

14. What is the net core area for three stepped transformers?

a) 0.45

b) 0.56

c) 0.6

d) 0.62

Answer: c

Explanation: 0.45 is the net core area for the square core transformers. 0.56 is the core area for cruciform or stepped core transformers.

15. What is the relationship between the number of steps to the area of circumscribing circle?

a) number of steps is directly proportional to the area of the circumscribing circle

b) number of steps is indirectly proportional to the area of the circumscribing circle

c) number of steps is directly proportional to square of the area of the circumscribing circle

d) number of steps is indirectly proportional to square of the area of the circumscribing circle

Answer: a

Explanation: As the number of steps increase, the area of the circumscribing circle also increases. But as the area of the circumscribing circle increases, the ratio of the net core area and gross core area to the area of circumscribing circle decreases.

This set of Design of Electrical Machines Questions and Answers for Freshers focuses on “Design of Core and Winding – 2”.

1. What is the formula for the number of turns in primary winding?

a) number of turns of primary winding = Voltage of primary windings * voltage per turn

b) number of turns of primary winding = Voltage of primary windings/voltage per turn

c) number of turns of primary winding = Voltage of secondary windings * voltage per turn

d) number of turns of primary winding = Voltage of secondary windings/voltage per turn

Answer: b

Explanation: For calculating the number of turns of primary windings first we calculate the voltage across the primary windings. Then the voltage per turn is calculated and the ratio gives the number of turns.

2. What is the formula for obtaining the current in the primary winding?

a) current in primary winding = kVA per turn * 10 3 * primary voltage

b) current in primary winding = kVA per phase * 10 3 * primary voltage

c) current in primary winding = kVA per turn * 10 3 / primary voltage

d) current in primary winding = kVA per phase * 10 3 / primary voltage

Answer: d

Explanation: For obtaining the current in primary winding, the kVA output per phase is obtained. Then the primary voltage is calculated, and the ratio of both gives the current in primary windings.

3. What does the area of conductors in primary and secondary windings depend on?

a) current

b) voltage

c) power

d) current density

Answer: d

Explanation: The area of the conductors is directly dependent on the current density. The area of the conductors are determined after choosing a suitable current density.

4. What does the permissible current density depend upon?

a) local heating

b) efficiency

c) output power

d) local heating and efficiency

Answer: d

Explanation: The permissible current density depends upon the local heating as the heating should not affect the output. It also depends on the efficiency of the transformer.

5. What is the relationship between temperature and the current density?

a) current density is directly proportional to the temperature

b) current density is directly proportional to the square of the temperature

c) current density is indirectly proportional to the square of the temperature

d) current density is indirectly proportional to the temperature

Answer: a

Explanation: As the current density increases, the temperature also increases. As the temperature increases, it can cause damage to the insulation.

6. What is the relationship between the losses and the maximum efficiency on the current density?

a) current density increases, losses decrease, efficiency increases

b) current density increases, losses increase, efficiency increases

c) current density decreases, losses decrease, efficiency increases

d) current density decreases, losses increase, efficiency increases

Answer: c

Explanation: As the current density decreases, the losses decrease. As the losses decrease the maximum efficiency increases.

7. What is the range of current density for small and medium power transformers?

a) 1-2 A per mm 2

b) 1-2.5 A per mm 2

c) 1.1-2.2 A per mm 2

d) 1.1-2.3 A per mm 2

Answer: d

Explanation: In small and medium power transformers, the lowest value of current density is 1.1. The highest permissible value is 2.3 for small and medium power transformers.

8. What is the range of current density for large power transformer with self oil cooled type?

a) 1-2 A per mm 2

b) 1.5-2.5 A per mm 2

c) 2.2-3.2 A per mm 2

d) 2-3 A per mm 2

Answer: c

Explanation: For large transformers with self oil cooled type, the highest permissible value of current density is 3.2. The minimum current density value required is 2.2.

9. What is the condition for minimum loss condition?

a) current density in primary < current density in secondary

b) current density in primary > current density in secondary

c) current density in primary = current density in secondary

d) current density in primary >= current density in secondary

Answer: c

Explanation: The condition for the minimum loss should be the current density in primary should be equal to the current density in secondary. Any different condition, could lead to high amount of loss.

10. The current density in relatively better cooled outer winding is made 10 percent greater than the inner winding.

a) true

b) false

Answer: b

Explanation: In practical case, the current density in relatively better cooled outer winding is made greater than that in the inner winding. It is usually made 5 percent greater in practical.

11. How many total high voltage windings are present?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: There are 3 high voltage windings present. They are i) Cylindrical winding, ii) Cross-over winding iii) Continuous disc type winding.

12. The low voltage windings are generally divided into 2 types.

a) true

b) false

Answer: a

Explanation: The low voltage windings are basically divided into 2 types. They are i) cylindrical winding ii) helical winding.

13. What is the rating for cylindrical type of winding with circular conductors?

a) 5000-10000 kVA

b) 5000-8000 kVA

c) 160-10000 kVA

d) 200-10000 kVA

Answer: a

Explanation: 5000-8000 kVA is used for rectangular conductors with cylindrical winding. 160-10000 kVA is used for helical winding. 200-10000 kVA is used for continuous disc type of winding.

14. What is the voltage for cross over type of winding?

a) upto 15 kV

b) upto 33 kV

c) upto 66 kV

d) upto 6 kV

Answer: b

Explanation: Helical windings have a voltage of upto 15 kV. Whereas, the cylindrical winding with rectangular conductors have a voltage of upto 6 kV.

15. What is the maximum current per conductor for helical winding?

a) from 12 A and above 12 A

b) from 300 A and above 300 A

c) upto 40 A

d) upto 80 A

Answer: b

Explanation: The maximum current per conductor for continuous disc winding is from 12 A and above 12 A. The maximum current per conductor for cross over winding is upto 40 A and the maximum current per conductor for cylindrical winding with circular conductors is upto 80 A.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Temperature Rise in Transformers”.

1. The problem of temperature rise and cooling of transformers is essentially the same as that of rotating machinery.

a) true

b) false

Answer: a

Explanation: There are problems of temperature rise and cooling of transformers which decreases the efficiency of the transformers. The same problems are also seen in the rotating machinery.

2. How are the losses in the transformer and rotating machines converted to?

a) the losses are converted to electrical energy

b) the losses are converted to electrical and mechanical energy

c) the losses are converted to mechanical energy

d) the losses are converted to thermal energy

Answer: d

Explanation: In both the transformer and the rotating machines the losses are converted to thermal energy. This thermal energy causes heating of the transformer parts.

3. In how many ways does heat dissipation occur in transformers?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: The heat dissipation takes place in 3 ways in transformers. They are radiation, convection and conduction.

4. What type of heat dissipation takes place when the heat flows from the outer surface of the transformer part to the oil that cools it?

a) conduction

b) convection

c) conduction and convection

d) radiation

Answer: b

Explanation: When heat flows from the outer surface of transformer part to the oil which cools it, it is convection. In transformers all 3 types of heat dissipation occurs.

5. What type of heat dissipation takes place when heat flows from oil to walls of a cooler?

a) conduction

b) convection

c) radiation

d) conduction and convection

Answer: b

Explanation: When heat flows from oil to walls of the cooler, the heat dissipation type is convection. In transformers all 3 types of heat dissipation takes place.

6. What type of heat dissipation takes place when heat flows from the walls of the cooler to the cooling medium?

a) convection

b) radiation

c) convection and radiation

d) conduction and radiation

Answer: c

Explanation: When the heat flows from the walls of the cooler to the cooling medium, it is both convection and radiation. In transformer all 3 types of heat dissipation occurs.

7. What is the range of the working temperature of oil determined by the tests?

a) 40-60° C

b) 30-60° C

c) 45-60° C

d) 50-60° C

Answer: d

Explanation: The minimum value of the working temperature of oil as cooling medium is determined to be 50°C. The maximum value of the working temperature of oil as cooling medium is determined to be 60°C.

8. What is the formula for specific heat dissipation due to convection of oil?

a) specific heat dissipation = 40.3* 1/4 W per m 2 – °C

b) specific heat dissipation = 40.3 /  1/4 W per m 2 – °C

c) specific heat dissipation = 40.3* 1/4 W per m 2 – °C

d) specific heat dissipation = 40.3* 1/4 W per m 2 – °C

Answer: a

Explanation: First the temperature difference of the surface relative to oil is calculated, then the height of the dissipating surface is also calculated. Substituting in the above formula provides the specific heat dissipation due to convection of oil.

9. What is the value of specific heat dissipation for convection due to air?

a) 8 W per m 2 – °C

b) 6 W per m 2 – °C

c) 9 W per m 2 – °C

d) 10 W per m 2 – °C

Answer: a

Explanation: The value of the specific heat dissipation for convection due to air is 8 W per m 2 – °C. The value of specific heat dissipation will vary for different medium.

10. The convection due to air is 10 times the convection due to oil.

a) true

b) false

Answer: b

Explanation: The convection due to oil is 10 times the convection due to air. This constitutes a major valuable property of oil as a cooling medium.

11. How do the walls of the transformer tank dissipate heat?

a) by radiation

b) by convection

c) by conduction

d) by convection and radiation

Answer: d

Explanation: The plain walled tanks of the transformer also dissipate heat through convection and radiation. The property is similar to that of the rotating machinery.

12. What is the specific heat dissipated by the plain walled tanks of the transformer by radiation and convection?

a) 6.5, 6 W per m 2 – °C

b) 6, 6.5 W per m 2 – °C

c) 6.5, 6.5 W per m 2 – °C

d) 6, 6 W per m 2 – °C

Answer: b

Explanation: The heat dissipated by the plain walled tanks of the transformer by radiation is 6 W per m 2 – °C. The heat dissipated by the plain walled tanks of the transformer by convection is 6.5 W per m 2 – °C.

13. What is the formula for the temperature rise of the transformers?

a) temperature rise = total loss * specific heat dissipation * surface temperature

b) temperature rise = total loss / 

c) temperature rise = total loss / specific heat dissipation / surface temperature

d) temperature rise = total loss * specific heat dissipation / surface temperature

Answer: b

Explanation: Firstly the surface temperature is calculated along with the specific heat dissipation which is nothing but 12.5. Then the losses are calculated and substituted in the above formula.

14. Can the plain walled tanks accommodate the transformer for both large and small outputs?

a) the plain walled tanks can accommodate for large outputs but cannot accommodate for small outputs

b) the plain walled tanks can accommodate for large outputs and small outputs

c) the plain walled tanks cannot accommodate for large outputs and small outputs

d) the plain walled tanks can accommodate for small outputs but cannot accommodate for large outputs

Answer: d

Explanation: The plain walled tanks are large enough to accommodate the transformer and oil has sufficient surface to keep the temperature rise within limits for small outputs. But the plain walled tanks cannot accommodate the transformers for large outputs.

15. How are the ratings of the transformer, losses and temperature rise related?

a) increase, decrease, increase

b) decrease, increase, increase

c) increase, increase, increase

d) decrease, increase, decrease

Answer: c

Explanation: As the rating of the transformer increases, the losses also increase. As the losses increase the heat dissipated increases and gives high temperature rise.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Tank”.

1. What is the usage of the tanks with tubes?

a) if the temperature rise with plain tank is very low

b) if the temperature rise with plain tank is very high

c) if the temperature rise is zero

d) if the temperature rise with plain tank exceeds the specific limits

Answer: d

Explanation: Temperature rise in transformers is calculated with plain walled tanks. If the limits is exceeded then the plain walled tank is replaced by tank with tubes.

2. What is the relation of the provision of tubes with respect to dissipation of heat?

a) the provision of tubes is directly proportional to the dissipation of heat

b) the provision of tubes is indirectly proportional to the dissipation of heat

c) the provision of tubes is directly proportional to square of the dissipation of heat

d) the provision of tubes is indirectly proportional to square of the dissipation of heat

Answer: b

Explanation: The provision of tubes increases the dissipating area. The increase in dissipation of heat is not proportional to area because tube screen some of the tank surface preventing radiation from there.

3. What is the relation of the transformer surface with respect to dissipation of heat?

a) transformer surface has no relation with respect to dissipation of heat

b) transformer surface has minor changes with respect to dissipation of heat

c) transformer surface has major changes with respect to dissipation of heat

d) transformer surface has no change with respect to dissipation of heat

Answer: d

Explanation: When the tanks with tubes are provided, the dissipation of heat increases. The dissipation of heat has no effect on the transformer surface.

4. How is the circulation of oil improved in tanks with tubes?

a) it can be improved by using dissipating heat

b) it can be improved by using more effective air circulation

c) it can be improved by using more effect power flow

d) it can be improved by using more effective heads of pressure

Answer: d

Explanation: The circulation of oil is improved in tanks with tubes. It takes place with the help of using more effective heads of pressure.

5. An addition of 35 percent should be made to tube area of the transformers.

a) true

b) false

Answer: a

Explanation: An addition of 35 percent should be made to tube area of the transformer. This should be done in order to take into account this improvement in dissipation of loss by convection.

6. What is the loss dissipated by tubes by convection, given area of the tubes = 3.5?

a) 12.3 W per °c

b) 2.51 W per °c

c) 5.3 W per °c

d) 30.8 W per °c

Answer: d

Explanation: Loss dissipated by tubes by convection = 8.8 * Area of tubes

Loss = 8.8 * 3.5 = 30.8 W per °c.

7. What is the formula for temperature rise with tubes?

a) temperature rise with tubes = total loss / dissipating surface*

b) temperature rise with tubes = total loss * dissipating surface*

c) temperature rise with tubes = total loss / dissipating surface / 

d) temperature rise with tubes = total loss + dissipating surface*

Answer: a

Explanation: The total losses in the transformers are obtained firstly the iron loss and copper loss. Next the dissipating surface temperature is obtained and substituting in the above formula gives the temperature rise.

8. What is the formula for number of tubes?

a) number of tubes =  * 

b) number of tubes =  * 

c) number of tubes =  / 

d) number of tubes =  + 

Answer: a

Explanation: First the temperature rise with tubes is obtained. Then the iron loss and copper loss are obtained and added. Area of each tube is also obtained. Substituting all the values in the above formula provides the number of tubes.

9. What is the range of the diameter of the tubes used?

a) 50-60 mm

b) 60-70 mm

c) 70-80 mm

d) 50-70 mm

Answer: d

Explanation: The minimum value of the diameter of tubes is derived to be around 50 mm. The maximum value of the diameter of tubes should be less than 70 mm.

10. Elliptical tubes with pressed radiators are increasingly been used.

a) true

b) false

Answer: a

Explanation: Elliptical tubes with pressed radiators are on high demand now a days. This is because they give a greater dissipating surface for the small volume of oil.

11. What is the formula for width of the tank for single phase transformers used?

a) width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank

b) width of tank = distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank

c) width of tank = 2*distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank

d) width of tank = distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank

Answer: b

Explanation: Width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank is the formula for three phase transformer. For single phase transformers, the distance between adjacent limbs is not multiplied.

12. What is the formula for the length of the tank?

a) length of the tank = external diameter of h.v winding + clearance on each side between the winding and tank along the width

b) length of the tank = external diameter of h.v winding * clearance on each side between the winding and tank along the width

c) length of the tank = external diameter of h.v winding + 2*clearance on each side between the winding and tank along the width

d) length of the tank = external diameter of h.v winding / 2*clearance on each side between the winding and tank along the width

Answer: c

Explanation: The external diameter of h.v winding is obtained. Next the clearance on each side between the winding and tank along the width is calculated and is substituted in the above formula.

13. What is the formula for height of transformer tank?

a) height of transformer tank = Height of transformer frame + clearance height between the assembled transformer and tank

b) height of transformer tank = Height of transformer frame * clearance height between the assembled transformer and tank

c) height of transformer tank = Height of transformer frame/clearance height between the assembled transformer and tank

d) height of transformer tank = Height of transformer frame – clearance height between the assembled transformer and tank

Answer: a

Explanation: Firstly, the height of the transformer frame is calculated. Next, the clearance height between the assembled transformer and tank is also calculated. Substitute the values to obtain the height of transformer tank.

14. What is the rating of the transformer for the voltage of about 11 kV?

a) 1000-2000 kVA

b) 100-3000 kVA

c) 1000-5000 kVA

d) 100-500 kVA

Answer: c

Explanation: The minimum value of the rating of the transformer for a voltage of about 11 kV should be 1000 kVA. The maximum value of the rating of the transformer for a voltage of about 11 kV should be about 5000 kVA.

15. What is the rating of the transformer for the voltage of above 11 kV upto 33 kV?

a) 1000-5000 kVA

b) less than 1000 kVA

c) above 1000 kVA

d) 100-500 kVA

Answer: b

Explanation: 1000-5000 kVA is the rating of the transformer for the voltage of about 11 kV. When the voltage rating is about 11-33 kV, then the rating of the transformer is less than 1000 kVA.

This set of Basic Design of Electrical Machines Questions and Answers focuses on “Methods of Cooling of Transformers”.

1. How many types of cooling methods are available for the transformer?

a) 3

b) 2

c) 1

d) 4

Answer: a

Explanation: There are 3 types of cooling methods available for transformers. They are natural cooling, air blast cooling, forced oil circulation.

2. How are the radiators cooled in the present time?

a) by natural cooling

b) by forced cooling using small fans

c) by forced cooling using large fans

d) by using external air

Answer: b

Explanation: At present time the radiators are cooled using forced cooling. The forced cooling takes place with the help of the small fans mounted on each radiator.

3. What type of cooling is being made use of in transformers having a capacity of less than 11MVA?

a) natural cooling

b) forced cooling

c) air blast cooling

d) forced cooling and air blast cooling

Answer: a

Explanation: For transformers having capacity less than 11MVA, natural cooling is made use of. For transformers having capacity more than 11MVA, air blast cooling is used.

4. Compared to the natural cooling, how much of heat dissipation is increased by air blast cooling?

a) 50-70%

b) 60-70%

c) 50-60%

d) 40-60%

Answer: c

Explanation: Air blast cooling helps in increased heat dissipation. The minimum value of increased heat dissipation is 50% and maximum value is 60%.

5. Increase in the velocity of oil circulation increases the transformer output.

a) true

b) false

Answer: a

Explanation: The increases in velocity of the air circulation increases the temperature. The temperature rise increases the transformer output.

6. What is the relation of the increase of the oil circulation rate with energy losses?

a) increase of the oil circulation rate is not depending with energy losses

b) increase of the oil circulation rate is directly proportional to the energy losses

c) increase of the oil circulation rate is directly proportional to the square of energy losses

d) increase of the oil circulation rate is indirectly proportional to energy losses

Answer: b

Explanation: The increase in the oil circulation rate is unsuitable because this increases the large energy losses In the pumping unit. To cool the oil, it is circulate through a special oil cooler.

7. What is the flow rate of the circulating oil in an air cooler with natural air cooling?

a) 12.5 litre per minute per KW of losses

b) 12 litre per minute per KW of losses

c) 14 litre per minute per KW of losses

d) 13 litre per minute per KW of losses

Answer: b

Explanation: When natural air cooling is used, the flow rate is 12 litres per minute per KW of losses. Even when the air blast cooling is used, the transformer output increases roughly to the same extent.

8. What is the range of the cooler surfaces per 1 KW of losses?

a) 0.1-0.25 m 2

b) 0.18-0.2 m 2

c) 0.1-0.2 m 2

d) 0.18-0.25 m 2

Answer: d

Explanation: The minimum value of the cooler surfaces per 1 KW of losses is 0.18 m 2 . The maximum value of the cooler surfaces per 1 KW of losses is 0.25 m 2 .

9. What is the range of the flow rate of circulating oil per KW of losses?

a) 6-7 liters per minute

b) 5-6 liters per minute

c) 6-8 liters per minute

d) 6-7 liters per minute

Answer: c

Explanation: The minimum value of the flow rate of circulating oil per KW of losses is derived to be 6 liters per minute. The maximum value of the flow rate of circulating oil per KW of losses is derived to be 8 liters per minute.

10. The temperature difference between the incoming and outgoing water is greater than 10°C.

a) true

b) false

Answer: b

Explanation: The water flow rate is about 1.5 litres per minute. The difference in temperature between the incoming water and outgoing water is 10°C.

11. What is the formula for width of the tank for single phase transformers used?

a) width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank

b) width of tank = distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank

c) width of tank = 2*distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank

d) width of tank = distance between adjacent limbs * external diameter of h.v windings + 2*clearance between h.v windings and tank

Answer: b

Explanation: Width of tank = 2*distance between adjacent limbs + external diameter of h.v windings + 2*clearance between h.v windings and tank is the formula for three phase transformer. For single phase transformers, the distance between adjacent limbs is not multiplied.

12. What is the formula for the length of the tank?

a) length of the tank = external diameter of h.v winding + clearance on each side between the winding and tank along the width

b) length of the tank = external diameter of h.v winding * clearance on each side between the winding and tank along the width

c) length of the tank = external diameter of h.v winding + 2*clearance on each side between the winding and tank along the width

d) length of the tank = external diameter of h.v winding / 2*clearance on each side between the winding and tank along the width

Answer: c

Explanation: The external diameter of h.v winding is obtained. Next, the clearance on each side between the winding and tank along the width is calculated and is substituted in the above formula.

13. What is the formula for the height of transformer tank?

a) height of transformer tank = Height of transformer frame + clearance height between the assembled transformer and tank

b) height of transformer tank = Height of transformer frame * clearance height between the assembled transformer and tank

c) height of transformer tank = Height of transformer frame/clearance height between the assembled transformer and tank

d) height of transformer tank = Height of transformer frame – clearance height between the assembled transformer and tank

Answer: a

Explanation: Firstly, the height of the transformer frame is calculated. Next, the clearance height between the assembled transformer and tank is also calculated. Substitute the values to obtain the height of transformer tank.

14. What is the rating of the transformer for the voltage of about 11 kV?

a) 1000-2000 kVA

b) 100-3000 kVA

c) 1000-5000 kVA

d) 100-500 kVA

Answer: c

Explanation: The minimum value of the rating of the transformer for a voltage of about 11 kV should be 1000 kVA. The maximum value of the rating of the transformer for a voltage of about 11 kV should be about 5000 kVA.

15. What is the rating of the transformer for the voltage of above 11 kV upto 33 kV?

a) 1000-5000 kVA

b) less than 1000 kVA

c) above 1000 kVA

d) 100-500 kVA

Answer: b

Explanation: 1000-5000 kVA is the rating of the transformer for the voltage of about 11 kV. When the voltage rating is about 11-33 kV, then the rating of the transformer is less than 1000 kVA.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Overall Dimensions – 1”.

1. What is the formula for the diameter of the single phase core type transformer?

a) D = diameter of circumscribing circle + Width of window

b) D = diameter of circumscribing circle – Width of window

c) D = diameter of circumscribing circle * Width of window

d) D = diameter of circumscribing circle / Width of window

Answer: a

Explanation: First the diameter of the circumscribing circle is obtained. Next, the width of the window is calculated, and the sum of both the data provides the diameter of the single phase core transformer.

2. What is the formula for height of the single phase core type transformer?

a) height = height of the window – height of the yoke

b) height = height of the window + height of the yoke

c) height = height of the window – 

d) height = height of the window + 

Answer: d

Explanation: The height of the window is first obtained. Next, the height of the yoke is calculated and it is multiplied by 2. Addition of both the values gives the height of the single phase core type transformer.

3. What is the formula for width of the single phase core type transformer?

a) width = Width of largest stamping / Diameter of the transformer

b) width = Width of largest stamping + Diameter of the transformer

c) width = Width of largest stamping – Diameter of the transformer

d) width = Width of largest stamping * Diameter of the transformer

Answer: b

Explanation: Firstly, the width of the largest stamping is calculated. Next, the diameter of the transformer is calculated and the sum of both the values gives the width of the transformer.

4. What is the formula for the width over 2 limbs?

a) width = Width of largest stamping + Diameter of the transformer

b) width = Diameter + outer diameter of hv windings

c) width = Diameter – outer diameter of hv windings

d) width = outer diameter of hv windings

Answer: b

Explanation: width = outer diameter of hv windings is the width over one limb. width = Width of largest stamping + Diameter of the transformer is the formula for width of the transformer.

5. The formula for single phase core type and three phase core type diameter and height are same.

a) true

b) false

Answer: a

Explanation: D = diameter of circumscribing circle + Width of window is the diameter of single phase and 3 phase core type transformers. height = height of the window +  is the height of the single and three phase core type transformers.

6. What is the formula for the width over one limb?

a) width over one limb = outer diameter of hv winding

b) width over one limb = 2*Diameter – outer diameter of hv winding

c) width over one limb = 2*Diameter + outer diameter of hv winding

d) width over one limb = Diameter + outer diameter of hv winding

Answer: a

Explanation: Width over one limb = 2*Diameter + outer diameter of hv winding is the formula for the width over 3 limbs. For one limb the width is equal to the outer diameter of hv winding.

7. What is the formula for the width of the single phase shell type transformer?

a) width = 2*Width of the window + width of the largest stamping

b) width = Width of the window + 4*width of the largest stamping

c) width = Width of the window + width of the largest stamping

d) width = 2*Width of the window + 4*width of the largest stamping

Answer: d

Explanation: First the width of the window is obtained. Next the width of the largest stamping is obtained. Substituting in the above formula provides the width of the single-phase shell type transformer.

8. What is the height of the single phase shell type transformer?

a) height = height of window + width of the largest stamping

b) height = 2*height of window + width of the largest stamping

c) height = height of window + 2* width of the largest stamping

d) height = height of window – width of the largest stamping

Answer: c

Explanation: First the height of the window is obtained. Then the width of the largest stamping is calculated and substituting in the above formula provides the height of the single phase shell type transformer.

9. What is the formula to calculate the voltage per turn of the transformer?

a) voltage per turn = space factor * square root of output power

b) voltage per turn = space factor / square root of output power

c) voltage per turn = space factor / square root of output power

d) voltage per turn = space factor * 2*square root of output power

Answer: a

Explanation: The corresponding space factor is obtained using the formula. Then the output power is obtained and square root of the output power is taken and substituted in the above formula to obtain the voltage per turn.

10. What is the formula for the net cross sectional area of the core of the transformer?

a) cross sectional area = voltage per turn * 4.44 * frequency * magnetic field

b) cross sectional area = voltage per turn / 4.44 * frequency * magnetic field

c) cross sectional area = voltage per turn * 4.44 / frequency * magnetic field

d) cross sectional area = voltage per turn * 4.44 * frequency / magnetic field

Answer: b

Explanation: For obtaining the cross sectional area, the voltage per turn is obtained. The frequency is always 50 Hz. Then the magnetic field is obtained and substituted in the above formula.

11. What is the formula for the diameter of the circumscribing circle of the transformer?

a) diameter of the circumscribing circle = 2*square root of ratio of cross sectional area of the core to the space factor

b) diameter of the circumscribing circle = 3*square root of ratio of cross sectional area of the core to the space factor

c) diameter of the circumscribing circle = square root of ratio of cross sectional area of the core to the space factor

d) diameter of the circumscribing circle = 4*square root of ratio of cross sectional area of the core to the space factor

Answer: c

Explanation: First the cross sectional area of the core is obtained by the formula cross sectional area = voltage per turn / 4.44 * frequency * magnetic field. Next the space factor is obtained. Substituting in the formula provides the diameter of the circumscribing circle.

12. What is the formula for the width of the window of the transformer?

a) width of the window = distance between core centers + diameter of the circumscribing circle

b) width of the window = distance between core centers – diameter of the circumscribing circle

c) width of the window = distance between core centers * diameter of the circumscribing circle

d) width of the window = distance between core centers / diameter of the circumscribing circle

Answer: b

Explanation: The diameter of the circumscribing circle is obtained from the formula, diameter of the circumscribing circle = square root of ratio of cross sectional area of the core to the space factor. After obtaining the distance between core centers, the width of the window is obtained.

13. What is the formula for window area of the transformer?

a) window area = output power * 2.22 * frequency * magnetic field * window space factor * current density * area of cross section of the core *10 3

b) window area = output power / 2.22 * frequency * magnetic field * window space factor * current density * area of cross section of the core *10 3

c) window area = output power / 3.33 * frequency * magnetic field * window space factor * current density * area of cross section of the core *10 3

d) window area = output power * 3.33 * frequency * magnetic field * window space factor * current density * area of cross section of the core*10 3

Answer: b

Explanation: The window space factor, the current density and the core cross sectional area are obtained by their respective formula. The frequency is 50Hz and then the magnetic field and the output power is calculated to obtain the window space factor.

14. What is the formula for the height of the window?

a) height of window = area of window * width of the window

b) height of window = area of window + width of the window

c) height of window = area of window – width of the window

d) height of window = area of window / width of the window

Answer: d

Explanation: First the area of the window is obtained. Next the window width is obtained. The ratio of both gives the height of the window.

15. The range of the ratio of the height of the window to the width of the window is 2-4.

a) true

b) false

Answer: a

Explanation: The ratio of the height of the window to the width of the window should be adjusted such that it is above 2. The ratio of the height of the window to the width of the window should be adjusted such that it is below 4.

This set of Design of Electrical Machines Interview Questions and Answers for freshers focuses on “Overall Dimensions – 2”.

1. What is the formula for the depth and height of the yoke for stepped core?

a) depth = width of largest stamping, height = 2* width of largest stamping

b) depth = 2*width of largest stamping, height = width of largest stamping

c) depth = width of largest stamping, height = width of largest stamping

d) depth = 2*width of largest stamping, height = 2* width of largest stamping

Answer: c

Explanation: The depth of the yoke of stepped core is equal to the width of the largest stamping. The height of the yoke for the stepped core is also equal to the width of the largest stamping.

2.The height and the width of the single phase and three phase core type transformers are equal.

a) true

b) false

Answer: b

Explanation: The height of both the single phase and three phase core type transformers are equal. The width of the single phase and three phase core type are not same.

3. What is the formula for the height and width of the single phase shell transformer?

a) width = 2*width of the window + 4*width of the largest stamping, height = height of the window + 2*width of the largest stamping

b) width = 2*width of the window – 4*width of the largest stamping, height = height of the window + 2*width of the largest stamping

c) width = 2*width of the window + 4*width of the largest stamping, height = height of the window – 2*width of the largest stamping

d) width = 2*width of the window – 4*width of the largest stamping, height = height of the window -2*width of the largest stamping

Answer: a

Explanation: First the width of the window is obtained. Next, the height of the window is obtained. Then, the width of the largest stamping is obtained and substituted in the above formula.

4. What is the formula to calculate the number of turns/phase?

a) number of turns = secondary voltage * voltage per turn

b) number of turns = secondary voltage / voltage per turn

c) number of turns = secondary voltage + voltage per turn

d) number of turns = secondary voltage – voltage per turn

Answer: b

Explanation: First the voltage across the secondary winding of the transformer is obtained. Next, the voltage across each turn is obtained. On substituting we get the number of turns.

5. What is the formula for the cross sectional area of the secondary conductor of the transformer?

a) cross sectional area = secondary current * current density

b) cross sectional area = secondary current + current density

c) cross sectional area = secondary current / current density

d) cross sectional area = secondary current – current density

Answer: c

Explanation: The current flowing through the secondary winding of the transformer is calculated. Next the current density is calculated and the ratio gives the cross sectional area of the secondary conductor.

6. What is the formula for the conductor dimensions in transformer?

a) conductor dimensions = conductor width * conductor thickness + 0.5 mm

b) conductor dimensions = conductor width / conductor thickness + 0.5 mm

c) conductor dimensions = conductor width + conductor thickness + 0.5 mm

d) conductor dimensions = conductor width – conductor thickness + 0.5 mm

Answer: a

Explanation: The width of the conductor is first calculated. Next, the thickness of the conductor is calculated. On obtaining these data the conductor dimensions can be obtained.

7. What is the formula for axial depth of low voltage winding?

a) axial depth = number of secondary turns / width of the conductor

b) axial depth = number of secondary turns * width of the conductor

c) axial depth = number of secondary turns + width of the conductor

d) axial depth = number of secondary turns – width of the conductor

Answer: b

Explanation: The number of secondary turns is calculated first. Then the width of the conductor is obtained. With the 2 data, the axial depth is obtained.

8. What is the formula for the window clearance of the transformer?

a) window clearance = /2

b) window clearance = 

c) window clearance = /2

d) window clearance = 

Answer: c

Explanation: First the height of the window is obtained. Then the axial depth is calculated using the formula axial depth = number of secondary turns * width of the conductor and substituting in the above formula provides the window clearance.

9. What is the formula to calculate the radial depth of low voltage windings?

a) radial depth of the lv windings = number of layers * radial depth of the conductors * insulation between layers

b) radial depth of the lv windings = number of layers * radial depth of the conductors – insulation between layers

c) radial depth of the lv windings = number of layers / radial depth of the conductors + insulation between layers

d) radial depth of the lv windings = number of layers * radial depth of the conductors + insulation between layers

Answer: d

Explanation: The number of layers is first taken note of. Then the radial depth of the conductors is calculated along with the insulation between layers. On substituting the values in the above formula the radial depth of the low voltage windings is obtained.

10. What is the formula for the inside diameter of the low voltage windings?

a) inside diameter = diameter of the circumscribing circle + pressboard thickness insulation between l.v winding and core

b) inside diameter = diameter of the circumscribing circle – pressboard thickness insulation between l.v winding and core

c) inside diameter = diameter of the circumscribing circle + 2*pressboard thickness insulation between l.v winding and core

d) inside diameter = diameter of the circumscribing circle – 2* pressboard thickness insulation between l.v winding and core

Answer: c

Explanation: For calculating the inner diameter, first the diameter of the circumscribing circle is obtained using the corresponding formula. Then the pressboard thickness insulation is calculated.

11. What is the assumption for width of the largest stamping for the stepped core transformer?

a) 0.9*d

b) 0.71*d

c) 0.85*d

d) 0.8*d

Answer: a

Explanation: If the width of the largest stamping is not provided, then for stepped core a = 0.9*d. For the cruciform it is a = 0.85*d and for the square core it is a = 0.71*d.

12. What is the range for the current density at HT side for a distribution transformer?

a) 2.4-3.5 Amp per mm 2

b) 2-2.5 Amp per mm 2

c) 1-3.5 Amp per mm 2

d) 2-3.5 Amp per mm 2

Answer: b

Explanation: 2.4-3.5 Amp per mm 2 is the range for the current density at HT side for a power transformer. 2-2.5 Amp per mm 2 is the range for the current density at HT side for a distribution transformer.

13. What is the relation of the height of the window with the winding height with respect to the rectangular conductors?

a) winding height = 60% * window height

b) winding height = 50% * window height

c) winding height = 80% * window height

d) winding height = 70% * window height

Answer: d

Explanation: In case of selection of the rectangular conductors, first the window height is obtained. Next the 70% of the window height provides the winding height.

14. What is the formula for number of turns/coil axially?

a) number of turns/coil axially = axial length / diameter of the insulated conductor

b) number of turns/coil axially = axial length * diameter of the insulated conductor

c) number of turns/coil axially = axial length – diameter of the insulated conductor

d) number of turns/coil axially = axial length + diameter of the insulated conductor

Answer: a

Explanation: First the axial length is calculated from its respective formula. Then the diameter of the insulated conductor is calculated, and the ratio gives the number of turns/coil axially.

15. The axial length of 16 coils = axial length of each coil * 16.

a) true

b) false

Answer: a

Explanation: The axial length of each coil is calculated initially from its corresponding formula. Then the value is multiplied by the number of coils present.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Main Dimensions”.

1. What are the parameters which come under the term “Main Dimensions”?

a) diameter

b) length

c) diameter and length

d) area

Answer: c

Explanation: The main dimensions entirely deal with the calculation of the diameter and length. They are obtained from the output equation of the machine.

2. What is the range of the ratio of the core length to pole pitch for minimum cost?

a) 1.5-2

b) 1-2

c) 1.0-1.25

d) 2-2.5

Answer: a

Explanation: 1.0-1.25 is the range of the ratio of the core length to pole pitch for good power factor. For minimum cost, 1.5 is the minimum range for the ratio and 2 is the maximum range for the ratio.

3. What is the value of the ratio of the core length to pole pitch for good efficiency?

a) 1

b) 1.5

c) 2

d) 3

Answer: b

Explanation: 1 is the value of the ratio of the core length to pole pitch for good overall design. 1.5 is the value of the ratio of the core length to pole pitch for good efficiency.

4. What is the relation between motors and ratio of core length to pole pitch?

a) for small motors high ratio is preferred

b) for big motors high ratio is preferred

c) for small motors small ratio is preferred

d) for big motors small ratio is preferred

Answer: c

Explanation: For small motors, high values of the ratio of core length to pole pitch which may not be able to accommodate even a small amount of slots. Thus small motors prefer only small ratio to accommodate slots.

5. What are the factors the value of core length to pole pitch depend upon?

a) area of the slots

b) size of machine

c) size of conductors

d) size of machine, minimum cost, good efficiency

Answer: d

Explanation: The value of core length to pole pitch varies from 0.6 to 2. It depends on the cost, efficiency, power factor, overall design.

6. What is the Relation between pole pitch and the core length in terms of the best power factor?

a) pole pitch =  3

b) pole pitch =  2

c) pole pitch =  1/2

d) pole pitch =  1/3

Answer: c

Explanation: For the best power factor, the pole pitch is equal to the square root of the product of 0.18 and core length. This value however is not dimensionally correct and is valid for the values in meters.

7. What is the range of the permissible peripheral speeds in the 3 phase induction machine?

a) 60-75 m per s

b) 60-70 m per s

c) 40-70 m per s

d) 50-70 m per s

Answer: a

Explanation: For the standard rotor construction can generally be used for peripheral speeds upto 60 m per s. For special rotor construction, the peripheral speeds upto 75 m per s are permissible.

8. What is the maximum permissible level for the peripheral speed for a normal design?

a) < 30 m per s

b) > 30 m per s

c) <=30 m per s

d) >=30 m per s

Answer: c

Explanation: For the normal design, the diameter value is chosen such that the peripheral speed value doesn’t exceed 30 m per s. The diameter value is directly proportional to the peripheral speed.

9. What is the range of the core length for which the stator is provided with the ventilating ducts?

a) 105-120 mm

b) 100-120 mm

c) 100-150 mm

d) 100-125 mm

Answer: d

Explanation: The stator is provided with the ventilating ducts if the core length exceeds 100-125 mm. If the value is less than 100 mm, then no ventilating ducts are provided.

10. The width of each duct is about 8 to 10 mm.

a) true

b) false

Answer: a

Explanation: The width of the duct is chosen such that, the minimum value of the duct is 8 mm. The maximum value of the duct is chosen to be less than 10 mm.

This set of Design of Electrical Machines Questions and Answers for Experienced people focuses on “Design of Rotor Bars and Slots”.

1. What is the formula for current in each of rotor bar?

a) current in rotor bar = 2 * slot pitch * window space factor * stator torque * stator current * power factor * rotor slots

b) current in rotor bar = 2 * slot pitch * window space factor / stator torque * stator current * power factor * rotor slots

c) current in rotor bar = 2 * slot pitch * window space factor * stator torque * stator current * power factor / rotor slots

d) current in rotor bar = 2 * slot pitch / window space factor * stator torque * stator current * power factor / rotor slots

Answer: c

Explanation: For the calculation of the current through each rotor bar, firstly we find out slot pitch, window space factor and rotor slots. Then the stator torque and stator current are obtained with respect to stator side.

2. What is the relation between rotor mmf and stator mmf?

a) rotor mmf = 0.85 * stator mmf

b) rotor mmf = 0.80 * stator mmf

c) rotor mmf = 0.75 * stator mmf

d) rotor mmf = 0.70 * stator mmf

Answer: a

Explanation: First the stator mmf is calculated. Then it is multiplied by 0.85 to obtain the rotor mmf.

3. What is the relation of the rotor resistance with respect to the starting torque?

a) rotor resistance is indirectly proportional to the starting torque

b) rotor resistance is directly proportional to the starting torque

c) rotor resistance is indirectly proportional to the square of the starting torque

d) rotor resistance is directly proportional to the square of the starting torque

Answer: b

Explanation: The rotor resistance is directly proportional to the starting torques. High resistance leads to high starting torque.

4. What is the relation of the rotor resistance to efficiency and losses?

a) as rotor resistance, losses increase, efficiency increases

b) as rotor resistance, losses increase, efficiency decreases

c) as rotor resistance, losses decrease, efficiency has no change

d) as rotor resistance, losses decrease, efficiency decreases

Answer: b

Explanation: As the rotor resistance increases, the I 2 R losses increases and cause heating effects. This increase in losses decreases efficiency.

5. What is the relationship between current density, conductor area and resistance?

a) higher the current density, higher the conductor area, higher the resistance

b) higher the current density, higher the conductor area, lower the resistance

c) higher the current density, lower the conductor area, higher the resistance

d) lower the current density, lower the conductor area, lower the resistance

Answer: c

Explanation: Higher the current density leads to lower conductor area, as current density is the ratio of current per area. As the conductor area decreases, resistance increases.

6. What is the formula for the calculation of rotor resistance?

a) rotor resistance = resistance of the bars + resistance of end rings

b) rotor resistance = resistance of the bars – resistance of end rings

c) rotor resistance = resistance of the bars * resistance of end rings

d) rotor resistance = resistance of the bars / resistance of end rings

Answer: a

Explanation: First the resistance of the bars are obtained. Next, the resistance of the end rings are calculated and the sum gives the rotor resistance.

7. What is the range of current density in rotor bars?

a) 4-9 A per mm 2

b) 4-6 A per mm 2

c) 4-7 A per mm 2

d) 5-6 A per mm 2

Answer: b

Explanation: The minimum value of the current density in the rotor bars is 4 A per mm 2 . The maximum value of the current density in the rotor bars is 6 A per mm 2 .

8. What is the formula for the area of each bar?

a) area of each bar = current of the rotor bars + current density in rotor bars

b) area of each bar = current of the rotor bars / current density in rotor bars

c) area of each bar = current of the rotor bars * current density in rotor bars

d) area of each bar = current of the rotor bars – current density in rotor bars

Answer: b

Explanation: For calculating the area of each bar, current flowing across the rotor bars should be first calculated. Then the current density in rotor bars should be calculated next and the ratio gives the area of each bar.

9. Closed slots are preferred for small machines.

a) true

b) false

Answer: a

Explanation: Closed slots are preferred for small machines. It is because the reluctance of the air gap is not large owing to absence of slot openings.

10. What is the relation of closed slots with leakage reactance?

a) closed slots give no leakage reactance

b) closed slots give high leakage reactance

c) closed slots give low leakage reactance

d) closed slots give negative leakage reactance

Answer: b

Explanation: It is an advantage that closed slots give large leakage reactance. If the leakage reactance is large, the current at the starting can be limited.

11. What is the relation of closed slots with leakage reactance and overload capacity?

a) closed slots give high leakage reactance, and increases the overload capacity

b) closed slots give high leakage reactance, and decreases the overload capacity

c) closed slots give low leakage reactance, and decreases the overload capacity

d) closed slots give low leakage reactance, and increases the overload capacity

Answer: b

Explanation: The closed slots have the main advantage of giving high leakage reactance. A high leakage reactance gives the advantage that the current at the starting can be limited.

12. What is the relation between surface of rotor and the operation?

a) smooth surface leads to the quiet operation

b) rough surface leads to the quiet operation

c) smooth surface leads to the noisy operation

d) rough surface leads to the noisy operation

Answer: a

Explanation: For the design of the rotor bars of the three phase induction machine, smooth surface is preferred. Smooth surface helps in the silent operation.

13. Rectangular shaped bars and slots are preferred to circular bars and slots.

a) true

b) false

Answer: a

Explanation: Rectangular shaped bars and slots are preferred to circular bars and slots. This is because while using the rectangular shaped bars, rotor resistance increases and this leads to the improvement of starting torque.

14. What is the relation between clearances and slots?

a) high clearances are provided for salient slots

b) low clearances are provided for skewed slots

c) low clearances are provided for salient slots

d) high clearances are provided for skewed slots

Answer: d

Explanation: When the skewed slots are being used, higher clearances are provided. High clearances can lead to the smooth and efficient operation of the machine for skewed slots.

15. What is the range of clearance that can be left between rotor bars and the core?

a) 0.1-0.4 mm

b) 0.2-0.4 mm

c) 0.15-0.4 mm

d) 0.4-0.6 mm

Answer: c

Explanation: The range of clearance is chosen based on whether the slots are skewed or not. The range is usually chosen between 0.15-0.4 mm.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of End Rings”.

1. How does the revolving field produce emf in the bars?

a) revolving field produces emf of fundamental frequency in the bars

b) revolving field produces emf of third frequency in the bars

c) revolving field produces emf of no frequency in the bars

d) revolving field produces emf of sinusoidal frequency in the bars

Answer: a

Explanation: The stator winding is 3 phase distributed winding and thus produces a revolving field. This revolving field produces emfs of fundamental frequency in the bars.

2. What happens if the resistance of the end rings is negligible?

a) resistance coming in each current path is resistance of three bars

b) resistance coming in each current path is resistance of four bars

c) resistance coming in each current path is resistance of two bars

d) resistance coming in each current path is resistance of five bars

Answer: c

Explanation: If the resistance of end rings is negligible then the resistance of combined bars are taken into account. Generally the resistance of two bars are taken into account.

3. What factors does the current in the bars depend on?

a) emfs, position of bars in magnetic field

b) instantaneous emfs, position of bars in magnetic field

c) emf

d) instantaneous emf

Answer: b

Explanation: The current that the bars carry are proportional to their instantaneous emfs. The instantaneous emfs are proportional to the position of the bars in the magnetic field.

4. The end resistance, if not negligible, will tend to distort the bar current distribution from being sinusoidal.

a) true

b) false

Answer: a

Explanation: The end resistance is proportional to the current distribution. If it is not negligible then the resistance will distort the bar current distribution.

5. What is the formula for the maximum current in end ring, if the current in all bars are maximum at the same time?

a) maximum current in the end ring= bars per pole * 2 * current per bar

b) maximum current in the end ring=  * current per bar

c) maximum current in the end ring= bars per pole / 2 / current per bar

d) maximum current in the end ring= bars per pole * 2 / current per bar

Answer: b

Explanation: First the bars per pole are obtained. Then the current per bar is calculated. Then substituting in the above formula, the maximum current in the end rings.

6. Given the bars per pole is 6 and the current per bar is 20 A, what is the value of the maximum current in the end rings?

a) 60 A

b) 80 A

c) 90 A

d) 70 A

Answer: a

Explanation: maximum current in the end ring= bars per pole / 2 * current per bar

Maximum current in the end ring = *20 = 3 * 20 = 60 A.

7. What is the formula for the maximum value of current through end ring, when the current is not maximum in all the bars under one pole at the same time?

a) maximum current in end ring=  /  * current per bar

b) maximum current in end ring=  *  / current per bar

c) maximum current in end ring=  *  * current per bar

d) maximum current in end ring=  /  / current per bar

Answer: c

Explanation: The bars per pole is first obtained. Then the no of poles is calculated along with the current per bar. On substituting in the formula the maximum current in the end ring with the current through all the bars under one pole is not maximum.

8. What is the formula for the maximum current through each bar?

a) maximum value of the current through each bar = 2 * current through each bar

b) maximum value of the current through each bar =  1/2

c) maximum value of the current through each bar =  2

d) maximum value of the current through each bar =  1/3

Answer: b

Explanation: Firstly, the current through each bar is calculated. Then it is multiplied by 2 and its square root provides the maximum value of current through each bar is obtained.

9. What is the formula for the rms value of the end ring current?

a) rms value of end ring current =  / 

b) rms value of end ring current =  * 

c) rms value of end ring current =  / 

d) rms value of end ring current =  / 

Answer: a

Explanation: First the bars per pole is obtained. Then the current per bar is calculated and the no of poles is calculated. Substituting in the above formula, the rms value of the end ring current is obtained.

10. The value of the current density is chosen for the end rings such that the desired value of rotor resistance is obtained.

a) true

b) false

Answer: a

Explanation: The value of current density is chosen for the end rings should be chosen such that the desired value of the rotor resistance is obtained. If not it can lead to the starting problems in the machine.

11. How is the current density of the rotor bars chosen with respect to the end rings?

a) current density of rotor bars < current density of end rings

b) current density of rotor bars > current density of end rings

c) current density of rotor bars = current density of end rings

d) current density of rotor bars <= current density of end rings

Answer: a

Explanation: The ventilation is generally better for the end rings. Thus, the current density of the end rings should be greater than that of the rotor bars.

12. What is the formula for the area of the ring?

a) area of the ring = depth of the end ring + thickness of the end ring

b) area of the ring = depth of the end ring – thickness of the end ring

c) area of the ring = depth of the end ring / thickness of the end ring

d) area of the ring = depth of the end ring * thickness of the end ring

Answer: d

Explanation: For calculation of the area of end rings, first the depth of end rings is calculated. Next, the thickness of the end ring is calculated and substituting in the above formula gives the area of the ring.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Wound Rotor”.

1. What is the rotor windings for wound rotor motors?

a) single phase windings

b) double phase windings

c) three phase windings

d) concentrated phase windings

Answer: c

Explanation: The rotor windings for wound rotor motors are 3 phase windings. The number of rotor slots should be such that a balanced winding is obtained.

2. How should the number of slots be in the case of fractional slot windings?

a) multiples of slots

b) multiple of phases

c) multiple of phases and pair of poles

d) multiples of pair of poles

Answer: c

Explanation: Fractional slot windings can be used in the wound rotor design. It is preferable to use a number of slots which are a multiple of phases and pair of poles in the case of fractional slot windings.

3. How many types of design of wound rotors are available?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: They are 4 types of design of wound rotors. They are Number of Rotor Slots, Number of Rotor Turns, Area of Rotor Conductors, Rotor Windings.

4. What should be done to keep the rotor voltage to an acceptable level?

a) rotor to stator turns must be properly adjusted

b) stator to rotor turns must be properly adjusted

c) stator turns must be adjusted

d) rotor turns must be adjusted

Answer: b

Explanation: The effective ratio of stator to rotor turns should be adjusted to keep the rotor voltage to an acceptable level. The choice of this turns ratio is arbitrary and is controllable by the designer.

5. The rotor voltage on open circuit between slip rings should not exceed 500 V for small machines.

a) true

b) false

Answer: a

Explanation: The rotor voltage on open circuit between slip rings should not exceed 500 V for the small machines. The voltage is limited to a small value in order to protect persons working the motor if the brush gear is not perfectly protected.

6. How should the rotor voltage be with respect to the high voltage and large machines?

a) low

b) moderate

c) high

d) very high

Answer: c

Explanation: The high voltage and large machines should have high rotor voltage. If the rotor voltage is kept low, the rotor current becomes large, involving use of large conductor sections.

7. What is the range of the rotor voltage for the large machines?

a) 1000-1500 V

b) 1000-1750 V

c) 500-1500 V

d) 1000-2000 V

Answer: d

Explanation: The minimum value of the rotor voltage should be 1000 V. The maximum value of the rotor voltage should not exceed 2000 V.

8. What is the formula for rotor turns per phase?

a) rotor turns per phase =  *  * Number of turns per phase for stator

b) rotor turns per phase =  /  * Number of turns per phase for stator

c) rotor turns per phase =  *  / Number of turns per phase for stator

d) rotor turns per phase =  /  / Number of turns per phase for stator

Answer: a

Explanation: Firstly, the winding factor for stator is obtained along with the winding factor for stator. Next the ratio of the rotor voltage per phase to the stator voltage per phase. Finally, the number of turns per phase for stator is also calculated to obtain the rotor turns per phase.

9. What is the formula for the full load rotor mmf?

a) 65% of stator mmf

b) 75% of stator mmf

c) 85% of stator mmf

d) 90% of stator mmf

Answer: c

Explanation: The full load rotor mmf is taken as 0.85 of stator mmf. Full load rotor mmf = 0.85 *

 / no of rotor turns.

10. The value of the current density of rotor is chosen almost equal to that in the stator.

a) true

b) false

Answer: a

Explanation: There occurs a lot of excessive copper loss in the rotors. The value of current density of rotor is almost equal to that in the stator.

11. What type of conductor is chosen for the small motors?

a) round

b) bar

c) skewed

d) rectangular

Answer: a

Explanation: Round conductors are used for the small motors. Bar conductors are being used for the large motors.

12. What type of winding is made use of for the small motors?

a) mush windings

b) cross windings

c) interconnected windings

d) rounded windings

Answer: a

Explanation: For the small motors, it is a normal practice to use mush windings. The mush windings should be housed in the semi-closed slots.

13. What type of winding is made use of for the large motors?

a) mush windings

b) bar type windings

c) cross windings

d) rounded windings

Answer: b

Explanation: For the small motors mush windings are made use of. For the large motors, a double layer bar type winding is made use of.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “No Load Current”.

1. How many methods are present to obtain all the machine performance characteristics?

a) 3

b) 2

c) 1

d) 4

Answer: b

Explanation: There are 2 methods in obtaining all the open circuit characteristics. They are no load characteristics and short circuit characteristics.

2. How many components does the no load current characteristics comprise of?

a) 2

b) 3

c) 4

d) 1

Answer: a

Explanation: There are 2 main components under the no load current. They are Magnetizing current and Loss component of current.

3. How is the Magnetizing component with respect to the voltage?

a) the magnetizing component is in phase with the voltage

b) the magnetizing component is 90° leading the voltage

c) the magnetizing component is 90° lagging the voltage

d) the magnetizing component is 90° out of phase with the voltage

Answer: d

Explanation: The magnetizing current component is 90° out of phase with the voltage. The loss component is in phase with the voltage.

4. How many parts does the flux produced by stator mmf passes through?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: The flux produced by stator mmf passes through 5 parts. They are air gap, rotor teeth, rotor core, stator teeth, stator core.

5. The flux is distributed sinusoidally and the mmf varies sinusoidally in a DC Machine.

a) true

b) false

Answer: b

Explanation: In a DC Machine, the flux is assumed to be uniform over any cross section and the same mmf for all the paths. But in an induction machine, the flux is distributed sinusoidally, and the mmf varies sinusoidally.

6. What factors does the value of magnetizing current depend on?

a) flux tube

b) output power

c) mean mmf

d) mean mmf and flux tube

Answer: d

Explanation: If the permeability of iron were constant this would cause no difficulty. The value of magnetizing current would be accurately obtained by considering the mean mmf and the flux tube where this mean occurs.

7. When maximum values of the design factors are considered, what is the relation between flux and the magnetizing current?

a) flux is directly proportional to the magnetizing current

b) flux is indirectly proportional to the magnetizing current

c) flux is directly proportional to square of the magnetizing current

d) flux is indirectly proportional to square of the magnetizing current

Answer: b

Explanation: The flux value is indirectly proportional to the magnetizing current. The flux is too small or rather the magnetizing current becomes high.

8. At what angle with respect to the interpolar axis does the flux tube gives a good approximation?

a) 30°

b) 45°

c) 60°

d) 90°

Answer: c

Explanation: The flux tube crossing the air gap at 60° from the interpolar axis will always give a good approximation. The calculation of the magnetizing mmf should be based upon the value of the flux density at 60° from the interpolar axis.

9. What is the formula for mmf for air gap?

a) mmf for air gap = 800000 * air gap flux density * air gap factor * length of air gap

b) mmf for air gap = 800000 / air gap flux density * air gap factor * length of air gap

c) mmf for air gap = 800000 * air gap flux density / air gap factor * length of air gap

d) mmf for air gap = 800000 * air gap flux density * air gap factor / length of air gap

Answer: a

Explanation: For calculating the mmf for air gap, the air gap flux density is first calculated. Next the air gap factor is calculated along with the length of air gap.

10. What is the formula for the mmf required for stator teeth?

a) mmf required for stator teeth = mmf per metre + depth of stator slots

b) mmf required for stator teeth = mmf per metre * depth of stator slots

c) mmf required for stator teeth = mmf per metre / depth of stator slots

d) mmf required for stator teeth = mmf per metre – depth of stator slots

Answer: b

Explanation: First the mmf per meter is obtained separately from its design equation. Then the depth of the stator slots is obtained and the product of both gives mmf required for stator teeth.

11. What is the formula for the mmf required for stator teeth?

a) stator teeth mmf = mmf per metre / length of flux path in rotor core

b) stator teeth mmf = mmf per metre + length of flux path in rotor core

c) stator teeth mmf = mmf per metre * length of flux path in rotor core

d) stator teeth mmf = mmf per metre – length of flux path in rotor core

Answer: c

Explanation: First the mmf per meter of stator slots is calculated by its equation. Then the length of the flux path in rotor core is obtained and the product of both gives the stator teeth mmf value.

12. What is the formula for the magnetizing current per phase?

a) magnetizing current per phase =  / stator winding factor * no of turns of stator slots

b) magnetizing current per phase =  / stator winding factor * no of turns of stator slots

c) magnetizing current per phase =  / stator winding factor * no of turns of stator slots

d) magnetizing current per phase =  * stator winding factor * no of turns of stator slots

Answer: a

Explanation: Firstly the total magnetizing mmf per pole is calculated. Then the no of poles and the stator winding factor is calculated. Next the no. of turns of stator slots is calculated and the magnetizing current per phase can be obtained.

13. What is the no load current percent of the full load current for the output of 0.75 KW?

a) 50%

b) 40%

c) 33%

d) 90%

Answer: a

Explanation: For output of 3 kW, the no load current is 40% of full load current. For output of 15 kW, the no load current is 33% of the full load current.

14. What is the no load current percent of the full load current for the output of 37 KW?

a) 50%

b) 30%

c) 27%

d) 67%

Answer: b

Explanation: For output of 0.75 kW, the no load current is 50% of full load current. For output of 75 kW and above, the no load current is 27% of the full load current.

15. The no load power factor is the ratio of full load current to no load current.

a) true

b) false

Answer: a

Explanation: For obtaining the no load power factor first the no load current value is obtained. Then the full load current value is obtained and the ratio gives the no load power factor.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Dispersion Coefficient”.

1. How many factors influence the power factor of an induction motor?

a) 3

b) 2

c) 1

d) 4

Answer: b

Explanation: There are 2 factors which influence the power factor of an induction motor. They are magnetizing current and ideal short circuit current.

2. What is the relation between the magnetizing current and power factor?

a) magnetizing current is directly proportional to the power factor

b) magnetizing current is indirectly proportional to the power factor

c) magnetizing current is directly proportional to the square of the power factor

d) magnetizing current is indirectly proportional to the square of the power factor

Answer: a

Explanation: Magnetizing current is indirectly proportional to the power factor. As the magnetizing current is large, the power factor is poor.

3. What is the relation between the leakage current and power factor?

a) leakage current is directly proportional to the power factor

b) leakage current is indirectly proportional to the power factor

c) leakage current is directly proportional to the square of the power factor

d) leakage current is indirectly proportional to the square of the power factor

Answer: b

Explanation: Leakage current is indirectly proportional to the power factor. A small leakage current means a very good power factor.

4. What is the formula for dispersion coefficient?

a) dispersion coefficient = magnetizing current / ideal short circuit current

b) dispersion coefficient = magnetizing current * ideal short circuit current

c) dispersion coefficient = magnetizing current + ideal short circuit current

d) dispersion coefficient = magnetizing current – ideal short circuit current

Answer: a

Explanation: First the magnetizing current is calculated. Next the ideal short circuit current is calculated. The ratio of both gives the value of dispersion coefficient.

5. What is the formula for dispersion coefficient?

a) dispersion coefficient = 0.838 * 10 6 * 3.14 / air gap length * effective specific permeance / pole pitch *  2 * number of slots per pole per phase

b) dispersion coefficient = 0.838 * 10 6 * 3.14 * air gap length / effective specific permeance / pole pitch *  2 * number of slots per pole per phase

c) dispersion coefficient = 0.838 * 10 6 * 3.14 * air gap length * effective specific permeance * pole pitch *  2 * number of slots per pole per phase

d) dispersion coefficient = 0.838 * 10 6 * 3.14 * air gap length * effective specific permeance / pole pitch *  2 * number of slots per pole per phase

Answer: d

Explanation: For the calculation of dispersion coefficient, first the air gap length, effective specific permeance is calculated. Next the pole pitch, window space factor and the number of slots per pole per phase.

6. The increase in number of poles, the dispersion coefficient increases and this gives a low power factor.

a) true

b) false

Answer: a

Explanation: The increase in number of poles increases the dispersion coefficient. The increases in dispersion coefficient gives a low power factor.

7. What is the relation between the number of poles and pole pitch with power factor?

a) number of poles increases, pole pitch increases, bad power factor

b) number of poles increases, pole pitch decreases, good power factor

c) number of poles increases, pole pitch decreases, good power factor

d) number of poles increases, pole pitch increases, bad power factor

Answer: c

Explanation: As the number of poles increases, the pole pitch decreases and the number of slots per pole per phase also decreases. This increases the dispersion coefficient and it leads to poor power factor.

8. What is the relation between the power factor and the air gap length?

a) small air gap length, dispersion coefficient increases, good power factor

b) small air gap length, dispersion coefficient decreases, bad power factor

c) small air gap length, dispersion coefficient increases, bad power factor

d) small air gap length, dispersion coefficient decreases, good power factor

Answer: d

Explanation: If the air gap length is small, the dispersion coefficient decreases. As the dispersion coefficient decreases, the power factor increases.

9. What is the relation between the dispersion coefficient and maximum power factor?

a) dispersion coefficient is directly proportional to the power factor

b) dispersion coefficient is indirectly proportional to the power factor

c) dispersion coefficient is directly proportional to the square of the power factor

d) dispersion coefficient is indirectly proportional to the square of the power factor

Answer: b

Explanation: The dispersion coefficient is indirectly proportional to the maximum power factor. As the dispersion coefficient increases, the power factor reduces drastically.

10. What is the value of the no. of poles for obtaining a dispersion coefficient = 0.5?

a) 5

b) 7

c) 10

d) 6

Answer: d

Explanation: The machines with 6 poles can result in a dispersion coefficient of 0.5. The dispersion coefficient of 0.5 can be obtained for 2 pole and 4 pole machines also.

11. What is the relation between the overload capacity and dispersion coefficient?

a) overload capacity is directly proportional to the dispersion coefficient

b) overload capacity is indirectly proportional to the dispersion coefficient

c) overload capacity is directly proportional to the square of the dispersion coefficient

d) overload capacity is indirectly proportional to the square of the dispersion coefficient

Answer: b

Explanation: Overload capacity is indirectly proportional to the dispersion coefficient. The overload capacity of induction motors decreases with an increase in the dispersion coefficient.

12. What is the relation between the overload capacity and magnetizing current?

a) overload capacity is directly proportional to the magnetizing current

b) overload capacity is indirectly proportional to the magnetizing current

c) overload capacity is directly proportional to the square of the magnetizing current

d) overload capacity is indirectly proportional to the square of the magnetizing current

Answer: a

Explanation: Overload capacity is directly proportional to the magnetizing current. Overload capacity increases the magnetizing current and this increases the dispersion coefficient and this gives a poor power factor.

13. What is the relation between the ideal short circuit current and the number of poles?

a) short circuit current is directly proportional to the number of poles

b) short circuit current is directly proportional to the square of the number of poles

c) short circuit current is indirectly proportional to the number of poles

d) short circuit current is indirectly proportional to the square of the number of poles

Answer: c

Explanation: The ideal short circuit current is indirectly proportional to the number of poles. As the number of poles increases, the ideal short circuit current decreases.

14. What is the relation between maximum power and the number of poles?

a) maximum power factor is directly proportional to the number of poles

b) maximum power factor is directly proportional to the square of the number of poles

c) maximum power factor is indirectly proportional to the number of poles

d) maximum power factor is indirectly proportional to the square of the number of poles

Answer: c

Explanation: Short circuit current is indirectly proportional to the number of poles. The short circuit current increases the dispersion coefficient. As the dispersion coefficient increases, the maximum power factor decreases.

15. The magnetizing current decreases as the number of poles is decreased.

a) true

b) false

Answer: a

Explanation: As the number of poles is reduced the magnetizing current is reduced. As the magnetizing current is reduced, the dispersion coefficient decreases and the power factor increases.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Losses and Efficiency”.

1. How many losses are present in induction motors?

a) 4

b) 3

c) 5

d) 2

Answer: c

Explanation: There are 5 losses present in the induction motor. They are i) Stator copper losses, ii) Rotor copper losses, iii) Stator iron losses, iv) Friction and winding losses, v) Additional losses.

2. What is the formula for efficiency at full load?

a) efficiency at full load = output / output + losses

b) efficiency at full load = output / output – losses

c) efficiency at full load = output / output * losses

d) efficiency at full load = output * output + losses

Answer: a

Explanation: First the various losses are calculated for the machine. Then the output obtained is observed and the substitution of the values in the formula gives the efficiency at full load.

3. How many types of additional losses are present?

a) 1

b) 4

c) 2

d) 3

Answer: c

Explanation: The additional losses are divided into 2 types. They are i) Additional copper loss ii) Additional iron losses.

4. What factor does the additional copper losses depend upon?

a) skin effect

b) mmf harmonics

c) machine design

d) mmf harmonics and skin effect

Answer: d

Explanation: With a sinusoidal voltage impressed over the terminal of the motor, the additional copper losses are caused. They are caused due to the higher order mmf harmonics and skin effect.

5. The additional losses owing to the higher order mmf harmonics occur mainly in windings of squirrel cage rotor.

a) true

b) false

Answer: a

Explanation: The additional losses are depending on the higher order mmf harmonics and skin effect. The losses occur mainly in the squirrel cage rotor.

6. How can the additional losses be decreased in the induction motor?

a) chording the stator winding

b) skewing the rotor

c) having a proper slot combination

d) chording the stator winding, skewing the rotor, having a proper slot combination

Answer: d

Explanation: There are 3 methods to decrease the additional losses in induction motor. They are chording the stator winding, skewing the rotor, having a proper slot combination.

7. What is the use of skin effects in the induction motor?

a) it helps in improving the efficiency

b) it helps in improving the stopping characteristics

c) it helps in improving the starting characteristics

d) it helps in improving the running characteristics

Answer: c

Explanation: The skin effect phenomenon is observed in stator and rotor windings in the induction motor. The effect may be used for improving starting characteristics.

8. What should be the maximum permissible level for frequency in normal operating conditions?

a) < 2 Hz

b) > 3 Hz

c) < 4 Hz

d) > 3 Hz

Answer: b

Explanation: The normal condition operation depends upon the frequency levels in the machine. It should not exceed 3 Hz.

9. How many types are the additional losses in iron classified into?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: The additional iron losses are classified into 2 types. They are i) pulsation losses and ii) surface losses.

10. The pulsation losses are caused by the direct axis pulsation of magnetic flux.

a) true

b) false

Answer: a

Explanation: The pulsation losses are one type of additional iron losses produced. They are produced by the direct axis pulsation of magnetic flux due to the variation of permeance caused by the continuous change in mutual positions of rotor and stator teeth during rotation of rotor.

11. How much does the addition iron losses relate with the supplied power?

a) additional iron losses = 0.5% of supplied power

b) additional iron losses = 0.6% of supplied power

c) additional iron losses = 0.8% of supplied power

d) additional iron losses = 0.9% of supplied power

Answer: a

Explanation: The additional iron losses are a small amount when compared with the supplied power. They are 0.5% of the supplied power.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Stator Winding”.

1. What type of winding is generally used for the stators?

a) double layer wave winding

b) double layer lap winding

c) single layer wave winding

d) single layer lap winding

Answer: b

Explanation: The double layer wave winding is generally used for stators. The wave winding is with diamond coils is used for stators.

2. What type of winding is made use of small motors?

a) single layer mush winding

b) single layer lap winding

c) single layer wave winding

d) double layer wave winding

Answer: a

Explanation: Small motors consisting of small number of slots have a large number of turns per phase. These small motors use single layer mush windings.

3. What class does the slot and phase insulation belong to?

a) B

b) Y

c) H

d) E

Answer: d

Explanation: The modern insulating materials for diamond coils belong to class E, B and F. The slot and phase insulation is Polyester foil coated with compressed fibre for Class E.

4. What class does the plastic foil baked with polyamide fibres belong to?

a) Y

b) B

c) F

d) H

Answer: c

Explanation: The modern insulating materials for diamond coils belong to classes E, B, and F. The plastic foil baked with polyamide fibres belong to class F.

5. What is the formula for flux per pole?

a) flux per pole = average magnetic flux * pole pitch * length

b) flux per pole = average magnetic flux / pole pitch * length

c) flux per pole = average magnetic flux * pole pitch / length

d) flux per pole = average magnetic flux * pole pitch + length

Answer: a

Explanation: First the average magnetic flux is calculated. Then the pole pitch is calculated and then the length of the pole is calculated.

6. What is the initial assumption for the value of winding factor?

a) 0.9

b) 0.95

c) 0.93

d) 0.92

Answer: b

Explanation: The winding factor may be initially assumed as 0.955. It is the value of winding factor for infinitely distributed winding with full pitch coils.

7. What is the formula for stator turns per phase?

a) stator turns per phase = Stator voltage per phase / 4.44 * f * maximum flux / stator winding factor

b) stator turns per phase = Stator voltage per phase * 4.44 * f * maximum flux * stator winding factor

c) stator turns per phase = Stator voltage per phase / 4.44 * f * maximum flux * stator winding factor

d) stator turns per phase = Stator voltage per phase * 4.44 * f * maximum flux / stator winding factor

Answer: c

Explanation: For the finding out of stator turns per phase, first the stator voltage per phase is obtained. Next, the maximum flux is calculated, then the stator winding factor is calculated.

8. What should be the range of current density in the stator windings?

a) 2-5 A per mm 2

b) 4-5 A per mm 2

c) 3-5 A per mm 2

d) 2-3 A per mm 2

Answer: c

Explanation: The minimum value for the current density in stator winding is 3 A per mm 2 . The maximum value of the current density in the stator windings should not exceed 5 A per mm 2 .

9. For the lower values of current, round conductors would be convenient to use.

a) true

b) false

Answer: a

Explanation: For lower values of current, round conductors would be the most convenient to use while for higher current bars. It should be less than 2 or 3 mm in diameter or else it is difficult to wind.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Types of Synchronous Machines”.

1. How many categories can the synchronous motors be divided into?

a) 4

b) 3

c) 5

d) 2

Answer: c

Explanation: The synchronous motors can be divided into 5 types of categories. They are hydro-generators, turbo-alternators, engine driven generators, motors, compensators.

2. How is the hydro-generator driven by?

a) water turbine

b) steam turbine

c) internal combustion engines

d) control of reactive power networks

Answer: a

Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The hydro-generators are also known as water wheel generators.

3. What is the rating of the hydro-generators?

a) 750 MW

b) 1000 MW

c) 20 MW

d) 700 MW

Answer: a

Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The rating of the synchronous generators is 750 MW.

4. What is the speed by which the hydro-generators are driven?

a) 100-1000 rpm

b) 5000 rpm

c) 1500 rpm

d) 3000 rpm

Answer: a

Explanation: The synchronous generators which are driven by the water turbines are called as hydro-generators. The hydro-generators are driven at a speed of 100-1000 rpm.

5. How is the turbo-alternators driven by?

a) water turbines

b) steam turbines

c) engine driven generators

d) compensators

Answer: b

Explanation: The turbo alternators are one category among the synchronous machines. They are driven by the steam turbines.

6. What is the speed of the turbo-alternators?

a) 100-1000 rpm

b) 5000 rpm

c) 1500 rpm

d) 3000 rpm

Answer: d

Explanation: The turbo alternators are one category among the synchronous machines. They are driven by steam turbines. The rating of the turbo-alternators is 3000 rpm.

7. What is the rating of the turbo-alternators?

a) 750 MW

b) 1000 MW

c) 20 MW

d) 700 MW

Answer: b

Explanation: The turbo alternators are one category among the synchronous machines. They are driven by the steam turbines. The speed of the turbo-alternators is 1000 MW.

8. How is the engine driven generators driven by?

a) water turbines

b) steam turbines

c) internal combustion engine

d) compensators

Answer: c

Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings.

9. What is the rating of the engine driven generators?

a) 750 MW

b) 1000 MW

c) 20 MW

d) 700 MW

Answer: c

Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings. These generators have ratings upto 20 MW.

10. What is the speed of the engine driven generators?

a) 100-1000 rpm

b) 5000 rpm

c) 1500 rpm

d) 3000 rpm

Answer: c

Explanation: The engine driven generators are driven by the different forms of internal combustion engine. These generators have higher speeds for higher power ratings. These generators have speed of 1500 rpm.

11. The synchronous motors are cheaper than the induction motors.

a) true

b) false

Answer: a

Explanation: The synchronous motors are cheaper than the induction motors. They have high power lower speed applications and are cheaper when compared to the induction motors.

12. Which among the following are the applications of synchronous motors?

a) compressors

b) blowers

c) fans

d) compressors, fans, blowers

Answer: d

Explanation: They are mainly used in the high power and low speed applications. They include compressors, fans, blowers and low head pumps.

13. What is the application of synchronous compensators?

a) control of real power

b) control of active power

c) control of reactive power

d) control of apparent power

Answer: c

Explanation: The main application of the synchronous compensators is to control the reactive powers. They are used in the power supply networks.

14. What is the rating of the synchronous generators?

a) 750 MVAr

b) 1000 MVAr

c) 100 MVAr

d) 700 MVAr

Answer: c

Explanation: The main application of the synchronous compensators is to control the reactive powers. They are designed for ratings upto 100 MVAr.

15. What is the speed of the engine driven generators?

a) 100-1000 rpm

b) 5000 rpm

c) 1500 rpm

d) 3000 rpm

Answer: d

Explanation: The main application of the synchronous compensators is to control the reactive powers. They are designed to speeds upto 3000 rpm.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Construction of Hydro-Generators – 1”.

1. How many factors are involved in the construction of hydro-generators?

a) 11

b) 10

c) 9

d) 8

Answer: b

Explanation: The hydro-generators are the synchronous generators that are driven by the water turbines. They are chosen based on 10 different categories.

2. Which factor does the constructional feature of the hydro-generators depend on?

a) speed

b) voltage

c) power

d) current

Answer: a

Explanation: The constructional feature of hydro-generators are basically dependent upon the mechanical considerations. The main mechanical consideration is speed of the machine.

3. What factors does the speed of the machines depend upon?

a) head

b) blades

c) type of turbine used

d) head and type of turbine used

Answer: d

Explanation: The hydro-generators are generally low speed machines. The speed depends upon the head and the type of turbines used.

4. Why is the stator core built up of laminations?

a) to reduce core loss

b) to reduce copper loss

c) to reduce iron loss

d) to reduce eddy current loss

Answer: d

Explanation: The stator core is built up of laminations in order to reduce the eddy current loss. The loss in the laminated core is usually the single largest loss and hence the choice of type and grade of steel is of utmost importance.

5. The modern synchronous machines make use of non-directional cold rolled steel.

a) true

b) false

Answer: a

Explanation: The modern synchronous machines make use of the non-directional cold rolled steel which has electrical characteristics similar to that of the hot rolled steel. But the non-directional cold rolled steel has much improved mechanical characteristics.

6. What is the thickness of the most commonly used grade for stator laminations?

a) 0.5 mm

b) 1 mm

c) 1.5 mm

d) 2 mm

Answer: a

Explanation: The most commonly used grade for the stator laminations is grade 230. The thickness of this most commonly used grade for the stator laminations is 0.5 mm.

7. What is the range of the outside diameter of the stator frame of the large hydro-generator?

a) 3-18 m

b) 2-18 m

c) 3.5-18 m

d) 4-18 m

Answer: c

Explanation: The minimum value for the outside diameter of the stator frame of the large hydro-generators is 3.5 m. The maximum value for the outside diameter of the stator frame of the large hydro-generators is 18 m.

8. How are the stator windings of all synchronous generator connected?

a) star-delta connection

b) star connection

c) star connection with neutral earthed

d) delta connected

Answer: c

Explanation: The stator windings of all the synchronous generator are connected in the star connection with neutral earthing. The main advantage that the winding has to be insulated to earth for the phase voltage and not the line voltage.

9. What among the following is the advantages of the star connection?

a) eliminates single frequency harmonics

b) eliminates double frequency harmonics

c) eliminates triple frequency harmonics

d) eliminates sinusoidal harmonics

Answer: c

Explanation: The hydro-generators stator windings are connected in the star connected with the neutral earthed. It eliminates the triple frequency harmonics from the line voltage.

10. The capacity of the pull out machines used for making the coils limits the pole pitch to less than 0.8 m.

a) true

b) false

Answer: a

Explanation: The capacity of the pull out machines used for making the coils limits the pole pitch to less than 0.8 m. The capacity of the pull out machines used for making coils limits the length of slot portion to about 3 m.

11. What is the main advantage of a winding with multi-turns coils?

a) reduce the choosing of the value of stator slots

b) allows greater flexibility in selecting the value of stator slots

c) increases the flexibility in selecting the number of stator slots

d) has no effect on the number of stator slots

Answer: b

Explanation: The main advantage of a winding with multi-turns coils is that it allows a greater flexibility in selecting the value of stator slots. It in turn helps in choosing the required number of turns per phase.

12. What happens to the current in the windings during the sudden short circuits at the line terminals?

a) the current reduces to 15 times the full load current

b) the current increases to 15 times the full load current

c) the current reduces to 10 times the full load current

d) the current increases to 10 times the full load current

Answer: b

Explanation: The current in the windings during short circuit at the line terminals increases to about 15 times the full load current. It can also increase to higher level depending on the direct axis sub-transient reactance.

13. What happens to the electromagnetic forces during the sudden short circuits at the line terminals?

a) the electromagnetic forces get increased by 250 times the force under normal full load condition

b) the electromagnetic forces get decreased by 250 times the force under normal full load condition

c) the electromagnetic forces get increased by 200 times the force under normal full load condition

d) the electromagnetic forces get decreased by 200 times the force under normal full load condition

Answer: a

Explanation: The electromagnetic forces are directly proportional to the square of the current. It increases by 250 times the force under normal full load conditions.

14. What should be done to the conductors in the overhang of the stator?

a) bracing

b) shaving

c) punching

d) compressing

Answer: a

Explanation: The conductors in the overhang of the stator must be braced. Bracing is the process of raising the mechanical strength of the conductors.

15. How many steel brackets are used along with the support to steel rings?

a) 5

b) 7

c) 4-6

d) 5-9

Answer: c

Explanation: During the process of bracing one or two circular steel rings are used to support the overhang. Along with the steel rings 4-6 steel brackets are also used.

This set of Design of Electrical Machines Question Bank focuses on “Construction of Hydro-Generators – 2”.

1. What factor does the rotor body depends upon in the construction of hydro-generators?

a) speed

b) voltage

c) peripheral voltage

d) power

Answer: c

Explanation: In the hydro-generators the salient poles are attached to the rotor body. The type of rotor body used depends in general on the peripheral speed.

2. Which machine makes use of the forged steel construction?

a) low speed

b) high speed

c) very high speed

d) medium speed

Answer: b

Explanation: The forged steel construction is made use for the high speed machines. The earliest construction particularly at relatively low outputs consisted of a body and shaft extension made as single forging.

3. In what type of generator is the thick rolled steel discs made use of?

a) generators running at 400 rpm

b) generators running at 500 rpm

c) generators running at 600 rpm

d) generators running at 600 rpm and above

Answer: d

Explanation: Another type of rotor construction made use of in the hydro-generators is the thick rolled steel discs. It is made use for the generators running at 600 rpm and above.

4. What is the range of the length of the thick rolled sheets used in the rotor design of the hydro generators?

a) 130-180 mm

b) 120-180 mm

c) 150-180 mm

d) 160-190 mm

Answer: b

Explanation: The minimum value of the length of the thick rolled sheets used in the rotor design of the hydro-generators is 120 mm. The maximum value of the length of the thick rolled sheets used in the rotor design of the hydro-generators is 180 mm.

5. The laminations used in the rotor body design is 1.8 mm.

a) true

b) false

Answer: a

Explanation: The laminations used in the rotor body design is 1.8 mm. They are in the form of the overlapping segments tightly bolted.

6. For what peripheral speed of the machine is the segmental rim on fabricated spider used?

a) >120 m per s

b) <120 m per s

c) >130 m per s

d) <130 m per s

Answer: d

Explanation: The segmental rim on fabricated spider is used for the machines whose peripheral speed is upto 130 m per s. The advantage is it is easy to transport and assemble at site.

7. For what peripheral speed is the poles bolted to the yoke?

a) 20 m per s

b) 30 m per s

c) 25 m per s

d) 27 m per s

Answer: c

Explanation: The poles are clamped or fixed to the rim in different ways. In case of the generators the poles are bolted to the yoke when the peripheral speed is upto 25 m per s.

8. What is the range of the peripheral speeds in the water wheel generators?

a) 20 to 80 m per s

b) 20 to 50 m per s

c) 50 to 80 m per s

d) 30 to 70 m per s

Answer: a

Explanation: The minimum peripheral speed in the water wheel generators is 20 m per sec. The maximum peripheral speed in the water wheel generators is 80 m per sec.

9. The length of the mean turn of the winding is smallest with circular poles.

a) true

b) false

Answer: a

Explanation: The circular poles have a lot of advantages. The length of the mean turn of the winding is smallest with circular poles and therefore cost is reduced. The copper losses are also less.

10. Why are damper windings used in the construction of hydro-generators?

a) to control losses

b) to increases efficiency

c) to reduce the oscillations

d) to reduces voltage surges

Answer: c

Explanation: The damper windings are generally used in the construction of hydro-generators. It is generally used to reduce the oscillations in the generator.

11. By how much is the rotor and the turbine runner and the hydraulic thrust requirements more than the dead weight of rotating masses?

a) thrice

b) twice

c) by 4 times

d) by 8 times

Answer: b

Explanation: In case of the vertical shaft generators special features have to be incorporated in the bearing set up. This is because the rotor and the turbine runner and the hydraulic thrust requirements is twice more than the dead weight of rotating masses.

12. How are the bearing cooled in the construction?

a) natural cooling

b) forced cooling

c) air cooling

d) oil cooling

Answer: d

Explanation: Oil is the material which is used for the cooling of the bearings. It is being supplied to the bearings by the pump.

13. By how many minutes is the complete energy of the rotating parts and the machine is brought to rest?

a) 2 minutes

b) 1 minute

c) 3 minutes

d) 4 minutes

Answer: c

Explanation: The brakes are so designed such that the complete energy of the rotating parts and the machine is brought to rest. It is brought to rest by 3 minutes.

14. What material is used to make the material of the pads of the brakes?

a) pads using asbestos

b) pads using metal wires

c) pads using copper

d) pads using the asbestos interlaced with metal wires.

Answer: d

Explanation: The brakes are made to stop the machines immediately to protect the bearings. The brakes have pads using the asbestos interlaced with metal wires.

15. Why are the slip rings made use of?

a) they are used to provide excitation to field windings

b) they are used to provide excitation to the armature windings

c) they are used to reduce the heating effects

d) they are used to reduce the losses

Answer: a

Explanation: The slip rings are the last construction material made use of in the hydro-generators construction. They are used to provide excitation to the field windings.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Construction of Turbo-Alternators”.

1. How many factors does the construction of the turbo alternators depend upon?

a) 2

b) 4

c) 3

d) 6

Answer: c

Explanation: There are 3 factors which decide the construction of the turbo-alternators. They are i) stator core ii) stator winding iii) rotor.

2. How many poles does the turbo-alternators have and what is the speed of the turbo alternators?

a) 4 poles, 3000 rpm

b) 3 poles, 6000 rpm

c) 2 poles, 5000 rpm

d) 2 poles, 3000 rpm

Answer: d

Explanation: The turbo-alternators are made up of 2 poles. The speed with which the turbo alternators operate are 3000 rpm.

3. What is the relation of the lengths and diameters with the turbo-alternators?

a) long length, long diameters

b) long length, short diameters

c) short lengths, short diameters

d) short lengths, long diameters

Answer: b

Explanation: The turbo-alternators are characterized by the long lengths. The turbo-alternators are also characterized by short diameters.

4. What is the core length and the shaft length of a 500 MW turbo alternator?

a) 5 m, 12 m

b) 6 m, 15 m

c) 3 m, 10 m

d) 4 m, 13 m

Answer: a

Explanation: The core length of a 500 MW turbo alternator is 5 m. The shaft length of the 500 MW turbo alternator is 12 m.

5. What is the outer diameter of the stator core and outer casing of 500 MW turbo alternator?

a) 5 m, 2 m

b) 6 m, 5 m

c) 3 m, 4 m

d) 4 m, 7 m

Answer: c

Explanation: For a 500 MW turbo alternator the outer diameter of the stator core is 3 m. The diameter of the outer casing is 4 m.

6. What type of lamination is used for the stator core of the turbo alternators?

a) stepped

b) smooth

c) interleaved

d) segmental

Answer: d

Explanation: The stator core is made up of segmental laminations. The grain oriented steel laminations are another type of stator laminations made use of.

7. What is the advantage of the grain oriented steel laminations?

a) it lowers the iron loss

b) it lowers the core loss

c) it lowers the heating effects

d) it lowers the harmonics

Answer: b

Explanation: The grain oriented steel laminations are also one type of laminations used for the stator core. They result in the lowering of the core loss.

8. What is the pulsational force produced for 500 MW machine?

a) 70 kN per m

b) 60 kN per m

c) 80 kN per m

d) 90 kN per m

Answer: c

Explanation: The generator designed with low voltage gives high currents which produce high pulsational forces. The pulsational forces can be as high as 80 kN per m.

9. What is the use of the laminated and transposed conductors in turbo alternators?

a) to decrease the harmonics

b) to decrease the heating effects

c) to decrease the iron and core loss

d) to decrease the eddy current loss

Answer: d

Explanation: The presence of large currents requires the use of multi circuit windings. The use of the laminated and transposed conductors in turbo alternators is to reduce the eddy current loss.

10. What is the voltage range for large turbo-alternators?

a) 15-20 kV

b) 18-23 kV

c) 20-23 kV

d) 20-25 kV

Answer: d

Explanation: The generation voltage normally used are 15 kV for 100-200 MW machines. The voltage range for the large turbo-alternators are 20-25 kV.

11. The overhang has to be highly reinforced in turbo alternators.

a) true

b) false

Answer: a

Explanation: The overhang has to be highly reinforced in turbo alternators. The overhang is distributed over a large number of sections which eliminates the concentration of conductors in the overhang.

12. What is the use of the slot in the rotor?

a) for inserting the field windings

b) for inserting the armature windings

c) for securing the field windings

d) for inserting and securing the field windings

Answer: d

Explanation: The rotors slots are used for inserting the field windings. The rotor slots are also used for securing the field windings also.

13. How many types are the rotor slots distinguished into?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are 2 types of rotor slots in the turbo-alternators. They are i) radial sort ii) parallel slot rotors.

14. The rotor is slotted for one-third of its periphery.

a) true

b) false

Answer: b

Explanation: The rotor is slotted for only two-thirds of its periphery. It is because to reduce the flux pulsations and the cost factors.

15. What is the mechanical strength of the end bells used?

a) 1120 MN per m 2

b) 1130 MN per m 2

c) 1150 MN per m 2

d) 1140 MN per m 2

Answer: c

Explanation: The end bells are made of a non-magnetic austenitic steel in order to reduce the leakage flux. The end bells have high mechanical strength of about 1150 MN per m 2 .

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Armature Design – 1”.

1. How many factors does the design of armature of synchronous machines depend upon?

a) 2

b) 4

c) 3

d) 5

Answer: d

Explanation: There are 5 factors that decide the design of the armature of synchronous machines. They are i) single or double layer winding, ii) number of armature slots, iii) coil span, iv) turns per phase, v) conductor section.

2. When are the double layer bar windings made use of during the armature design?

a) large values of flux per pole and small number of turns per phase

b) large values of flux per pole and large number of turns per phase

c) small values of flux per pole and small number of turns per phase

d) small values of flux per pole and large number of turns per phase

Answer: a

Explanation: The double layer bar windings are made use of when there is large values of flux per poles. The double layer bar windings are made use of when there are small number of turns per phase.

3. Which type of machines have a large number of poles per phase?

a) high voltage machines and machines with high value of flux per pole

b) high voltage machines and machines with small value of flux per pole

c) small voltage machines and machines with high value of flux per pole

d) small voltage machines and machines with low value of flux per pole

Answer: b

Explanation: The high voltage machines have a large number of poles per phase. The small value of flux per pole per phase also results in large number of poles per phase.

4. Which among the following makes double layer windings advantageous than the single layer windings?

a) ease in the manufacture of coils and lower cost of winding

b) less number of coils are required as spare in the case of winding repairs

c) fractional slot windings can be employed

d) ease in the manufacture of coils and lower cost of winding, fractional slot windings can be employed, less number of coils are required as spare in the case of winding repairs

Answer: d

Explanation: The double layer windings have an advantage over single layer windings because of ease in the manufacture of coils and lower cost of winding, less number of coils are required as spare in the case of winding repairs, fractional slot windings can be employed.

5. The single layer windings have higher efficiency and quieter operation because of narrow slot openings.

a) true

b) false

Answer: a

Explanation: The single layer winding has high efficiency and quiet operation because of narrow slot openings. They also have higher space factor owing to the absence of inter layer separator.

6. When is the double layer bar or wave windings made use of?

a) when single turns coils are necessary as with turbo alternators and unipolar low voltage machines

b) when single turns coils are necessary as with turbo alternators and bipolar low voltage machines

c) when single turns coils are necessary as with turbo alternators and multipolar low voltage machines

d) when double turns coils are necessary as with turbo alternators and unipolar low voltage machines

Answer: c

Explanation: The double layer bar or wave windings are used when the single turns coils are necessary. They are also made use of with the multipolar low voltage machines.

7. How many factors are related in the selection of the armature slots?

a) 5

b) 6

c) 7

d) 4

Answer: b

Explanation: There are 6 factors associated with the selection of the armature slots. They are i) Balanced windings ii) Cost iii) Hot spot temperature iv) Leakage reactance v) Tooth ripples vi) flux density in iron.

8. How is the number of armature slots associated with the armature windings?

a) number of slots should be such that unbalanced winding is obtained

b) number of slots should be such that balanced winding is obtained

c) number of slots should be so low as possible

d) number of slots should be high as possible

Answer: b

Explanation: The number of slots should be such that balanced winding is obtained. Balanced windings should be obtained because it may lead to losses and heating effects.

9. How is the number of armature slots associated with the cost factor?

a) small number of slots leads to less cost

b) small number of slots leads to high cost

c) large number of slots leads to high cost

d) large number of slots leads to low cost

Answer: a

Explanation: The smaller the number of slots, the less will be the cost. This is because there are fewer coils to wind, form insulate, place into slots, and connect.

10. How is the number of armature slots associated with the hotspot temperature?

a) small number of slots leads to less hotspot temperature

b) small number of slots leads to high hotspot temperature

c) large number of slots leads to high hotspot temperature

d) large number of slots leads to low hotspot temperature

Answer: b

Explanation: The small number of slots leads to an increase in hotspot temperature. The small number of slots results in bunching of conductors leaving small space for circulation of air.

11. How is the number of armature slots associated with the leakage reactance?

a) small number of slots leads to less leakage reactance

b) small number of slots leads to high leakage reactance

c) large number of slots leads to high leakage reactance

d) large number of slots leads to low leakage reactance

Answer: a

Explanation: When the number of slots is small the leakage flux is increased. As the leakage flux is increased, the leakage reactance is increased owing to conductors lying near each other.

12. How is the number of armature slots associated with the tooth ripples?

a) tooth ripples are increased, if the number of slots are increased

b) tooth ripples are decreased, if the number of slots are increased

c) tooth ripples are increased, if the number of slots are decreased

d) tooth ripples are decreased, if the number of slots are decreased

Answer: b

Explanation: The tooth ripples in the field form and the consequent pulsation losses in pole face decrease if a large number of slots are used. Also, the waveform of generated voltage is free from ripples.

13. How is the number of armature slots associated with the flux densities in iron?

a) tooth ripples are increased, if the number of slots are increased

b) tooth ripples are decreased, if the number of slots are increased

c) tooth ripples are increased, if the number of slots are decreased

d) tooth ripples are decreased, if the number of slots are decreased

Answer: a

Explanation: The large number of slots a greater space is taken up by the insulation. This results in the narrower teeth giving flux densities which may go beyond acceptable limits.

14. The value of slot pitch depends upon the voltage of the machine.

a) true

b) false

Answer: a

Explanation: The value of the slot pitch serves as a guide when choosing the number of armature slots. The value of the slot pitch depends upon the voltage of the machine.

15. What is the value of the slot pitch for the low voltage machines?

a) slot pitch < 25 mm

b) slot pitch = 25 mm

c) slot pitch less than equal to 25 mm

d) slot pitch greater than equal to 25 mm

Answer: c

Explanation: The value of the slot pitch depends upon the voltage of the machine. The slot pitch is less than equal to 25 mm for low voltage machines.

This set of Tough Design of Electrical Machines Questions and Answers focuses on “Armature Design – 2”.

1. What is the range for the stator slot pitch for the large hydro-electric generators?

a) 50-60 mm

b) 50-70 mm

c) 50-80 mm

d) 50-90 mm

Answer: d

Explanation: The minimum value for the stator slot pitch for the large hydro-electric generators is 50 mm. The maximum value for the stator slot pitch for the large hydro-electric generators is 90 mm.

2. When is the range of the number of slots per pole per phase in the salient pole machines?

a) 2-3

b) 3-4

c) 2-4

d) 2-6

Answer: c

Explanation: The minimum value of the number of slots per pole per phase is 2. The maximum value of the number of slots per pole per phase is 4.

3. Fractional windings are invariably used in synchronous machines.

a) true

b) false

Answer: a

Explanation: Fractional slot windings reduces the distribution factor for higher harmonics thus reducing their corresponding generated emfs and making the voltage nearly sinusoidally. Fractional slot windings are invariably used in synchronous machines.

4. What is the relation between coil span and harmonics?

a) low coil span decreases harmonics to less amount

b) low coil span decreases the harmonics drastically

c) high coil span decreases the harmonics drastically

d) high coil span decreases the harmonics by small amount

Answer: b

Explanation: The coil span is kept low in order to decreases the harmonics. The advantage of having lower coil spans is that it reduces the harmonics drastically.

5. The coil span should be 8.33 percent of pole pitch to obtain the maximum reduction of harmonics.

a) true

b) false

Answer: a

Explanation: The coil span adjustment indirectly affects the harmonics reduction. The coil span should be minimum of about 8.33 percent of pole pitch to obtain the maximum reduction of harmonics.

6. When is the formula for the flux per pole?

a) flux per pole = average magnetic field * pole pitch * length of the core

b) flux per pole = average magnetic field / pole pitch * length of the core

c) flux per pole = average magnetic field * pole pitch / length of the core

d) flux per pole = 1 / average magnetic field * pole pitch * length of the core

Answer: a

Explanation: To obtain the value of flux per pole first the average magnetic field is obtained. Then the pole pitch and the length of the core is obtained to obtain the flux per pole.

7. What is the formula for the turns per phase in the armature design?

a) turns per phase = voltage per phase * parallel paths per phase / 4.44 * flux density * frequency * winding space factor

b) turns per phase = voltage per phase / parallel paths per phase * 4.44 * flux density * frequency * winding space factor

c) turns per phase = voltage per phase * parallel paths per phase * 4.44 * flux density * frequency * winding space factor

d) turns per phase = voltage per phase * parallel paths per phase * 4.44 * flux density / frequency * winding space factor

Answer: a

Explanation: For obtaining the turns per phase, the voltage per phase is obtained along with the parallel paths per phase. Next the winding space factor is calculated and the substitution in the formula gives the turns per phase.

8. What is the formula for current in each conductor?

a) current in each conductor = kVA * 10 3 * 3 * voltage per phase

b) current in each conductor = kVA / 10 3 * 3 * voltage per phase

c) current in each conductor = kVA * 10 3 / 3 * voltage per phase

d) current in each conductor = kVA * 10 3 * 3 / voltage per phase

Answer: c

Explanation: The kVA output is first obtained from the operation of the machine. Next the voltage per phase is calculated to obtain the current in each conductor.

9. What is the permissible current density in the armature conductors?

a) 3-4 A per mm 2

b) 3-6 A per mm 2

c) 4-6 A per mm 2

d) 3-5 A per mm 2

Answer: d

Explanation: The minimum permissible value of the current density in the armature conductors is 3 A per mm 2 . The maximum allowed value of the current density in the armature conductors is 6 A per mm 2 .

10. What is the formula for the area of cross section of armature conductors?

a) area of cross section = current per conductor * current density in the armature conductors

b) area of cross section = current per conductor + current density in the armature conductors

c) area of cross section = current per conductor – current density in the armature conductors

d) area of cross section = current per conductor / current density in the armature conductors

Answer: d

Explanation: For obtaining the area of the cross section the current per conductor is calculated. Next the current density is calculated and the ratio of both gives the current density of the area of cross section.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Field Winding – 1”.

1. What type of coils are made use of for machines with small number of poles?

a) iron wound coils

b) wire wound coils

c) rectangular coils

d) square coils

Answer: b

Explanation: Different type of coils are being made use of for the machines as per the quantity of number of poles. For small number of poles the wire wound coils are made use of.

2. What type of strips is made use of for field coils of small alternators?

a) wood covered rectangular strips

b) bare copper strips

c) glass covered rectangular strips

d) iron strips

Answer: c

Explanation: The glass covered rectangular strips are made use of for the field coils of small alternator. The bare copper strips are being used are made use of for the field coils of large alternator.

3. What should be the maximum width of the edge conductors used in the large alternators?

a) 6 mm

b) 5 mm

c) 4 mm

d) 3 mm

Answer: a

Explanation: The field coils of the large alternators use strip on edge winding wherein the bare copper strips are insulated from each other by interturn insulation. The width of the edge conductors does not exceed 6 mm.

4. For machines with Class B insulation, how many layers of inter turn insulation is made use of and what is the distance between the layers?

a) 4, 0.18 mm

b) 3, 0.25 mm

c) 2, 018 mm

d) 2, 0.25 mm

Answer: c

Explanation: The machines using Class B insulation makes use of 2 layers of inter turn insulation. The distance between the insulation is given to be 0.18 mm.

5. What material is the paper strips stuck with?

a) synthetic resin varnish

b) shellac

c) synthetic resin varnish and shellac

d) synthetic resin varnish or shellac

Answer: d

Explanation: The paper strips used in the Class B insulation machines are stuck on with shellac. They are also made of the synthetic resin varnish materials.

6. What is the thickness of the flanges and what material is used in the flanges?

a) 10 mm thick, resins

b) 10 mm thick, asbestos

c) 15 mm thick, asbestos

d) 10 mm thick, bakelized asbestos

Answer: d

Explanation: The flanges used in the Class B insulation is made up of 10 mm thickness. The flanges are made up of the bakelized asbestos.

7. Current is passed simultaneously through the conductors to raise the temperature of the field coil.

a) true

b) false

Answer: a

Explanation: The current is passed simultaneously through the conductors to raise the temperature of the field coil. The temperature should be high enough so that polymerization of interturn insulation is complete.

8. During the pressing and consolidation by how much is the thickness of the interturn insulation reduced to?

a) 0.36 mm to 0.26 mm

b) 0.36 mm to 0.25 mm

c) 0.30 mm to 0.25 mm

d) 0.32 mm to 0.25 mm

Answer: a

Explanation: The thickness of the interturn insulation before the pressing and consolidation is 0.36 mm. After the process of pressing and consolidation the thickness of the interturn insulation is reduced to 0.26 mm.

9. How many layers does the machine with Class F insulation consists of?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: The machines with Class B insulation consists of 2 layers. The machines with Class F insulation consists of 3 layers.

10. What is the thickness of the layers of Class F insulation and what material is layers made of?

a) 0.18 mm, asbestos paper

b) 0.10 mm, asbestos paper

c) 0.18 mm, thick epoxy treated asbestos paper

d) 0.10 mm, thick epoxy treated asbestos paper

Answer: c

Explanation: The thickness of the layers is 0.18 mm. The layers are made up of thick epoxy treated asbestos paper.

11. What is the lamination material of the pole body and the thickness of the pole body insulation?

a) epoxy resin, 5 mm thick

b) epoxy resin. 4 mm thick

c) asbestos, 4 mm thick

d) asbestos 5 mm thick

Answer: b

Explanation: The pole body insulation is made up of the epoxy resin laminations. The thickness of the pole body insulation is 4 mm thick.

12. What is the range of the pressure under which the field coils are consolidated?

a) 4-10 MN per m 2

b) 3-10 MN per m 2

c) 4-12 MN per m 2

d) 4-15 MN per m 2

Answer: c

Explanation: The minimum pressure under which the field coils are consolidated is 4 MN per m 2 . The maximum value of the pressure under which the field coils are 12 MN per m 2 .

13. What is the range of the exciter voltage in the field coils?

a) 50-100 V

b) 150-300 V

c) 200-400 V

d) 50-400 V

Answer: d

Explanation: The minimum value of the exciter voltage across the field coils is 50 V. The maximum value of the exciter voltage is 400 V.

14. The field winding should be designed for a voltage from 15-20% less than the exciter voltage.

a) true

b) false

Answer: a

Explanation: The field winding should be designed for a voltage from 15-20% less than the exciter voltage. This is because to allow for the drop in voltage between field and exciter and to allow for variations in the reluctance of the magnetic field.

15. What is the formula for the voltage across each field coil?

a) voltage across each field coil = *exciter voltage/number of poles

b) voltage across each field coil = *exciter voltage*number of poles

c) voltage across each field coil = /exciter voltage*number of poles

d) voltage across each field coil = /exciter voltage/number of poles

Answer: a

Explanation: The exciter voltage value is first set up for the particular machine. Then with the number of poles, the voltage across each field coil is calculated.

This set of Tricky Design of Electrical Machines Questions and Answers focuses on “Design of Field Winding – 2”.

1. What is the formula for the winding height in the design of the field windings?

a) winding height = height of the pole – height of shoe + space taken by the spool, flanges, etc

b) winding height = height of the pole + height of shoe + space taken by the spool, flanges, etc

c) winding height = height of the pole + height of shoe – space taken by the spool, flanges, etc

d) winding height = height of the pole – height of shoe – space taken by the spool, flanges, etc

Answer: d

Explanation: The height of the pole and the height f the shoe is calculated. Next the space taken by the spool, flanges is calculated and the winding height is obtained.

2. What is the approximate value of the space taken by spools, flanges, etc?

a) 15 mm

b) 10 mm

c) 12 mm

d) 20 mm

Answer: d

Explanation: The value of the space taken by spools, flanges are required for the calculation of the winding height. The approximate value of the space taken by spools, flanges, etc. is 20 mm.

3. What is the winding depth for the pole pitch of 0.1 mm?

a) 25 mm

b) 35 mm

c) 45 mm

d) 50 mm

Answer: a

Explanation: The winding depth is 25 mm for the pole pitch of 0.1 mm. The winding depth is 35 mm for the pole pitch of 0.2 mm. The winding depth is 45 mm for the pole pitch is 0.3 mm.

4. What is the formula for the voltage across each field coil?

a) voltage across each field coil = field current * resistance of each field at 75°C

b) voltage across each field coil = field current / resistance of each field at 75°C

c) voltage across each field coil = field current + resistance of each field at 75°C

d) voltage across each field coil = field current – resistance of each field at 75°C

Answer: a

Explanation: The field current is first calculated from the machine. Next the resistance value of each field at 75°C is calculated and this gives the voltage across each field coil.

5. What is the range of the current density in the field conductors?

a) 3 to 5 A per mm 2

b) 3 to 4 A per mm 2

c) 4 to 5 A per mm 2

d) 3 to 6 A per mm 2

Answer: b

Explanation: The minimum range of the current density in the field conductors is 3 A per mm 2 . The maximum value of the current density in the field conductors is 4 A per mm 2 .

6. What is the formula for the field current of the synchronous machines?

a) field current = current density * area of conductors

b) field current = current density / area of conductors

c) field current = current density – area of conductors

d) field current = current density + area of conductors

Answer: a

Explanation: For the calculation of the field current, first the current density is calculated. Next, the area of conductors is calculated and the field current is calculated.

7. What is the formula for the number of field turns of the field windings?

a) number of field turns = field mmf per pole at full load * field current

b) number of field turns = field mmf per pole at full load / field current

c) number of field turns = field mmf per pole at full load + field current

d) number of field turns = field mmf per pole at full load – field current

Answer: b

Explanation: The field mmf per pole at full load is calculated from the voltage across each field coil. Next, the field current is calculated and from these values the number of field turns is calculated.

8. What is the relation between winding space and the depth?

a) winding space is directly proportional to the depth

b) winding space is indirectly proportional to the depth

c) winding space is directly proportional to the square of the depth

d) winding space is indirectly proportional to the square of the depth

Answer: b

Explanation: The winding space is indirectly proportional to the depth. If the winding space is less, then the depth is increased.

9. What is the formula of the resistance of the winding is calculated at 75°C?

a) resistance of the winding =  / area of the field conductors

b) resistance of the winding =  * area of the field conductors

c) resistance of the winding =  / area of the field conductors

d) resistance of the winding =  / area of the field conductors

Answer: a

Explanation: The number of field turns is calculated along with the pole proportion. The length of mean turns of the coil and the area of the field conductors is calculated and on substituting the values the resistance of the winding is obtained.

10. What is the formula of the dissipating surface of the coil?

a) dissipating surface of the coil = 2*length of mean turns of the coil*

b) dissipating surface of the coil = 2*length of mean turns of the coil*

c) dissipating surface of the coil = 2*length of mean turns of the coil*

d) dissipating surface of the coil = 2*length of mean turns of the coil/

Answer: c

Explanation: First the length of the mean turns of the coil is calculated. Then the winding height and the diameter of the winding is calculated to obtain the dissipating surface of the coil.

11. What is the formula for the cooling co-efficient to the rotating field coils?

a) cooling coefficient of rotating field coils = 0.05 to 0.08 / 1 + armature voltage

b) cooling coefficient of rotating field coils = 0.05 to 0.08 / 1 – armature voltage

c) cooling coefficient of rotating field coils = 0.08 to 0.12 / 1 + armature voltage

d) cooling coefficient of rotating field coils = 0.08 to 0.12 / 1 – armature voltage

Answer: c

Explanation: The armature voltage is first calculated for the calculation of the cooling coefficient of rotating field coils. The cooling coefficient is used to calculate the temperature rise.

12. What is the formula for the temperature rise in the design of field windings?

a) temperature rise = 1 / copper loss in each field coil at 75°C * cooling coefficient of rotating field coils * dissipating surface of the coil

b) temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils * dissipating surface of the coil

c) temperature rise = copper loss in each field coil at 75°C / cooling coefficient of rotating field coils * dissipating surface of the coil

d) temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils / dissipating surface of the coil

Answer: d

Explanation: The copper loss in each field coil is first calculated using its formula. Next, the cooling coefficient of rotating field coils is calculated. Finally dissipating surface of the coil is calculated and this gives the temperature rise.

13. If the temperature increases beyond the acceptable limits the depth of the winding should be decreased.

a) true

b) false

Answer: b

Explanation: The temperature rise = copper loss in each field coil at 75°C * cooling coefficient of rotating field coils / dissipating surface of the coil is calculated. If the temperature rise crosses the specified limits, the depth of the winding is increased.

14. The increase in the depth of the winding increases the heat dissipating surface.

a) true

b) false

Answer: a

Explanation: The increase in the depth of the windings increase the heat dissipating surface. The increase in the heat dissipation decreases the temperature rise.

15. What is the minimum clearance between adjacent field coils and pole drawing?

a) 14 mm

b) 15 mm

c) 13 mm

d) 12 mm

Answer: b

Explanation: The final step in the design of field windings is the checking of the clearance between the adjacent field coils and pole drawing. The minimum value of clearance should be 15 mm.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Rotor – 1”.

1. How many factors does the design of rotor of synchronous machines depend upon?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: There are 4 factors which are associated with the design of rotor in the synchronous machines. They are height of pole, design of damper windings, height of pole shoe, pole profile drawing.

2. What is the formula for the flux in pole body?

a) flux in pole body = leakage coefficient * useful flux per pole

b) flux in pole body = leakage coefficient / useful flux per pole

c) flux in pole body = leakage coefficient – useful flux per pole

d) flux in pole body = leakage coefficient + useful flux per pole

Answer: a

Explanation: The leakage coefficient is obtained first from its formula. Next, the value of useful flux per pole is calculated and this gives the flux in pole body value.

3. What is the range of the permissible values of the flux densities in pole body?

a) 1.4-1.7 Wb per m 2

b) 1.5-1.7 Wb per m 2

c) 1.4-1.6 Wb per m 2

d) 1.5-1.6 Wb per m 2

Answer: b

Explanation: The minimum value of the flux density in the pole body is given to be 1.5 Wb per m 2 .The maximum permissible value of the flux density in the pole body is given to be 1.7 Wb per m 2 .

4. What is the range of the leakage coefficient in the pole body?

a) 1.1 to 1.2

b) 1.00 to 1.5

c) 1.15 to 1.2

d) 0.75 to 2.3

Answer: c

Explanation: The minimum value of the leakage coefficient in the pole body is 1.15. The maximum value of the leakage coefficient in the pole body is 1.2.

5. What is the formula for the area of cross-section of pole body for rectangular poles?

a) area of cross section of pole body = 0.98 * axial length of the pole * breadth of the pole

b) area of cross section of pole body = 0.98 / axial length of the pole * breadth of the pole

c) area of cross section of pole body = 0.98 * axial length of the pole / breadth of the pole

d) area of cross section of pole body = 1/0.98 * axial length of the pole * breadth of the pole

Answer: a

Explanation: The axial length of the pole and the breadth of the pole are calculated. Next by multiplying the two values with the stacking factor, we get the area of cross section of pole body.

6. What is the formula for the copper area of the field windings?

a) copper area = full load field mmf * current density in the field winding

b) copper area = full load field mmf / current density in the field winding

c) copper area = full load field mmf + current density in the field winding

d) copper area = full load field mmf – current density in the field winding

Answer: b

Explanation: For the calculation of the copper area, first the current density in the field winding is calculated. Next the full load field mmf is calculated and the ratio gives the copper area of field windings.

7. What is the formula for the total space required for the winding?

a) total space = copper area + space factor

b) total space = copper area – space factor

c) total space = copper area / space factor

d) total space = copper area * space factor

Answer: c

Explanation: The copper area is calculated from its respective formula. Then the space factor is calculated and the ratio gives the value of total space.

8. What is the value of space factor for the strip on edge winding?

a) 0.8-0.9

b) 0.4

c) 0.65

d) 0.75

Answer: a

Explanation: The space factor for the strip on edge winding is 0.8-0.9. The space factor for small round wires is 0.4 and for large round wires it is 0.65. The space factor for large rectangular conductors is 0.75.

9. What is the formula for the height of winding?

a) height of winding = total winding area / depth of winding

b) height of winding = total winding area * depth of winding

c) height of winding = total winding area + depth of winding

d) height of winding = total winding area – depth of winding

Answer: a

Explanation: The total winding area is first calculated. Next the depth of the winding is calculated. The ratio of both gives the height of winding.

10. What is the formula for the radial length of the pole shoe?

a) radial length of the pole shoe = height of winding – height of pole shoe – 0.02

b) radial length of the pole shoe = height of winding + height of pole shoe – 0.02

c) radial length of the pole shoe = height of winding – height of pole shoe + 0.02

d) radial length of the pole shoe = height of winding + height of pole shoe + 0.02

Answer: d

Explanation: First the height of the winding is calculated from its formula. Next the height of pole shoe is calculated. Both the values are added with 0.02 to give the radial length of the pole shoe.

11. What is the formula for the height of pole body?

a) height of pole body = height of the winding + 0.02

b) height of pole body = height of the winding * 0.02

c) height of pole body = height of the winding – 0.02

d) height of pole body = height of the winding / 0.02

Answer: a

Explanation: The height of the pole body is one of the design factors in the design of rotor. It is obtained by adding the value of the height of winding with 0.02, which is the approximate space occupied by flanges.

12. What is the range of the ratio of radial length of pole to pole pitch?

a) 0.3-1

b) 0.3-1.5

c) 0.7-1

d) 0.7-1.5

Answer: b

Explanation: The minimum value of the ratio of radial length of pole to pole pitch is given to be 0.3. The maximum value of the ratio of radial length of pole to pole pitch is given to be 1.5.

13. The damper windings are made use of in synchronous generators to reduce the oscillations and to prevent hunting.

a) true

b) false

Answer: a

Explanation: The purpose of the damper windings is to reduce the oscillations and to prevent the hunting in synchronous generators. Next, the damper windings are used to suppress the negative sequence field in the synchronous generator.

14. The mmf of the damper windings depends on the pole pitch value.

a) true

b) false

Answer: a

Explanation: The mmf of the damper windings depends on the pole pitch value. The value for the mmf of the damper windings = 0.143 * specific electric loading * pole pitch.

15. What is the formula for the area per pole of damper pass provided?

a) area per pole of damper pass = 0.2 * specific electric loading * pole pitch * current density in damper bars

b) area per pole of damper pass = 0.2 * specific electric loading * pole pitch / current density in damper bars

c) area per pole of damper pass = 0.2 * specific electric loading – pole pitch / current density in damper bars

d) area per pole of damper pass = 0.2 + specific electric loading * pole pitch / current density in damper bars

Answer: b

Explanation: The specific electric loading and the pole pitch is calculated first. Next the current density in damper bars is next calculated. Substituting in the above formula gives the area per pole of damper pass provided.

This set of Advanced Design of Electrical Machines Questions and Answers focuses on “Design of Rotor – 2”.

1. What is the range of current density in the damper bars?

a) 3-4 A per mm 2

b) 3-5 A per mm 2

c) 3-6 A per mm 2

d) 4-6 A per mm 2

Answer: a

Explanation: The minimum value of the current density in the damper bars is given to be 3 A per mm 2 . The maximum value of the current density in the damper bars is given to be 4 A per mm 2 .

2. What is the percentage of the damper windings slot pitch with respect to stator slot pitch?

a) 30%

b) 40%

c) 20%

d) 60%

Answer: c

Explanation: The slot pitch of damper windings is 20% of the stator slot pitch. This is because to reduce the current induced in damper windings by tooth ripples.

3. What is the formula for the pole arc?

a) pole arc = number of bars per pole * stator slot pitch * 0.8

b) pole arc = number of bars per pole / stator slot pitch * 0.8

c) pole arc = number of bars per pole * stator slot pitch / 0.8

d) pole arc = 1 / number of bars per pole * stator slot pitch * 0.8

Answer: a

Explanation: The number of bars per pole is calculated along with the stator slot pitch. Next, all the values are multiplied by 0.8 and the pole arc value is obtained.

4. What is the formula for the length of each damper bar for small machines?

a) length of each damper bar = 1.1 * axial length

b) length of each damper bar = axial length + 0.1

c) length of each damper bar = axial length – 0.1

d) length of each damper bar = 1.1 / axial length

Answer: a

Explanation: Length of each damper bar = 1.1 * axial length is the formula for the length of each damper bar for small machines. length of each damper bar = axial length + 0.1 is the formula for the length of each damper bar for large machines.

5. What is the formula for the area of cross-section of each damper bar?

a) area of cross section of each damper bar = total area of bars per pole – number of damper bars per pole

b) area of cross section of each damper bar = total area of bars per pole + number of damper bars per pole

c) area of cross section of each damper bar = total area of bars per pole / number of damper bars per pole

d) area of cross section of each damper bar = total area of bars per pole * number of damper bars per pole

Answer: c

Explanation: The total area of bars per pole is first obtained. Next the number of damper bars per pole is calculated and the ratio gives the area of cross section of each damper bar.

6. What is the formula of the area of each ring short-circuiting the bars?

a) area of each ring short-circuiting the bars =  * area of damper bar

b) area of each ring short-circuiting the bars =  * area of damper bar

c) area of each ring short-circuiting the bars =  * area of damper bar

d) area of each ring short-circuiting the bars =  * area of damper bar

Answer: b

Explanation: First the area of damper bars is calculated first. Next the area of damper bars is multiplied by a value between 0.8-1 to obtain the area of each ring short-circuiting the bars.

7. Given : total area = 473 mm 2 and Number of bars = 8 for a rotor design, what is the value of area of each damper bar?

a) 59 mm 2

b) 455 mm 2

c) 475 mm 2

d) 3784 mm 2

Answer: a

Explanation: Area of each bar = total area / number of bars

Area of each bar = 473/8 = 59 mm 2 .

8. What is the formula for the height of pole shoe sufficient to accommodate the damper windings?

a) height of pole shoe = diameter of damper bars

b) height of pole shoe = 2 * diameter of damper bars

c) height of pole shoe = diameter of damper bars/2

d) height of pole shoe = 3 * diameter of damper bars/2

Answer: b

Explanation: First the diameter of the damper bars is calculated. It is then multiplied by 2 to get the height of pole shoe.

9. Pole profile drawing helps in obtaining the various dimensions of the pole.

a) true

b) false

Answer: a

Explanation: The pole profile drawing is essential to obtain the design characteristics of the rotor. It helps in the process of obtaining the various dimensions of the pole.

10. The pole shoe drawing is completed by fixing the height of pole shoe.

a) true

b) false

Answer: a

Explanation: The pole shoe surface can be drawn with the armature surface being fixed. The pole shoe drawing is completed by fixing the height of pole shoe.

This set of Design of Electrical Machines Problems focuses on “Losses and Temperature Rise”.

1. How many types of losses are present in synchronous machines?

a) 7

b) 3

c) 4

d) 5

Answer: a

Explanation: There are 7 losses in the synchronous machines. They are i) iron loss due to main field, ii) iron loss due to parasitic field, iii) I2R loss in the armature winding, iv) eddy current loss in armature conductors, v) stray load loss, vi) loss in field windings, vii) friction and windage loss.

2. What is the classification of the iron loss due to the main field?

a) hysteresis loss

b) eddy current loss

c) hysteresis loss or eddy current loss

d) hysteresis loss and eddy current loss

Answer: d

Explanation: The iron loss due to main field is due to the hysteresis loss. The eddy current loss also contribute to the iron losses due to main field.

3. What are the factors the pole face loss depends upon?

a) slot opening

b) air gap length

c) number of slots and speed of machines

d) slot opening, air gap length, number of slots and speed of machines

Answer: d

Explanation: The pole face loss depends upon the slot opening and the air gap length. It also depends on the number of slots and speed of machines.

4. What is the range of the pole face loss in the synchronous machines?

a) 40-60 % of iron loss

b) 20-60 % of iron loss

c) 25-70 % of iron loss

d) 40-80 % of iron loss

Answer: c

Explanation: The pole face loss has a minimum value of 25% of the iron loss. The pole face loss has a maximum value of 70% of the iron loss.

5. What is the formula for the copper loss in the synchronous machine?

a) copper loss per phase = current per phase * dc resistance

b) copper loss per phase = current per phase2 * dc resistance 2

c) copper loss per phase = current per phase2 * dc resistance

d) copper loss per phase = current per phase * dc resistance 2

Answer: c

Explanation: First the current per phase is calculated and the value is squared. Next the dc resistance is calculated and the sum of the square of the current per phase and dc resistance gives the copper loss per phase.

6. What is the formula for the total eddy current loss in conductors?

a) total copper loss = 3 * average value of the eddy current constant * current per phase 2 * dc resistance

b) total copper loss = 3 / average value of the eddy current constant * current per phase 2 * dc resistance

c) total copper loss = 3 * average value of the eddy current constant / current per phase 2 * dc resistance

d) total copper loss = 3 * average value of the eddy current constant * current per phase 2 / dc resistance

Answer: a

Explanation: The average value of the eddy current constant is obtained. Next the I 2 R loss values are calculated and multiplying with 3 gives the total copper loss.

7. What is the cause of the stray load losses in the synchronous machine?

a) stray field

b) stray armature

c) stray field and stray armature

d) stray field or stray armature

Answer: a

Explanation: The stray load loss occurs due to stray fields. They are formed when the machine is being loaded.

8. What is the voltage drop in the carbon and graphite brushes?

a) 1 V

b) 0.3 V

c) 0.6 V

d) 0.75 V

Answer: a

Explanation: The voltage drop in the carbon and graphite brushes is 1 V. The voltage drop in the brushes containing metal is 0.3 V.

9. What factors does the friction and windage loss depend upon?

a) construction of the machine

b) speed of the machine

c) rating of the machine

d) construction, speed, rating of the machine

Answer: d

Explanation: This loss consists of the bearing friction and rotor windage loss. The loss depends upon the type of construction, speed and ratings of the machines.

10. What is the reduction in the total friction loss with the hydrogen cooling?

a) 0.3-0.5 % of kVA rating

b) 0.2-0.3 % of kVA rating

c) 0.3-0.4 % of kVA rating

d) 0.3-0.6 % of kVA rating

Answer: c

Explanation: The friction loss depends upon the type of construction, speed and ratings of the machines. The hydrogen cooling reduces the total friction loss by 0.3-0.4% of the kVA rating.

11. What is the formula to obtain the temperature rise of the surface?

a) temperature rise of the surface = Surface area * cooling coefficient * dissipating surface

b) temperature rise of the surface = Surface area / cooling coefficient * dissipating surface

c) temperature rise of the surface = Surface area * cooling coefficient / dissipating surface

d) temperature rise of the surface =1 / Surface area * cooling coefficient * dissipating surface

Answer: c

Explanation: The surface area is first calculated from its formula. Next, the cooling coefficient and the dissipating surface are obtained and on substitution gives the temperature rise of the surface.

12. What factor/s does the cooling coefficient depend upon?

a) speed of the cooling medium

b) configuration of the surface

c) speed of the machine and configuration of the surface

d) speed of the machine or configuration of the surface

Answer: c

Explanation: The cooling coefficient depends upon the speed of the machine. The cooling coefficient also depends upon the configuration of the surface.

13. The value of the cooling coefficient varies from 0.025 to 0.04 in the back of the stator core.

a) true

b) false

Answer: a

Explanation: The cooling coefficient value is required in the calculation of the temperature rise of the surface. The value varies from 0.025 to 0.04 for the back of the stator core.

14. The peripheral speed is the armature peripheral speed in the stationary field coils.

a) true

b) false

Answer: a

Explanation: There are various peripheral speeds in various parts of the machine. In the stationary field coils the peripheral speed is nothing but the armature peripheral speed.

15. What all factors does the heat to be dissipated by cooling surfaces depend upon?

a) hysteresis loss

b) eddy current loss

c) heating loss

d) hysteresis, eddy and heating losses

Answer: d

Explanation: The heat to be dissipated by the cooling surface of the armature core would consist of the hysteresis loss and the eddy current loss. It also consists of the heating loss or the I2R in the active part of the armature.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Synchronous Machines Output Equation”.

1. What is the formula for output equations in synchronous machines?

a) kVA output = output coefficient * diameter 2 * length * synchronous speed

b) kVA output = output coefficient / diameter 2 * length * synchronous speed

c) kVA output = output coefficient * diameter 2 / length * synchronous speed

d) kVA output = output coefficient * diameter 2 * length / synchronous speed

Answer: a

Explanation: The output equation is found out by first calculating the output coefficient. Next, the diameter and length are obtained, and the synchronous speed is calculated using the tacho-generator to obtain the kVA output.

2. What is the formula of the output coefficient?

a) output coefficient = 11 * specific magnetic loading / specific electrical loading * winding space factor * 10 -3

b) output coefficient = 11 / specific magnetic loading * specific electrical loading * winding space factor * 10 -3

c) output coefficient = 11 * specific magnetic loading * specific electrical loading * winding space factor * 10 -3

d) output coefficient = 11 * specific magnetic loading * specific electrical loading / winding space factor * 10 -3

Answer: c

Explanation: The output coefficient is one of the terms required in the calculation of the output of the machine. The specific magnetic and electrical loading terms are first calculated along with the winding space factor.

3. What is the formula for the output equation with respect to the peripheral speed?

a) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor * 10 -3 * peripheral speed 2 *Length * synchronous speed

b) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor * 10 -3 * peripheral speed 2 *Length / synchronous speed

c) output = 1.11* specific magnetic loading * specific electrical loading * winding space factor / 10 -3 * peripheral speed 2 *Length * synchronous speed

d) output = 1.110 / specific magnetic loading * specific electrical loading * winding space factor * 10 -3 * peripheral speed 2 *Length * synchronous speed

Answer: b

Explanation: The output equation with respect to the peripheral speed depends on the square of the peripheral speed of the machine. It doesn’t consist of the diameter term in the output equation.

4. How many factors does the choice of specific magnetic loading depend upon?

a) 4

b) 2

c) 5

d) 8

Answer: c

Explanation: The choice of specific magnetic loading depends upon 5 factors basically. They are a) Iron Loss, b) Voltage, c) Transient short circuit current, d) Stability, e) Parallel Operation.

5. How is the iron loss related with the choice of specific magnetic loading?

a) choice of magnetic loading is directly proportional to the iron loss

b) choice of magnetic loading is indirectly proportional to the iron loss

c) choice of magnetic loading is directly proportional to the square of the iron loss

d) choice of magnetic loading is indirectly proportional to the square of the iron loss

Answer: a

Explanation: The choice of specific magnetic loading is directly proportional to the iron loss. The iron loss increases with the increase in the air gap density.

6. How is the voltage related with the air gap density?

a) air gap density is directly proportional to the voltage

b) air gap density is indirectly proportional to the voltage

c) air gap density is directly proportional to the square of the voltage

d) air gap density is indirectly proportional to the square of the voltage

Answer: b

Explanation: The air gap density is indirectly proportional to the voltage. High voltage machine should have low air gap density, to avoid excessive values of flux density in the teeth and core.

7. How is the transient short circuit current related with the air gap density?

a) air gap density is directly proportional to the short circuit current

b) air gap density is indirectly proportional to the short circuit current

c) air gap density is directly proportional to the square of the short circuit current

d) air gap density is directly proportional to the square of the short circuit current

Answer: a

Explanation: The air gap density is directly proportional to the short circuit current. The air gap density should be kept low in order to reduce the initial electromagnetic forces under short circuit condition.

8. How is the steady state stability related with the air gap density?

a) air gap density is directly proportional to the steady state stability

b) air gap density is indirectly proportional to the steady state stability

c) air gap density is directly proportional to the square of the steady state stability

d) air gap density is directly proportional to the square of the steady state stability

Answer: a

Explanation: The air gap density is directly proportional to the steady state stability. The steady state stability is improved if the air gap density is high.

9. The machines having high air gap density operates poorly when connected in synchronism.

a) true

b) false

Answer: b

Explanation: The machines having high air gap density allows high amount of synchronizing power. Thus the machines having high air gap density provides high synchronism.

10. What is the range of the air gap density for salient pole machines?

a) 0.52-0.65 Wb per m 2

b) 0.5-0.6 Wb per m 2

c) 0.54-0.65 Wb per m 2

d) 0.44-0.65 Wb per m 2

Answer: a

Explanation: The range of air gap density for salient pole machines is 0.52-0.65 Wb per m 2 . The range of air gap density for turbo-alternators is 0.54-0.65 Wb per m 2 .

11. How many factors influence the choice of specific electric loading?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: There are 4 factors that influence the choice of specific electric loading. They are a) Copper loss and temperature rise, b) voltage, c) synchronous reactance, d) stray load loss.

12. How is the specific electric loading related to copper losses and temperature rise?

a) high specific electric loading gives high copper losses and high temperature rise

b) high specific electric loading gives low copper losses and high temperature rise

c) high specific electric loading gives high copper losses and low temperature rise

d) high specific electric loading gives low copper losses and low temperature rise

Answer: a

Explanation: The specific electric loading is directly proportional to the copper losses and the temperature rise. The high specific electric loading gives high copper losses and high temperature rise.

13. High value of the specific electric loading can be used for low voltage machines.

a) true

b) false

Answer: a

Explanation: High value of specific electric loading can be used for low voltage machines. This is because the space required for insulation is small.

14. How is the specific electric loading related to the synchronous reactance of the machines?

a) specific electric loading is high, leakage reactance is high, giving low synchronous reactance

b) specific electric loading is high, leakage reactance is low, giving low synchronous reactance

c) specific electric loading is high, leakage reactance is high, giving high synchronous reactance

d) specific electric loading is low, leakage reactance is high, giving high synchronous reactance

Answer: c

Explanation: The specific electric loading is directly proportional to the synchronous reactance. If the specific electric loading is high, the synchronous reactance becomes high.

15. What is the value of specific electric loading for the salient pole alternators?

a) 20,000-40,000 A per m

b) 50,000-75,000 A per m

c) 25,000-40,000 A per m

d) 20,000-45,000 A per m

Answer: a

Explanation: The value of specific electric loading for the salient pole alternators is 20,000-40,000 A per m. The value of specific electric loading for the turbo alternators is 50,000-75,000 A per m.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Slot Dimensions”.

1. What is the range of the flux density in the teeth at no load?

a) 1.7-1.8 Wb per mm 2

b) 1.3-1.5 Wb per mm 2

c) 1.3-1.6 Wb per mm 2

d) 1.4-1.6 Wb per mm 2

Answer: a

Explanation: The minimum value of the flux density in the teeth is given to be 1.7 Wb per mm 2 . The maximum value of the flux density in the teeth is given to be 1.8 A per mm 2 .

2. What is the formula for the minimum width of the tooth?

a) minimum width of tooth = flux * pole proportion *  * length * 1.8

b) minimum width of tooth = flux / pole proportion *  * length * 1.8

c) minimum width of tooth = flux * pole proportion /  * length * 1.8

d) minimum width of tooth = flux * pole proportion *  * length / 1.8

Answer: b

Explanation: The flux value, pole proportion and the length values are first obtained. Then the ratio of the number of stator slots to number of poles is obtained and on substitution gives the minimum width of tooth.

3. Name the slots that are commonly used.

a) parallel sided

b) square sided

c) rectangular

d) circular

Answer: a

Explanation: The most commonly used type of slots are parallel sided. The other type of slots may be used for the purposes required.

4. How is the teeth and the minimum width designed in the machines?

a) teeth is tapered and minimum width is across the medium

b) teeth is sharpened and minimum width occurs across the air gap

c) teeth is widened and minimum width occurs across the air gap

d) teeth is reduced and minimum width occurs across the medium

Answer: a

Explanation: Parallel sided slots are made use of. Hence, the teeth is tapered and their minimum width occurs at the air gap surface.

5. What is the formula for the maximum permissible width of slot?

a) maximum permissible width = slot pitch * minimum width of the teeth

b) maximum permissible width = slot pitch + minimum width of the teeth

c) maximum permissible width = slot pitch / minimum width of the teeth

d) maximum permissible width = slot pitch – minimum width of the teeth

Answer: d

Explanation: The slot pitch is first calculated with its respective formula. Next, the minimum width of the teeth is calculated and the difference between both gives the maximum permissible width.

6. By how much should the depth of slot not exceed the width?

a) two times

b) three times

c) four times

d) six times

Answer: b

Explanation: The depth of the slot depends upon the width of the slot. The depth should not exceed three times the width of the slot.

7. Why are slot made deeper in the machine?

a) to increase the short circuit current

b) to reduce the short circuit current

c) to increase the open circuit current

d) to reduce the open circuit current

Answer: b

Explanation: The slots used in the machine are basically deeper slots. The slots are made deeper even more to increase the leakage reactance and to limit the short circuit current.

8. What is the formula for the height of length of mean turn of armature?

a) length of mean turn = 2*length + 2.5*pole pitch + 0.06 kV + 0.2

b) length of mean turn = 2*length + 2*pole pitch + 0.06 kV + 0.2

c) length of mean turn = 2*length + 2.5*pole pitch – 0.06 kV – 0.2

d) length of mean turn = 2*length – 2.5*pole pitch – 0.06 kV – 0.2

Answer: a

Explanation: The length of the slots is obtained along with the pole pitch. The output kV is calculated and on substituting we get the length of mean turn.

9. The flux density in the armature core of salient pole machines lies between 1-1.2 Wb per m 2 .

a) true

b) false

Answer: a

Explanation: The value of depth of core can be calculated by assuming a suitable value of flux density. The value of the flux density varies from 1-1.2 Wb per m 2 .

10. What is the formula for the depth of armature core?

a) depth of armature core = flux / length of the iron core * flux density

b) depth of armature core = flux * length of the iron core * flux density

c) depth of armature core = flux / 2 * length of the iron core * flux density

d) depth of armature core = flux * 2 * length of the iron core * flux density

Answer: c

Explanation: The flux value is calculated along with the length of the iron core. Next, the suitable flux density is chosen and the depth of armature core is calculated.

11. What is the formula for the outer diameter of the stator?

a) outer diameter = inner diameter + depth of the slots + depth of armature core

b) outer diameter = inner diameter + 2*depth of the slots + depth of armature core

c) outer diameter = inner diameter + 2*

d) outer diameter = inner diameter + depth of the slots + 2*depth of armature core

Answer: c

Explanation: The depth of the armature core is calculated and the depth of the slots is also calculated. The inner diameter is calculated and substituting the outer diameter is obtained.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design Preliminaries”.

1. What does the copper factor in PMDC motors represent?

a) it represents the armature circular area for conductors

b) it represents the field circular area for conductors

c) it represents the fraction of the armature circular area for conductors

d) it represents the fraction of the field circular area for conductors

Answer: c

Explanation: The copper factor represents the fraction of the armature circular area for conductors. It is represented by the letter K.

2. What is the range of the copper factor in PMDC motors?

a) 0.1-0.3

b) 0.1-0.2

c) 0.1-0.4

d) 0.2-0.4

Answer: b

Explanation: The copper factor represents the fraction of the armature circular area for conductors. The range of the copper factor is between 0.1-0.2.

3. What is the formula for the armature resistance in PMDC motor?

a) armature resistance = *total number of armature conductors/1.2 * 10 4 * number of parallel paths in the armature 2

b) armature resistance = *total number of armature conductors*1.2 * 10 4 * number of parallel paths in the armature 2

c) armature resistance = *total number of armature conductors/1.2 * 10 4 + number of parallel paths in the armature 2

d) armature resistance = +total number of armature conductors/1.2 * 10 4 * number of parallel paths in the armature 2

Answer: a

Explanation: First the diameter, length and the total number of armature conductors are obtained. Next the number of parallel paths in the armature is calculated and on substitution it provides the armature resistance.

4. What happens to the diameter when the poles are more than 2?

a) diameter = 2 * diameter * 

b) diameter = 2.32 * diameter * 

c) diameter = 2.32 * diameter * 

d) diameter = 2 * diameter / 

Answer: b

Explanation: The diameter is the exact calculated value for 2 pole motors. But when the poles are more than 2, the above formula is made use of to calculate the armature resistance.

5. What factor does the permeance coefficient depend upon?

a) geometry of the magnet

b) geometry of the magnet, airgap, associated non-portions of the magnetic circuit

c) airgap

d) associated non-portions of the magnetic circuit

Answer: b

Explanation: The permeance coefficient depends upon the geometry of the magnet and the airgap. It also depends on the associated non-portions of the magnetic circuit.

6. What is the range of the permeance coefficient in the PMDC motors?

a) 3-5

b) 4-9

c) 4-8

d) 3-9

Answer: c

Explanation: The minimum value of the permeance coefficient used in the PMDC motors is 4. The maximum value of the permeance coefficient used in the PMDC motor is 8.

7. What is the usual value of the permeance coefficient of the PMDC motor?

a) 4

b) 5

c) 6

d) 7

Answer: c

Explanation: The range of the permeance coefficient for the PMDC motor is 4-8. The value is usually around 6 for most of the applications.

8. The field current flowing in the conductor’s acts as demagnetizing force on the fraction tips of the magnet.

a) true

b) false

Answer: b

Explanation: The armature current flowing in the conductors acts as demagnetizing force. Its acts on the fraction tips of the magnets present.

9. What is the value of the demagnetizing coefficient if the total number of teeth is greater than 10 7 ?

a) d = angle/360

b) d = angle/240

c) d = angle/540

d) d = angle/720

Answer: d

Explanation: If the total number of teeth is greater than 10 7 then the demagnetizing coefficient become the ratio of the angle and 720. Otherwise d is one half the ratio of the maximum number of teeth that can be situated within the angle to the total number of teeth.

10. What is the value of the reluctance factor in the calculation of the intensity of magnetic field?

a) 1

b) 2

c) 1.15

d) 1,45

Answer: c

Explanation: The reluctance factor is one of the factors made use of in the calculation of the intensity of magnetic field. The value of the reluctance factor is around 1.15 generally.

11. What is the formula of the magnetic to electrical boarding ratio?

a) magnetic to electrical boarding ratio = number of poles * permeance coefficient * flux per pole/number of conductors * armature current

b) magnetic to electrical boarding ratio = number of poles / permeance coefficient * flux per pole*number of conductors * armature current

c) magnetic to electrical boarding ratio = number of poles + permeance coefficient * flux per pole/number of conductors * armature current

d) magnetic to electrical boarding ratio = number of poles * permeance coefficient / flux per pole*number of conductors * armature current

Answer: a

Explanation: The permeance coefficient is first calculated along with the number of poles and the flux per pole. Then the number of conductors are noted and the armature current is calculated to give the magnetic to electrical boarding ratio.

12. How is the value of the magnetic to electrical boarding ratio related with the volume of iron and volume of copper?

a) high magnetic to electrical boarding ratio gives high copper volume and high iron volume

b) high magnetic to electrical boarding ratio gives low copper volume and high iron volume

c) low magnetic to electrical boarding ratio gives low copper volume and low iron volume

d) low magnetic to electrical boarding ratio gives low copper volume and high iron volume

Answer: b

Explanation: The high value of magnetic to electrical boarding ratio gives a high volume of iron. But the high value of magnetic to electrical boarding ratio gives low copper volume.

13. For good performance the small dc motor should have magnetic to electrical boarding ratio greater than 70.

a) true

b) false

Answer: b

Explanation: The performance of the small DC motor depends on the magnetic to electrical boarding ratio. The magnetic to electrical boarding ratio should be greater than 50 for good performance.

14. What is the formula for the flux density for the PM motors?

a) flux density = residual flux density / 1 + 

b) flux density = residual flux density * 1 + 

c) flux density = residual flux density / 1 + 

d) flux density = residual flux density * 1 + 

Answer: a

Explanation: The residual flux density is calculated first along with the permeance coefficient to obtain the flux density of the PMDC motor. The flux density is 0.85 times the residual flux density.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design Steps and Considerations – 1”.

1. How many design steps are present in the design of PMDC motors?

a) 8

b) 9

c) 10

d) 11

Answer: d

Explanation: There are 11 steps involves in the design of the PMDC motors. They are minimum sum of air gap volume and magnet volume, ratio of magnetic to electric loading, area of magnet, length of magnet, value of flux, number of turns per coil, running armature resistance, armature diameter, axial dimensions, wire cross section and radial thickness.

2. What happens to the armature diameter and the volume of air gap and magnet when the angle is lower in value?

a) volume of air gap and magnet increases, armature diameter increases

b) volume of air gap and magnet increases, armature diameter decreases

c) volume of air gap and magnet decreases, armature diameter decreases

d) volume of air gap and magnet decreases, armature diameter increases

Answer: d

Explanation: The lower values of angle, reduces the volume of air gap and magnet. The reduction of volume of air gap and magnet, increases the armature diameter.

3. What should be the range of the product of the magnetic field and magnetic flux density?

a) 4-4.5 * 10 6

b) 4-4.3 * 10 6

c) 4.3-4.6 * 10 6

d) 4.2-4.5 * 10 6

Answer: c

Explanation: The product of the magnetic field and magnetic flux density has a minimum value of 4.3 * 10 6 . The product of the magnetic field and magnetic flux density has a minimum value of 4.6 * 10 6 .

4. What should be the minimum value of the ratio of the magnetic to electric loading?

a) 40

b) 30

c) 50

d) 60

Answer: c

Explanation: The calculation of the ratio of the magnetic to electric loading is the second step in the design of the PMDC motors. It should have a minimum value of 50.

5. What is the formula for the area of the magnet in the design of PMDC motors?

a) area of magnet = flux * 4.95 * residual flux density

b) area of magnet = flux / 4.95 * residual flux density

c) area of magnet = flux * 4.95 / residual flux density

d) area of magnet = 1/flux * 4.95 * residual flux density

Answer: b

Explanation: First the residual flux density is calculated. Next, the flux is calculated and substitution in the formula gives the area of magnet.

6. What is the range of length of the magnet in the PMDC motors?

a) 2.5-4 cm

b) 2-3 cm

c) 2.5-3 cm

d) 1.5-4 cm

Answer: a

Explanation: The minimum value of the length of the magnet in the PMDC motor is 2.5 cm. The maximum value of the length of the magnet in the PMDC motor is 4 cm.

7. What is the formula of the length of the magnet?

a) length of the magnet = sum of the volume of air gap and magnet * Area of the magnet + 0.06

b) length of the magnet = sum of the volume of air gap and magnet / Area of the magnet + 0.06

c) length of the magnet = sum of the volume of air gap and magnet / Area of the magnet – 0.06

d) length of the magnet = sum of the volume of air gap and magnet * Area of the magnet – 0.06

Answer: c

Explanation: The sum of the volume of the air gap and magnet is first calculated. Next, the area of the magnet is calculated from its formula and on substitution gives the length of the magnet.

8. What is the relation between the flux and the no local speed?

a) flux is directly proportional to the no local speed

b) flux is indirectly proportional to the no local speed

c) flux is directly proportional to the square of the no local speed

d) flux is indirectly proportional to the square of the no local speed

Answer: b

Explanation: The calculation of the flux value is one of the design steps. The flux is indirectly proportional to the no local speed calculated.

9. What is the formula of the number of turns per coil?

a) number of turns per coil = number of conductors/2*coils/slot*number of armature teeth

b) number of turns per coil = number of conductors*2*coils/slot*number of armature teeth

c) number of turns per coil = number of conductors*2*coils/slot/number of armature teeth

d) number of turns per coil = number of conductors/2*coils/slot/number of armature teeth

Answer: a

Explanation: The number of conductors is calculated along with the coils per slot is calculated. Next, the number of armature teeth is calculated, and on substitution gives the number of turns per coil.

10. What is the formula for the armature resistance?

a) armature resistance = running armature resistance / 1.0 to 1.0

b) armature resistance = running armature resistance * 1.3 to 1.5

c) armature resistance = running armature resistance * 1.4 to 1.5

d) armature resistance = running armature resistance / 1.3 to 1.3

Answer: d

Explanation: The running armature resistance is first calculated in the PMDC motor. It is divided by 1.3 and that gives the armature resistance of the machine.

11. What is the relation between axial dimension and the area of the magnet?

a) area of the magnet is directly proportional to the axial dimension

b) area of the magnet is indirectly proportional to the axial dimension

c) area of the magnet is directly proportional to the square of the axial dimension

d) area of the magnet is indirectly proportional to the square of the axial dimension

Answer: a

Explanation: The calculation of the axial dimension is one of the steps in the PMDC motors. The axial dimension is directly proportional to area of the magnet.

12. What is the relation of the wire cross-section with respect to the armature resistance?

a) wire section is directly proportional to the armature resistance

b) wire section is indirectly proportional to the armature resistance

c) wire section is directly proportional to the square of the armature resistance

d) wire section is indirectly proportional to the square of the armature resistance

Answer: a

Explanation: The 10th design step of the PMDC motor is the calculation of the wire cross section. The wire cross section is directly proportional to the armature resistance.

13. The radial thickness of the joke directly proportional to the flux.

a) true

b) false

Answer: a

Explanation: The last design step in the PMDC motor is the calculation of the radial thickness of the joke. The flux value is directly proportional to the radial thickness of the joke.

14. The radial thickness of the joke is directly proportional to the length of the stator slots.

a) true

b) false

Answer: b

Explanation: The last design step in the PMDC motor is the calculation of the radial thickness of the joke. The radial thickness of the joke is indirectly proportional to the length of the stator slots.

15. What is the formula for the length of the stator slots?

a) length of the stator slots = 2 * perimeter of one magnet

b) length of the stator slots = 1/2 * perimeter of one magnet

c) length of the stator slots = 1/3 * perimeter of one magnet

d) length of the stator slots = 3 * perimeter of one magnet

Answer: b

Explanation: The length of the stator slots is required in the calculation of the radial thickness of the joke. The length of the stator slots is equal to half the perimeter of one magnet.

This set of Design of Electrical Machines Assessment Questions and Answers focuses on “Design Steps and Considerations”.

1. What is the relation between number of poles and total volume of magnet?

a) number of poles is directly proportional to the total volume of the magnet

b) number of poles is indirectly proportional to the total volume of the magnet

c) number of poles is directly proportional to the square of the total volume of the magnet

d) number of poles is indirectly proportional to the square of the total volume of the magnet

Answer: b

Explanation: The first design consideration in the PMDC motor is the number of poles. The volume of the magnet is indirectly proportional to the number of poles.

2. What is the relation between number of poles and flux reversal in the armature?

a) number of poles is directly proportional to the flux reversal in the armature

b) number of poles is indirectly proportional to the flux reversal in the armature

c) number of poles is directly proportional to the square of the flux reversal in the armature

d) number of poles is indirectly proportional to the square of the flux reversal in the armature

Answer: a

Explanation: The first design consideration in the PMDC motor is the number of poles. The flux reversal in the armature is directly proportional to the number of poles.

3. How many number of poles should be used for large motors of relatively low speed?

a) should be equal to 2

b) should be lesser than 2

c) should be greater than 2

d) should be more than 4

Answer: c

Explanation: The number of poles should be greater than 2 for large motors with relatively low speed. The number of poles is equal to 2 for small motors.

4. In the PMDC motors the brush shift should be approached with considerable caution.

a) true

b) false

Answer: a

Explanation: The brush shift in the PMDC motor should be approached with considerable caution. This is because as flux shift in the ceramic magnet will be found to be almost negligible.

5. What is the relation of the brush shift with the demagnetization effect?

a) brush shift is directly proportional to the demagnetization effect

b) brush shift is indirectly proportional to the demagnetization effect

c) brush shift is directly proportional to the square of the demagnetization effect

d) brush shift is indirectly proportional to the square of the demagnetization effect

Answer: a

Explanation: The second design consideration is the brushes in the PMDC motor. The brush shift increases the demagnetizing effect.

6. How many primary reasons are present for the thermal failure?

a) 3

b) 4

c) 2

d) 5

Answer: c

Explanation: There are 2 primary reasons for the thermal failure. The first one is an increase of the resistance of motor winding and second one is the inability of the motor to dissipate the heat generated.

7. What does the increase of the resistance of the motor winding cause?

a) high starting current

b) low motor torque

c) low starting current

d) high motor torque

Answer: b

Explanation: The increase of the resistance of the motor winding is one of the cause of thermal failure. This produces lower motor torque in the machine.

8. What does the inability of the motor to dissipate the heat cause?

a) causes high starting current

b) insulation failure

c) causes low starting current

d) causes high starting current

Answer: b

Explanation: The inability of the motor to dissipate the heat is one of the causes of thermal failure. The inability of the motor to dissipate heat causes insulation failure.

9. What is the solution to prevent the increase of resistance of motor windings?

a) higher ventilation arrangement

b) reduction in the number of poles

c) increase the coil windings

d) insulate the windings

Answer: a

Explanation: The increase of resistance of the motor winding is one of the cause of thermal failure. They can be reduced by providing a higher ventilation arrangement in the machine.

10. How many types of gears are made use of in the PMDC motors?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: Generally 3 types of gears are made use of in the PMDC motors. They are spur gears, helical gears, and worm gears.

11. What type of gears are used in the small loads and low inertia motors?

a) spur gears

b) helical gears

c) worm gears

d) worm gears and helical gears

Answer: a

Explanation: Three types of gears are made use of in the PMDC motors. The spur gears are made use of in the low inertia and small load motors.

12. What type of gear is made use of in the high inertia motors?

a) spur gears

b) helical gears

c) worm gears

d) spur and helical gears

Answer: c

Explanation: There are three types of gears made use of in the PMDC motors. The worm gears are used in the high inertia load.

13. What type of gear is made use for the silent operation?

a) spur gears

b) helical gears

c) worm gears

d) spur gears and worm gears

Answer: b

Explanation: There are three types of gears made use of in the gearing system of the PMDC motors. The helical gears is made use for the silent operation.

14. How many types of bearings are made use of in the PMDC motors?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are two types of bearings used in the PMDC motors. They are bell bearing and the journal bearing.

15. How many principle types of lubricants are available in the PMDC motor?

a) 2

b) 3

c) 5

d) 7

Answer: b

Explanation: There are 3 principle types of lubricants in use. They are oil, dry film lubricants and grease.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Types of Electromagnets”.

1. How many types of electromagnets are present?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are 2 types of electromagnets present. They are i) Tractive type and ii) Portative type.

2. What is the other name of the Tractive electromagnet and what is the means of movement of the armature?

a) solenoidal, electrical movement

b) solenoidal, mechanical movement

c) traction, electrical movement

d) traction, mechanical movement

Answer: b

Explanation: The other name for the tractive electromagnet is solenoids. They are designed to produce mechanical.

3. What is the supply given to the tractive electromagnets?

a) only dc supply

b) only ac supply

c) ac and dc supply

d) ac or dc supply

Answer: d

Explanation: The other name for the tractive electromagnet is solenoids. The tractive electromagnets are operated either from ac or dc supply.

4. Among the following what are the applications of the tractive electromagents?

a) track switches

b) electric bells

c) buzzers

d) track switches, bells, buzzers

Answer: d

Explanation: The tractive electromagnets have a large number of applications. They are made use of in the track switches, electric bells and buzzers.

5. How does the portative electromagnet work as?

a) holding magnet

b) connecting magnet

c) repulsion magnets

d) attraction magnets

Answer: a

Explanation: The second type of electromagnet is the portative electromagnet. They usually function as a holding magnet.

6. What type of supply is being provided to the portative electromagnet?

a) only ac supply

b) only dc supply

c) ac and dc supply

d) ac or dc supply

Answer: b

Explanation: Portative electromagnets are one type of electromagnet, which function as a holding magnet. They operated usually from dc supply only.

7. How many most commonly used electromagnets are present?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 most commonly used electromagnets are present. They are I) Flat-faced armature type, II) Horse shoe type, III) Flat-faced plunger type.

8. What type of magnet is made use of to produce large force through a relatively small distance?

a) flat-faced armature type

b) horse shoe type

c) flat-faced plunger type

d) flat-faced plunger type and horse shoe type

Answer: a

Explanation: There are 3 types of most commonly used electromagnets. The flat faced armature type electromagnet is made use of to produce large force through a relatively small distance.

9. What material is the flat faced armature type made of?

a) hard steel

b) cast steel

c) cast iron

d) soft steel

Answer: b

Explanation: The flat faced armature type is made up of cast steel. It is used for lifting scrap iron, sheet iron and iron castings.

10. How are the air gaps arranged in the flat faced armature type?

a) magnetic in series and mechanical in parallel

b) magnetic in series and parallel

c) mechanical in series and parallel

d) magnetic in parallel and mechanical in series

Answer: a

Explanation: The flat faced armature types have 2 air gaps within them. They are magnetic in series and are mechanical in parallel and hence produce a holding surface of large effective area.

11. Which among the following are the application of portative electromagnets?

a) lifting magnets

b) magnetic clutches

c) magnetic chucks

d) lifting magnets, magnetic clutches, magnetic chucks

Answer: d

Explanation: The portative electromagnets generally function as holding magnets. The lifting magnets, magnetic chucks, magnetic clutches are all applications of portative electromagnets.

12. What is the relation between force and the air gap length in the flat-faced armature type?

a) force is directly proportional to the air gap length

b) force is indirectly proportional to the air gap length

c) force is directly proportional to the square of the air gap length

d) force is indirectly proportional to the square of the air gap length

Answer: c

Explanation: The force is directly proportional to the square of the air gap length. This condition exists under ideal conditions wherein the effects of saturation and magnetic leakage are negligible.

13. Horse shoe is usually employed for the small magnets.

a) true

b) false

Answer: a

Explanation: The horse shoe is usually employed for the small magnets. It is because of the mechanical adaptability and the ease with which it can be constructed.

14. How many air gaps are present in the flat-faced plunger type?

a) 1

b) 2

c) 3

d) 4

Answer: a

Explanation: The magnetic circuit in the flat faced plunger type is usually short and heavy. It has only one air gap present.

15. What is the relation of the force and the air gap length in the flat faced plunger type?

a) force is directly proportional to the air gap length

b) force is indirectly proportional to the air gap length

c) force is directly proportional to the square of the air gap length

d) force is indirectly proportional to the square of the air gap length

Answer: c

Explanation: Force is directly proportional to the square of the air gap length. The characteristics are similar to that of the flat-faced armature type.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Construction of Electromagnets”.

1. What type of core does the electromagnetic consist of?

a) paramagnetic

b) diamagnetic

c) ferromagnetic

d) paramagnetic and diamagnetic

Answer: c

Explanation: Electromagnets consists of a ferromagnetic core. It carries the flux and a winding which produces a flux when excited by an external source.

2. What material is used for the construction of core of electromagnets?

a) soft magnetic materials

b) hard magnetic materials

c) non-magnetic materials

d) anti-ferromagnetic materials

Answer: a

Explanation: The electromagnets consist of ferromagnetic core. The soft magnetic materials are made use of in the construction of core of electromagnets.

3. What are the ferromagnetic elements used in the electromagnets?

a) iron

b) nickel

c) cobalt

d) iron, nickel, cobalt

Answer: d

Explanation: The electromagnets consist of ferromagnetic core. The ferromagnetic materials include iron. nickel and cobalt.

4. What are the non-magnetic materials being used in the electromagnets?

a) silicon

b) molybdenum

c) silicon, chromium, molybdenum

d) chromium

Answer: c

Explanation: The electromagnets consists of the ferromagnetic core. Some times the non-ferro-magnetic materials are being made use of like silicon, chromium and molybdenum.

5. Coils are being made use of in electromagnets as an exciting source for production of magnetic field.

a) true

b) false

Answer: a

Explanation: The coils are being made use of in electromagnets. They are used as an exciting source for the production of magnetic field.

6. What is the insulation material being used in the electromagnets?

a) paper

b) wood

c) brass

d) copper

Answer: a

Explanation: Insulation material is being used in between the coils of the electromagnet to provide insulation. The material used in the electromagnet is paper.

7. What is the conductor material being used in the electromagnet?

a) copper

b) zinc

c) bronze

d) aluminum

Answer: a

Explanation: Conductor materials are being used in the electromagnet to conduct the current in the electromagnet. Copper is the material used in the electromagnet.

8. What type of conductors are being used in the coils made of heavy wire?

a) circular

b) rounded

c) conical

d) rectangular

Answer: d

Explanation: The cross-section of coils is generally rectangular and the cross-section of conductors is usually rounded. The coils made of heavy wires, rectangular conductors with rounded corners are used.

9. The coil insulation used in the electromagnets is of sheet form.

a) true

b) false

Answer: a

Explanation: Insulation material used in the machine is usually paper. The insulation is arranged in the form of sheets.

10. What materials are used along with paper for insulation?

a) varnish

b) glass

c) synthetic resin

d) varnish, glass, synthetic resin

Answer: d

Explanation: The insulation used in the electromagnets is paper. It is being insulated along with the varnish, glass and synthetic resin and treated into the form of sheets to form proper insulation.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Magnet Coils”.

1. What is the formula for the mean diameter of the magnet coils?

a) mean diameter = inside diameter of coil + outer diameter of coil / 2

b) mean diameter = inside diameter of coil – outer diameter of coil / 2

c) mean diameter = inside diameter of coil * outer diameter of coil / 2

d) mean diameter = inside diameter of coil / outer diameter of coil / 2

Answer: a

Explanation: First the inner diameter of coil is calculated. Secondly, the outer diameter of coil is calculated. On substitution, we finally get the mean diameter.

2. What is the formula for the outside diameter of the magnet coils?

a) outside diameter = mean diameter + 2*depth of winding

b) outside diameter = mean diameter + depth of winding

c) outside diameter = mean diameter – 2*depth of winding

d) outside diameter = mean diameter – depth of winding

Answer: b

Explanation: The mean diameter is found out from its respective formula. Next, the depth of the winding is calculated and on substitution gives the outside diameter.

3. What is the formula for depth of winding of the magnet coils?

a) depth of winding = mean diameter of coil – inner diameter

b) depth of winding = mean diameter of coil + inner diameter

c) depth of winding = mean diameter of coil – 2* inner diameter

d) depth of winding = mean diameter of coil + 2*inner diameter

Answer: a

Explanation: The mean diameter of coil is calculated first from its respective formula. The inner diameter is next calculated and on substitution gives the depth of winding.

4. What is the formula of the cross winding area of the magnet coils?

a) cross winding area = axial length of coil + depth of winding

b) cross winding area = axial length of coil – depth of winding

c) cross winding area = axial length of coil * depth of winding

d) cross winding area = axial length of coil / depth of winding

Answer: c

Explanation: First the axial length of coil is calculated. Next, the depth of winding is calculated and on substitution gives the cross winding area of the magnet coils.

5. What is the formula for the length of mean turn of magnet coils?

a) length of mean turns = 3.14 * 

b) length of mean turns = 3.14 / 

c) length of mean turns = 3.14 * 

d) length of mean turns = 3.14 + 

Answer: a

Explanation: The inside diameter of the coil is first calculated. Next, the depth of windings is next calculated and on substitution gives the length of mean turns.

6. What is the formula for the total heat dissipating surface of the magnet coils?

a) total heat dissipating surface = length of mean turn * depth of winding * axial length of coil

b) total heat dissipating surface = length of mean turn * depth of winding + axial length of coil

c) total heat dissipating surface = 2 * length of mean turn * 

d) total heat dissipating surface = 2 * length of mean turn * depth of winding * axial length of coil

Answer: c

Explanation: The length of mean turn is calculated first. Next, the depth of winding and axial length of coil is next calculated and on substitution gives the total heat dissipating surface.

7. What is the formula for the outer cylindrical heat dissipating surface of the magnet coils?

a) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil + axial length of coil

b) outer cylindrical heat dissipating surface = 3.14 + outer diameter of coil + axial length of coil

c) outer cylindrical heat dissipating surface = 3.14 / outer diameter of coil + axial length of coil

d) outer cylindrical heat dissipating surface = 3.14 * outer diameter of coil * axial length of coil

Answer: d

Explanation: The outer diameter of the coil is first calculated. Next, the axial length of the coil is next calculated and on substitution gives the outer cylindrical heat dissipating surface of the magnet coils.

8. What is the formula of the inner cylindrical heat dissipating surface?

a) inner cylindrical heat dissipating surface = length of mean turn * axial length of coil

b) inner cylindrical heat dissipating surface = 2 *length of mean turn * axial length of coil

c) inner cylindrical heat dissipating surface = length of mean turn / axial length of coil

d) inner cylindrical heat dissipating surface =1 / length of mean turn * axial length of coil

Answer: b

Explanation: The length of mean turn is first calculated. Next, the axial length of coil is calculated and on substitution gives the inner cylindrical heat dissipating surface.

9. What is the ambient temperature of the magnet coils?

a) 10°C

b) 15°C

c) 20°C

d) 25°C

Answer: c

Explanation: The temperature is one of the factors which is used in the efficient operation of the magnet coils. The ambient temperature of the magnet coils is 20°C.

10. What is the formula for the area of the conductors of the magnet coils?

a) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn * terminal voltage

b) area of the conductors = mmf per coil / resistivity of conductor * length of mean turn * terminal voltage

c) area of the conductors = mmf per coil * resistivity of conductor * length of mean turn / terminal voltage

d) area of the conductors = mmf per coil * resistivity of conductor / length of mean turn * terminal voltage

Answer: c

Explanation: For calculating the area of the conductors, first the mmf per coil is calculated along with the resistivity of conductors. The length of mean turn and terminal voltage is calculated and on substitution gives the area of the conductors.

11. What is the value of the resistivity temperature coefficient of copper?

a) 0.017 ohm per m per mm 2

b) 0.0173 ohm per m per mm 2

c) 0.01734 ohm per m per mm 2

d) 0.0175 ohm per m per mm 2

Answer: c

Explanation: The resistivity temperature coefficient of copper is first calculated at a temperature of 20°C. The resistivity temperature coefficient of copper is 0.01734 ohm per m per mm 2 .

12. What is the value of the resistance temperature coefficient of copper?

a) 0.00393 per °C

b) 0.0040 per °C

c) 0.00383 per °C

d) 0.00373 per °C

Answer: a

Explanation: The resistance temperature coefficient of copper is calculated at a temperature of 20°C. The resistance temperature coefficient of copper is 0.00393 per °C.

13. What is the formula for total number of turns in the magnet coils?

a) total number of turns = mmf per coil * current

b) total number of turns = mmf per coil / current

c) total number of turns = mmf per coil – current

d) total number of turns = mmf per coil + current

Answer: b

Explanation: The mmf per coil is first calculated. Next, the current flowing through the coils is measured and on substitution gives the total number of turns.

14. What is the formula for the total winding area?

a) total winding area = number of turns * area of each conductor * space factor

b) total winding area = number of turns / area of each conductor * space factor

c) total winding area = number of turns * area of each conductor / space factor

d) total winding area = 1/number of turns * area of each conductor * space factor

Answer: c

Explanation: First the number of turns is calculated along with the area of each conductor. Next, the space factor is calculated and on substitution gives the total winding area.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Errors”.

1. What is the definition of current transformer?

a) it is used for measuring high voltage

b) it is used for measuring low voltage

c) it is used for measuring high current

d) it is used for measuring low current

Answer: c

Explanation: For the measuring of high currents, the current transformer is made use of. The measured current is scaled down to lower values equivalently.

2. How many classifications are present for the current transformers?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: The current transformers is divided into 2 types. They are I) measuring current transformer and II) protective current transformer.

3. What is the definition of the ideal current transformer?

a) the primary and secondary windings are in exact ratio and same phase relationship

b) the primary winding and secondary winding ratio is greater than 1 and same phase relationship

c) the primary and secondary winding ratio is lesser than 1 are in exact ratio and different phase relationship

d) the primary and secondary windings ratio is greater than 1 and different phase relationship

Answer: a

Explanation: The primary and secondary winding ratio are same. The phase relationship of the windings are also same.

4. How many types of errors are present in the current transformers?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: There are 2 types of errors present in the current transformers. They are ratio error and phase angle error.

5. What is the formula of the angle between secondary induced voltage and secondary current?

a) phase angle = tan -1 *[ / ]

b) phase angle = tan -1 *[ / ]

c) phase angle = tan -1 *[ / ]

d) phase angle = tan -1 *[ / ]

Answer: d

Explanation: The reactance of the secondary windings and the external burden is first calculated. Next, the resistance of the secondary windings and external burden is calculated and on substitution gives the value of the phase angle.

6. What is the formula of the phase angle of the secondary load circuit?

a) phase angle of secondary load circuit = tan -1 * 

b) phase angle of secondary load circuit = tan -1 * 

c) phase angle of secondary load circuit = tan -1 * 

d) phase angle of secondary load circuit = tan -1 * 

Answer: a

Explanation: The reactance and resistance of the external burden is first calculated. Next, the value is taken tan inverse to obtain the phase angle of secondary load circuit.

7. What is the formula of the ratio error in the current transformers?

a) ratio error = turns ratio – regulation / regulation

b) ratio error = turns ratio + regulation / regulation

c) ratio error = turns ratio * regulation / regulation

d) ratio error = 1 / turns ratio * regulation

Answer: a

Explanation: First the turns ratio is calculated. Next the regulation of the current transformer is obtained and on substitution gives the ratio error.

8. The ratio of active conductor section to total conductor section is called space factor.

a) true

b) false

Answer: a

Explanation: The space factor is a term that is used in the design of the magnet coils. It is the ratio of the active conductor section to the total conductor section.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Current Transformers Construction”.

1. How many types are the current transformers classified into?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: The current transformers are classified into 2 types. They are wound type and bar type.

2. What is the wound type current transformer?

a) primary winding having one full turn wound on core

b) primary winding having more than one full turn wound on core

c) secondary winding having one full turn wound on core

d) secondary winding having more than one full turn wound on core

Answer: b

Explanation: The wound type current transformer is one of the classifications of the current transformers. The primary winding has more than one full turn wound on core.

3. What is the bar type current transformer?

a) primary winding consists of a rod of suitable size and material

b) primary winding consists of a bar of suitable size and material

c) secondary winding consists of a rod of suitable size and material

d) secondary winding consists of a bar of suitable size and material

Answer: b

Explanation: The bar type current transformer is one of the classifications of the current transformers. In bar type winding, primary winding consists of a bar of suitable size and material.

4. How many commonly used shapes of current transformer are present?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: The current transformers consist of 3 commonly used shapes. They are stadium, circular, rectangular orifices.

5. What material is made use of for the lamination in the current transformer?

a) cold rolled steels

b) hot rolled steels

c) copper

d) hot iron

Answer: b

Explanation: The current transformer consists of stacks of laminations. The lamination used in the current transformer is hot rolled steel.

6. What is the insulation material used in the current transformer?

a) elephantide

b) presspahn

c) elephantide and presspahn

d) elephantide or presspahn

Answer: d

Explanation: The insulation in current transformer is by means of end collars and circumferential wraps. The insulation material used in elephantide or presspahn.

7. What is the additional usage of the presspahn material used as insulation material?

a) lamination

b) to reduce the losses

c) to protect secondary winding conductor from mechanical damage

d) to protect secondary winding conductor from electrical damage

Answer: c

Explanation: The presspahn is used as insulating material in the current transformer. In addition to that the presspahn is also used to protect the secondary winding conductor from mechanical damage.

8. What is the other name of the ring type current transformer?

a) brush transformer

b) cloud transformer

c) circular transformer

d) bushing transformer

Answer: d

Explanation: The ring type current transformer is one type of current transformer. It is also known as the bushing transformer.

9. How many faces are present in the split core current transformer?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: The split core transformer consists of a split core. The split half consists of 2 finely grounded or lapped gap faces.

10. The current transformers are assembled on to the secondary conductors “on site” for either permanent or temporary duty.

a) true

b) false

Answer: b

Explanation: The current transformers are assembled on to the primary conductors. They are assembled “on site” for the permanent or temporary duty.

11. What is the insulation material on the primary conductor?

a) bakelized paper tube

b) resin

c) bakelized paper tube and resin

d) bakelized paper tube or resin

Answer: d

Explanation: The insulation material on the primary conductors is generally made up of the bakelized paper tube. It can also be made use of resin directly moulded on the bar.

12. How is the reluctance of the interleaved corner related with the magnetizing current?

a) reluctance of the interleaved corner is directly proportional to the magnetizing current

b) reluctance of the interleaved corner is indirectly proportional to the magnetizing current

c) reluctance of the interleaved corner is directly proportional to the square of the magnetizing current

d) reluctance of the interleaved corner is indirectly proportional to the square of the magnetizing current

Answer: a

Explanation: The reluctance of the interleaved corner is directly proportional to the magnetizing current. As the reluctance is being reduced it in turn reduces the magnetizing current.

13. To reduce the peak voltage between layers, the secondary winding is being sectionalized.

a) true

b) false

Answer: a

Explanation: The large number of secondary turns requiring more than 1 winding layer, the secondary winding is sectionalized. This is because to reduce the peak voltage between layers.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design Principles”.

1. How many design principles are present in the current transformers?

a) 2

b) 3

c) 4

d) 5

Answer: d

Explanation: There are 5 design principles present in the current transformers. They are core design, secondary current rating, primary current rating, windings and behavior of the transformer under short circuit current.

2. What should be done in order to reduce the errors in the core?

a) armature mmf is to kept low

b) field mmf to be kept high

c) the exciting mmf is to be kept low

d) the field mmf is to be kept high

Answer: c

Explanation: The errors in the core are reduced by keeping the exciting mmf low. This can take place with the core having a low reluctance and low iron loss.

3. How many classifications are the magnetic alloys used in the current transformers classified into?

a) 3

b) 2

c) 4

d) 5

Answer: a

Explanation: The magnetic alloys used in the current transformers are divided into 3 types. They are hot rolled silicon steel, cold rolled grain oriented silicon steel and nickel iron alloys.

4. What is the material used in the transformer when the transformer errors should be small?

a) mumetal cores

b) steel cores

c) permender cores

d) presshamn cores

Answer: a

Explanation: The mumetal cores are commonly used when it is essential that transformer errors shall be small. Mumetal has the properties of high permeability, low loss and small retentivity.

5. What is the relation of the secondary winding leakage reactance and secondary circuit impedance?

a) secondary winding leakage reactance is directly proportional to the secondary circuit impedance

b) secondary winding leakage reactance is indirectly proportional to the secondary circuit impedance

c) secondary winding leakage reactance is directly proportional to the square of the secondary circuit impedance

d) secondary winding leakage reactance is indirectly proportional to the square of the secondary circuit impedance

Answer: a

Explanation: The secondary winding leakage reactance is directly proportional to the secondary circuit impedance. In secondary winding the leakage reactance is reduced and in turn the secondary circuit impedance is reduced.

6. The ring shaped cores are made use of in the reduction of the secondary winding leakage reactance and secondary impedance.

a) true

b) false

Answer: a

Explanation: The secondary winding leakage reactance is directly proportional to the secondary impedance. The ring shaped cores around which the toroidal secondary windings of one or more layers are uniformly distributed.

7. What type of core is employed when the performance standard required is not so high?

a) rectangular strips

b) c-shaped sections

c) rectangular strips or c-shaped sections

d) rectangular strips and c-shaped sections

Answer: c

Explanation: Ring core type is used only for the high performance operation. The rectangular strips or c-shaped sections are used when the standard of performance required is not so high.

8. What should the magnetic path be in order to reduce the core reluctance?

a) length of the magnetic path in core should be low

b) length of the magnetic path in core should be medium

c) length of the magnetic path in core should be high

d) length of the magnetic path in core should be very high

Answer: a

Explanation: The length of the magnetic path in core should be as small as possible. This reduces the core reluctance of the current transformer.

9. What is the value of the rated secondary current?

a) 1 A

b) 2 A

c) 3 A

d) 5 A

Answer: d

Explanation: The rating of the secondary current is the maximum current that can be passed through the secondary windings. It is 5 A for the current transformer.

10. What are the disadvantages of the low rated secondary current transformer?

a) high cost

b) high voltages

c) high voltages or high cost

d) high voltages and high cost

Answer: d

Explanation: When there is a low secondary current rating in the current transformers they produces high voltages if the secondary is left open. It is also costly to produce the windings because of the extra time involved in the making.

11. What is the ideal condition with respect to the primary current rating?

a) ratio of secondary mmf to primary mmf should be high

b) ratio of secondary mmf to primary mmf should be low

c) ratio of excitation mmf to primary mmf should be high

d) ratio of excitation mmf to primary mmf should be low

Answer: d

Explanation: The primary current rating depends on exciting mmf and primary mmf. The ratio of the exciting mmf to the primary mmf should be low.

12. What is the rating of the primary current in the current transformer?

a) 200 A

b) 300 A

c) 400 A

d) 500 A

Answer: d

Explanation: The rating of the primary current is minimum 500 A. If the rating is less than 500 A, then multiturn primary windings and rating is above than 500 A, then single turn winding is enough.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Turns Compensation”.

1. What is the use of the turns compensation in current transformer?

a) to increases the ratio error

b) to reduce the ratio error

c) to increase the phase angle error

d) to reduce the phase angle error

Answer: b

Explanation: There are 2 types of errors in the current transformer. The turns compensation is used to reduce the ratio error.

2. What is the formula of the actual ratio?

a) actual ratio = turns ratio + load current * secondary current

b) actual ratio = turns ratio * load current * secondary current

c) actual ratio = turns ratio + load current / secondary current

d) actual ratio = turns ratio / load current * secondary current

Answer: c

Explanation: The turns ratio is first calculated. Next the load current and secondary current is calculated and on substitution gives the actual ratio.

3. What happens if the number of secondary turns is reduced?

a) the primary turns is reduced

b) the output is reduced

c) the efficiency is reduced

d) the transformation ratio is reduced

Answer: d

Explanation: The reduction of the number of secondary turns reduces the transformation ratio. If the number of secondary turns reduces by 1 percent the actual transformation ratio reduces by equal percentage.

4. What is the best number of secondary turns of the current transformer?

a) 1

b) 2

c) 1 or 2 less than the number such that the turns ratio is equal to the nominal current ratio

d) 1 or 2 more than the number such that the turns ratio is equal to the nominal current ratio

Answer: c

Explanation: The best number of secondary turns of the current transformer is 1 or 2 less than the number such that the turns ratio is equal to the nominal current ratio. For example in a 1000/5 current transformer, the secondary turns number would be 198 or 199 rather than 200.

5. The phase angle error is significantly affected by the small change in secondary turns.

a) true

b) false

Answer: b

Explanation: There are 2 types of errors which is the ratio error and phase angle error. The phase angle error is not significantly effected by a small change in secondary turns.

6. What is the dimension of the round copper wire made use of in the windings of current transformer?

a) 3 cm 2

b) 3 mm 2

c) 3 m 2

d) 3 cm

Answer: b

Explanation: Copper strip is used for primary windings. The dimension of the round copper wire made use of in the windings of current transformer is 3 mm 2 .

7. What is the range of current density in the windings of the current transformer?

a) 1-3 A per mm 2

b) 2-3 A per mm 2

c) 1-2 A per mm 2

d) 0.5-2 A per mm 2

Answer: c

Explanation: The minimum value of the current density in the windings I A per mm 2 . The maximum value of the current density is 2 A per mm 2 .

8. How many factors are present in the behavior of transformer under short circuit conditions?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 factors present in the behavior of the transformer under short circuit conditions. They are temperature rise, current density, mechanical forces.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Shaft”.

1. What is the consideration for the determination of the diameter of shaft?

a) stiffness

b) voltage

c) current

d) rigidity

Answer: c

Explanation: The main aspect for the design of the diameter of the shaft is the stiffness. The diameter of the shaft depends on the stiffness of the machine.

2. What is the meaning of stiffness?

a) ability to transmit the power

b) ability to withstand the weight of the rotor

c) ability to withstand unbalanced magnetic pull

d) ability to withstand the weight of rotor and unbalanced magnetic pull

Answer: d

Explanation: The diameter of shaft for an electrical machine is determined by considerations of stiffness. The stiffness is the ability to withstand the weight of rotor and unbalanced magnetic pull.

3. What should be the first property of the shaft design?

a) the shaft design should be such that the shaft must have enough corrosion resistance

b) the shaft design should be such that the shaft must have enough mechanical strength

c) the shaft design should be such that the shaft has enough tensile strength

d) the shaft design should be able to withstand the voltage fluctuations

Answer: b

Explanation: The shaft design should be such that the shaft must have enough mechanical strength. The strength should be such that it should withstand all loads without causing much residual strain.

4. What is the second property of the shaft design?

a) the shaft design should be such that it has high rigidity

b) the shaft design should be such that it should have high tensile strength

c) the shaft design should be such that it should have high corrosion resistance

d) the shaft design should be such that it should withstand voltage fluctuations

Answer: a

Explanation: The second property of the shaft design such that it should have high rigidity. The rigidity should be such that the deflection of shaft under operation of machine does not reach such a dangerous value as to cause the rotor to touch the stator.

5. The critical speeds of rotation should be different from running speed of machine.

a) true

b) false

Answer: a

Explanation: The critical speed relation is the third property of the shaft design. The critical speeds of rotation should be different from the running speed of the machine.

6. What is the formula of the diameter of the shaft?

a) diameter of the shaft = 5.5 +  1/3 mm

b) diameter of the shaft = 5.5 –  1/3 mm

c) diameter of the shaft = 5.5 *  1/3 mm

d) diameter of the shaft = 5.5 /  1/3 mm

Answer: c

Explanation: The output is first calculated from the operation of the machine. Next, the tachogenerator is used to calculate the speed of the machine and on the substitution of the values gives the diameter of the shaft.

7. What is the relation of the diameter of the shaft in the bearings to the diameter under the armature?

a) diameter of the shaft is very much greater than the diameter under the armature

b) diameter of the shaft is greater than the diameter under the armature

c) diameter of the shaft is equal to the diameter under the armature

d) diameter of the shaft is lesser than the diameter under the armature

Answer: d

Explanation: There are certain rules in the design of the shaft. The diameter of the shaft in the bearings is less than the diameter under the armature.

8. What happens when the diameter under armature is 150 mm or above?

a) diameter of the shaft in bearing is 100 mm smaller than the maximum diameter

b) diameter of the shaft in bearing is 90 mm smaller than the maximum diameter

c) diameter of the shaft in bearing is 70 mm smaller than the maximum diameter

d) diameter of the shaft in bearing is 50 mm smaller than the maximum diameter

Answer: d

Explanation: The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter of the shaft in bearing is 50 mm smaller than the maximum diameter.

9. What happens in the case of the small shafts?

a) the diameter in the bearings should be about 1/3 of the maximum diameter

b) the diameter in the bearing should be about 2/3 of the maximum diameter

c) the diameter in the bearing should be about 2/5 of the maximum diameter

d) the diameter in the bearing should be about 1/5 of the maximum diameter

Answer: b

Explanation: The diameter of the shaft in the bearings is less than the diameter under the armature. The diameter in the bearing should be 2/3 of the maximum diameter.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Bearings”.

1. What bearing is made use of in the horizontal shaft machines?

a) plain bearing

b) thrust bearing

c) push bearing

d) throw bearing

Answer: a

Explanation: The plain bearing is made use of in the horizontal shaft machine. The thrust bearing are used for vertical shaft machine.

2. What is the plain bearing used?

a) sleeve bearings

b) anti friction bearing

c) sleeve or anti friction bearing

d) sleeve and anti friction bearing

Answer: c

Explanation: The plain bearing can be the sleeve bearings made use of. It can also be friction bearings used which is either ball or roller bearings.

3. In horizontal shaft machines, the forces acting in which direction is prominent?

a) circular

b) vertical

c) horizontal

d) radial

Answer: d

Explanation: Plain bearings are made use of for the horizontal shaft machines. The forces acting in the radial direction are most prominent.

4. In vertical shaft machines, which load is taken up by the thrust bearings?

a) perpendicular load acting upwards

b) perpendicular load acting downwards

c) axial load acting upwards

d) axial load acting downwards

Answer: d

Explanation: The thrust bearings are made use of for the vertical shaft machines. In vertical shaft machines, the axial load acting downwards is taken up by the thrust bearings.

5. How are the radial loads caused?

a) dynamic unbalance of rotor

b) unbalanced magnetic pull of rotor

c) dynamic unbalance of rotor or unbalanced magnetic pull of rotor

d) dynamic unbalance of rotor and unbalanced magnetic pull of rotor

Answer: c

Explanation: The radial loads are caused by the dynamic unbalance of rotor. It can also be caused by the unbalanced magnetic pull of rotor.

6. What is the additional setup provided to the simple thrust bearing when it is not able to take up radial loads?

a) ball bearing

b) gear bearing

c) steel bearing

d) guide bearing

Answer: d

Explanation: Sometimes the simple thrust bearing wont be able to pull the radial loads. Therefore an additional bearing called guide bearing is provided.

7. How many guide bearings are used along with the simple thrust bearings to pick up radial loads?

a) 2

b) 3

c) 2 or 3

d) 4

Answer: c

Explanation: The simple thrust bearing cannot pull the radial load and hence guide bearings are used in addition to it. Usually 2-3 guide bearings are used in vertical shaft machines depending upon the load.

8. Where are the bearings for the horizontal shaft placed?

a) outside the machine

b) in the end shields of the machine

c) outside the machine or in the end shields of the machine

d) outside the machine and in the end shields of the machine

Answer: c

Explanation: The bearings for the horizontal shaft machines are being placed outside the machine. It can also be placed along the end shields of the machine.

9. What is the other name of the bearings outside the machine?

a) ball bearing

b) pedestal bearing

c) guide bearing

d) gear bearing

Answer: b

Explanation: The bearings for the horizontal shaft machines are placed outside the machine or in the end shields of the machine. The other name for the bearings outside the machine is pedestal bearing.

10. The phosphor bronze sleeve bearings are used for the small electrical machines.

a) true

b) false

Answer: a

Explanation: The phosphor bronze sleeve bearings are used for the small electrical machines. The diameter of the machines are between 50-60 mm.

11. Anti-friction bearings are lubricated by charcoal.

a) true

b) false

Answer: b

Explanation: Anti-friction bearings are one type of bearing used to take radial loads. They are always lubricated by grease.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Frames of D.C. & A.C. Machines”.

1. What is the work of the frame of dc machines?

a) to reduce the voltage

b) to reduce the flux

c) to carry the flux

d) to carry the current

Answer: c

Explanation: The main function of the frame of dc machines is to carry the flux. Thus the frame must be large enough to carry flux.

2. Why is the length of the yoke made larger?

a) to protect the armature windings

b) to cover the armature windings

c) to cover the field windings

d) to cover and protect the field windings

Answer: d

Explanation: The length of the yoke is usually made larger than the pole cores. It is because to cover and protect the field windings.

3. What is the formula for the depth of the yoke?

a) depth of yoke = thickness/2

b) depth of yoke = thickness

c) depth of yoke = 2*thickness

d) depth of yoke = 3*thickness

Answer: b

Explanation: The depth of yoke is equal to the thickness of the yoke. It is calculated to give the required cross-section for the magnetic circuit.

4. In large machines, the thickness is relatively larger to the diameter.

a) true

b) false

Answer: b

Explanation: The thickness is used in the calculation of the depth of the yoke. In large machines, the thickness is relatively smaller to the diameter.

5. What is the formula in order to check the rigidity?

a) moment of inertia ≥ (weight of magnetic frame * radius 2 * 10 -6 ) / 225

b) moment of inertia ≤ (weight of magnetic frame * radius 2 * 10 -6 ) / 225

c) moment of inertia = (weight of magnetic frame * radius 2 * 10 -6 ) / 225

d) moment of inertia < (weight of magnetic frame * radius 2 * 10 -6 ) / 225

Answer: a

Explanation: The moment of inertia, the weight of magnetic frame and the radius is calculated first. The machine is highly rigid if the moment of inertia is greater than or equal to the product of weight of magnetic frame and square of radius divided by 225.

6. What is the formula for the thickness of the ac machines?

a) thickness = 40 * inner diameter of frame/12

b) thickness = 40 + inner diameter of frame/12

c) thickness = 40 – inner diameter of frame/12

d) thickness = 40 * inner diameter of frame*12

Answer: a

Explanation: The thickness of the ac machines depend upon the inner diameter of the frame. On obtaining the inner diameter of frame and on substitution gives the thickness of ac machines.

7. What is the formula for the breadth of the ac machine?

a) breadth = 6 + 0.01 * inner diameter of frame

b) breadth = 6 – 0.01 * inner diameter of frame

c) breadth = 6 * 0.01 * inner diameter of frame

d) breadth = 6 / 0.01 * inner diameter of frame

Answer: a

Explanation: The breadth of the ac machines also depends upon the inner diameter of the frame. On substituting the values the breadth is calculated.

8. What is the formula for the checking of rigidity of induction machines?

a) moment of inertia ≥ radius / length of stator core * 90

b) moment of inertia ≥ radius * length of stator core * 90

c) moment of inertia ≥ radius * length of stator core / 90

d) moment of inertia ≤ radius / length of stator core * 90

Answer: c

Explanation: The radius and length of the stator core along with the moment of inertia is calculated. If the moment of inertia is greater than or equal to the product of length and radius divided by 90, the machine is more rigid.

9. What is the formula for the radius at the centre of gravity?

a) radius at the centre of gravity = inner diameter 1.5 /6.3

b) radius at the centre of gravity = inner diameter 2 /6.3

c) radius at the centre of gravity = outer diameter 1.5 /6.3

d) radius at the centre of gravity = outer diameter 2 /6.3

Answer: c

Explanation: The outer diameter of stator core is first calculated. On substituting the values the radius at the centre of gravity is obtained.

10. What is the formula of the centrifugal force?

a) centrifugal force = weight of revolving body * 39.43 * speed 2 * radius of circular path

b) centrifugal force = weight of revolving body / 39.43 * speed 2 * radius of circular path

c) centrifugal force = weight of revolving body * 39.43 / speed 2 * radius of circular path

d) centrifugal force = weight of revolving body * 39.43 * speed 2 / radius of circular path

Answer: a

Explanation: The weight of revolving body, speed, radius of circular path is calculated. On substitution the centrifugal force is obtained.

This set of Design of Electrical Machines Questions and Answers for Aptitude test focuses on “Bracing of Rotor Windings”.

1. What is the use of the wire bands of rotor?

a) used for bracing the rotor windings

b) used for circulating the current in the rotor windings

c) used for the encircling of the rotor windings

d) used for the protecting the rotor windings

Answer: a

Explanation: Bands used on the rotors of electrical machines are intended for bracing the rotor windings. This is done against their shift in the radial direction under action of centrifugal forces.

2. Where are the wire bands placed?

a) active portions of rotor conductors

b) inactive portions of rotor conductors

c) active or inactive portions of rotor conductors

d) active and inactive portions of the rotor conductors

Answer: d

Explanation: The wire bands are placed on the active portions of the rotor conductors. They are also placed in the inactive portions of the rotor conductors.

3. What are the factors on which the sizes of bands placed on depend?

a) length of air gap

b) method of cooling of armatures

c) length of air gap and method of cooling of armatures

d) method of cooling of armatures or length of air gap

Answer: c

Explanation: The sizes of bands placed on the active portions of the conductors depend upon the length of air gap. They also depend upon the method of cooling of armatures.

4. In what machines are the wire bands along the active length of windings placed?

a) dc or ac machines

b) dc and ac machines

c) dc machines

d) ac machines

Answer: c

Explanation: Wire bands are generally placed on both the active and inactive portions of rotor conductors. The wire bands along the active length of windings are placed along the dc machines.

5. What is the range of the width of each band that should not be exceeded?

a) 10-15 mm

b) 15-20 mm

c) 20-25 mm

d) 18-23 mm

Answer: b

Explanation: Bands placed along the active length of windings are housed in the ring slots. The width of each band should not exceed 15 to 20 mm.

6. What is the maximum value above which the total width of the bands should not exceed?

a) 25-35% of the axial length of armature core

b) 30-35% of the axial length of armature core

c) 25-30% of the axial length of armature core

d) 35-40% of the axial length of armature core

Answer: a

Explanation: The total width should not exceed 25% of the axial length of the armature core. The total width should not exceed maximum of 35% of the axial length of the armature core.

7. What is the formula for the breadth of the ring slot?

a) breadth of the ring slot = *diameter of band wire – 2*constant

b) breadth of the ring slot = *diameter of band wire + 2*constant

c) breadth of the ring slot = *diameter of band wire * 2*constant

d) breadth of the ring slot = *diameter of band wire / 2*constant

Answer: b

Explanation: The number of turns in a band is first calculated along with the diameter of band wire. The value of constant is just fixed and on substitution gives the breadth of the ring slot.

8. What is the value of the constant used in the calculation of the breadth of the ring slot for the diameter of band wire < 1.5 mm?

a) 1 mm

b) 1.5 mm

c) 2 mm

d) 3 mm

Answer: a

Explanation: The value of the constant used in the calculation of the breadth of the ring slot is 1 mm for the diameter of band wire < 1.5 mm. The value of the constant used in the calculation of the breadth of the ring slot is 1.5 mm for the diameter of band wire > 1.5 mm.

9. What is the maximum width of the bands placed on the end windings of induction machines and high speed dc machines?

a) 30 mm

b) 35 mm

c) 40 mm

d) 45 mm

Answer: c

Explanation: The maximum width are obtained for the bands placed on the end windings of induction machines and high speed dc machines. The maximum width is 40 mm.

10. What is the diameter of the wire bands made of tin, steel or bronze wire?

a) 2 mm

b) 1 mm

c) 4 mm

d) 3 mm

Answer: d

Explanation: The wire bands are generally made up of tin, steel or bronze wires. The diameter of those wire bands are 3 mm.

11. What is the function of the bands when it is placed on overhang?

a) used to reduce the centrifugal forces

b) used to increase the centrifugal forces

c) used to balance the centrifugal forces

d) used to withstand the centrifugal forces

Answer: d

Explanation: The bands when placed on overhead only are used to withstand the centrifugal forces. The centrifugal forces are due to the weight of the overhang.

12. What is the function of the bands when they are distributed along the axial length of armature?

a) used to reduce the centrifugal forces

b) used to increase the centrifugal forces

c) used to decrease the centrifugal forces

d) used to withstand the centrifugal forces

Answer: d

Explanation: The bands are distributed along the axial length of the armature and they withstand the centrifugal forces. The centrifugal forces are due to the weight of both the active and inactive parts of armature.

13. What is the formula of the mean diameter at the position of centre of gravity?

a) mean diameter at the position of centre of gravity = Inner diameter + diameter of stator wires

b) mean diameter at the position of centre of gravity = Inner diameter * diameter of stator wires

c) mean diameter at the position of centre of gravity = Inner diameter / diameter of stator wires

d) mean diameter at the position of centre of gravity = Inner diameter – diameter of stator wires

Answer: d

Explanation: The inner diameter and the diameter of the stator wires is first calculated. Then on substitution gives the mean diameter at the position of centre of gravity.

14. What is the value of permissible stress for bronze wire for the diameter of branding wire of 1 mm?

a) 350 NM per m 2

b) 250 NM per m 2

c) 300 NM per m 2

d) 450 NM per m 2

Answer: a

Explanation: The permissible stress for bronze wire for the diameter of branding wire of 1 mm is 350 NM per m 2 . The permissible stress for bronze wire for the diameter of branding wire of 1.5 mm is 300 NM per m 2 .

15. What is the value of permissible stress for steel wire for the diameter of branding wire of 0.5-1.2 mm?

a) 570 NM per m 2

b) 600 NM per m 2

c) 650 NM per m 2

d) 700 NM per m 2

Answer: b

Explanation: The value of permissible stress for steel wire for the diameter of branding wire of 0.5-1.2 mm is 600 NM per m 2 . The value of permissible stress for steel wire for the diameter of branding wire of 1.5-2 mm is 570 NM per m 2 .

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Fan”.

1. What is the formula for the fundamental relationship for the design of the ventilation system?

a) head of air inside the machine = hydrodynamic resistance * volume of air passing 2

b) head of air inside the machine = hydrodynamic resistance + volume of air passing 2

c) head of air inside the machine = hydrodynamic resistance – volume of air passing 2

d) head of air inside the machine = hydrodynamic resistance / volume of air passing 2

Answer: a

Explanation: First the hydrodynamic resistance is calculated along with the volume of air passing. On substitution in the formula gives the head of air inside the machine.

2. What is the formula for the total head produced?

a) total head produced = ∑ coefficient of hydrodynamic resistance + volume of air passing per second 2

b) total head produced = ∑ coefficient of hydrodynamic resistance – volume of air passing per second 2

c) total head produced = ∑ coefficient of hydrodynamic resistance * volume of air passing per second 2

d) total head produced = ∑ coefficient of hydrodynamic resistance / volume of air passing per second 2

Answer: c

Explanation: The various coefficient of hydrodynamic resistance is calculated along with the volume of air passing per second. On substituting the various values and on addition gives the total head produced.

3. What are the ventilating parts in the ventilating circuits?

a) sharp or projecting inlet edges

b) inlet corners

c) variations in cross-sections of air paths

d) sharp or projecting inlet edges, inlet corners, variations in cross-sections of air paths

Answer: d

Explanation: There are various ventilation parts provided in the ventilation circuits. They are sharp or projecting inlet edges, inlet corners, variations in cross-sections of air paths.

4.What is the range of the coefficients of hydrodynamic resistances for the protruding edges at inlet?

a) 40-50 * 10 -3

b) 40-60 * 10 -3

c) 30-50 * 10 -3

d) 30-40 * 10 -3

Answer: b

Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For protruding edges the range is about 40-60 * 10 -3 .

5. What is the range of the coefficients of hydrodynamic resistances for the rectangular edges at inlet?

a) 10-20 * 10 -3

b) 30 * 10 -3

c) 20-30 * 10 -3

d) 20-25 * 10 -3

Answer: b

Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For rectangular edges the range is about 30 * 10 -3 .

6. What is the range of the coefficients of hydrodynamic resistances for the rounded edges at inlet?

a) 12-20 * 10 -3

b) 10-20 * 10 -3

c) 15-20 * 10 -3

d) 12-30 * 10 -3

Answer: a

Explanation: The coefficients of hydrodynamic resistances are used in the calculation of the total head. For rounded edges the range is about 12-20 * 10 -3 .

7. What factor/factors are required to evaluate the hydrodynamic resistance?

a) area of cross section

b) hydrodynamic coefficients

c) area of cross section or hydrodynamic coefficients

d) area of cross section and hydrodynamic coefficients

Answer: d

Explanation: The hydrodynamic resistance is calculated from the hydrodynamic coefficients. It is also calculated from the area of cross section.

8. How many data are required for the design of fan?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: There are 3 data required in the design of fan. They are outside diameter of fan, volume of air, hydrodynamic resistance.

9. How many steps are required in the design of the fan?

a) 7

b) 8

c) 9

d) 6

Answer: a

Explanation: There are 7 steps involved in the design of fan. They are maximum air passing per second, peripheral speed of fan, width of fan, static head of fan under rated duty, inside diameter of fan, number of blades, power input of fan.

10. What is the formula for the volume of air?

a) volume of air = 0.9 * losses in kW / difference of air temperature at inlet and outlet

b) volume of air = 0.9 * losses in kW * difference of air temperature at inlet and outlet

c) volume of air = 0.9 / losses in kW * difference of air temperature at inlet and outlet

d) volume of air = 1 / 0.9 * losses in kW * difference of air temperature at inlet and outlet

Answer: a

Explanation: The losses in kW is calculated along with the difference of air temperatures at inlet and outlet. On substitution the volume of air can be obtained.

11. What is the range of the difference of air temperature at inlet and outlet?

a) 11-15 0 C

b) 10-13 0 C

c) 12-16 0 C

d) 14-18 0 C

Answer: c

Explanation: The difference of air temperatures at inlet and outlet has a minimum value of 12 0 C. The difference of air temperatures at inlet and outlet has a maximum value of 16 0 C.

12. What is the formula for the area of outlet opening?

a) area of outlet opening = maximum air passing per second / 0.42 * peripheral speed

b) area of outlet opening = maximum air passing per second * 0.42 * peripheral speed

c) area of outlet opening = maximum air passing per second * 0.42 / peripheral speed

d) area of outlet opening = 1/maximum air passing per second * 0.42 * peripheral speed

Answer: a

Explanation: The maximum air passage per second along with the peripheral speed is calculated. On substitution the area of outlet opening is obtained.

13. What is the formula of the width of fan?

a) width of fan = area of outlet opening * 2.88 * outside diameter */ coefficient of utilization

b) width of fan = 1/ area of outlet opening * 2.88 * outside diameter * coefficient of utilization

c) width of fan = area of outlet opening * 2.88 * outside diameter * coefficient of utilization

d) width of fan = area of outlet opening / 2.88 * outside diameter * coefficient of utilization

Answer: d

Explanation: The area of outlet opening and the outside diameter is calculated. After fixing the coefficient of utilization and on substituting the width of fan is obtained.

14. What is the formula for the number of blades?

a) number of blades = 3.14 * outside diameter * * width of fan

b) number of blades = 3.14 / outside diameter * * width of fan

c) number of blades = 3.14 * outside diameter / * width of fan

d) number of blades = 3.14 * outside diameter *  / width of fan

Answer: c

Explanation: The outside diameter and the width of fan is first calculated. Then the range is fixed according to the diameter and on substitution gives the number of blades.

15. What is the formula for the maximum air passing per second at maximum efficiency?

a) maximum air passing = 2 * volume of air passing per second

b) maximum air passing = volume of air passing per second

c) maximum air passing = 2 / volume of air passing per second

d) maximum air passing = volume of air passing per second / 2

Answer: a

Explanation: The volume of air passing per second is first calculated and multiplying it by 2 gives the maximum air passing per second. The maximum air passing per second is used in the calculation of the area of the outlet opening.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Types of Motor”.

1. What is the special feature of single phase induction motor?

a) high starting torque

b) low starting torque

c) average starting torque

d) zero starting torque

Answer: d

Explanation: The single phase induction motor has no inherent starting torque. Thus special means should be used to make it self starting.

2. How many methods are present in the self starting of the single phase induction motor?

a) 1

b) 2

c) 3

d) 4

Answer: c

Explanation: There are 3 methods involved in the self starting of the single phase induction motor. They are split phase starting, shaded pole starting, repulsion motor starting.

3. What are the names of the windings used in the split phase starting?

a) starting windings

b) auxiliary windings

c) starting or auxiliary windings

d) starting and auxiliary windings

Answer: c

Explanation: The single phase induction motor is not self starting. The starting or auxiliary windings are used along with the running or main windings.

4. What is the displacement of the running and the starting windings used?

a) running winding displaces the starting winding by 180°

b) running winding displaces the starting winding by 90°

c) starting winding displaces the running winding by 90°

d) starting winding displaces the running winding by 180°

Answer: b

Explanation: The split phase starting makes use of the starting windings along with the running windings. The running winding displaces the starting windings by 900.

5. How is the required phase displacement between the current in the running and starting windings obtained?

a) by connecting a suitable resistor

b) by connecting a suitable capacitor

c) by connecting a suitable inductor

d) by connecting a suitable impedance

Answer: d

Explanation: The running winding displaces the starting winding by 90°. The required phase displacement is obtained by connecting a suitable impedance in series with any of the windings.

6. When is the starting winding cut out of the circuit in the split phase motor?

a) when the motor speed reaches 65 % of the full load speed

b) when the motor speed reaches 75 % of the full load speed

c) when the motor speed reaches 50 % of the full load speed

d) when the motor speed reaches 85 % of the full load speed

Answer: b

Explanation: The single phase induction motor is not a self starting machine and hence starting windings are connected in series with the running winding. The starting windings are cut out when the motor speed reaches 75 % of the full load speed.

7. What is the shaded pole starting method?

a) part of the pole is shaded by open circuited copper ring

b) part of the pole is shaded by short circuited copper ring

c) the pole is shaded by open circuited copper ring

d) the pole is shaded by short circuited copper ring

Answer: b

Explanation: One of the starting methods of the single phase induction motor is the shaded pole starting method. Here the part of the pole is shaded by the short circuited copper ring.

8. What happens in the shaded pole starting method according to the displacement?

a) displacement between shaded and unshaded portion varies between 20°-25°

b) displacement between shaded and unshaded portion varies between 20°-35°

c) displacement between shaded and unshaded portion varies between 20°-30°

d) displacement between shaded and unshaded portion varies between 30°-45°

Answer: c

Explanation: The shaded pole starting method is that the part of the pole is shaded by short circuited copper ring. The displacement between shaded and unshaded portion varies between 200-300.

9. For what type of machines is the shaded pole starting method suitable?

a) for outputs below 60 watt

b) for output below 50 watt

c) for output below 40 watt

d) for output above 50 watt

Answer: a

Explanation: The efficiency of the shaded pole starting method is very low. The shaded pole starting method is used for outputs below 60 watts.

10. When is the repulsion motor starting method used?

a) when low starting torque is required

b) when high starting torque is required

c) when high running torque is required

d) when low running torque is required

Answer: b

Explanation: The repulsion motor starting method is one of the 3 methods of starting the single phase induction motor. It is used when the high starting torque is required.

11. What is the specialty in the repulsion motor starting method?

a) cage winding is replaced by armature windings

b) cage winding is replaced by field windings

c) cage winding is replaced by commutator windings

d) cage winding is replaced by bearings

Answer: c

Explanation: The repulsion motor starting method is one of the methods used in the starting of the single phase induction motor. Here the cage winding is replaced by the commutator windings.

12. What happens in the repulsion motor starting method?

a) the cage windings is dominant

b) the commutator windings are dominant

c) the rotor windings are dominant

d) the stator windings are dominant

Answer: b

Explanation: The repulsion motor starting method is one of the starting methods of single phase induction motor. The commutator windings are dominant and hence gives good starting torque.

13. What is the range of output watt for the shaded pole induction machine?

a) 0.37-50

b) 90-750

c) 90-3700

d) 7.5-370

Answer: a

Explanation: The range of output watt for capacitor type induction motor is 90-750 and that of the repulsion start motor is 90-3700. The range of output watt for resistor type induction motor is 7.5-370 and that of the shaded pole type is 0.37-50.

14. What is the range of the starting current of capacitor type induction motor?

a) 5-7

b) 4–6

c) 2-6

d) 2-3

Answer: b

Explanation: The range of starting current is 5-7 for resistor type induction motor and that of the repulsion start motor is 2-3. The range of the starting current of capacitor induction motor is 4-6.

15. What is the range of the starting torque of shaded pole induction motor?

a) 2-4

b) 2-3.5

c) 0.2-0.3

d) 0.25-0.5

Answer: c

Explanation: The range of starting torque in the capacitor induction motor is 2-3.5 and that of the repulsion start induction motor is 2-4. The range of the starting torque of the shaded pole induction motor is 0.2-0.3.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Single Phase Induction Motor Construction”.

1. How many steps are involved in the construction of single phase induction motor?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: There are 5 steps in the construction of the single phase induction motor. They are stator, stator windings, rotor, starting switches, electrolytic capacitor.

2. What is the lamination used for the stator?

a) cast iron

b) die cast aluminium alloy frame

c) cast iron or die cast aluminium alloy frame

d) cast iron and die cast aluminium alloy frame

Answer: c

Explanation: The stator is made up of a block of laminations. The block of laminations are made up of cast iron or die cast aluminium alloy frame.

3. What type of coils are used for winding the single phase induction motor generally?

a) rectangular coils

b) square coils

c) cruciform coils

d) circular coils

Answer: d

Explanation: The slots house the starting and running windings. The single phase induction motors are generally wound with concentric coils.

4. How many kinds of single phase windings are present?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are basically 3 kinds of single phase windings. They are concentric, progressive and skein.

5. How are the poles and pitches in the concentric windings?

a) single pole, different pitches

b) different pole, different pitches

c) different pole, single pitch

d) single pole, single pitch

Answer: a

Explanation: The concentric windings have a single pole for a common centre. They have different pitches for each individual coil.

6. What is the form of the progressive windings?

a) double layer diamond coil windings

b) single layer diamond coil windings

c) multi layer diamond coil windings

d) three layer diamond coil windings

Answer: b

Explanation: The progressive windings is one kind of the stator windings. They are in the form of the single layer diamond coil windings.

7. When is the skein winding made use of?

a) when small amount of relatively small size wire is used

b) when large amount of relatively small size wire is used

c) when large amount of relatively large size wire is used

d) when small amount of relatively large size wire is used

Answer: a

Explanation: Skein winding is one of the 3 kinds of single phase windings used. It is used when small amount of relatively small size wire is used.

8. What kind of motor employs the skein winding made use of?

a) maximum horse power single phase induction motor

b) fractional horse power single phase induction motor

c) minimum horse power single phase induction motor

d) zero horse power single phase induction motor

Answer: b

Explanation: The skein winding is one of the 3 kinds of single phase induction motor. The skein winding is used when fractional horse power single phase induction motor is used.

9. Which winding is mostly used winding in the single phase induction motor?

a) circular winding

b) concentric winding

c) progressive winding

d) skein winding

Answer: b

Explanation: The concentric winding is the most widely used winding. It is also the most flexible winding of the windings used in the single phase induction motor.

10. What is/are the advantages of the skein winding?

a) low cost to wind

b) low cost to insert

c) permits some freedom of choice of distribution

d) low cost to wind, low cost to insert, permits some freedom of choice of distribution

Answer: d

Explanation: The skein winding is the low cost to wind and to insert. It also permits some freedom of choice of distribution.

11. What material is used in the tunnel of the rotor of the single phase induction motor?

a) aluminium

b) copper

c) steel

d) wood

Answer: a

Explanation: The rotor consists of a block of slotted laminations. The slots form a series of tunnels that are filled with aluminium in its molten state.

12. What type of operations are used in the starting switches?

a) mechanical operation

b) electrical operation

c) centrifugal operation and mechanical operation

d) centrifugal operation

Answer: c

Explanation: The starting switch is used to cut the auxillary winding when the motor attains 75% of the full load speed. The switches operate in both the centrifugal as well as mechanical operation.

13. The ac electrolytic capacitor is formed by winding two sheets of etched aluminium foil.

a) true

b) false

Answer: a

Explanation: Modern capacitor start motors employ ac electrolytic capacitors. The ac electrolytic capacitor is formed by winding two sheets of etched aluminium foil, separated by two layers of insulating paper, into a cylindrical shape.

14. The electrolytic capacitor and insulator unit is impregnated using ethylene glycol or a derivative.

a) true

b) false

Answer: a

Explanation: The electrolytic capacitor and insulator unit is impregnated using the ethylene glycol. It is also impregnated using the derivative of ethylene glycol.

15. What is the range of the power factor of electrolytic capacitors?

a) 2-4

b) 4-6

c) 6-8

d) 7-9

Answer: c

Explanation: The minimum power factor of the electrolytic capacitor is 6. The maximum power factor of the electrolytic capacitor is 8.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Output Equation”.

1. What is the formula of the output equation of ac machines?

a) kVA input Q = output coefficient * diameter 2 * length * synchronous speed

b) kVA input Q = output coefficient / diameter 2 * length * synchronous speed

c) kVA input Q = output coefficient * diameter 2 / length * synchronous speed

d) kVA input Q = output coefficient * diameter 2 * length / synchronous speed

Answer: a

Explanation: The output coefficient, diameter, length and synchronous speed are first calculated. Then on substitution gives the kVA input and in turn gives the output equation.

2. What is the formula for the output coefficient of the output equation?

a) output coefficient = 11 * winding space factor * specific magnetic loading / specific electric loading * 10 -3

b) output coefficient = 11 * winding space factor * specific magnetic loading * specific electric loading * 10 -3

c) output coefficient = 11 * winding space factor / specific magnetic loading * specific electric loading * 10 -3

d) output coefficient = 11 / winding space factor * specific magnetic loading * specific electric loading * 10 -3

Answer: b

Explanation: The winding space factor, specific magnetic loading, specific electric loading is calculated. On substitution gives the output coefficient used for the calculation of output equation.

3. What is the formula of the kVA input if the rating of the machine is given in horse power?

a) kVA input = horse power / 0.746 * efficiency * power factor

b) kVA input = horse power * 0.746 * efficiency * power factor

c) kVA input = horse power * 0.746 / efficiency * power factor

d) kVA input = horse power * 0.746 * efficiency / power factor

Answer: c

Explanation: The horse power rating of the machine along with the efficiency and power factor is calculated. On substitution the kVA input can be obtained.

4. What is the ratio of the efficiency for 75 watt to 750 watt motor?

a) 4:7

b) 5:7

c) 6:7

d) 3:7

Answer: b

Explanation: The efficiency of the 75 watt motor is 50%. The efficiency of the 750 watt motor is 70%.

5. What is the ratio of power factor of the 75 watt to 750 watt motor?

a) 0.55 : 0.60

b) 0.50 : 0.60

c) 0.55 : 0.65

d) 0.50 : 0.65

Answer: c

Explanation: The power factor of the 75 watt motor is 0.55. The power factor of 750 watt motor is 0.60.

6. The smaller values are applicable for lower rating machines.

a) true

b) false

Answer: a

Explanation: The smaller values are applicable for lower rating machines. The power factor is 0.55 for 75 watt motor and the efficiency is 50% for the 75 watt motor.

7. What is the efficiency for the output watt of 180?

a) 0.38

b) 0.48

c) 0.57

d) 0.65

Answer: c

Explanation: The efficiency of the output watt of 37 is 0.38 and the efficiency of output watt of 90 is 0.48. The efficiency of the output watt of 180 is 0.57.

8. What is the power factor of output watt of 90?

a) 0.46

b) 0.51

c) 0.56

d) 0.62

Answer: b

Explanation: The power factor of output watt of 37 is 0.46 and the power factor for 180 output watt is 0.56. The power factor of the output watt of 90 is 0.51.

9. What factor does the output coefficient depend upon?

a) specific magnetic loading

b) specific electric loading

c) specific electric loading or specific magnetic loading

d) specific electric loading and specific magnetic loading

Answer: b

Explanation: The output coefficient depends upon the specific electric loading. The output coefficient also depends upon the specific magnetic loading.

10. What is the range of the average flux density used in the output equation?

a) 0.30-0.55 weber per m 2

b) 0.30-0.50 weber per m 2

c) 0.35-0.45 weber per m 2

d) 0.35-0.55 weber per m 2

Answer: d

Explanation: The minimum value of the average flux density is 0.35 weber per m 2 . The maximum value of the average flux density is 0.55 weber per m 2 .

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Stator”.

1. What type is the stator windings of the single phase induction motor?

a) hollow

b) cylindrical

c) concentric

d) rectangular

Answer: c

Explanation: The stator windings are also known as the running winding or the main winding. The type of stator winding used is concentric type

2. How many coils are present in the stator windings?

a) 2

b) 3

c) 2 or more

d) 3 or more

Answer: d

Explanation: The stator windings of single phase induction motors are concentric type. There are usually 3 or more coils per pole each having same or different number of turns.

3. How much of the total slots are used for the reduction of the mmf wave harmonics?

a) 60%

b) 65%

c) 70%

d) 80%

Answer: c

Explanation: 70% of the total slots are used for the reduction of the mmf wave harmonics. The remaining 30% are used for accommodating the starting windings.

4. How can the small single phase motor reduce the harmonics still much further?

a) removing the winding

b) insulating the winding

c) grading the winding

d) shading the winding

Answer: c

Explanation: 70% of the total slots are used for the reduction of the mmf wave harmonics. The mmf wave harmonics can be still further reduced by grading the winding.

5. What is the formula for the mean pitch factor?

a) mean pitch factor = pitch factor of each coil per pole group + turns in the coil / total number of turns

b) mean pitch factor = pitch factor of each coil per pole group / turns in the coil * total number of turns

c) mean pitch factor = pitch factor of each coil per pole group * turns in the coil * total number of turns

d) mean pitch factor = pitch factor of each coil per pole group * turns in the coil / total number of turns

Answer: d

Explanation: The pitch factor of each coil per pole group, turns in the coil and total number of turns are obtained. On substitution, it gives the mean pitch factor.

6. What is the range of the winding factor for the usual windings distribution?

a) 0.70-0.80

b) 0.75-0.85

c) 0.70-0.85

d) 0.70-0.75

Answer: b

Explanation: The minimum value of the winding factor of the usual winding distribution is 0.75. The maximum value of the winding factor of the usual winding distribution is 0.85.

7. What is the formula of the maximum flux in the running winding?

a) maximum flux = flux * pole

b) maximum flux = flux/pole

c) maximum flux = flux / turns

d) maximum flux = flux * turns

Answer: b

Explanation: First the flux is calculated along with the number of poles used. On substituting the values the maximum flux value is obtained.

8. What is the value of the stator induced voltage with respect to the supply voltage?

a) stator induced voltage = 95% of supply voltage

b) stator induced voltage = 90% of supply voltage

c) stator induced voltage = 85% of supply voltage

d) stator induced voltage = 80% of supply voltage

Answer: a

Explanation: The winding factor is assumed to be 0.75-0.85 for the running winding. The stator induced voltage is 95% of the supply voltage.

9. How many design data are present in the design of the stator?

a) 6

b) 7

c) 8

d) 9

Answer: c

Explanation: There are 8 design data available in the design of the stator. The design data are running winding, number of turns In running winding, running winding conductors, number of stator slots, size of stator slot, stator teeth, stator core, length of mean turn.

10. What is the range of the current density for the open type motors split phase, capacitor and repulsion start motors?

a) 4-5 A per mm 2

b) 3-4 A per mm 2

c) 2-4 A per mm 2

d) 1-4 A per mm 2

Answer: b

Explanation: The minimum value of the current density for the open type motors split phase, capacitor and repulsion start motors is 3 A per mm 2 . The maximum value of the current density for the open type motors split phase, capacitor and repulsion start motors is 4 A per mm 2 .

11. What is the relation of the number of slots with the leakage reactance?

a) small number of slots, high leakage reactance

b) large number of slots, high leakage reactance

c) large number of slots, small leakage reactance

d) small number of slots, small leakage reactance

Answer: c

Explanation: The number of slots is indirectly proportional to the leakage reactance. The larger the number of slots, the lower will be the leakage reactance.

12. What is the formula for the area required for the insulated conductors?

a) area required for the insulated conductors = total number of conductors per slot * 0.785 / diameter of insulated conductor 2

b) area required for the insulated conductors = total number of conductors per slot / 0.785 * diameter of insulated conductor 2

c) area required for the insulated conductors = total number of conductors per slot * 0.785 * diameter of insulated conductor 2

d) area required for the insulated conductors = 1/total number of conductors per slot * 0.785 * diameter of insulated conductor 2

Answer: c

Explanation: The total number of conductors per slot and the diameter of insulated conductors are calculated. On substitution the area required for the insulated conductors are calculated.

13. The flux density of the high torque machines is 1.8 weber per m 2 .

a) true

b) false

Answer: a

Explanation: The flux density of the general purpose machine is 1.45 weber per m 2 . The flux density of the high torque machines is 1.8 weber per m 2 .

14. The flux density of the stator core should not exceed 1.3 weber per m 2 .

a) true

b) false

Answer: b

Explanation: The flux density of the stator core should not exceed 1.5 weber per m 2 . The range lies between 0.9 – 1.4 weber per m 2 .

15. What is the formula for the flux density in stator core?

a) flux density in stator core = maximum flux / length of the iron * depth of stator core

b) flux density in stator core = maximum flux * length of the iron * depth of stator core

c) flux density in stator core = maximum flux / 2 *length of the iron * depth of stator core

d) flux density in stator core = maximum flux * length of the iron / depth of stator core

Answer: c

Explanation: The maximum flux, length of iron and depth of stator core is calculated. On substitution it provides the flux density in stator core.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Rotor”.

1. How many design steps are available for the design of rotor?

a) 5

b) 6

c) 7

d) 8

Answer: b

Explanation: There are 6 design steps involved in the design of the rotor. They are number of rotor slots, area of rotor bars, area of end rings, rotor resistance, rotor teeth, rotor core.

2. What is the main motive while choosing the number of rotor slots?

a) increasing the efficiency

b) decreasing the losses

c) no noise is produced

d) high output is produced

Answer: c

Explanation: There are basically 6 steps involved in the rotor design. The number of slots is chosen such that no noise is produced.

3. What is the formula for the harmonic poles due to slots?

a) harmonic poles due to slots = 2 * 

b) harmonic poles due to slots = 2 / 

c) harmonic poles due to slots = 2 * 

d) harmonic poles due to slots = 1/ 2 * 

Answer: a

Explanation: First the number of slots and number of poles are first calculated. On substitution we get the harmonic poles due to the slots.

4. What factors are used fixing the number of stator slots?

a) winding arrangement

b) number of poles

c) winding arrangement or number of poles

d) winding arrangement and number of poles

Answer: d

Explanation: The number of poles are fixed according to the winding arrangement. The number of poles are also fixed according to the number of poles.

5. Which condition satisfies the quiet operation in machines?

a) number of stator slots is divisible by number of pairs of poles

b) number of rotor slots differs from the number of stator slots by more than the number of poles

c) number of rotor slots is not divisible by number of pairs of poles

d) number of stator slots differs from the number of rotor slots by more than the number of poles

Answer: b

Explanation: The number of rotor slots are decided for quieter operation of the machine. The number of rotor slots differs from the number of stator slots by more than the number of poles.

6. What among the following are considered for the selection of number of rotor slots?

a) magnetic locking

b) cusps

c) magnetic locking or cusps

d) magnetic locking and cusps

Answer: d

Explanation: The selection of number of rotor slots depends on the magnetic locking. The selection of number of rotor slots depends on the cusps also.

7. What is the formula for the total stator copper section for main winding?

a) total stator copper section for main winding = number of turns in the running winding * area of the running winding conductor

b) total stator copper section for main winding = 2 * number of turns in the running winding * area of the running winding conductor

c) total stator copper section for main winding = number of turns in the running winding / area of the running winding conductor

d) total stator copper section for main winding = 2* number of turns in the running winding / area of the running winding conductor

Answer: b

Explanation: First the number of turns in the running winding is calculated along with the area of the running winding conductor. On substitution it gives the total stator copper section for main winding.

8. What is the formula for the total cross section of rotor bars?

a) total cross section of rotor bars = number of rotor slots * area of each bar

b) total cross section of rotor bars = number of rotor slots / area of each bar

c) total cross section of rotor bars = number of rotor slots + area of each bar

d) total cross section of rotor bars = number of rotor slots – area of each bar

Answer: a

Explanation: The number of rotor slots and area of each bar is first calculated. On substitution it gives the total cross section of rotor bars.

9. What is the range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding for copper?

a) 0.4-0.8

b) 0.3-0.7

c) 0.5-0.8

d) 0.8-0.9

Answer: c

Explanation: The minimum value of range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding is 0.5. The maximum value range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding is 0.8.

10. What is the formula of the end ring current?

a) end ring current = number of rotor slots * bar current * 3.14 * number of poles

b) end ring current = number of rotor slots * bar current * 3.14 / number of poles

c) end ring current = number of rotor slots / bar current * 3.14 * number of poles

d) end ring current = number of rotor slots * bar current / 3.14 * number of poles

Answer: d

Explanation: The number of rotor slots and the bar current along with the number of poles is calculated. On substitution it gives the end ring current value.

11. What is the range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding for aluminium?

a) 1-1.3

b) 1-1.4

c) 1-1.6

d) 1.2-1.5

Answer: c

Explanation: The minimum value of range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding is 1. The maximum value range of the ratio of the total cross section of rotor bars to the total stator copper section for main winding is 1.6.

12. What is the formula for the area of each bar?

a) area of each bar = current through each bar / current density through each bar

b) area of each bar = current through each bar * current density through each bar

c) area of each bar = current density through each bar / current through each bar

d) area of each bar = current density through each bar * current through each bar

Answer: a

Explanation: The current through each bar and the current density through each bar is calculated. On substitution the area of each bar is obtained.

13. What is the formula of the area of each end ring?

a) area of each end ring = 0.32 * total cross section of rotor bars * number of poles

b) area of each end ring = 0.32 / total cross section of rotor bars * number of poles

c) area of each end ring = 0.32 * total cross section of rotor bars / number of poles

d) area of each end ring = 1/0.32 * total cross section of rotor bars * number of poles

Answer: c

Explanation: First the total cross section of rotor bars along with the number of poles are calculated. On substitution the area of each end ring is obtained.

14. What is the formula of the rotor teeth flux density?

a) flux density of rotor teeth = maximum flux /  * length of the teeth * depth of rotor core

b) flux density of rotor teeth = maximum flux *  * length of the teeth * depth of rotor core

c) flux density of rotor teeth = 1/maximum flux *  * length of the teeth * depth of rotor core

d) flux density of rotor teeth = maximum flux /  * length of the teeth * depth of rotor core

Answer: a

Explanation: The maximum flux, the number of rotor slots per pole and the length of teeth along with the depth of rotor core is calculated. On substitution the flux density of the rotor teeth is obtained.

15. What is the range for the ratio of the resistance to reactance in the split phase motors?

a) 0.40-0.55

b) 0.45-0.55

c) 0.45-0.8

d) 0.45-0.6

Answer: b

Explanation: The range for the ratio of the resistance to reactance in the split phase motors is 0.45-0.55. The range for the ratio of the resistance to reactance in the capacitor start motors is 0.45-0.8.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Parameters”.

1. What is the formula for the resistance of running winding?

a) resistance of running winding = 0.021 * no of turns in running winding * length of mean turns of running winding * area of running winding conductor

b) resistance of running winding = 0.021 / no of turns in running winding * length of mean turns of running winding * area of running winding conductor

c) resistance of running winding = 0.021 * no of turns in running winding / length of mean turns of running winding * area of running winding conductor

d) resistance of running winding = 0.021 * no of turns in running winding * length of mean turns of running winding / area of running winding conductor

Answer: d

Explanation: First the no of turns in running winding along with length of mean turns of running winding and area of running winding conductor is calculated. On substitution the resistance of running winding is calculated.

2. How many parameters are present in the single phase induction motor?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 parameters in the single phase induction motor. They are running winding resistance, rotor resistance and leakage reactance calculations of single phase motor.

3. How many parameters are present under the leakage reactance calculations?

a) 6

b) 5

c) 7

d) 4

Answer: c

Explanation: There are 7 parameters present under the leakage reactance calculations. They are slot leakage reactance, rotor slot leakage reactance, zigzag leakage reactance, overhang leakage reactance, skew leakage reactance, magnetizing reactance, total leakage reactance.

4. How is the winding arrangement and how is the conductors in each slot?

a) circular winding and same conductors in each slot

b) circular winding and different conductor in each slot

c) concentric winding and same conductor in each slot

d) concentric winding and different conductor in each slot

Answer: d

Explanation: The winding of the induction motors are concentric type. The different conductors are being used in each slot.

5. What is the relation of the total slot leakage reactance with number of stator slots?

a) slot leakage reactance is directly proportional to the number of stator slots

b) slot leakage reactance is indirectly proportional to the number of stator slots

c) slot leakage reactance is directly proportional to the square of the number of stator slots

d) slot leakage reactance is indirectly proportional to the square of the number of stator slots

Answer: b

Explanation: The slot leakage reactance is one of the parameters used in the leakage reactance calculations. The slot leakage reactance is indirectly proportional to the number of stator slots.

6. What is the relation between slot leakage reactance and specific slot permeance?

a) slot leakage reactance is directly proportional to the specific slot permeance

b) slot leakage reactance is indirectly proportional to the specific slot permeance

c) slot leakage reactance is directly proportional to the square of the specific slot permeance

d) slot leakage reactance is indirectly proportional to the square of the specific slot permeance

Answer: a

Explanation: Specific slot permeance is one of the parameters present in the leakage reactance calculation. It is directly proportional to the slot leakage reactance.

7. What is the relation of the total slot leakage reactance with number of stator slots?

a) slot leakage reactance is directly proportional to the number of rotor slots

b) slot leakage reactance is indirectly proportional to the number of rotor slots

c) slot leakage reactance is directly proportional to the square of the number of rotor slots

d) slot leakage reactance is indirectly proportional to the square of the number of rotor slots

Answer: b

Explanation: The slot leakage reactance is one of the parameters in the leakage reactance calculation. The slot leakage reactance is indirectly proportional to the number of rotor slots.

8. What is the relation of the zigzag reactance with the specific permeance for zigzag leakage?

a) zigzag reactance is directly proportional to the specific permeance for zigzag leakage

b) zigzag reactance is indirectly proportional to the specific permeance for zigzag leakage

c) zigzag reactance is directly proportional to the square of the specific permeance for zigzag leakage

d) zigzag reactance is indirectly proportional to the square of the specific permeance for zigzag leakage

Answer: a

Explanation: Zigzag reactance is one of the parameters used in the leakage reactance calculation. The zigzag reactance is directly proportional to the specific permeance for zigzag leakage.

9. What is the relation of the stator slot leakage factor with the skew leakage reactance?

a) skew leakage reactance is directly proportional to the stator slot leakage factor

b) skew leakage reactance is indirectly proportional to the stator slot leakage factor

c) skew leakage reactance is directly proportional to the square of stator slot leakage factor

d) skew leakage reactance is indirectly proportional to the square of stator slot leakage factor

Answer: a

Explanation: Skew leakage reactance is one of the parameters used in the leakage reactance calculation. The skew leakage reactance is directly proportional to the stator slot leakage reactance.

10. What is the formula for the rotor bar skew angle?

a) rotor bar skew angle = 3.14 / rotor slot pitches through which bars are skewed * 

b) rotor bar skew angle = 3.14 * rotor slot pitches through which bars are skewed * 

c) rotor bar skew angle = 3.14 * rotor slot pitches through which bars are skewed / 

d) rotor bar skew angle = 3.14 * rotor slot pitches through which bars are skewed / 

Answer: d

Explanation: The rotor slot pitches through which bars are skewed, number of rotor slots and number of poles are calculated. On substitution the rotor bar skew angle is obtained.

11. What is the value of the stator slot leakage factor?

a) 0.90

b) 0.80

c) 0.95

d) 0.85

Answer: c

Explanation: Stator slot leakage factor is used in the calculation of the skew leakage reactance. The value of the stator slot leakage factor is 0.95 in the single phase induction motor.

12. What is the relation of the overhang leakage reactance with the average coil span in slots?

a) overhang leakage reactance is directly proportional to the square of the average coil span in slots

b) overhang leakage reactance is indirectly proportional to the square of the average coil span in slots

c) overhang leakage reactance is directly proportional to the average coil span in slots

d) overhang leakage reactance is indirectly proportional to the average coil span in slots

Answer: c

Explanation: The overhang leakage reactance is one of the parameters used in the leakage reactance calculation. The overhang leakage reactance is directly proportional to the average coil span in slots.

13. What is the relation between pole pitch and the magnetizing reactance?

a) magnetizing reactance is directly proportional to the square of the pole pitch

b) magnetizing reactance is directly proportional to the pole pitch

c) magnetizing reactance is indirectly proportional to the pole pitch

d) magnetizing reactance is indirectly proportional to the square of the pole pitch

Answer: b

Explanation: The magnetizing reactance is one of the parameters in the leakage reactance calculation. The magnetizing reactance is directly proportional to the pole pitch.

14. The magnetizing reactance is directly proportional to the saturation factor.

a) true

b) false

Answer: b

Explanation: The magnetizing reactance is one of the components used in the leakage reactance calculations. The magnetizing reactance is indirectly proportional to the saturation factor.

15. What is the formula of the saturation factor?

a) saturation factor = total mmf required for magnetic circuit * mmf required for air gap

b) saturation factor = total mmf required for magnetic circuit – mmf required for air gap

c) saturation factor = total mmf required for magnetic circuit / mmf required for air gap

d) saturation factor = total mmf required for magnetic circuit + mmf required for air gap

Answer: c

Explanation: The total mmf required for the magnetic circuit and the mmf required for air gap is calculated. On substitution of the values the saturation factor is calculated.

16. What factor is the core length made equal to in theoretical conditions?

a) pole length

b) pole proportion

c) pole length

d) number of poles

Answer: c

Explanation: The core length is made equal to the pole length in theoretical conditions. But in practical the exact dimensions are governed by manufacturing conditions.

17. What is the output equation of a single phase induction motor developed by P.H Tricky?

a) diameter 2 * length = 16.5 / H.P * output coefficient * frequency constant * motor type constants * rpm * 10 6

b) diameter 2 * length = 16.5 * H.P * output coefficient * frequency constant * motor type constants /rpm * 10 6

c) diameter 2 * length = 16.5 * H.P / output coefficient * frequency constant * motor type constants * rpm * 10 6

d) diameter 2 * length = 16.5 * H.P * output coefficient / frequency constant * motor type constants * rpm * 10 6

Answer: b

Explanation: The output coefficient, horse power, frequency constant, motor type constant along with the speed is calculated. On substitution we get the diameter 2 * length value.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Operating Characteristics”.

1. How many factors are present in the operating characteristics?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: There are 4 operating characteristics present in the single phase induction motor. They are mmf for air gap, saturation factor, iron loss, friction and windage loss.

2. How many parts does the stator mmf passes through?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: The stator mmf produced in the motor passes through 5 parts. They are air gap, stator teeth, stator core, rotor teeth, rotor core.

3. What is the angle at which the value of the flux density should be for the calculation of mmf?

a) 40°

b) 60°

c) 80°

d) 70°

Answer: b

Explanation: The calculation of mmf should be based upon the value of the flux density. The value of flux density at 60° from the interpolar axis as far as gap and teeth are concerned.

4. What is the value of the flux density with respect to average flux density?

a) value of flux density = 1.67 times of average flux density

b) value of flux density = 1.70 times of average flux density

c) value of flux density = 1.60 times of average flux density

d) value of flux density = 1.50 times of average flux density

Answer: a

Explanation: The value of flux density at 60° from the interpolar axis as far as gap and teeth are used in the calculation of mmf. The value of flux density = 1.67 times of average flux density.

5. What is the formula for the mmf required for air gap?

a) mmf required for air gap = 8,00,000 * air gap flux density * air gap constant / air gap length

b) mmf required for air gap = 8,00,000 * air gap flux density * air gap constant * air gap length

c) mmf required for air gap = 8,00,000 * air gap flux density / air gap constant * air gap length

d) mmf required for air gap = 8,00,000 / air gap flux density * air gap constant * air gap length

Answer: b

Explanation: The air gap flux density, air gap constant, air gap length are calculated first. On substitution, the mmf required for air gap can be obtained.

6. What is the formula for the saturation factor?

a) saturation factor = total mmf required for the magnetic circuit/mmf required for air gap

b) saturation factor = total mmf required for the magnetic circuit + mmf required for air gap

c) saturation factor = total mmf required for the magnetic circuit – mmf required for air gap

d) saturation factor = total mmf required for the magnetic circuit * mmf required for air gap

Answer: a

Explanation: The total mmf required for the magnetic circuit and the mmf required for air gap is calculated. On substitution, the saturation factor is obtained.

7. What is the range of the saturation factor in the single phase induction motor?

a) 1.1-1.3

b) 1.0-1.2

c) 1.1-1.35

d) 1.2-1.6

Answer: c

Explanation: The minimum value of the saturation factor in the single phase induction motor is 1.1. The maximum value of the saturation factor in the single phase induction motor is 1.35.

8. What is the relation between flux densities with respect to saturation factor?

a) flux density is indirectly proportional to the square of the saturation factor

b) flux density is directly proportional to the square of the saturation factor

c) flux density is indirectly proportional to the saturation factor

d) flux density is directly proportional to the saturation factor

Answer: d

Explanation: The saturation factor is kept low if the flux densities in the teeth and core are low. The saturation factor is kept high if the flux densities in the teeth and core are high.

9. What factors are considered while calculating iron loss in stator teeth and core?

a) flux densities

b) weights

c) flux densities or weights

d) flux densities and weights

Answer: d

Explanation: The iron loss in stator teeth and core are found by calculating their flux densities. The iron loss in stator teeth and core are found by calculating their weights.

10. What is the relation between total iron loss for induction motors and the sum of stator tooth and core loss?

a) total iron loss for induction motors = 1.3-2.3 times the sum of stator tooth and core loss

b) total iron loss for induction motors = 1.4-2.4 times the sum of stator tooth and core loss

c) total iron loss for induction motors = 1.5-2.5 times the sum of stator tooth and core loss

d) total iron loss for induction motors = 1.3-2 times the sum of stator tooth and core loss

Answer: c

Explanation: The total iron loss for induction motors is 1.5-2.5 times the sum of stator tooth and core loss. The total iron loss is due to fundamental frequency flux.

11. What is the range of the multiplying factor when test data is not available?

a) 1.7-2

b) 1.75-2.2

c) 1.6-2.3

d) 1.5-2

Answer: b

Explanation: The multiplying factor should be obtained from tests of motors of similar design. When test data is not available, a value of 1.75 to 2 may be used.

12. What are the factors which result in the bearing friction and windage loss?

a) ball bearings

b) sleeve bearing

c) ball bearing and sleeve bearing

d) ball bearing or sleeve bearing

Answer: d

Explanation: The bearing friction and windage loss will depend upon the ball bearings. The bearing friction and windage loss will also depend upon the sleeve bearings.

13. What is the friction and windage loss for a 1500 rpm machine?

a) 3-7% of watt output

b) 3-9% of watt output

c) 4-8% of watt output

d) 1-5% of watt output

Answer: c

Explanation: The friction and windage loss and a speed of 1500 rpm, it is usually minimum 4% of the watt output. The friction and windage loss and a speed of 1500 rpm, it is usually maximum of 8% of the watt output.

14. The high values actually apply for the small motors below 150 W.

a) true

b) false

Answer: b

Explanation: The friction and windage loss and a speed of 1500 rpm, it is usually 4-8% of the watt output. The high values apply to small motors below 180 W.

15. The loss for the sleeve bearing having stator outer diameter 150 mm and 1000 rpm is 3.7 W.

a) true

b) false

Answer: a

Explanation: The sleeve bearing having a stator outer diameter of 150 mm has losses at different running speed. The 1000 rpm machine gives loss of 3.7 W.

16. What is the range of the frequency constant?

a) 0.86-1.0

b) 0.82-1.0

c) 0.9-1.0

d) 0.5-1.0

Answer: a

Explanation: The minimum value of the frequency constant used in the output equation of P.H Tricky is 0.86. The maximum value of the frequency constant used in the output equation of P.H Tricky is 1.0.

17. What is the range of the motor type constant?

a) 1.1-1.3

b) 1.0-1.4

c) 1.1-1.42

d) 1.1-1.5

Answer: c

Explanation: The minimum value of the motor type constants is 1.1. The maximum value of the motor type constants is 1.42.

18. What is the formula of the most economical relation between D and L?

a) length = 0.6 * diameter

b) length = 0.5 / diameter

c) length = 0.6 / diameter

d) length = 0.63 * diameter

Answer: d

Explanation: The length and the diameter are the 2 main dimensions of the single phase induction motor. The most economical relation between length and diameter is length = 0.63 * diameter.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of a Small Reluctance Motor”.

1. How is the reluctance motor with respect to a synchronous motor and are the field windings?

a) small synchronous motor with field windings

b) small synchronous motor without field windings

c) large synchronous motor with field windings

d) large synchronous motor without field windings

Answer: b

Explanation: Reluctance motor is nothing but a simple small synchronous motor with salient pole rotor. They are without field windings in which the field flux is produced.

2. Why is the three phase reluctance motor preferred over single phase reluctance motor?

a) single phase reluctance motors have the phenomenon of hunting

b) single phase reluctance motors have the phenomenon of over voltage

c) single phase reluctance motors have high losses

d) single phase reluctance motors have low output

Answer: a

Explanation: The reluctance motor is a small synchronous motor with salient pole rotor. The single phase reluctance motors have the phenomenon of hunting.

3. What is the relation of the input voltage with the magnetic flux?

a) if the input voltage is constant, the magnetic flux increases

b) if the input voltage is constant, the magnetic flux decreases

c) if the input voltage is constant, the magnetic flux is constant

d) if the input voltage is constant, the magnetic flux is zero

Answer: c

Explanation: The input voltage is given constant, which results in the constant magnetic flux. The magnetic flux is independent of the excitation.

4. What is the power factor in the reluctance motor and the range of efficiency?

a) leading power factor, 60-75%

b) lagging power factor, 50-75%

c) zero power factor, 55-80%

d) lagging power factor, 55-75%

Answer: d

Explanation: The power factor in the reluctance motor is lagging power factor. The efficiency of the machine is about 55-75%.

5. What is the angle at which the electromagnetic torque is maximum?

a) 30°

b) 45°

c) 60°

d) 90°

Answer: b

Explanation: The electromagnetic torque is maximum at the angle of 45°. The range of operation of the reluctance motor lies in the range of 0-45°.

6. What is the range of the ratio of the direct axis reactance to the quadrature axis reactance?

a) 1.5-2.3

b) 1.6-2.7

c) 1.6-2.2

d) 1.2-2.0

Answer: c

Explanation: The minimum value of the ratio of the direct axis reactance to the quadrature axis reactance is 1.6. The maximum value of the ratio of the direct axis reactance to the quadrature axis reactance is 2.2.

7. How many design dimension are present in the design of the small reluctance motor?

a) 3

b) 4

c) 5

d) 6

Answer: c

Explanation: There are 5 design dimensions present in the design of the small reluctance motors. They are the design of the main dimensions, design of stator windings, design of the rotor of the reluctance motor, design of performance parameters, design of losses and efficiency.

8. What is the range of the constant used in the calculation of the active power of reluctance motor?

a) 0.3-0.4

b) 0.35-0.55

c) 0.40-0.50

d) 0.35-0.60

Answer: b

Explanation: The minimum value of the range of the constant used in the calculation of the active power of reluctance motor is 0.35. The maximum value of the range of the constant used in the calculation of the active power of reluctance motor is 0.55.

9. How many steps are present in the calculation of the determination of main dimensions?

a) 5

b) 4

c) 3

d) 2

Answer: a

Explanation: There are 5 steps present in the calculation of the determination of main dimensions. They are electromagnetic power of reluctance motor, output coefficient, pole pitch, pole arc, peripheral velocity.

10. How many steps are present in the calculation of the design of stator windings?

a) 10

b) 11

c) 9

d) 12

Answer: b

Explanation: There are 11 steps involved in the calculation of the design of stator windings. They are input current to motor, number of stator slots, stator winding pitch, winding factor, useful flux, number of turns per stator winding, cross sectional area of the stator winding, slot area, mean length for conductor, active resistance of stator winding, specific permeance of leakage flux.

11. How many steps are present in the calculation of the design of rotor of reluctance motors?

a) 4

b) 5

c) 3

d) 2

Answer: a

Explanation: There are 4 steps involved in the design of rotor of reluctance motor. They are rotor diameter calculation, height of rotor core, mmf for magnetic circuit, saturation coefficient of motor.

12. How many steps are involved in the design of performance parameters?

a) 6

b) 5

c) 7

d) 8

Answer: c

Explanation: There are 7 steps involved in the design of the performance parameters. They are no load current, height of steel stator teeth, weight of steel in the stator core, copper loss in the stator winding under no load, active resistance and leakage reactance, active component of no load current, starting torque of 3 phase reluctance motor.

13. How many design steps are involved in the determination of the losses and efficiency?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 steps involved in the determination of the losses and efficiency. They are copper loss in stator winding, iron loss in stator steel, mechanical loss in the motor.

14. What is the formula for the slot pitch factor in design of rotors?

a) slot pitch factor = 3.14*rotor diameter*number of rotor slots

b) slot pitch factor = 3.14/rotor diameter*number of rotor slots

c) slot pitch factor = 3.14*rotor diameter/number of rotor slots

d) slot pitch factor = 1/3.14*rotor diameter*number of rotor slots

Answer: c

Explanation: First the rotor diameter and the number of rotor slots are first calculated. On substitution the slot pitch factor can be obtained.

15. The active resistance of the stator winding is calculated at the temperature of 45° C.

a) true

b) false

Answer: b

Explanation: The active resistance of the stator winding determination is one of the steps in the design of stator windings. The value is calculated at the temperature of 45° C.

This set of Design of Electrical Machines Interview Questions and Answers for Experienced people focuses on “Design of Small Universal Commutator Motors”.

1. What are the applications of the small universal commutator motors?

a) industry

b) medicine

c) domestic sector

d) industry, medicine and domestic sector

Answer: d

Explanation: Small universal commutator motors have power outputs varying from few watts to hundreds of watts. They have lots of application in industry, medicine, domestic sector.

2. What type of excitation is used in the small universal commutator motors and what type of supply is provided?

a) parallel excitation, dc supply

b) series excitation, dc or ac supply

c) series excitation, ac supply

d) parallel excitation, dc supply

Answer: b

Explanation: The excitation which is provided is the series excitation in the small universal commutator motors. The type of supply provided is dc or ac supply.

3. What is the material used in the lamination of the magnetic poles of small universal commutator motor?

a) copper

b) aluminium

c) gold

d) sheet steel

Answer: d

Explanation: Unlike dc motors universal commutator motors are having laminated poles. The laminated poles are laminated using the sheet steel.

4. What is the thickness of the laminations of magnetic poles?

a) 0.3-0.5 mm

b) 0.2-0.4 mm

c) 0.35-0.5 mm

d) 0.4-0.5 mm

Answer: c

Explanation: The minimum value of the thickness of the laminations of magnetic poles is 0.35 mm. The maximum value of the thickness of the laminations of magnetic poles is 0.5 mm.

5. How many type of excitations does the universal commutator motors have?

a) 1

b) 2

c) 3

d) 4

Answer: b

Explanation: There are two kinds of excitation present for the universal commutator motors. One winding is intended to run the motor to ac supply voltage. The other winding is connected in series with the first winding when the motor is operated from dc supply voltage.

6. The number of turns of field winding in the motor must be considerably less than the number of turns in the armature winding.

a) true

b) false

Answer: a

Explanation: The number of turns of field winding in the motor must be considerably less than the number of turns in the armature winding. The speed of the machine is irrespective of the supply.

7. What is the relation of the copper loss and brush contacts with the total loss in small dc motors?

a) copper loss and brush contacts = 2 * total loss

b) copper loss and brush contacts = 2/3 * total loss

c) copper loss and brush contacts = 1/3 * total loss

d) copper loss and brush contacts = total loss

Answer: b

Explanation: The copper loss and brush contact loss is being compared with the total loss in order to deduce an equation. The copper loss and brush contact loss is 2/3 times the total loss.

8. What is the relation of the copper loss and brush contacts with the total loss in universal commutator motors?

a) copper loss and brush contacts = total loss

b) copper loss and brush contacts = total loss/2

c) copper loss and brush contacts = total loss * 2

d) copper loss and brush contacts = total loss * 3

Answer: b

Explanation: The copper loss and brush contact loss is being compared with the total loss in order to deduce equations. The copper loss and brush contact loss is half the total loss.

9. What is the range of the transformation ratio in the pole machines?

a) 0.05-0.1

b) 0.1-0.2

c) 0.1-0.25

d) 0.3-0.4

Answer: c

Explanation: The transformation ratio in 2 pole motors is 0.1-0.25. The transformation ratio in the 4 pole motors is 0.05-0.1.

10. For what outputs are the 2 pole machines made use of?

a) output > 200 W

b) output < 200 W

c) output > 300 W

d) output < 300 W

Answer: b

Explanation: The 2 pole machines are made use of when the output is below 200 W. The 4 pole machines are made use of when the output is above 200 W.

11. What is the range of the power factor for the 4 pole motors?

a) 0.6-0.85

b) 0.75-0.95

c) 0.6-0.8

d) 0.7-0.9

Answer: a

Explanation: The power factor for the 4 pole motors is 0.6-0.85. The power factor for the 2 pole motors is 0.75-0.95.

12. What is the value of the specific electric loading for the continuous duty motor type?

a) 8000-11000 A per m

b) 6000-9000 A per m

c) 12000-20000 A per m

d) 15000-25000 A per m

Answer: b

Explanation: The specific electric loading for continuous duty motor type is 6000-9000 A per m. The specific electric loading for power rating above 100 W but below 200 W is 8000-11000 A per m and the specific electric loading for power rating above 200 W but below 750 W is 12000-20000 A per m.

13. What is the specific magnetic loading for the motors having output less than 100 W?

a) 0.25-0.30 T

b) 0.3-0.4 T

c) 1.3-1.5 T

d) 0.25-0.35 T

Answer: d

Explanation: The specific magnetic loading for the output below 100 W is 0.25-0.35 T and the specific magnetic loading for the continuous duty motors is 0.3-0.4 T and the specific magnetic loading for the short time duty motors is 1.3-1.5 T.

14. What is the formula for the pole pitch in the universal commutator motor?

a) pole pitch = 3.14 * diameter * 2 * no. of poles

b) pole pitch = 3.14 / diameter * 2 * no. of poles

c) pole pitch = 3.14 * diameter / 2 * no. of poles

d) pole pitch = 3.14 * diameter * 2 / no. of poles

Answer: c

Explanation: The diameter and the number of poles are calculated. On substitution, the pole pitch of the universal commutator motor is obtained.

15. What is the formula of the pole arc of the universal commutator motor?

a) pole arc = ratio of armature axial length to armature diameter * pole pitch

b) pole arc = ratio of armature axial length to armature diameter + pole pitch

c) pole arc = ratio of armature axial length to armature diameter – pole pitch

d) pole arc = ratio of armature axial length to armature diameter / pole pitch

Answer: b

Explanation: Firstly the ratio of armature axial length to armature diameter is calculated. Next, the pole pitch is calculated and on addition of both the terms the pole arc is obtained.

This set of Design of Electrical Machines test focuses on “Motor Starters, Calculation of Resistance Steps & Design of Field Regulators”.

1. What is the function of the motor starter with respect to current?

a) to slow the low current flow

b) to prevent the low current flow

c) to allow the large current flow

d) to prevent the large current flow

Answer: d

Explanation: There are varied type of starters which vary among themselves according to the function. The main function is to prevent the excessive current at the starting.

2. What is the work of the starter with respect to the mechanical stress?

a) to allow large mechanical stress

b) to restrict large mechanical stress

c) to allow small mechanical stress

d) to restrict small mechanical stress

Answer: b

Explanation: The main function of the starter is to prevent the excessive current at the starting. The other work of the starter is to restrict the large mechanical stress from acting on the machines.

3. What is the relation of the current with the starting torque in the starter concept?

a) the starter should restrict current to prevent low starting torque

b) the starter should restrict current to produce high starting torque

c) the starter should send current to prevent low starting torque

d) the starter should send current to produce high starting torque

Answer: d

Explanation: The main function of the starter is to prevent excessive current flow. At the same time it should allow current to produce good or high starting torque.

4. When does the starter take up liquid rheostat?

a) when the resistance can be varied heavily

b) when the resistance can be varied gradually

c) when the resistance cannot be varied

d) when the resistance should not be varied

Answer: b

Explanation: The starter actually picks up either liquid rheostat or metallic resistance during its operation. The starter takes up liquid rheostat during the situation where the resistance is gradually varied.

5. The starter take up metallic resistance when the resistance should not be varied in steps.

a) true

b) false

Answer: b

Explanation: The starter actually picks up either liquid rheostat or metallic resistance during its operation. The starter takes up metallic resistance when the resistance is to be varied in steps.

6. What happens when the starter takes up metallic resistance?

a) voltage fluctuates from high to low

b) voltage fluctuates between fixed upper and lower limits

c) current fluctuates from high to low

d) current fluctuates between fixed upper and lower limits

Answer: d

Explanation: The starter actually picks up either liquid rheostat or metallic resistance during its operation. The starter takes up metallic resistance when the resistance is to be varied in steps. When taken with resistance steps, the current fluctuated between upper and lower limits.

7. What is the product of the ratio of the current and the useful flux per pole?

a) product of ratio of current and useful per pole =  / 

b) product of ratio of current and useful per pole =  / 

c) product of ratio of current and useful per pole =  / 

d) product of ratio of current and useful per pole =  / 

Answer: d

Explanation: First the product of useful flux per pole due to upper limit current and lower limit current and then the product of useful flux per pole due to lower limit current and upper limit current. On substitution the product of ratio of current and useful per pole is obtained.

8. How many machines are considered in the calculation of the resistance steps?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 machines considered in the calculation of the resistance steps. They are starters for dc shunt motors, starters for dc series motors, starters for three phase slip ring induction motor.

9. What is the concept of notching operation?

a) process of decreasing the voltage

b) process of increasing the efficiency

c) process of cutting out the resistance

d) process of adding on the resistance

Answer: c

Explanation: The process of notching operation occurs In the dc shunt motors. The concept of notching operation means the cutting out the resistance.

10. What happens in the dc shunt motor when the notching process occurs?

a) flux remains constant

b) speed remains constant

c) voltage remains constant

d) current remains constant

Answer: b

Explanation: Notching operation means cutting out the resistance. During the process of notching speed remains constant.

11. What is the formula of the ratio of the lower limit to upper limit of current with respect to the resistance?

a) lower limit of current / upper limit of current = motor resistance / resistance to limit the starting current

b) lower limit of current / upper limit of current =  1/number of resistance

c) lower limit of current / upper limit of current =  1/number of resistance

d) lower limit of current / upper limit of current = motor resistance * resistance to limit the starting current

Answer: b

Explanation: The motor resistance, resistance to limit the starting current and number of resistance is calculated. On substitution the ratio of lower limit to upper limit of current is obtained.

12. What is the relation of the ratio of rotor current limits and the ratio of lower limit to upper limit current?

a) ratio of rotor current limits = ratio of lower limit to upper limit current

b) ratio of rotor current limits > ratio of lower limit to upper limit current

c) ratio of rotor current limits < ratio of lower limit to upper limit current

d) no relation between ratio of rotor current limits and ratio of lower limit to upper limit current

Answer: a

Explanation: The ratio of rotor current limits and the ratio of lower limit to upper limit current are first calculated. The ratio of rotor current limits is approximately equal to the ratio of lower limit to upper limit current.

13. How many machines are considered for the design of field regulators for dc machines?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: Two machines are considered for the design of field regulators for dc machines. They are the shunt generators and shunt motor.

14. What is the first step in the design of the field regulators for dc machines?

a) calculation of the resistance of section

b) calculation of the total field circuit resistance

c) calculation of the field circuit resistance

d) resistance to be inserted

Answer: b

Explanation: 2 machines are considered in the design of field regulators for dc machines. They are the shunt generators and shunt motor. The first step is the calculation of the field circuit resistance.

15. What is the second step in the design of the field regulators for dc machines?

a) calculation of the resistance of section

b) calculation of the total field circuit resistance

c) calculation of the field circuit resistance

d) resistance to be inserted

Answer: d

Explanation: 2 machines are considered in the design of field regulators for dc machines. They are the shunt generators and shunt motor. The second step is the calculation of the resistance to be inserted.

16. What is the last step involved in the design of field regulators for shunt generators?

a) calculation of the resistance of section

b) calculation of the total field circuit resistance

c) calculation of resistance of section

d) resistance to be inserted

Answer: c

Explanation: 2 machines are considered in the design of field regulators for dc machines. They are the shunt generators and shunt motor. The last step is the calculation of resistance of section.

17. What is the first step in the design of the field regulators for dc motor?

a) number of sections

b) shunt field circuit resistance

c) shunt field resistance

d) resistance of step

Answer: a

Explanation: 2 machines are considered in the design of field regulators for dc machines. They are the shunt generators and shunt motor. The first step is the calculation of number of sections.

18. What is the plot of the magnetization curve?

a) field current in the y axis vs voltage in x axis

b) field current in the x axis vs voltage in y axis

c) armature current in the y axis vs voltage in x axis

d) armature current in the x axis vs voltage in y axis

Answer: b

Explanation: The magnetization curve is the curve which is used to obtain the various values required in the design of field regulators. The curve is the plot of field current in x axis and voltage in y axis.

This set of Design of Electrical Machines Quiz focuses on “Computer Aided Design & Computer Aided Design of Transformers”.

1. When was the computer aided design introduced and who was the founder?

a) 1950, Heroz

b) 1959, Heroz

c) 1959, Veinott

d) 1956, Veinott

Answer: b

Explanation: The commonly accepted papers to machine design were introduced in the year 1959. It was devised by the author Heroz et al.

2. How many commonly accepted papers are present in the machine design?

a) 3

b) 2

c) 4

d) 5

Answer: b

Explanation: There are 2 commonly accepted papers devised by Heroz et al. They are analytic method and synthesis method.

3. What is the concept of analysis method?

a) the choice of dimension alone is made by designer and provided to computer

b) the choice of dimension and materials are made by designer and provided to computer

c) the choice of dimension, materials and types of construction are made by designer and provided to computer

d) the choice of types of construction are made by designer and provided to computer

Answer: c

Explanation: The analysis method is one of the 2 commonly accepted approaches. The choice of dimension, materials, and types of construction are made by designer and provided to computer.

4. How many different approaches are present in the computer aided design?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: There are 4 types of different approaches present in the computer aided design. They are analysis method, synthesis method, hybrid method, optimisation method.

5. It is fairly easy to program and to use and understand the analysis method.

a) True

b) False

Answer: a

Explanation: Out of the 4 different approaches, analysis method is one commonly used methodology. It is fairly easy to program and to use and understand the analysis method.

6. What happens in the synthesis method?

a) The computer takes its own values

b) Desired performance is given as input to the computer

c) Logical instructions are incorporated in the program

d) Desired performance is given as input along with the logical instruction being incorporated in the program

Answer: d

Explanation: The synthesis method is one of the 4 different approaches in the computer aided design. The desired performance is given as input along with the logical instruction being incorporated in the program.

7. What is the hybrid method of computer aided design?

a) advanced analysis method

b) advanced synthesis method

c) combination of advanced and synthesis method

d) different method

Answer: c

Explanation: They hybrid method is one of the 4 different approaches in the computer aided design. The hybrid method is the combination of advanced and synthesis method.

8. How many transformers are considered in the power system?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are 2 types of transformers used in the power system. They are power transformer and distribution transformer.

9. How many design procedures are present in the design of transformers?

a) 10

b) 9

c) 11

d) 8

Answer: c

Explanation: There are 11 design steps present in the computer aided design of transformers. They are core design, window dimension design, yoke design, overall design, low voltage winding design, high voltage winding design, resistance calculation, leakage reactance calculation, loss calculation, efficiency calculation, no load current calculation.

10. What is the symbol used for the maximum flux density in computer aided designing?

a) Bm

b) Bf

c) Bfd

d) Bmf

Answer: a

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for maximum flux density in computer aided design is Bm.

11. What is the symbol used for the number of turns in the secondary winding?

a) T

b) Tsw

c) Ts

d) Tws

Answer: c

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for number of turns in secondary winding in computer aided design is Ts.

12. What is the symbol used for the mean diameter of the HV/LV winding?

a) Dm

b) Dmp

c) Dms

d) Dmp/Dms

Answer: d

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for mean diameter of the HV/LV winding in computer aided design is Dmp/Dms.

13. What is the symbol used for the resistance referred to HV winding?

a) RR

b) RRw

c) RRp

d) RRhw

Answer: c

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for resistance referred to the HV winding in computer aided design is RRp.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Dc Motor & Synchronous Machine”.

1. How many design factors are involved in the CAD design of DC motors?

a) 5

b) 6

c) 7

d) 4

Answer: b

Explanation: There are 6 types of design factors involved in the CAD design of DC motors. They are main dimension, armature winding, design of field winding, design of commutator, design of interpoles, losses and efficiency.

2. What is the symbol used for the terminal voltage in dc motor design?

a) V

b) Vs

c) Vt

d) Tv

Answer: c

Explanation: Computer aided design is one of the design algorithms made use of in the design of various machines. The symbol used for the terminal voltage is Vt.

3. What is the symbol used for the slot per pole arc in dc motor design?

a) sppa

b) spa

c) ssa

d) sp

Answer: b

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for slot per pole arc in dc motor design is spa.

4. What is the symbol used for the flux density produced in the pole body of dc motor design?

a) fp

b) fdp

c) fyp

d) fypd

Answer: c

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for flux density produced in the pole body of dc motor design is fyp.

5. What is the symbol used for the total mmf required for the teeth?

a) AT

b) ATt

c) Tm

d) Tmt

Answer: b

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for total mmf required for the teeth is ATt.

6. What is the symbol used for the pitch of commutator segments in the dc motor?

a) Pc

b) Pcs

c) BETAc

d) Ret

Answer: c

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for pitch of commutator segments is BETAc.

7. How many design steps are involved in the design of main dimensions of synchronous machine?

a) 10

b) 9

c) 7

d) 8

Answer: a

Explanation: There are 10 steps in the flowchart of design of the main dimensions of synchronous machines. There are 8 design data in the design of the synchronous machine using CAD.

8. What is the number of design steps involved in the length of air gap?

a) 7

b) 8

c) 9

d) 10

Answer: d

Explanation: There are 8 design data involved in the design of synchronous machine using CAD. There are 10 steps in the flowchart of the length of air gap in the CAD design of synchronous machine.

9. What is the number of design steps involved in the stator?

a) 6

b) 7

c) 8

d) 9

Answer: b

Explanation: There are 8 design data involved in the design of synchronous machine using CAD. There are 7 steps in the flowchart of the stator design in the CAD design of synchronous machine.

10. What is the number of design steps involved in the number of slots?

a) 4

b) 5

c) 6

d) 7

Answer: c

Explanation: There are 8 design data involved in the design of synchronous machine using CAD. There are 6 steps in the flowchart of the number of slots in the CAD design of synchronous machine.

11. What is the symbol used for the armature mmf per pole?

a) AT

b) ATM

c) Amp

d) ATa

Answer: d

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for the armature mmf per pole is Ata.

12. What is the symbol used for the width of the ventilating duct?

a) Wv

b) Nw

c) Wvd

d) Wvds

Answer: b

Explanation: Computer aided design in one of the design algorithms made use of to design various devices. The symbol used for width of the ventilating ducts is Nw.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Design of Machines, Limitations in Design”.

1. What does design mean?

a) creative physical relationship of theoretical concepts

b) producing the hardware

c) idea made into a drawing

d) idea laid out as a thesis

Answer: a

Explanation: Design is one of the important steps in the manufacturing of a product. It involves the process of producing the creative physical relationship of theoretical concepts.

2. What are the factors which are employed in the engineering design?

a) application of science

b) application of science and technology

c) application of science and innovation

d) application of science, technology and innovation

Answer: d

Explanation: Design is the process of producing the creative physical relationship of theoretical concepts. The application of science, technology and innovation is what called as engineering design.

3. How many considerations are present in the design?

a) 6

b) 4

c) 5

d) 3

Answer: d

Explanation: There are 3 considerations present in the design of the machine. They are cost, durability, compliance with performance criteria as laid down in specifications.

4. What is the condition of the good design?

a) machine having operating life of 5-10 years

b) machine having operating life of 10-15 years

c) machine having operating life of 15-20 years

d) machine having operating life of 20-30 years

Answer: d

Explanation: The machine having operating life span of 20-30 years is one of the signs of good design. This type of machines have initial cost to be low.

5. What are the factors which are important in design when it comes to the machines used in power systems?

a) reliability

b) durability

c) cost

d) reliability and durability

Answer: d

Explanation: The machines used in the power system are given weightage to the reliability and durability. Thus the cost of the machines are not taken into account.

6. How many parts are used in the design of the machines?

a) 2

b) 3

c) 4

d) 5

Answer: d

Explanation: There are 5 parts involved in the design of the machine. They are magnetic circuit, electric circuit, dielectric circuit, thermal circuit, mechanical parts.

7. How many limitations are present in the design of the machine?

a) 7

b) 8

c) 9

d) 10

Answer: c

Explanation: There are basically 9 limitations in the design of the machines. They are saturation, temperature rise, insulation, efficiency, mechanical parts, commutation, power factor, consumer specification, standard specification.

8. How does the saturation levels limit the design?

a) saturation levels decrease the flux density

b) saturation levels increase the flux density

c) saturation levels provides no flux density

d) saturation levels provide very high flux density

Answer: d

Explanation: The maximum allowable flux density to be used is determined by the saturation levels of ferromagnetic materials. The saturation levels provides high flux density which results in high cost.

9. What happens if the insulating material is operated beyond the maximum allowable temperature?

a) the insulation gets damaged

b) the insulation is boosted

c) the insulating materials peels off

d) the lifetime is drastically reduced

Answer: d

Explanation: The temperature rise is one of the limitations present in the design of machines. The insulating material when operated beyond the maximum temperature leads to the lifetime being drastically reduced.

10. How many types of stresses are present in the machine?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are basically 3 stresses available in the machine. They are electrical stress, mechanical stress and thermal stress.

11. What is the relationship between efficiency and the operating costs?

a) efficiency is directly proportional to the square of the operating cost

b) efficiency is directly proportional to the operating costs

c) efficiency is indirectly proportional to the operating costs

d) efficiency is indirectly proportional to the square of the operating costs

Answer: c

Explanation: Efficiency is one of the limitations in the design of the machines. The efficiency should be as high as possible to reduce the initial cost.

12. The mechanical parts design is very much important in which type of machines?

a) low speed machines

b) high speed machines

c) medium speed machines

d) low voltage machines

Answer: b

Explanation: The design of mechanical parts is very important in the high speed machines. It is so important because to reduce the mechanical stresses which occurs more in high speed machines.

13. What is the relationship between power factor, current and conductor sizes?

a) poor power factor leads to small amount of current and hence high conductor sizes should be used

b) high power factor leads to small amount of current and hence small conductor sizes should be used

c) poor power factor leads to high amount of current and hence high conductor sizes should be used

d) high factor leads to small amount of current and hence high conductor sizes should be used

Answer: c

Explanation: The power factor is one of the limitations in the design of the machine. The poor power factor leads to high amount of current and in turn large conductor sizes have to be used.

This set of Design of Electrical Machines MCQs focuses on “Modern Trends in Machine Design & Magnetic Leakage”.

1. What are the subjects to which the design of electrical machines is compared to?

a) science and maths

b) maths and art

c) maths

d) science and art

Answer: d

Explanation: The design of electrical machines is basically compared to as a science. It is also compared to as an art because of the designing.

2. How many design problems are present according to the modern trends in design of electrical machines?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 design problems involved in the modern trends in the design. They are electromagnetic design, mechanical design and thermal design.

3. What is one of the major aspects in the modern day design?

a) design machines to form a group

b) design machines to provided integrated systems

c) design machines to provide a single system

d) design machine to form multiple group with interconnection

Answer: c

Explanation: The major aspect of the modern trends is that the design of machines should be many in number. At the same time, all the machines should form a part of a single system.

4. How are the machines sometimes designed with respect to ratings?

a) design 2 machines with different rating

b) design a series of machines with different ratings to fit into a single frame size

c) design a series of machines with same rating to fit into a single frame size

d) design 2 machines with the same rating

Answer: b

Explanation: The machines are sometimes being designed as a series of machines with different ratings. They are done so to fit into a single frame size.

5. What are the factors which are considered when the optimal solution involves iterations wherein the values of variables are changed?

a) performance

b) cost constraint

c) performance or cost constraint

d) performance and cost constraint

Answer: d

Explanation: The optimal solution involves iteration wherein the values of variables are changed. This is done to satisfy both the performance and cost constraints.

6. The computer aided design is one of the modern techniques which is used to provide accurate and comprehensive design.

a) true

b) false

Answer: a

Explanation: The computer aided design is one of the most modern techniques of designing various machines. It is very easy, accurate, easily changeable and a comprehensive design procedure.

7. What is the relation between reluctance, flux and mmf of the machine?

a) low reluctance, less flux leakage, high mmf

b) low reluctance, high flux leakage, high mmf

c) high reluctance, high flux leakage, low mmf

d) high reluctance, less flux linkage, high mmf

Answer: a

Explanation: The designer’s problem is to provide a path of low reluctance to that of comparatively little flux leaks away. In order to compensate for the flux leakage, high mmf is maintained.

8. How should the air gaps be present in the magnetic circuit according to length and cross section?

a) low length, low cross section

b) high length, high cross section

c) high length, low cross section

d) low length, high cross section

Answer: d

Explanation: The air gaps should be present in the magnetic circuit in the low length. The air gap should have maximum cross section to reduce the reluctance.

9. What is the function of the leakage flux?

a) contributes to the transfer of energy

b) contributes to the conversion of energy

c) contributes to both the conversion of energy and transfer of energy

d) does not contributes to the conversion or transfer of energy

Answer: d

Explanation: The leakage flux produced due to the low reluctance path does not transfer the energy produced. The leakage flux also does not convert the energy in the system.

10. How many factors does the leakage flux affect?

a) 6

b) 7

c) 8

d) 9

Answer: c

Explanation: There are 8 factors which are affected by the leakage flux. They are performance of rotating machines and transformers, excitation demands of salient pole machines, leakage reactance of windings, forces between windings, voltage regulation of ac generator and transformers, commutation condition in dc machines, stray load losses, circulating current in transformer tank walls.

11. In the B-H magnetization curve, the flux density occupies the x axis.

a) true

b) false

Answer: b

Explanation: In B-H curve, the flux density occupies the y axis always. In the magnetization curve, magnetic field intensity is plotted.

12. What is the formula of the leakage coefficient?

a) leakage coefficient = total flux * useful flux

b) leakage coefficient = total flux / useful flux

c) leakage coefficient = useful flux / total flux

d) leakage coefficient = total flux + useful flux

Answer: b

Explanation: The total flux is the addition of the useful flux and the leakage flux is first calculated. On substitution the leakage coefficient can be obtained.

This set of Design of Electrical Machines Questions and Answers for Campus interviews focuses on “Determination of Iron Losses & Slot Leakage”.

1. What is the other name for the iron loss?

a) field loss

b) armature loss

c) winding loss

d) core loss

Answer: d

Explanation: When ferromagnetic materials are subjected to a flux in a fixed direction in space and having a magnitude varying in time, losses are produced in the machine. This is called iron loss or core loss.

2. How many types of iron losses are present?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are basically 2 losses present under the core loss or iron loss. They are hysteresis loss and the eddy current loss.

3. What is the formula to obtain the hysteresis loss devised by Steinmetz?

a) hysteresis loss = hysteresis coefficient / frequency of magnetization *  steinmetz coefficient

b) hysteresis loss = hysteresis coefficient * frequency of magnetization /  steinmetz coefficient

c) hysteresis loss = hysteresis coefficient * frequency of magnetization *  steinmetz coefficient

d) hysteresis loss = 1/hysteresis coefficient * frequency of magnetization *  steinmetz coefficient

Answer: c

Explanation: The hysteresis coefficient along with the frequency of magnetization and maximum flux density is calculated first. Next the Steinmetz coefficient is calculated and on substitution it gives the hysteresis loss.

4. How are the eddy current losses in the machine reduced?

a) by using conducting materials

b) by using magnetic materials

c) by using insulating materials

d) by laminating the core

Answer: d

Explanation: Eddy current loss is one type of loss present under the iron or core loss. It can be reduced by laminating the core.

5. What is the relation between resistivity, magnetizing mmf and magnetizing current?

a) high resistivity, low magnetizing mmf, high magnetizing current

b) high resistivity, high magnetizing mmf, high magnetizing current

c) low resistivity, high magnetizing mmf, high magnetizing current

d) low resistivity, high magnetizing mmf, low magnetizing current

Answer: b

Explanation: The high resistivity materials are made use of to reduce the eddy current loss. The high resistivity materials have high magnetizing mmf which in turn leads to high magnetizing current.

6. What must be done in order to take additional iron losses into account in dc machines?

a) iron loss obtained from iron loss curve is multiplied by 1.2-1.4

b) iron loss obtained from iron loss curve is multiplied by 1.4-1.6

c) iron loss obtained from iron loss curve is multiplied by 1.3-1.5

d) iron loss obtained from iron loss curve is multiplied by 1.15-1.25

Answer: b

Explanation: In order to take additional losses into account the iron loss obtained from the iron loss curves is multiplied into 1.4-1.6 in dc machines and synchronous machines. The iron loss obtained from the iron loss curves is multiplied into 1.2-1.4 in induction motors.

7. What is the value of constant ‘a’ in the core part of the ac machines?

a) 4.7

b) 6.5

c) 6.7

d) 2.3

Answer: a

Explanation: The value of constant ‘a’ in the core part of the ac machines is 4.7 and that in the teeth part is 6.5. The value of constant ‘a’ in the core part of the dc machines is 6.5 and that in the teeth part is 2.3.

8. What are the equations that are to be satisfied by flux patterns?

a) Poisson equation

b) Laplace equation

c) Laplace and Poisson equation

d) Gauss equation

Answer: c

Explanation: The flux patterns must satisfy the Poisson equation within the section of the conductors. The flux patterns must satisfy the Laplace equation in the other parts.

9. How many assumptions can be used for the slot leakage calculation?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 assumption made in the slot leakage calculation. They are the current in the slot conductors is uniformly distributed, the leakage path is straight across the slot and around the iron at the bottom and the permeance of air paths is only considered.

10. What are the factors the slot leakage permeance will depend upon?

a) shape of the slot

b) arrangement of winding in the slot

c) number of windings

d) shape of the slot and arrangement of winding in the slot

Answer: d

Explanation: The slot leakage permeance depends upon the shape of the slots. The slot leakafe permeance also depends upon the arrangement of winding in the slot.

11. What is the formula of the permeance of the strip in the conductor portion?

a) permeance of the strip = permeability in air * area of the flux path * length of flux path

b) permeance of the strip = permeability in air / area of the flux path * length of flux path

c) permeance of the strip = permeability in air * area of the flux path / length of flux path

d) permeance of the strip =1 / permeability in air * area of the flux path * length of flux path

Answer: c

Explanation: The permeability in air, area of the flux path and the length of flux path is first calculated. On substitution, the permeance of the strip is obtained.

12. What is the formula of the effective permeance of conductor portion?

a) effective permeance of conductor portion = 1/total flux linkages * total turns * total mmf

b) effective permeance of conductor portion = total flux linkages / total turns * total mmf

c) effective permeance of conductor portion = total flux linkages * total turns * total mmf

d) effective permeance of conductor portion = total flux linkages * total turns / total mmf

Answer: b

Explanation: The total flux linkages, total turns and total mmf is first calculated. On substitution the effective permeance of conductor portion is obtained.

13. Which machine incorporates the usage of the closed slots?

a) dc motors

b) synchronous motors

c) induction motors

d) special motors

Answer: c

Explanation: The closed slots are being made use often in the small induction motors. The closed slots are used to increase the leakage reactance which in turn decreases the starting current.

14. The leakage flux on the top of the slot is through iron, which is called ‘bridge’?

a) true

b) false

Answer: a

Explanation: The leakage flux on the top of the slot is through iron, which is called ‘bridge’. Therefore the specific permeance of this path depends upon the permeance of iron which in turn depends upon the degree of saturation and hence upon the relative permeability of iron.

This set of Design of Electrical Machines Multiple Choice Questions & Answers  focuses on “Pulsation Losses & Types of Leakages”.

1. What happens in the rotating electrical machine?

a) armatures are slotted

b) armatures are slotted and results in the movement of rotor

c) the rotor remains stationary

d) the rotor slots are rotating

Answer: b

Explanation: In the rotating electrical machine, the armatures are slotted and there is movement of rotor. Due to the movement of the rotor there are rapid changes of local gap reluctance.

2. What is pulsation losses?

a) flux pulsations are caused due to the slotted armature

b) flux pulsations are caused due to the rotation of machine

c) flux pulsation occurs due to rotor slots are rotating

d) flux pulsations occur due to the change in reluctance

Answer: d

Explanation: In the rotating electrical machine, the armatures are slotted and there is movement of rotor which leads to changes of local gap reluctance. This change leads to flux pulsations which causes additional losses called pulsating losses.

3. In which machine part/parts does the pulsation loss occurs?

a) teeth

b) pole face

c) conductors

d) teeth and pole faces

Answer: d

Explanation: The pulsation losses occurs in the teeth of the machine. The pulsation losses also occur in the pole faces.

4. How are the pulsation losses aggravated?

a) if the air gap is small compared with slot openings

b) if the air gap is reduced

c) if the air gap is increased

d) if the air gap is made larger than the slot openings

Answer: a

Explanation: The pulsation losses occurs in the pole faces and teeth. The pulsation losses are aggravated if the air gap is small compared with slot openings.

5. The slotting produces harmonic fields which cause high frequency losses near the gap surface.

a) true

b) false

Answer: a

Explanation: The pulsation losses are aggravated if the length of air gap is small than the slot openings. The slotting produces harmonic fields which cause high frequency losses near the gap surface.

6. In which machine is the pulsation losses considerable?

a) synchronous motors

b) induction motors

c) dc shunt motors

d) dc series motors

Answer: b

Explanation: The pulsation losses are aggravated if the length of air gap is small than the slot openings. The pulsation losses although are considerable in the induction motors.

7. What are the factors the permeance depends upon in the zigzag leakage?

a) relative position of stator

b) relative position of rotor

c) relative position of stator and rotor

d) stored energy at any position

Answer: c

Explanation: The permeance for the path of zigzag leakage will depend upon the relative position of stator. The permeance for the path of zigzag leakage will depend upon the relative position of rotor.

8. What is the formula for the stored energy at any position?

a) stored energy at any position = mmf per slot 2 * permeance in a particular position

b) stored energy at any position = 2 * mmf per slot 2 * permeance in a particular position

c) stored energy at any position = 1/2 * mmf per slot 2 * permeance in a particular position

d) stored energy at any position = 1/3 * mmf per slot 2 * permeance in a particular position

Answer: c

Explanation: First the mmf per slot along with the permeance in a particular position is calculated. On substitution, the stored energy at any position is calculated.

9. What is the formula for the zigzag permeance?

a) zigzag permeance = average width of the rotor tooth / (1/2 * mmf per slot 2 )

b) zigzag permeance = average width of the rotor tooth * (1/2 * mmf per slot 2 )

c) zigzag permeance = 1/average width of the rotor tooth *(1/2 * mmf per slot 2 )

d) zigzag permeance = average width of the rotor tooth *(1/2 / mmf per slot 2 )

Answer: a

Explanation: The average width of the rotor tooth is first calculated along with the mmf per slot. On substitution, the zigzag permeance is found out.

10. What is the formula of the zigzag specific permeance?

a) zigzag specific permeance = average width of the rotor tooth * length / (1/2 * mmf per slot 2 )

b) zigzag specific permeance = average width of the rotor tooth / length * (1/2 * mmf per slot 2 )

c) zigzag specific permeance = average width of the rotor tooth * length * (1/2 * mmf per slot 2 )

d) zigzag specific permeance =1/ average width of the rotor tooth * length * (1/2 * mmf per slot 2 )

Answer: b

Explanation: The average width of the rotor tooth along with the length and mmf per slot is calculated. On substitution the zigzag specific permeance is obtained.

11. What are the factors the overhang leakage reactance is obtained?

a) length of the overhang

b) diameter of the overhang

c) shape of the overhang

d) length of the overhang along with the shape of the overhang

Answer: d

Explanation: The overhang leakage reactance depends upon the length of the overhang. It also depends upon the shape of the overhang.

12. The overhang leakage reactance depends on the degree of saturation in the ferromagnetic parts.

a) true

b) false

Answer: a

Explanation: The overhang leakage reactance depends on the length of the overhang and the shape of the overhang. The overhang leakage reactance also depends upon the degree of saturation in the ferromagnetic parts.

13. What is the relation between the overhang specific permeance and the slot pitch?

a) overhang specific permeance is directly proportional to the slot pitch

b) overhang specific permeance is indirectly proportional to the slot pitch

c) overhang specific permeance is directly proportional to the square of the slot pitch

d) overhang specific permeance is indirectly proportional to the square of the slot pitch

Answer: b

Explanation: The overhang specific permeance relation obtained is an empirical relation. In the empirical relation, the overhang specific permeance is indirectly proportional to the slot pitch.

This set of Design of Electrical Machines Multiple Choice Questions & Answers focuses on “Specific Permeance & Calculation of Magnetizing Current”.

1. What is specific permeance?

a) specific permeance is product of permeance of unit length and depth of field

b) specific permeance is ratio of permeance of unit length and depth of field

c) specific permeance is the permeance per unit length

d) specific permeance is the permeance per unit pole

Answer: c

Explanation: Specific permeance is defined as permeance per unit length. It is also known as the depth of field.

2. What is the formula of the specific permeance?

a) specific permeance = permeability in air * ∫small change in width + length

b) specific permeance = permeability in air * ∫small change in width/length

c) specific permeance = permeability in air * ∫small change in width * length

d) specific permeance = 1/permeability in air * ∫small change in width * length

Answer: b

Explanation: The permeability in air, small change in width and length is calculated first. On substitution the specific permeance is calculated.

3. What is the assumption made in the calculation of the specific permeance?

a) voltage is kept constant

b) current is kept constant

c) mmf is kept constant

d) speed is kept constant

Answer: c

Explanation: The mmf is kept constant over all the flux tubes. The mmf should be kept constant when the integration is carried out during the calculation of specific permeance.

4. What is the formula of the effective permeance?

a) effective permeance = effective flux/total mmf

b) effective permeance = effective flux/mmf of air gap

c) effective permeance = effective flux * total mmf

d) effective permeance = effective flux * mmf of air gap

Answer: a

Explanation: The effective flux and the total mmf is first calculated. On substitution of the values the effective permeance is calculated.

5. What is the formula of the flux dividing into infinitesimal parts?

a) flux dividing into infinitesimal parts = mmf producing the flux / permeance of infinitesimal part

b) flux dividing into infinitesimal parts = mmf producing the flux * permeance of infinitesimal part

c) flux dividing into infinitesimal parts = mmf producing the flux + permeance of infinitesimal part

d) flux dividing into infinitesimal parts = mmf producing the flux – permeance of infinitesimal part

Answer: b

Explanation: The mmf producing the flux and the permeance of infinitesimal part is calculated. On substitution the flux dividing into infinitesimal parts is calculated.

6. What is the relation between the specific permeance of a differential path and the length?

a) specific permeance of a differential path is directly proportional to the length

b) specific permeance of a differential path is indirectly proportional to the length

c) specific permeance of a differential path is directly proportional to the square of the length

d) specific permeance of a differential path is indirectly proportional to the square of the length

Answer: b

Explanation: The specific permeance is defined as the permeance per unit length. It is indirectly proportional to the length.

7. How many factors does the value of the magnetizing current depends upon?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 factors upon which the magnetizing current depends upon. They are total mmf required, number of turns in the exciting winding and upon the way in which the winding is distributed.

8. What is the formula for the magnetizing current?

a) magnetizing current = total mmf * number of turns

b) magnetizing current = total mmf / number of turns

c) magnetizing current = total mmf + number of turns

d) magnetizing current = total mmf – number of turns

Answer: b

Explanation: First the total mmf is calculated along with the number of turns of the magnetizing winding. Then on substitution the magnetizing current is obtained.

9. What is the formula for the rms value of the magnetizing current?

a) rms value of the magnetizing current = maximum magnetizing current / peak factor

b) rms value of the magnetizing current = maximum magnetizing current * peak factor

c) rms value of the magnetizing current = maximum magnetizing current + peak factor

d) rms value of the magnetizing current = maximum magnetizing current – peak factor

Answer: a

Explanation: The maximum value of the magnetizing current and the peak factor are calculated first. On substitution, the rms value of the magnetizing current is obtained.

10. What is the relation of the type of winding with the flux linkage?

a) in distributed windings the flux does not link with all the turns

b) in distributed windings the flux links with all the turns

c) in concentrated windings the flux links with all the turns

d) in concentrated windings the flux does not link with all the turns

Answer: a

Explanation: The magnetizing current is actually calculated for the concentrated windings and the distributed windings. In the distributed windings, the flux does not link with all the turns.

11. What is the relation of the magnetizing current with the turns per phase?

a) magnetizing current is directly proportional to the turns per phase

b) magnetizing current is directly proportional to the square of the turns per phase

c) magnetizing current is indirectly proportional to the turns per phase

d) magnetizing current is indirectly proportional to the turns per phase

Answer: c

Explanation: The magnetizing current is actually calculated for the concentrated windings and the distributed windings. In the distributed windings, magnetizing current is indirectly proportional to the turns per phase.

12. The plot of the flux density distribution curve is between the interpolar axis consisting of the flux density and the angle difference between phases.

a) true

b) false

Answer: a

Explanation: The flux density distribution curve is used to calculate the magnetizing current in the non sinusoidal flux distribution of the distributed windings. The curve is between the flux density and the phase angle.

This set of Design of Electrical Machines online test focuses on “Field Form, Harmonic Analysis of Flux Distribution Curve”.

1. How is the flux distributed in the field form?

a) to reduce the high voltage

b) to reduce the high current

c) to reduce the harmonics

d) to keep the total reluctance low

Answer: d

Explanation: The flux in passing from poles into the armature, does not confine itself over the pole arc but spreads out over the entire pole pitch. The flux will distribute itself in the air gap in such a way that the total reluctance is minimum.

2. What does the flux distribution curve determine in the ac machine?

a) waveshape of voltage

b) waveshape of current

c) waveshape of power

d) commutation conditions

Answer: a

Explanation: The flux distribution curve in the ac machine determines the waveshape of the voltage. In the dc machine the flux distribution curve determines the commutation conditions.

3. How many techniques are used to plot the field form in salient pole machines?

a) 2

b) 3

c) 4

d) 5

Answer: a

Explanation: There are 2 techniques which are used to plot the field form in salient pole machines. They are carter’s fringe curves and the flux plotting by method of curvilinear squares.

4. What is the formula of the flux density in the gap at a distance ‘x’ from the centre of the pole?

a) flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap

b) flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole / length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap

c) flux density in the gap at a distance ‘x’ from the centre of the pole = length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole / maximum flux density in air gap

d) flux density in the gap at a distance ‘x’ from the centre of the pole = 1/ length of air gap at the centre of pole * length of air gap at a distance ‘x’ from the centre of the pole * maximum flux density in air gap

Answer: b

Explanation: The length of air gap at the centre of pole, the length of air gap at a distance ‘x’ from the centre of the pole and the maximum flux density in air gap is calculated. On substitution the flux density in the gap at a distance ‘x’ from the centre of the pole.

5. The plot between carter’s coefficient and the relative flux density is the carter’s fringe curve.

a) true

b) false

Answer: a

Explanation: Carter’s fringe curves is one of the techniques used in the plotting of the field form in the salient pole machines. The plot between carter’s coefficient and the relative flux density is called as carter’s fringe curve.

6. What Is the formula of the permeance of the flux tube considering unit depth in the flux plotting technique?

a) permeance of the flux tube = permeability in the air * mean width of flux tube * mean length of flux tube

b) permeance of the flux tube = permeability in the air / mean width of flux tube * mean length of flux tube

c) permeance of the flux tube = permeability in the air * mean width of flux tube / mean length of flux tube

d) permeance of the flux tube =1/ permeability in the air * mean width of flux tube * mean length of flux tube

Answer: c

Explanation: The permeability in the air, mean width of the flux tube and the mean length of flux tube is calculated first. On substitution the permeance of the flux tube is obtained.

7. How many rules are to be followed while the flux plotting by method of curvilinear squares?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: The flux lines leave and enter from surfaces, bounding the gap, at right angles if it is assumed that iron has infinite permeability as compared with air and then the flux and equipotential lines intersect at right angles. The flux and the equipotential lines are so drawn that each flux is divided into equal number of curvilinear squares.

8. How many factors does the flux distribution in the rotating machines depend on?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: The flux distribution in the rotating machines depends upon 3 factors. They are shape of pole, the distribution of field windings and the load condition.

9. How should the flux distribution be in the case of ac machines?

a) sinusoidal

b) rectangular

c) square

d) circular

Answer: a

Explanation: The flux distribution in the case of ac machines should be sinusoidal. The flux distribution in the case of dc machines should be rectangular.

10. How should the air gap and the fringing effects be if the field form of a salient pole machine is rectangular?

a) air gap under the pole arc is not constant, fringing effects are considered

b) air gap under the pole arc is constant, fringing effects are considered

c) air gap under the pole arc is constant, fringing effects are not considered

d) air gap under the pole arc is not constant, fringing effects are not considered

Answer: c

Explanation: The field form of a salient pole machine is rectangular if the air gap under the pole arc is not constant. The field form of a salient pole machine is rectangular if the fringing effects are not considered.

11. What series is used to analyze the field form?

a) z-series

b) fourier series

c) fourier transform

d) z-transform

Answer: b

Explanation: The field form of a salient pole is rectangular is the air gap under pole arc is uniform and if fringing effects are neglected. The field form can be analyzed for its harmonic contents with the help of Fourier series.

12. What happens if the field form is symmetrical about the pole axis?

a) north and south pole of a machine are similar

b) no harmonics

c) no cosine terms

d) north and south pole of a machine are similar, no harmonics, no cosine terms

Answer: d

Explanation: If the field form is symmetrical about the pole axis the north and south pole of a machine are similar. As the poles are similar, no harmonics and no cosine terms and the constant term is zero.

13. What Is the formula for the amplitude of the fundamental curve?

a) amplitude of the fundamental curve = 1.27 * flux density in the air gap * cosine 

b) amplitude of the fundamental curve = 1.27 / flux density in the air gap * cosine 

c) amplitude of the fundamental curve = 1.27 * flux density in the air gap / cosine 

d) amplitude of the fundamental curve = 1/1.27 * flux density in the air gap * cosine 

Answer: a

Explanation: The flux density in the air gap is calculated along with the cosine of the phase angle divided by 2. On substitution, the amplitude of the fundamental curve is calculated.

This set of Design of Electrical Machines online quiz focuses on “Relation Between Rating, Dimensions, Variation of Output of Rotating Machines”.

1. How many factors are present in the relation between rating and dimension of rotating machines?

a) 6

b) 7

c) 8

d) 9

Answer: c

Explanation: There are 8 factors present in the relation between rating and dimension of rotating machines. They are symbols, main dimensions, total loadings, specific loadings, output equation, factors affecting size of rotating machine, choice of specific magnetic loading, choice of specific electric loading.

2. What is the formula of the total magnetic loading?

a) total magnetic loading = total flux around the armature periphery at the air gap

b) total magnetic loading = total number of ampere conductors around the armature for periphery

c) total magnetic loading = number of poles/armature total flux

d) total magnetic loading = total flux/number of poles

Answer: a

Explanation: There are 2 types of loadings in the total loadings which are total magnetic loading and total electric loading. The total magnetic loading is the total flux around the armature periphery at the air gap.

3. What is the formula of the specific electric loading?

a) specific electric loading = total armature ampere conductors * armature periphery at air gap

b) specific electric loading = total flux around the air gap / area of flux path at the air gap

c) specific electric loading = total armature ampere conductors / armature periphery at air gap

d) specific electric loading = total flux around the air gap * area of flux path at the air gap

Answer: c

Explanation: The total flux around the air gap and the area of flux path at the air gap is calculated. On substitution, the specific electric loading is calculated.

4. How many terms can be used to express the output of a machine?

a) 3

b) 4

c) 5

d) 6

Answer: b

Explanation: There are 4 terms that can be used to express the output of a machine. They are main dimension, specific magnetic loading, specific electric loading, speed.

5. How many factors affecting the size of rotating machines?

a) 3

b) 4

c) 5

d) 2

Answer: d

Explanation: There are 2 factors that affect the size of rotating machines. They are speed of the machines and the output coefficient of the machines.

6. How many factors are used to determine the specific magnetic loading?

a) 2

b) 3

c) 4

d) 5

Answer: b

Explanation: There are 3 factors are used to determine the specific magnetic loading. They are maximum flux density in iron parts of machine, magnetizing current and core losses.

7. How many factors are used to determine the specific electric loading?

a) 2

b) 3

c) 4

d) 5

Answer: c

Explanation: There are 4 factors used to determine the specific electric loading. They are temperature rise, cooling coefficient, size of machine, current density.

8. What is the relation between the output and the flux density?

a) flux density is directly proportional to the output

b) flux density is indirectly proportional to the output

c) flux density is directly proportional to the square of the output

d) flux density is indirectly proportional to the square of the output

Answer: a

Explanation: Flux density is one of the factors used in the determination of the specific magnetic loading. The flux density is directly proportional to the output.

9. What is the relation between output and the specific electric loading?

a) specific electric loading is directly proportional to the output

b) specific electric loading is indirectly proportional to the output

c) specific electric loading is directly proportional to the square of the output

d) specific electric loading is indirectly proportional to the square of the output

Answer: a

Explanation: The choice of specific electric loading is one of the factors under the relation between rating and dimensions of rotating machines. The specific electric loading is directly proportional to the output.

10. What is the relation of the specific electric loading and the diameter?

a) specific electric loading is directly proportional to the diameter

b) specific electric loading is indirectly proportional to the diameter

c) specific electric loading is directly proportional to the square of the diameter

d) specific electric loading is indirectly proportional to the square of the diameter

Answer: b

Explanation: The specific electric loading is one of the factors under the relation between rating and dimension of rotating machines. The specific electric loading is indirectly proportional to the diameter.

11. What is the formula of the I 2 R loss?

a) I 2 R loss = number of conductors / copper loss in each conductor

b) I 2 R loss = number of conductors + copper loss in each conductor

c) I 2 R loss = number of conductors – copper loss in each conductor

d) I 2 R loss = number of conductors * copper loss in each conductor

Answer: d

Explanation: The number of conductors and copper loss in each conductor is first calculated. On substitution the I 2 R loss is obtained.

12. What is the formula of the efficiency of the machine?

a) efficiency = output / output + losses

b) efficiency = output * output + losses

c) efficiency = output – output + losses

d) efficiency = output + output + losses

Answer: a

Explanation: The output is first calculated from the running of the machine and the different losses are calculated. On substitution the efficiency of the machine is calculated.

13. The fractional horsepower motors have efficiency of order of 98%.

a) true

b) false

Answer: b

Explanation: The fractional horse power motors have efficiency of order of 60% or less. The large turbo-alternators have efficiency of order of 98% because efficiency increases with an increase in dimensions.

This set of Design of Electrical Machines Questions and Answers for Entrance exams focuses on “Separation of D and L, Standard Frames”.

1. What is the formula of the core length?

a) core length = pole length * pole pitch

b) core length = pole length / pole pitch

c) core length = pole length + pole pitch

d) core length = pole length – pole pitch

Answer: b

Explanation: Core length is one of the factors used in the separation of D and L of the DC machines. It is the ratio of the pole length to the pole pitch.

2. What is the range of the ratio of the pole length to pole pitch?

a) 0.60-0.70

b) 0.64-0.72

c) 0.65-0.75

d) 0.70-0.80

Answer: b

Explanation: The minimum value of the ratio of the pole length to pole pitch is 0.64. The maximum value of the ratio of the pole length to pole pitch is 0.72.

3. What is the maximum value of the peripheral speed that should not exceed?

a) 25 m per sec

b) 20 m per sec

c) 30 m per sec

d) 35 m per sec

Answer: c

Explanation: The peripheral speed of the machine should not exceed 30 m per sec. If it exceeds it leads to the damage of the windings. The banding wires on the overhang have to be made so strong if the peripheral speed exceeds 30 m per sec.

4. What is the range of the pole length to pole pitch ratio for obtaining good power factor in induction motors?

a) 1.5-2

b) 1.3–1.8

c) 1.0-1.25

d) 1.1-1.6

Answer: c

Explanation: The range of the pole length to pole pitch ratio should be between 1.0-1.25 for the good power factor. The range of the pole length to pole pitch ratio should be between 1.5-2 for minimum cost.

5. What is the relationship between the diameter and number of poles?

a) diameter is directly proportional to the number of poles

b) diameter is indirectly proportional to the number of poles

c) diameter is directly proportional to the square of the number of poles

d) diameter is indirectly proportional to the square of the number of poles

Answer: a

Explanation: The number of poles is one of the factors involved in the separation of D and L in the synchronous machines. The number of poles is directly proportional to the diameter.

6. The core length should be high to obtain high short circuit ratio.

a) true

b) false

Answer: a

Explanation: The core length is directly proportional to the short circuit ratio. If high short circuit ratio is required, high core length should be used.

7. What is the factor which is by modern motors which are into a series of standard frames?

a) current rating

b) voltage rating

c) power rating

d) output rating

Answer: c

Explanation: The modern motors for industrial applications is concentrated into a series of standard frames. The standard frames are used to cover a wide range of power ratings.

8. What is the frame used in the standard frames?

a) set of standard values

b) mechanical structure required to house a rotor of given outside diameter

c) mechanical structure required to house a stator of given outside diameter

d) mechanical structure required to house a stator of given outside length

Answer: c

Explanation: The frame is a mechanical structure required to house a stator of given outside diameter. The frame also includes housing the bearings, end covers, terminal box and maximum length.

9. How can the variation in ratings be obtained?

a) alternative windings

b) alternative core lengths

c) alternative pole pitch

d) alternative pole length

Answer: b

Explanation: The frame is a mechanical structure required to house a stator of given outside diameter, housing the bearings, end covers, terminal box and maximum length. The variation in the ratings can be obtained using alternative core lengths.

10. What is the value of the alternative core lengths below which variation in rating can be obtained?

a) 0.7 * length

b) 0.4 * length

c) 0.3 * length

d) 0.7 * length or 0.5 * length

Answer: d

Explanation: The variation in the ratings can be obtained using alternative core lengths. The value of the alternative core lengths below 0.5 * length or 0.7 * length provides the variation in the ratings.

11. What is the range of the central heights used in the standard frames?

a) 50-1000 mm

b) 90-1000 mm

c) 100-1000 mm

d) 56-1000 mm

Answer: d

Explanation: The modern motors of small and medium sizes are built with standard frame sizes as specified in IEC 72. The IEC 72 lists a coherent range of main dimensions with the central heights ranging between 56-1000 mm.

12. The frame size is designated by a number which is its centre height expressed in mm.

a) true

b) false

Answer: a

Explanation: The IEC 72 lists a coherent range of main dimensions with the central heights ranging between 56-1000 mm. The frame size is designated by a number which is its centre height expressed in mm. The frame designated 180 has a centre height of 180 mm.