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# Multiple choice question for engineering

## Set 1

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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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= (bars per pole / 2) * 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 [Reason:] 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 [Reason:] maximum current in the end ring= bars per pole / 2 * current per bar Maximum current in the end ring = (6/2)*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= (2*3.14) / (bars per pole/2*no of poles) * current per bar
b) maximum current in end ring= (2/3.14) * (bars per pole/2*no of poles) / current per bar
c) maximum current in end ring= (2*3.14) * (bars per pole/2*no of poles) * current per bar
d) maximum current in end ring= (2*3.14) / (bars per pole/2*no of poles) / current per bar

Answer: c [Reason:] 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 = (2 * current through each bar)1/2
c) maximum value of the current through each bar = (2 * current through each bar)2
d) maximum value of the current through each bar = (2 * current through each bar)1/3

Answer: b [Reason:] 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 = (bars per pole * current per bar) / (3.14*no of poles)
b) rms value of end ring current = (bars per pole * current per bar) * (3.14*no of poles)
c) rms value of end ring current = (bars per pole * current per bar) / (no of poles)
d) rms value of end ring current = (bars per pole * current per bar) / (3.14+no of poles)

Answer: a [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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.

## Set 2

1. What is the formula for the fundamental relationship for the design of ventilation system?
a) head of air inside the machine = hydrodynamic resistance * volume of air passing2
b) head of air inside the machine = hydrodynamic resistance + volume of air passing2
c) head of air inside the machine = hydrodynamic resistance – volume of air passing2
d) head of air inside the machine = hydrodynamic resistance / volume of air passing2

Answer: a [Reason:] 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 second2
b) total head produced = ∑ coefficient of hydrodynamic resistance – volume of air passing per second2
c) total head produced = ∑ coefficient of hydrodynamic resistance * volume of air passing per second2
d) total head produced = ∑ coefficient of hydrodynamic resistance / volume of air passing per second2

Answer: c [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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) 11-150C
b) 10-130C
c) 12-160C
d) 14-180C

Answer: c [Reason:] The difference of air temperatures at inlet and outlet has a minimum value of 120C. The difference of air temperatures at inlet and outlet has a maximum value of 160C.

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 [Reason:] 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 [Reason:] 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 * (1.25 – 1.5)* width of fan
b) number of blades = 3.14 / outside diameter * (1.25 – 1.5)* width of fan
c) number of blades = 3.14 * outside diameter / (1.25 – 1.5)* width of fan
d) number of blades = 3.14 * outside diameter * (1.25 – 1.5) / width of fan

Answer: c [Reason:] 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 [Reason:] 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.

## Set 3

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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 m2
b) 3-10 MN per m2
c) 4-12 MN per m2
d) 4-15 MN per m2

Answer: c [Reason:] The minimum pressure under which the field coils are consolidated is 4 MN per m2. The maximum value of the pressure under which the field coils are 12 MN per m2.

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 [Reason:] 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 [Reason:] 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 = (0.8-0.85)*exciter voltage/number of poles
b) voltage across each field coil = (0.8-0.85)*exciter voltage*number of poles
c) voltage across each field coil = (0.8-0.85)/exciter voltage*number of poles
d) voltage across each field coil = (0.8-0.85)/exciter voltage/number of poles

Answer: a [Reason:] 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.

## Set 4

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 thesis

Answer: a [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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.

## Set 5

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 [Reason:] 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 [Reason:] 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 [Reason:] 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 [Reason:] 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 * (inside diameter of coil + depth of windings)
b) length of mean turns = 3.14 / (inside diameter of coil + depth of windings)
c) length of mean turns = 3.14 * (inside diameter of coil * depth of windings)
d) length of mean turns = 3.14 + (inside diameter of coil + depth of windings)

Answer: a [Reason:] 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 * (depth of winding + axial length of coil)
d) total heat dissipating surface = 2 * length of mean turn * depth of winding * axial length of coil

Answer: c [Reason:] 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 [Reason:] 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 [Reason:] 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) 100C
b) 150C
c) 200C
d) 250C

Answer: c [Reason:] 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 200C.

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 [Reason:] 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 mm2
b) 0.0173 ohm per m per mm2
c) 0.01734 ohm per m per mm2
d) 0.0175 ohm per m per mm2

Answer: c [Reason:] The resistivity temperature coefficient of copper is first calculated at a temperature of 200C. The resistivity temperature coefficient of copper is 0.01734 ohm per m per mm2.

12. What is the value of the resistance temperature coefficient of copper?
a) 0.00393 per 0C
b) 0.0040 per 0C
c) 0.00383 per 0C
d) 0.00373 per 0C

Answer: a [Reason:] The resistance temperature coefficient of copper is calculated at a temperature of 200C. The resistance temperature coefficient of copper is 0.00393 per 0C.

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