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51.

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 – °CI had been asked this question in unit test.My query is from Temperature Rise in Transformers topic in section Design of Transformers of Design of Electrical Machines

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The correct choice is (a) 8 W per m^2 – °C

The best I can explain: 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.

52.

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 °cThe question was asked during an interview.My doubt stems from Design of Tank in chapter Design of Transformers of Design of Electrical Machines

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The CORRECT OPTION is (d) 30.8 W per °c

To elaborate: LOSS DISSIPATED by tubes by convection = 8.8 * Area of tubes

Loss = 8.8 * 3.5 = 30.8 W per °c.

53.

What is the formula for temperature rise with tubes?(a) temperature rise with tubes = total loss / dissipating surface*(12.5 + 8.8x)(b) temperature rise with tubes = total loss * dissipating surface*(12.5 + 8.8x)(c) temperature rise with tubes = total loss / dissipating surface / (12.5 + 8.8x)(d) temperature rise with tubes = total loss + dissipating surface*(12.5 + 8.8x)This question was addressed to me in a job interview.I would like to ask this question from Design of Tank in section Design of Transformers of Design of Electrical Machines

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54.

An addition of 35 percent should be made to tube area of the transformers.(a) true(b) falseI have been asked this question in an interview for job.Query is from Design of Tank topic in section Design of Transformers of Design of Electrical Machines

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Correct answer is (a) true

The 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.

55.

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 pressureThe question was asked in class test.This intriguing question comes from Design of Tank topic in division Design of Transformers of Design of Electrical Machines

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Right OPTION is (d) it can be improved by USING more effective HEADS of pressure

To elaborate: The circulation of oil is improved in tanks with tubes. It takes PLACE with the HELP of using more effective heads of pressure.

56.

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 heatThis question was posed to me in an internship interview.This intriguing question originated from Design of Tank topic in portion Design of Transformers of Design of Electrical Machines

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Right OPTION is (d) transformer SURFACE has no CHANGE with respect to dissipation of heat

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

57.

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 heatThe question was posed to me in examination.My question is taken from Design of Tank in portion Design of Transformers of Design of Electrical Machines

Answer» CORRECT answer is (b) the PROVISION of tubes is indirectly proportional to the dissipation of heat

Best 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.
58.

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 limitsThe question was asked in an interview for job.Query is from Design of Tank topic in portion Design of Transformers of Design of Electrical Machines

Answer» CORRECT choice is (d) if the temperature rise with plain TANK exceeds the SPECIFIC limits

The best I can explain: 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.
59.

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, decreaseThe question was posed to me in an online interview.My question comes from Temperature Rise in Transformers in portion Design of Transformers of Design of Electrical Machines

Answer»

Correct option is (c) increase, increase, increase

Best 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.

60.

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 outputsThe question was asked during an interview.This key question is from Temperature Rise in Transformers in chapter Design of Transformers of Design of Electrical Machines

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Correct option is (d) the plain walled tanks can accommodate for SMALL OUTPUTS but cannot accommodate for large outputs

To EXPLAIN: 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.

61.

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 – °CThis question was posed to me by my school teacher while I was bunking the class.This key question is from Temperature Rise in Transformers topic in portion Design of Transformers of Design of Electrical Machines

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Correct OPTION is (B) 6, 6.5 W per m^2 – °C

The explanation is: 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.

62.

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 / (specific heat dissipation * surface temperature)(c) temperature rise = total loss / specific heat dissipation / surface temperature(d) temperature rise = total loss * specific heat dissipation / surface temperatureI had been asked this question in homework.My doubt is from Temperature Rise in Transformers in chapter Design of Transformers of Design of Electrical Machines

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63.

What is the formula for specific heat dissipation due to convection of oil?(a) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil/height of the dissipating surface)^1/4W per m^2 – °C(b) specific heat dissipation = 40.3 / (temperature difference of the surface relative to oil/height of the dissipating surface)^1/4W per m^2 – °C(c) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil * height of the dissipating surface)^1/4W per m^2 – °C(d) specific heat dissipation = 40.3*(temperature difference of the surface relative to oil + height of the dissipating surface)^1/4W per m^2 – °CI had been asked this question in examination.I need to ask this question from Temperature Rise in Transformers topic in portion Design of Transformers of Design of Electrical Machines

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64.

The convection due to air is 10 times the convection due to oil.(a) true(b) falseI had been asked this question at a job interview.This question is from Temperature Rise in Transformers in division Design of Transformers of Design of Electrical Machines

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65.

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° CThis question was addressed to me in my homework.This interesting question is from Temperature Rise in Transformers in chapter Design of Transformers of Design of Electrical Machines

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66.

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 radiationThe question was posed to me at a job interview.This is a very interesting question from Temperature Rise in Transformers topic in portion Design of Transformers of Design of Electrical Machines

Answer» RIGHT choice is (c) CONVECTION and radiation

Explanation: When the HEAT flows from the walls of thecooler to the cooling MEDIUM, it is both convection and radiation. In transformer all 3 types of heat dissipation occurs.
67.

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) radiationThis question was posed to me in a job interview.Query is from Temperature Rise in Transformers topic in division Design of Transformers of Design of Electrical Machines

Answer»

The CORRECT answer is (B) convection

The explanation is: 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.

68.

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 convectionI had been asked this question in final exam.Question is from Temperature Rise in Transformers topic in division Design of Transformers of Design of Electrical Machines

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The correct choice is (B) CONVECTION

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.

69.

In how many ways does heat dissipation occur in transformers?(a) 2(b) 3(c) 4(d) 5I got this question in unit test.My doubt is from Temperature Rise in Transformers in chapter Design of Transformers of Design of Electrical Machines

Answer» RIGHT option is (B) 3

Easiest explanation: The HEAT dissipation takes PLACE in 3 ways in transformers. They are RADIATION, convection and conduction.
70.

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 energyI got this question in an interview.I'm obligated to ask this question of Temperature Rise in Transformers topic in portion Design of Transformers of Design of Electrical Machines

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Correct ANSWER is (d) the LOSSES are converted to thermal energy

The BEST I can EXPLAIN: In both the transformer and the rotating machines the losses are converted to thermal energy. This thermal energy causes heating of the transformer parts.

71.

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 kVThe question was asked in a national level competition.Asked question is from Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

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Correct choice is (b) upto 33 kV

The best explanation: Helical windings have a VOLTAGE of upto 15 kV. WHEREAS, the cylindrical WINDING with rectangular conductors have a voltage of upto 6 kV.

72.

The problem of temperature rise and cooling of transformers is essentially the same as that of rotating machinery.(a) true(b) falseThis question was posed to me at a job interview.My question is taken from Temperature Rise in Transformers in division Design of Transformers of Design of Electrical Machines

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Correct answer is (a) true

To explain: 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.

73.

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 AI have been asked this question during a job interview.I want to ask this question from Design of Core and Winding in division Design of Transformers of Design of Electrical Machines

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74.

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 kVAThis question was addressed to me by my college professor while I was bunking the class.I need to ask this question from Design of Core and Winding topic in chapter Design of Transformers of Design of Electrical Machines

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The correct choice is (a) 5000-10000 kVA

Easy 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.

75.

The low voltage windings are generally divided into 2 types.(a) true(b) falseThe question was asked in examination.My question comes from Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

Answer» RIGHT CHOICE is (a) true

For explanation I would say: The low voltage windings are basically divided into 2 TYPES. They are i) cylindrical winding ii) helical winding.
76.

How many total high voltage windings are present?(a) 1(b) 2(c) 3(d) 4This question was posed to me in an internship interview.My enquiry is from Design of Core and Winding in portion Design of Transformers of Design of Electrical Machines

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Right option is (c) 3

For explanation I would say: There are 3 high VOLTAGE windings present. They are i) Cylindrical WINDING, ii) Cross-over winding iii) CONTINUOUS DISC type winding.

77.

The current density in relatively better cooled outer winding is made 10 percent greater than the inner winding.(a) true(b) falseThis question was posed to me by my school principal while I was bunking the class.This intriguing question comes from Design of Core and Winding in portion Design of Transformers of Design of Electrical Machines

Answer» CORRECT CHOICE is (b) false

For 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.
78.

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 secondaryI had been asked this question in an international level competition.This interesting question is from Design of Core and Winding topic in section Design of Transformers of Design of Electrical Machines

Answer» RIGHT answer is (C) current density in PRIMARY = current density in secondary

For explanation I would say: 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.
79.

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^2The question was posed to me in a national level competition.Origin of the question is Design of Core and Winding in division Design of Transformers of Design of Electrical Machines

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The correct ANSWER is (c) 2.2-3.2 A per mm^2

The explanation is: 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.

80.

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^2I had been asked this question by my college director while I was bunking the class.I would like to ask this question from Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

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Correct choice is (d) 1.1-2.3 A PER mm^2

Easy 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.

81.

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 increasesThe question was posed to me by my college director while I was bunking the class.I want to ask this question from Design of Core and Winding in portion Design of Transformers of Design of Electrical Machines

Answer» RIGHT option is (c) current density DECREASES, LOSSES DECREASE, efficiency increases

Explanation: As the current density decreases, the losses decrease. As the losses decrease the MAXIMUM efficiency increases.
82.

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 temperatureI got this question in class test.This interesting question is from Design of Core and Winding in division Design of Transformers of Design of Electrical Machines

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The correct choice is (a) current DENSITY is directly proportional to the temperature

The EXPLANATION: As the current density increases, the temperature also increases. As the temperature increases, it can CAUSE DAMAGE to the insulation.

83.

What does the permissible current density depend upon?(a) local heating(b) efficiency(c) output power(d) local heating and efficiencyI have been asked this question by my school principal while I was bunking the class.The doubt is from Design of Core and Winding in division Design of Transformers of Design of Electrical Machines

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Right ANSWER is (d) local heating and efficiency

To ELABORATE: 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.

84.

What does the area of conductors in primary and secondary windings depend on?(a) current(b) voltage(c) power(d) current densityThis question was posed to me by my college professor while I was bunking the class.This interesting question is from Design of Core and Winding topic in chapter Design of Transformers of Design of Electrical Machines

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The CORRECT answer is (d) current DENSITY

For 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.

85.

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 voltageThe question was posed to me in homework.Question is taken from Design of Core and Winding in portion Design of Transformers of Design of Electrical Machines

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The CORRECT choice is (d) current in primary WINDING = KVA per phase * 10^3 / primary voltage

Easiest 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.

86.

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 turnI have been asked this question in quiz.Query is from Design of Core and Winding topic in portion Design of Transformers of Design of Electrical Machines

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Right answer is (b) NUMBER of turns of primary winding = Voltage of primary WINDINGS/voltage per turn

To elaborate: 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.

87.

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 circleThis question was addressed to me in an interview for internship.My question is based upon Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

Answer» RIGHT ANSWER is (a) number of steps is directly PROPORTIONAL to the area of the circumscribing CIRCLE

The best I can explain: 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.
88.

What is the net core area for three stepped transformers?(a) 0.45(b) 0.56(c) 0.6(d) 0.62This question was posed to me during an online interview.I'm obligated to ask this question of Design of Core and Winding topic in portion Design of Transformers of Design of Electrical Machines

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The correct CHOICE is (c) 0.6

The BEST I can explain: 0.45 is the net CORE AREA for the square core transformers. 0.56 is the core area for cruciform or STEPPED core transformers.

89.

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) 0.91I got this question by my college professor while I was bunking the class.Query is from Design of Core and Winding topic in portion Design of Transformers of Design of Electrical Machines

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The correct choice is (B) 0.79

To explain I WOULD say: 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.

90.

The laminations are manufactured in standard size to minimize the wastage of steel during punching of laminations.(a) true(b) falseI had been asked this question in examination.Question is from Design of Core and Winding in section Design of Transformers of Design of Electrical Machines

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The correct answer is (a) true

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.

91.

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) 0.80This question was addressed to me during an interview for a job.The above asked question is from Design of Core and Winding topic in division Design of Transformers of Design of Electrical Machines

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The CORRECT answer is (a) 0.58

For 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.

92.

Circular coils are preferred because of their electrical characteristics.(a) true(b) falseI have been asked this question in an internship interview.This intriguing question comes from Design of Core and Winding topic in division Design of Transformers of Design of Electrical Machines

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The correct answer is (B) false

The explanation is: 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.

93.

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 distributionI had been asked this question by my school teacher while I was bunking the class.Asked question is from Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

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Correct CHOICE is (c) when CIRCULAR coils are REQUIRED for high voltage distribution

Best explanation: Circular coils are required for high voltage distribution and power TRANSFORMER. When circular coils are required square and STEPPED cores are used.

94.

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) 1.5-2I have been asked this question during an interview.Enquiry is from Design of Core and Winding in portion Design of Transformers of Design of Electrical Machines

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The correct ANSWER is (c) 1.4-2

The 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.

95.

What type of core section is used for shell type transformer?(a) rectangular(b) square(c) stepped(d) cruciformI had been asked this question in quiz.My question is based upon Design of Core and Winding topic in section Design of Transformers of Design of Electrical Machines

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Right answer is (a) rectangular

The best I can explain: Shell type transformers PREFER only rectangular core section. Shell type transformer are MODERATE and LOW VOLTAGE transformer which use only rectangular core section.

96.

How many types of cores are available for core type of transformer?(a) 2(b) 3(c) 4(d) 5I had been asked this question during an online interview.My enquiry is from Design of Core and Winding in section Design of Transformers of Design of Electrical Machines

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Right answer is (b) 3

Best 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.

97.

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 increasesThe question was asked in an online quiz.This is a very interesting question from Design of Core and Winding topic in chapter Design of Transformers of Design of Electrical Machines

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The correct option is (b) core area increases and the volt/turns increases

The BEST I can explain: 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.

98.

What is the optimum number of steps for small and large transformers?(a) 5, 10(b) 10, 15(c) 6, 15(d) 1, 10I have been asked this question during an internship interview.Asked question is from Design of Core and Winding in chapter Design of Transformers of Design of Electrical Machines

Answer»

The correct OPTION is (c) 6, 15

For 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.

99.

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 stepsI have been asked this question during an online exam.The query is from Design of Core and Winding topic in division Design of Transformers of Design of Electrical Machines

Answer»

The correct ANSWER is (a) UTILIZATION factor is directly proportional to the number of core steps

Easiest explanation: The utilization factor increases with the INCREASE in the number of core steps used. This eventually increases the MANUFACTURING COST.

100.

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 coreI had been asked this question during a job interview.Question is from Design of Core and Winding topic in section Design of Transformers of Design of Electrical Machines

Answer»

Right choice is (d) utilization factor= cross sectional area / gross area of the CORE

The 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.