InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 51. |
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^2I got this question during an interview.Question is taken from Design of Rotor in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct option is (a) 3-4 A PER mm^2 |
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| 52. |
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 barsI have been asked this question during an internship interview.Query is from Design of Rotor topic in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (b) AREA per pole of DAMPER pass = 0.2 * specific electric loading * pole pitch / current density in damper bars |
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| 53. |
The mmf of the damper windings depends on the pole pitch value.(a) true(b) falseI had been asked this question by my college director while I was bunking the class.My query is from Design of Rotor in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (a) true |
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| 54. |
The damper windings are made use of in synchronous generators to reduce the oscillations and to prevent hunting.(a) true(b) falseI got this question in final exam.My question is based upon Design of Rotor in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct option is (a) true |
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| 55. |
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) 0.7-1.5I have been asked this question in an interview for job.My question is from Design of Rotor in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right answer is (b) 0.3-1.5 |
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| 56. |
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 / 0.02This question was addressed to me during an interview for a job.My question is from Design of Rotor topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» RIGHT CHOICE is (a) height of pole body = height of the WINDING + 0.02 To elaborate: 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. |
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| 57. |
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 + 0.02I got this question by my college professor while I was bunking the class.Question is taken from Design of Rotor topic in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right option is (d) RADIAL length of the POLE shoe = HEIGHT of WINDING + height of pole shoe + 0.02 |
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| 58. |
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 windingI have been asked this question during an interview.Asked question is from Design of Rotor in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (a) HEIGHT of winding = total winding AREA / depth of winding |
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| 59. |
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) 0.75I had been asked this question by my school principal while I was bunking the class.This interesting question is from Design of Rotor in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct option is (a) 0.8-0.9 |
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| 60. |
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 factorThis question was posed to me during an interview for a job.I'd like to ask this question from Design of Rotor topic in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct choice is (c) total SPACE = copper area / space FACTOR |
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| 61. |
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 windingI have been asked this question in examination.The origin of the question is Design of Rotor topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT option is (b) copper area = full load FIELD mmf / current density in the field winding The 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. |
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| 62. |
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 poleThis question was posed to me during an interview.Question is taken from Design of Rotor in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right CHOICE is (a) area of cross SECTION of pole body = 0.98 * AXIAL LENGTH of the pole * breadth of the pole |
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| 63. |
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 2.3This question was addressed to me by my school principal while I was bunking the class.This intriguing question comes from Design of Rotor in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT option is (c) 1.15 to 1.2 |
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| 64. |
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^2This question was addressed to me at a job interview.I would like to ask this question from Design of Rotor topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right answer is (b) 1.5-1.7 WB per m^2 |
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| 65. |
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 poleI have been asked this question in exam.My query is from Design of Rotor topic in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (a) flux in pole body = leakage coefficient * useful flux PER pole |
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| 66. |
What is the minimum clearance between adjacent field coils and pole drawing?(a) 14 mm(b) 15 mm(c) 13 mm(d) 12 mmThis question was addressed to me in exam.My doubt stems from Design of Field Winding in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct CHOICE is (b) 15 mm |
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| 67. |
If the temperature increases beyond the acceptable limits the depth of the winding should be decreased.(a) true(b) falseI have been asked this question in my homework.Question is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (B) false |
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| 68. |
The increase in the depth of the winding increases the heat dissipating surface.(a) true(b) falseI have been asked this question in an online interview.My question comes from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right answer is (a) true |
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| 69. |
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 coilThe question was posed to me in an online interview.The question is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (d) temperature rise = COPPER loss in each FIELD coil at 75°C * cooling coefficient of rotating field COILS / dissipating SURFACE of the coil |
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| 70. |
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 voltageThis question was addressed to me in semester exam.I would like to ask this question from Design of Field Winding in section Design of Synchronous Machines of Design of Electrical Machines |
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| 71. |
What is the formula of the resistance of the winding is calculated at 75°C?(a) resistance of the winding = (Number of field turns * pole proportion * length of mean turns of the coil) / area of the field conductors(b) resistance of the winding = (Number of field turns * pole proportion * length of mean turns of the coil) * area of the field conductors(c) resistance of the winding = (Number of field turns / pole proportion * length of mean turns of the coil) / area of the field conductors(d) resistance of the winding = (Number of field turns * pole proportion / length of mean turns of the coil) / area of the field conductorsThe question was posed to me in an interview for job.I want to ask this question from Design of Field Winding topic in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right option is (a) RESISTANCE of the winding = (Number of field TURNS * POLE proportion * length of mean turns of the coil) / AREA of the field conductors |
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| 72. |
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*(winding height * diameter of winding)(b) dissipating surface of the coil = 2*length of mean turns of the coil*(winding height / diameter of winding)(c) dissipating surface of the coil = 2*length of mean turns of the coil*(winding height + diameter of winding)(d) dissipating surface of the coil = 2*length of mean turns of the coil/(winding height * diameter of winding)I had been asked this question in exam.This question is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT option is (C) DISSIPATING surface of the coil = 2*length of mean turns of the coil*(winding height + diameter of winding) The EXPLANATION is: 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. |
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| 73. |
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 depthThis question was posed to me in final exam.This key question is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT answer is (B) winding space is indirectly PROPORTIONAL to the depth To ELABORATE: The winding space is indirectly proportional to the depth. If the winding space is LESS, then the depth is increased. |
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| 74. |
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 currentThis question was posed to me by my school principal while I was bunking the class.This is a very interesting question from Design of Field Winding in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (b) number of field TURNS = field mmf per POLE at full load / field current |
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| 75. |
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 conductorsI got this question by my college professor while I was bunking the class.I'd like to ask this question from Design of Field Winding topic in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (a) field CURRENT = current density * area of CONDUCTORS |
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| 76. |
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^2I had been asked this question during an interview.This interesting question is from Design of Field Winding in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT answer is (b) 3 to 4 A per mm^2 For explanation I would SAY: 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. |
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| 77. |
What is the approximate value of the space taken by spools, flanges, etc?(a) 15 mm(b) 10 mm(c) 12 mm(d) 20 mmI had been asked this question in semester exam.I would like to ask this question from Design of Field Winding in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT answer is (d) 20 MM Best 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. |
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| 78. |
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°CThis question was posed to me in class test.I'd like to ask this question from Design of Field Winding in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right ANSWER is (a) voltage across each FIELD coil = field current * resistance of each field at 75°C |
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| 79. |
What is the winding depth for the pole pitch of 0.1 mm?(a) 25 mm(b) 35 mm(c) 45 mm(d) 50 mmThe question was asked during an online exam.Query is from Design of Field Winding in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» CORRECT option is (a) 25 mm For 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. |
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| 80. |
The field winding should be designed for a voltage from 15-20% less than the exciter voltage.(a) true(b) falseI have been asked this question in an online interview.This intriguing question originated from Design of Field Winding in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (a) true |
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| 81. |
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, etcI have been asked this question during an interview for a job.Question is from Design of Field Winding topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right CHOICE is (d) winding HEIGHT = height of the pole – height of shoe – SPACE taken by the spool, FLANGES, etc |
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| 82. |
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 VI have been asked this question by my school principal while I was bunking the class.This interesting question is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT choice is (d) 50-400 V |
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| 83. |
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^2I have been asked this question in class test.Question is taken from Design of Field Winding topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct option is (c) 4-12 MN per m^2 |
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| 84. |
How many layers does the machine with Class F insulation consists of?(a) 2(b) 3(c) 4(d) 5The question was posed to me in examination.Asked question is from Design of Field Winding topic in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct choice is (b) 3 |
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| 85. |
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 paperThe question was asked in an interview for job.This intriguing question originated from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT choice is (c) 0.18 MM, thick epoxy treated ASBESTOS paper |
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| 86. |
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 thickI got this question in a national level competition.My doubt is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (b) epoxy resin. 4 mm THICK |
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| 87. |
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 mmThe question was asked during an online interview.Origin of the question is Design of Field Winding topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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| 88. |
Current is passed simultaneously through the conductors to raise the temperature of the field coil.(a) true(b) falseThe question was posed to me in examination.The question is from Design of Field Winding topic in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct choice is (a) true |
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| 89. |
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 mmI have been asked this question by my school teacher while I was bunking the class.I want to ask this question from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right answer is (c) 2, 018 mm |
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| 90. |
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 asbestosI had been asked this question by my college director while I was bunking the class.My doubt stems from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» RIGHT answer is (d) 10 mm thick, bakelized asbestos To explain I WOULD say: The flanges used in the Class B insulation is made up of 10 mm thickness. The flanges are made up of the bakelized asbestos. |
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| 91. |
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 shellacThis question was posed to me in an internship interview.Query is from Design of Field Winding in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct ANSWER is (d) synthetic resin VARNISH or shellac |
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| 92. |
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 mmThe question was posed to me in class test.This question is from Design of Field Winding topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right answer is (a) 6 mm |
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| 93. |
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 stripsThis question was posed to me in exam.My question is taken from Design of Field Winding topic in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT option is (c) glass covered rectangular STRIPS |
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| 94. |
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 coilsThis question was addressed to me in an online quiz.Query is from Design of Field Winding topic in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right choice is (B) wire WOUND coils |
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| 95. |
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 conductorsThis question was posed to me in an online interview.Asked question is from Armature Design in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT answer is (d) AREA of CROSS section = current per conductor / current density in the armature conductors |
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| 96. |
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^2The question was posed to me during an interview for a job.My enquiry is from Armature Design in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The correct answer is (d) 3-5 A per mm^2 |
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| 97. |
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 phaseI got this question by my college professor while I was bunking the class.This intriguing question comes from Armature Design topic in division Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right option is (C) current in each conductor = kVA * 10^3 / 3 * voltage per PHASE |
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| 98. |
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 factorI got this question during a job interview.The above asked question is from Armature Design in section Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Correct choice is (a) turns PER phase = VOLTAGE per phase * parallel paths per phase / 4.44 * flux density * frequency * winding space factor |
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| 99. |
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 coreThe question was posed to me during an interview for a job.Enquiry is from Armature Design in portion Design of Synchronous Machines of Design of Electrical Machines |
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Answer» The CORRECT answer is (a) flux PER pole = AVERAGE magnetic field * pole pitch * length of the core |
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| 100. |
The coil span should be 8.33 percent of pole pitch to obtain the maximum reduction of harmonics.(a) true(b) falseI had been asked this question in an interview for internship.I need to ask this question from Armature Design in chapter Design of Synchronous Machines of Design of Electrical Machines |
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Answer» Right option is (a) true |
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