Related InterviewSolutions

• What is the formula for the total winding area?(a) total winding area = number of turns * area of each conductor * space factor(b) total winding area = number of turns / area of each conductor * space factor(c) total winding area = number of turns * area of each conductor / space factor(d) total winding area = 1/number of turns * area of each conductor * space factorThe question was asked during a job interview.My question is taken from Design of Magnet Coils topic in division Design of Electromagnets of Design of Electrical Machines
• 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 + currentThis question was posed to me during an interview.Query is from Design of Magnet Coils topic in chapter Design of Electromagnets of Design of Electrical Machines
• What is the value of the resistance temperature coefficient of copper?(a) 0.00393 per °C(b) 0.0040 per °C(c) 0.00383 per °C(d) 0.00373 per °CThe question was posed to me by my school teacher while I was bunking the class.The above asked question is from Design of Magnet Coils in chapter Design of Electromagnets of Design of Electrical Machines
• 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 voltageThis question was posed to me during an interview.I want to ask this question from Design of Magnet Coils topic in section Design of Electromagnets of Design of Electrical Machines
• What is the value of the resistivity temperature coefficient of copper?(a) 0.017 ohm per m per mm^2(b) 0.0173 ohm per m per mm^2(c) 0.01734 ohm per m per mm^2(d) 0.0175 ohm per m per mm^2I have been asked this question in homework.This intriguing question originated from Design of Magnet Coils in section Design of Electromagnets of Design of Electrical Machines
• What is the ambient temperature of the magnet coils?(a) 10°C(b) 15°C(c) 20°C(d) 25°CThe question was asked in a national level competition.I'm obligated to ask this question of Design of Magnet Coils topic in chapter Design of Electromagnets of Design of Electrical Machines
• 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 coilI had been asked this question in an interview for internship.Question is taken from Design of Magnet Coils in section Design of Electromagnets of Design of Electrical Machines
• 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 coilThe question was posed to me in final exam.My question is based upon Design of Magnet Coils in portion Design of Electromagnets of Design of Electrical Machines
• 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 windingI got this question in an interview for internship.I need to ask this question from Design of Magnet Coils topic in portion Design of Electromagnets of Design of Electrical Machines
• 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 coilThis question was addressed to me during an interview.The doubt is from Design of Magnet Coils in division Design of Electromagnets of Design of Electrical Machines