Design Electrical Machines Design Magnet Coils Online Exam Quiz
Design Electrical Machines Design Magnet Coils GK Quiz. Question and Answers related to Design Electrical Machines Design Magnet Coils. MCQ (Multiple Choice Questions with answers about Design Electrical Machines Design Magnet Coils
What is the value of the resistance temperature coefficient of copper?
Options
A : 0.00393 per °C
B : 0.0040 per °C
C : 0.00383 per °C
D : 0.00373 per °C
What is the value of the resistivity temperature coefficient of copper?
Options
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
What is the ambient temperature of the magnet coils?
Options
A : 10°C
B : 15°C
C : 20°C
D : 25°C
What is the formula for the area of the conductors of the magnet coils?
Options
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
What is the formula of the cross winding area of the magnet coils?
Options
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
What is the formula for depth of winding of the magnet coils?
Options
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
What is the formula of the inner cylindrical heat dissipating surface?
Options
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
What is the formula for the length of mean turn of magnet coils?
Options
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)
What is the formula for the mean diameter of the magnet coils?
Options
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
What is the formula for the outer cylindrical heat dissipating surface of the magnet coils?
Options
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
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