# Design of Electrical Machines Questions and Answers – Design of Magnet Coils

This set of Design of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “Design of Magnet Coils”.

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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: b
Explanation: 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
View Answer

Answer: a
Explanation: 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.
advertisement
advertisement

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
View Answer

Answer: c
Explanation: 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)
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: d
Explanation: 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.
advertisement

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
View Answer

Answer: b
Explanation: 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) 10°C
b) 15°C
c) 20°C
d) 25°C
View Answer

Answer: c
Explanation: 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 20°C.
advertisement

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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: c
Explanation: The resistivity temperature coefficient of copper is first calculated at a temperature of 20°C. 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 °C
b) 0.0040 per °C
c) 0.00383 per °C
d) 0.00373 per °C
View Answer

Answer: a
Explanation: The resistance temperature coefficient of copper is calculated at a temperature of 20°C. The resistance temperature coefficient of copper is 0.00393 per °C.

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
View Answer

Answer: b
Explanation: The mmf per coil is first calculated. Next, the current flowing through the coils is measured and on substitution gives the total number of turns.

14. 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 factor
View Answer

Answer: c
Explanation: First the number of turns is calculated along with the area of each conductor. Next, the space factor is calculated and on substitution gives the total winding area.

Sanfoundry Global Education & Learning Series – Design of Electrical Machines.

To practice all areas of Design of Electrical Machines, here is complete set of 1000+ Multiple Choice Questions and Answers.

advertisement
advertisement
Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.