MCQOPTIONS
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MCQOPTIONS
This section includes 18 Mcqs, each offering curated multiple-choice questions to sharpen your Design Electrical Machines knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
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 |
| Answer» B. 0.0040 per °C | |
| 2. |
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 |
| Answer» D. 0.0175 ohm per m per mm2 | |
| 3. |
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 |
| Answer» D. area of the conductors = mmf per coil * resistivity of conductor / length of mean turn * terminal voltage | |
| 4. |
What is the ambient temperature of the magnet coils? |
| A. | 10°C |
| B. | 15°C |
| C. | 20°C |
| D. | 25°C |
| Answer» D. 25°C | |
| 5. |
WHAT_IS_THE_AMBIENT_TEMPERATURE_OF_THE_MAGNET_COILS??$ |
| A. | 100C |
| B. | 150C |
| C. | 200C |
| D. | 250C |
| Answer» D. 250C | |
| 6. |
What is the value of the resistivity temperature coefficient of copper?$ |
| A. | 0.017 ohm per m per mm<sup>2</sup> |
| B. | 0.0173 ohm per m per mm<sup>2</sup> |
| C. | 0.01734 ohm per m per mm<sup>2</sup> |
| D. | 0.0175 ohm per m per mm<sup>2</sup> |
| Answer» D. 0.0175 ohm per m per mm<sup>2</sup> | |
| 7. |
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 |
| Answer» D. area of the conductors = mmf per coil * resistivity of conductor / length of mean turn * terminal voltage | |
| 8. |
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 |
| Answer» D. total winding area = 1/number of turns * area of each conductor * space factor | |
| 9. |
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 |
| Answer» C. total number of turns = mmf per coil ‚Äö√Ñ√∂‚àö√ë‚àö¬® current | |
| 10. |
What_is_the_value_of_the_resistance_temperature_coefficient_of_copper? |
| A. | 0.00393 per 0C |
| B. | 0.0040 per 0C |
| C. | 0.00383 per 0C |
| D. | 0.00373 per 0C |
| Answer» B. 0.0040 per 0C | |
| 11. |
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 |
| Answer» C. inner cylindrical heat dissipating surface = length of mean turn / axial length of coil | |
| 12. |
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 |
| Answer» E. | |
| 13. |
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 |
| Answer» D. total heat dissipating surface = 2 * length of mean turn * depth of winding * axial length of coil | |
| 14. |
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) |
| Answer» B. length of mean turns = 3.14 / (inside diameter of coil + depth of windings) | |
| 15. |
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 |
| Answer» D. cross winding area = axial length of coil / depth of winding | |
| 16. |
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 |
| Answer» B. depth of winding = mean diameter of coil + inner diameter | |
| 17. |
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 |
| Answer» C. outside diameter = mean diameter ‚Äö√Ñ√∂‚àö√ë‚àö¬® 2*depth of winding | |
| 18. |
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 |
| Answer» B. mean diameter = inside diameter of coil ‚Äö√Ñ√∂‚àö√ë‚àö¬® outer diameter of coil / 2 | |