MCQOPTIONS
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MCQOPTIONS
This section includes 19 Mcqs, each offering curated multiple-choice questions to sharpen your Electrical Machines knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
For obtaining maximum current when we conduct the ‘Slip Test’ on a synchronous machine, its armature field will align along __________ |
| A. | d-axis |
| B. | q-axis |
| C. | 45° to d-axis |
| D. | 45° to q-axis |
| Answer» C. 45° to d-axis | |
| 2. |
for a synchronous motor, its power factor __________ |
| A. | improves with increase in excitation and may even become leading at higher excitations |
| B. | is independent of its excitation |
| C. | decreases with increase in excitation |
| D. | increases with loading for a given excitation |
| Answer» B. is independent of its excitation | |
| 3. |
While finding the parameters for an alternator, its voltage regulation is found to be negative. Then the load connected to it must be __________ |
| A. | capacitive nature |
| B. | inductive nature |
| C. | resistive nature |
| D. | can not be judged based on voltage regulation |
| Answer» B. inductive nature | |
| 4. |
For a 400 V, 3 phase alternator gives an open circuit voltage of 380V and armature current of 38A at a field current of 20A. Then the synchronous reactance of the machine is? |
| A. | 10 ohm |
| B. | 38/√3 |
| C. | 19 ohm |
| D. | 1.9 ohm |
| Answer» B. 38/√3 | |
| 5. |
Mark the correct expression for alternator working at a lagging power factor.a) tanψ = (IaXq – Vtsinθ)/(Vtcosθ – Ia*ra)b) tanψ = (IaXq + Vtsinθ)/(Vtcosθ – Ia*ra)c) tanψ = (IaXq + Vtsinθ)/(Vtcosθ + Ia*ra)d) tanψ = (IaXq – Vtsecθ)/(Vtcosecθ – Ia*r |
| A. | tanψ = (IaXq – Vtsinθ)/(Vtcosθ – Ia*ra) |
| B. | tanψ = (IaXq + Vtsinθ)/(Vtcosθ – Ia*ra) |
| C. | tanψ = (IaXq + Vtsinθ)/(Vtcosθ + Ia*ra) |
| D. | tanψ = (IaXq – Vtsecθ)/(Vtcosecθ – Ia*ra) |
| Answer» B. tanψ = (IaXq + Vtsinθ)/(Vtcosθ – Ia*ra) | |
| 6. |
A salient pole synchronous generator has following per-unit parameter Xd = 1.2 pu, Xq = 0.8pu, ra = 0.25 ohms. Then the armature current at 0.8 lagging is? |
| A. | 1,-36.9° |
| B. | 1, 36.9° |
| C. | 1, 73° |
| D. | 1, 27° |
| Answer» B. 1, 36.9° | |
| 7. |
If the internal pf angle of a synchronous motor is 30 degree. Then the quadrature axis component of the armature is? |
| A. | 0.866 pu |
| B. | 0.5 pu |
| C. | 1.73 pu |
| D. | 0 |
| Answer» B. 0.5 pu | |
| 8. |
If the internal power factor angle of synchronous motor is 30 degree. Then the direct-axis component of the armature current will be __________ |
| A. | 0.5 pu |
| B. | 0.866 pu |
| C. | 1.73 pu |
| D. | 0 |
| Answer» B. 0.866 pu | |
| 9. |
In salient-pole machines, the armature mmf cannot be accounted for by introducing one equivalent reactance due to __________ |
| A. | non-uniform airgap |
| B. | variable resistance |
| C. | non-uniform air gap |
| D. | any of the mentioned |
| Answer» B. variable resistance | |
| 10. |
FOR_A_SYNCHRONOUS_MOTOR,_ITS_POWER_FACTOR?$ |
| A. | improves with increase in excitation and may even become leading at higher excitations |
| B. | is independent of its excitation |
| C. | decreases with increase in excitation |
| D. | increases with loading for a given excitation |
| Answer» B. is independent of its excitation | |
| 11. |
For_obtaining_maximum_current_when_we_conduct_the_‘Slip_Test’_on_a_synchronous_machine,_its_armature_field_will_align_along$# |
| A. | d-axis |
| B. | q-axis |
| C. | 45° to d-axis |
| D. | 45° to q-axis |
| Answer» C. 45¬¨¬®‚Äö√†√ª to d-axis | |
| 12. |
While finding the parameters for an alternator, its voltage regulation is found to be negative. Then the load connected to it must b? |
| A. | capacitive nature |
| B. | inductive nature |
| C. | resistive nature |
| D. | can not be judged based on voltage regulation |
| Answer» B. inductive nature | |
| 13. |
For a 400 V, 3 phase alternator gives an open circuit voltage of 380V and armature current of 38A at a field current of 20A. Then the synchronous reactance of the machine is |
| A. | 10 ohm |
| B. | 38/‚àö3 |
| C. | 19 ohm |
| D. | 1.9 ohm |
| Answer» B. 38/‚Äö√Ñ√∂‚àö‚Ć‚àö‚àÇ3 | |
| 14. |
Mark the correct expression for alternator working at a lagging power factor |
| A. | tanψ = (IaXq – Vtsinθ)/(Vtcosθ – Ia*ra) |
| B. | tanψ = (IaXq + Vtsinθ)/(Vtcosθ – Ia*ra) |
| C. | tanψ = (IaXq + Vtsinθ)/(Vtcosθ + Ia*ra) |
| D. | tanψ = (IaXq – Vtsecθ)/(Vtcosecθ – Ia*ra) |
| Answer» B. tan‚âà√¨‚àö‚Ć = (IaXq + Vtsin‚âà√≠‚Äö√†√®)/(Vtcos‚âà√≠‚Äö√†√® ‚Äö√Ñ√∂‚àö√ë‚àö¬® Ia*ra) | |
| 15. |
The internal power factor angle is given for a lagging load of a 3- phase alternator. |
| A. | ψ = δ – θ |
| B. | ψ = δ + θ |
| C. | ψ = -δ -θ |
| D. | ψ = δ -θ |
| Answer» C. ‚âà√¨‚àö‚Ć = -‚âà√≠¬¨‚Ä¢ -‚âà√≠‚Äö√†√® | |
| 16. |
A salient pole synchronous generator has following per-unit parameter Xd = 1.2 pu, Xq = 0.8pu, ra = 0.25 ohms. |
| A. | |
| B. | 1,-36.9° |
| C. | 1, 36.9° |
| Answer» B. 1,-36.9¬¨¬®‚Äö√†√ª | |
| 17. |
If the internal pf angle of a synchronous motor is 30 degree. Then the quadrature axis component of the armature is |
| A. | 0.866 pu |
| B. | 0.5 pu |
| C. | 1.73 pu |
| D. | 0 |
| Answer» B. 0.5 pu | |
| 18. |
If the internal power factor angle of synchronous motor is 30 degree. Then the direct-axis component of the armature current will be |
| A. | 0.5 pu |
| B. | 0.866 pu |
| C. | 1.73 pu |
| D. | 0 |
| Answer» B. 0.866 pu | |
| 19. |
In salient-pole machines, the armature mmf cannot be accounted for by introducing one equivalent reactance due to |
| A. | non-uniform airgap |
| B. | variable resistance |
| C. | non-uniform air gap |
| D. | any of the mentioned |
| Answer» B. variable resistance | |