1. | A three-phase generating system has an efficiency of around: |
2. | The windings of a three-phase alternator are separated by: |
3. | Figure shows the stator winding arrangement of a 4 pole, three phase alternator: The alternator has a 24-slot stator and is wound with 24 coils in sets of: A. | 4 coils per pole per phase |
B. | 3 coils per pole per phase |
C. | 2 coils per pole per phase |
D. | 1 coil per pole per phase |
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4. | One of advantages of a three-phase system over a single phase system is that the power is more constant so that the torque of a rotating machine is more constant. The more constant torque results in: A. | much more vibration from the machine |
B. | no vibration from the machine |
C. | excessive load current from the machine |
D. | much less vibration from the machine |
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5. | Figure shows the waveform for a three-phase AC supply: The displacement between the voltage is: |
6. | In a three-phase system, where the three voltage sources are, connected to feed a three-phase load, the phase sequence is important for rotating machinery, because the: A. | direction of current flow will increase |
B. | direction of rotation will be affected |
C. | machinery could draw excessive load current |
D. | power factor will be leading instead of lagging |
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7. | One method of forming a three-phase system is to connect the three similar ends of the windings together at one point. This type of connection is called a: |
8. | Look at the following diagram: The three-phase winding shown in figure have been connected in the: B. | double series configuration |
C. | parallel configuration |
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9. | Look at the following diagram: The three-phase winding shown in figure has: |
10. | In a three-phase star-connected system the line current equals: A. | Ö3 x the phase current |
C. | 1.732 x the phase current |
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11. | In a three-phase star-connected system the line voltage equals: B. | the phase voltage / 1.732 |
C. | Ö3 x the phase voltage |
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12. | Figure shows one method of determining the line voltage of a three-phase star connected system: In the diagram, the line voltage phasor VA-B is equal to: |
13. | In a star connected three-phase system, the line voltages: A. | lead the phase voltages by 30°E |
B. | lag the phase voltages by 30°E |
C. | lead the phase voltages by 60°E |
D. | lag the phase voltages by 60°E |
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14. | Figure shows the phasor diagram method of determining the line voltage of a three-phase star connected system: In the diagram, the line voltage phasor VA-B: A. | lags the phase voltage VA by 60°E |
B. | leads the phase voltage VA by 30°E |
C. | leads the phase voltage VA by 60°E |
D. | lags the phase voltage VA by 30°E |
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15. | Figure shows the phasor diagram of a three-phase star connected system in which phase “C” phase has mistakenly been connected with the connections reversed: The waveforms of the three phase voltages then have a displacement of 120º E between A and B, and: A. | 90º E between A and C and also between C and B |
B. | 180º E between A and C and also between C and B |
C. | 60º E between A and C and also between C and B |
D. | 120º E between A and C and also between C and B |
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16. | One method of forming a three-phase system is to connect the dissimilar similar ends of the windings together. This type of connection is called a: |
17. | Look at the following diagram: The three-phase winding shown in figure have been connected in the: B. | double series configuration |
C. | parallel configuration |
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18. | In a three-phase delta-connected system the line current equals: B. | Ö3 x the phase current |
C. | 1.732 x the phase voltage |
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19. | In a three-phase delta-connected system the line voltage equals: A. | Ö3 x the phase voltage |
B. | the phase voltage / 1.732 |
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20. | Figure shows one a Delta connected 400 V three-phase system in which one phase has been mistakenly connected with reversed connections: In the diagram, the voltage between the open connections will be: |
21. | Figure shows a connection diagram for a three-phase delta connected load: The current in line 2, equals the: A. | phasor difference of phase currents IA and IB |
B. | phasor sum of phase currents IA and IB |
C. | arithmetic sum of phase currents IA and IB |
D. | arithmetic difference of phase currents IA and IB |
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22. | Figure shows a phasor diagram for determining the line current in a three-phase delta connected load: In the diagram, the line current phasor IL2: A. | leads the phase current IA by 60°E |
B. | leads the phase current IA by 30°E |
C. | lags the phase voltage VA by 60°E |
D. | lags the phase voltage VA by 30°E |
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23. | Electrical power can be transmitted using low voltage and high currents. Higher currents result in: A. | lower transmission losses |
B. | very efficient transmission |
C. | higher voltage outputs |
D. | higher transmission losses |
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24. | In economic terms, the higher the voltage used for power transmission systems, the: A. | lower the cost of installing and maintaining the transmission lines |
B. | higher the cost of installing and maintaining the transmission lines |
C. | higher the cost of the current along the transmission lines |
D. | lower the cost of installing underground transmission lines |
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25. | Look at the following diagram: Transmission system voltages are far higher than the voltages required by the average consumer, therefore the voltage is: A. | stepped up to a suitable value using transformers |
B. | stepped down to a suitable value using transformers |
C. | kept the same right through to the customer’s circuits |
D. | converted to DC for the customer’s machines |
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26. | The distribution system shown in figure is a: A. | switched wiring electrical reticulation system |
B. | sub-transmission electrical receiving system |
C. | single wire earth return distribution system |
D. | sixteen-kilovolt wiring earthed resister system |
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27. | Figure shows a 4-wire distribution system with the star point of the phase windings connected to earth: One advantage of this system is that: A. | only one voltage is available to the consumer |
B. | the neutral conductor is not used by the consumer |
C. | the active conductors are connected to the earth |
D. | two voltages are available to the consumer |
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28. | For a balanced load, the current in a neutral conductor of a three-phase, four-wire system is equal to: B. | the current in any phase |
C. | the arithmetic sum of the phase currents |
D. | the phasor difference of any two line currents |
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29. | For any type of load, the current in a neutral conductor of any three-phase, four-wire system is equal to: A. | the phaser difference of the line currents |
B. | minus the phasor sum of the line currents |
C. | minus the arithmetic sum of the line currents |
D. | the arithmetic difference of the line currents |
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30. | The currents in the lines of a three-phase four-wire system are: A phase—8 amperes at unity power factor B phase—10 amperes at 0.866 leading power factor C phase—10 amperes at 0.866 lagging power factor. In this case the current flowing in the neutral conductor will be approximately: |
31. | Figure shows the readings obtained when checking the voltages of a faulty 230/400 V three-phase four-wire system: The most likely cause of the fault will be a: A. | faulty three-phase alternator |
B. | faulty three-phase pole-top transformer |
C. | high resistance load on one phase |
D. | broken neutral conductor |
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32. | The power consumed by a three-phase load can be determined using the formula: |
33. | A three-phase 400 V motor draws 10 A at 0.8 lagging power factor. The power is consumed by this motor will be approximately: |
34. | Figure shows the one wattmeter method of measuring three-phase power on a three-phase four-wire system: One advantage of this method is that: A. | only one reading is needed |
B. | it is suitable for fluctuating loads |
C. | only one wattmeter is required |
D. | a neutral connection is not required |
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35. | Figure shows the two-wattmeter method of measuring power in a three-phase three-wire system: One advantage of using this method is that: A. | a neutral conductor is available for the meters |
B. | the meters only need to measure the phase voltages |
C. | three-terminal wattmeters can be used |
D. | the power factor can be obtained for balanced loads |
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36. | The power consumed by a three-phase motor has been measured using the two-wattmeter method. At full load, the two wattmeters gave readings of 6 kW and 3 kW. At this load, the power factor of the motor will be: |
37. | Figure shows the three-wattmeter method of measuring three-phase power in a three-phase four-wire system: To find the total power, the following calculation must be made using the readings: PTOTAL = |
38. | Figure shows the three-wattmeter method of measuring three-phase power in a three-phase four-wire system. This method is suitable for: C. | balanced and unbalanced loads |
D. | measuring the power factor with balanced loads |
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39. | One advantage of using three-phase digital wattmeters is as well as the power, they can show the value of such things as: A. | the current in each phase |
B. | the VAR value for each phase |
C. | frequency and total harmonic distortion |
D. | all of the given answers |
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40. | In power stations operating under normal conditions, the values of voltage and current are generally too high and unwieldy to be used directly with portable instruments. To avoid these high-energy circuits when measuring voltage, current and power, in these cases: A. | potential transformers and current transformers must be used |
B. | only high voltage ammeter and voltmeters must be used |
C. | digital voltmeters can be used with a high voltage extension arm |
D. | 275 kV ammeters and voltmeters must be used for 11 kV circuits |
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