| 1. | A three-phase generating system has an efficiency of around:
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| 2. | The windings of a three-phase alternator are separated by:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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:
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| 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 =
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| 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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