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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