# power EE test question



## sd3232 (Jan 14, 2011)

please look at attached file for question and answer, i dont get the answer, can someone explain this one to me? answer says phase to ground is equal to phase to phase, VBG=VBA=13.2kV, i thought phase to ground was 7.62kV? why is C a right answer? thanks


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## VolInGA (Jan 14, 2011)

sd3232 said:


> please look at attached file for question and answer, i dont get the answer, can someone explain this one to me? answer says phase to ground is equal to phase to phase, VBG=VBA=13.2kV, i thought phase to ground was 7.62kV? why is C a right answer? thanks


Draw it out. You should see that the phase-phase voltage will be the same as the phase-ground voltage for that leg.


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## Flyer_PE (Jan 14, 2011)

If you ground one leg of a delta system (or ungrounded wye for that matter), the magnitude of the voltage between either of the other two legs and ground will be the line voltage (VL-L).


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## Jonjo (Jan 14, 2011)

sd3232 said:


> please look at attached file for question and answer, i dont get the answer, can someone explain this one to me? answer says phase to ground is equal to phase to phase, VBG=VBA=13.2kV, i thought phase to ground was 7.62kV? why is C a right answer? thanks


sd3232 , The system is "ungrounded" the grounding point is floating for this system there is not going to draw any current also


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## sd3232 (Jan 14, 2011)

thanks guys, but its still fuzzy to me, is it 13.2kV because on delta phase to ground voltage is equal phase to phase? if some one could provide a sketch, i will keep reading the response and may be i will get it, need some time


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## bambooi (Jan 15, 2011)

sd3232 said:


> thanks guys, but its still fuzzy to me, is it 13.2kV because on delta phase to ground voltage is equal phase to phase? if some one could provide a sketch, i will keep reading the response and may be i will get it, need some time


The explaination is in the answer. There is no ground path.just like some one said above the system is floating.


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