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robertplant22

Camara Morning Exam #18

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The question goes as follows:

A 208V wye-connected, ungrounded sytem suffers a ground at pahse A. Waht is most nearly the line voltage at phase B?

The answer to the question is 208*SQRT(3) = 360V

however I tought that for a wye connected transformer:

VB= VBN - VCN

The explanation in the answer indicates that because phase A is grounded VB becomes

VB= VAN - VBN which results in a magnitude of 208*SQRT(3).

Can someone elavorate on this? Also if phase A is grounded would it not make sense to say that the voltage VAN = 0 ?

Thanks in advance for any input.

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The question goes as follows:

A 208V wye-connected, ungrounded sytem suffers a ground at pahse A. Waht is most nearly the line voltage at phase B?

The answer to the question is 208*SQRT(3) = 360V

however I tought that for a wye connected transformer:

VB= VBN - VCN

The explanation in the answer indicates that because phase A is grounded VB becomes

VB= VAN - VBN which results in a magnitude of 208*SQRT(3).

Can someone elavorate on this? Also if phase A is grounded would it not make sense to say that the voltage VAN = 0 ?

Thanks in advance for any input.

I also don't understand the Camara solution you've shown either. If you rewrite part of their solution and change subscripts/signs you get V = VAN + VNB (note I've changed sign and subscript order) so it appears they are defining VAB as their line voltage.

Since VAN and VNB have not changed, this seems to me like it would be 208V. Maybe I'm missing something as well but I don't see where they are going. Have you searched on here for any errata?

Can someone elavorate on this? Also if phase A is grounded would it not make sense to say that the voltage VAN = 0 ?

You have to remember that this is ungrounded system so grounding phase A doesn't change VAN. If the neutral point was originally grounded and you then grounded the A phase, you would obviously have a fault.

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