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:

V_{B}= V_{BN} - V_{CN }

The explanation in the answer indicates that because phase A is grounded V_{B }becomes

V_{B}= V_{AN} - V_{BN }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 V_{AN }= 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 = V

_{AN} + V

_{NB} (note I've changed sign and subscript order) so it appears they are defining V

_{AB} as their line voltage.

Since V

_{AN} and V

_{NB} 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 V_{AN }= 0 ?

You have to remember that this is

ungrounded system so grounding phase A doesn't change V

_{AN}. If the neutral point was originally grounded and you then grounded the A phase, you would obviously have a fault.