I just noticed if I don't use the generator in the MVA method, I get the same answer as my PU method answer, which is an answer of 882. I'm totally confused now.
Yes, the generator counts. You have to sum all impedances up to the point of the fault. However, as previously mentioned, the generator and XFMR are on different power bases. This has to be resolved, whereby the Znew for G = 0.05, not 0.1. This comes from the change of base formula, Znew=Zold(Vold/Vnew)^2(Snew/Sold)=0.1(11/11)^2(25/50). Get everything in terms of the XFMR base cause it has a "let through" of 25 MVA, and the line is connected to the XFMR of the same base (i.e. Zbase=(33kV)^2/(25 MVA)).
Why is Vnew 11? Shouldn't it be 33 to match the voltage base of the of the transformer and line?
The Vnew/old would come into affect if you had a generator producing at, say, 13.2 kV, but the XFMR was rated at 12.47 kV on that side (i.e. the primary side connected to the generator). In this case, we aren't given any info about the voltage of the generator, therefore, it's safe to assume it is producing voltage at the rated input of the XFMR.
820 is the closest and it wouldn't be unheard of to see something similar on the test. Remember, this exam will test your gut (and comprehension) as much or more than your subject knowledge and engineering competency. That said, I didn't have/use the Camara text when I took it, and it seems the rest of you should be cautious about some topics it presents. Ideally, this very problem, for one, and I've seen other erroneous representations on this very forum (power factor correction comes to mind... how the hell do you mess that up? - these are the softballs)
Good luck! Just use what works for you, and as I stated previously, I prefer the pu method cause errors will jump off the page at you. If you have fractional impedances but end up with an odd ball 10 pu, you know you've made a mistake somewhere. With that said, it wouldn't be unheard of to have a fault magnitude less than the full load current of the XFMR, depending on impedances, fault location and fault type. However, you most likely won't see it, but it is possible and fair game.
