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Blasphemy!  A quarter of the way there... And if you're in CA or PA, the clock hasn't even started yet.  

So serious! I guess you've never had Memphis BBQ....?  Dodged that one quicker than Hillary being quizzed about her rogue emails....

I love bacon!....and sausage, porkchops, ribs, bbq...

Admittedly, when I originally replied, I didn't notice your proclamation about it being an approximation.  Mea culpa.  That said, I entertained the rest cause your tones through text seemed condescending in nature.  I'm not proclaiming to know it all, nor am I the best in the world of power, but...

I stand by what I said, and nothing I've intimated is incorrect.  Why would you bring up an approximation on this board that isn't valid for this exam or for practice in industry?  I'd advise others reading your posts to take them with a pinch of salt.  Be sure to include this approximation on...

Nothing I stated was incorrect, other than my recollection from memory and the 0.85 PF approximation approach (read where I said I didn't have a copy in front of me).  It is never ok to scalar add reactance and resistance for any other reason than approximating, and even then, I'd only find it...

I do not have a copy of the NEC in front of me, but if memory serves, one of the "approximations" is only valid at 0.85 PF.  Also, this test is not about approximating! Work the problem legitimately and let the answer be "approximated" off your work, not the other way around!!  Good way to screw...

Not exactly sure what line of reasoning you used (best I can guess is a very general, inaccurate approximation), but you never add reactance and resistance in the manner you presented above.  It is a vector sum, not a scalar sum.  Hence, the Pythagorean theorem should be dinging like a doorbell.

Means nothing.  Don't get your hopes up! I'd wager they haven't even began the "grading" process.

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...

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 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...

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...

All banks are generally connected in parallel to the grid (usually Y-GND).  I may be misunderstanding you, but in AC circuits, all values of impedance (reactance or vector sum of reactance and resistance) are combined exactly like resistors in DC circuits.  Recall that impedance and...

I got 800.5 A using the pu method.  Keep in mind, the generator and XFMR are not on the same power base.  This has to be corrected, in turn, adjusting the impedance of the generator.  It seems to be somewhat of a consensus here that the answer in the book is incorrect.  I'd agree.  If someone...

No, what @TNSparky said is not exactly true, but I know what he means.  Without a source, the fault current would be goose eggs!  You have to get the generator and transformer on the same power base, then all other pu values will adjust accordingly. After obtaining all pu values, use the...

Customarily, an electromechanical 51 element (reasonable assumption) is a time over current device, hence B is the appropriate answer for the given question/scenario.

I believe you are right on all except #3.  A 51 element is a time over current device (does not look in both directions), therefore, I feel confident in B, not C, being the correct answer.  However, the 67 element is indicated to be a non-time delay (i.e. instantaneous) and is a device used for...

This problem can be solved using PF, but there is certainly no need to complicate it.  However, I will illustrate this below so that you can use this approach if you get hung up: From our power equations, we know P=VIcos(theta) Conversely, Q=VIsin(theta) and S=VI or VI* We are interested in...

Spent dang near $100 to frame mine.  Still awaiting the certificate. But hey, I'll take this wait over the one endured from about 4:00PM Oct. 28, 2016 till 7:57AM Dec. 9, 2016.   :hung-037: