TSM, Beanland, et al...
Boy, is my face red! I erred as well! Yes, it is ‘B’. My error was accepting the X/R-rule given in the PE Exam solution… that is, per-unit impedances were proportional to power… instead, the ratio must be in per-unit! (See Note 2 after the solution!)
Now, here is the correct solution using my Gi.Fi.E.S. method (pronounced Jiffy’s), where:
o Gi means ‘Given’
o Fi “ ‘Find’
o E “ ‘Equation’
o S “ ‘Solution’ The procedure follows:
Find:
Short-circuit current, ISC, at transformer secondary bus!
Equation(s):
If Capacity and Impedance were known for both the Gen’r and Xfmr, then the problem is easily solved, using the MVA, Ohm, or per-unit method! Unfortunately, for this problem, only ZT is known. But, ZG can be found using a little-known relationship that I call the X/R-Rule (see Note 2)! It is properly stated as Eq 1:
o Eq 1: Zg = Sb / Sg
o Eq 2, Ssc = Sbase / (Zg + Zt)
Then, using Per-Unit method,
o Eq 3, Isc = Ssc / ( Sqrt(3) x Vbase )
Solution:
o Zg = 2.0/20 = 1/10 or 0.10 (pu)
o Ssc = 2.0 x 10⁶ / (0.10 +0.05) = 13 x 10⁶
o Isc = 13.3 x 10E6 / Sqrt(3) x 460 = 16,729 A. Closest is B, 20 kA!
Note 1.
Since 460 was the only given voltage, I called it Vbase! Why? Because I perceived the series-circuit to be a generator connected to a 1:1 Xfmr or reactor!
Note 2:
To TheSmoothestCriminal… you asked for some reference on the X/R rule! No, I couldn’t find any! But, a search of my archives revealed where it was! I did numerous fault-calcs, mostly at client-sites, using a hand-held HP-35 calculator! The X/R rule was pre-programmed! But, I’d forgotten the restrictions involved! It works, because the variables must be converted to base figures, i.e., base volts and base-MVA! In addition, having equal X/R ratio’s insure their vectors have a magnitude, and their angles are the same! BTW, it can also be applied to problems having reactance as the magnitude, and the angle is 90º!
(Caveat… because of recent events, your moniker might get you into trouble in the future. LOL!)