Question from inbox:

This question is for problem 7 on page 96 of the Cram for the PE guide. The problem deals with a 150 MVA generator connected to a 13.8 kV bus. I think this is a very good problem, but I am having trouble getting the same answer mathematically. I think I am getting lost in the complex math at the bottom of page 96. Where does the -j component of 6.66 go? Also why are we taking the square root of the denominator.

This is an excellent question. For the three phase fault, I get the answer -j6.66. However, when I set the three phase fault equal to the line to ground fault, I suddenly drop the “-j” part. By disregarding the “-j” I am considering magnitude only. This “-j” should not be carried over since it really is in reference to V1. See below:

I did not put in in the problem, but V1 is taken as the reference and is 0°. It should be clear. This 90° angle has no business being placed in the new equation since the reference to V1 is no longer valid. I only care about magnitude.

The reason I take the square root of the sums in the denominator (j0.4+3Rn) , is the same reason I dropped the “-j”. I only care about magnitude of the line-to-ground short circuit. I CAN NOT manipulate j0.4+3Rn without first taking the magnitude. This is because the vector j0.4+3Rn is in reference to the “V” of the Isc calculated in the line-to-ground short circuit. This “V” is taken as 0°.
Once I set the three phase short circuit equal to the line to ground short circuit, these angles no longer make any sense. Remember, in real life these vectors are rotating. When we solve for this angles , we always take an arbitrary reference point.

Imagine it another way. Lets say for the three phase fault, I made V1 be referenced at +90°. Than my three-phase Isc would be:
jV/jZ1 = j1/j0.15 = 6.66 a pu.
This makes a current with no “j”. It is typical when doing fault calculations to assume this +90° for volts to cancel out the j’s.

Side Note: For load calculation it is preferable to use the voltage at 0° so that the angle of the current I represents its actual lag or lead value.