EasyPower considers the source impedance when performing a short circuit analysis. Below, we demonstrate two cases that generate the same short circuit currents.

**Case 1: ****Utility @ 115 kv**
Calculating the I_sc from the MVA values, we get:

3Ph = MVA / (SQRT(3)* I_sc_3ph*kV) = 12.551kA

1Ph = MVA / (SQRT(3)* I_sc_1Ph*kV) = 10.041kA

Assuming 2500MVA and 115kV as base values, we get Xp = 1pu. EasyPower assumes Xp=Xn

- p = Positive sequence
- n = Negative sequence
- z = Zero sequence

Using the following sequence formula, we find that Xz=1.75pu.

I_f (SLG) = 3 *{V/(Xp+Xn+Xz)}

**Case 2: ****Utility with Zero Impedance @ 115kV.**
Here, we have increased the MVA to the maximum that EasyPower allows. However, it still has a small impedance value:

Using the same method, and assuming 2500MVA and 115kV as base values with the current utility value, we get:

Xp = Xn = .005pu

Xp + Xn + Xz = .015 => Xz = .005pu

To match the impedance values of the utility in the first case, we need to add an impedance, where the value plus the above values gives us an equivalent of the utility in the first case. Using a CL reactor, we can add the reactance (X/R in the utility is chosen to be very large, so the impedance is all inductive).

For 3 phase the Xp of the reactor should be:

1 - .005 = 0.995pu or 5.26355 Ohm. (Z_base = 115^2/2500 = 5.29Ohms)

Xz = 1.75 – 0.05 = 1.745 pu or 9.23105 Ohms, where Z_base = 5.29 Ohms:

Below, you can see the exact matches for the 3 phase and SLG faults: