MepEngineer ?
Your problem is the classic ?Fault-Reduction? question that?s occasionally presented in the PE Exam! Using my GI.FI.E.S (pronounced Jiffys) method, here is my solution:
Part I:
GIven: Vs in Volts, I4, and I4A in Amperes
FInd: Reactance, Xr, to reduce I4 to I4A (At this point cable parameters are unnecessary!)
Equation: Xr = ( Vs/Sqrt3 )*[ (1/I4A ? (1/I4) ] Ohm/ph
Solution: Xr = (208/Sqrt3)*[ (1/12,367)-(1/40,000) ] = 0.00670 Ohm/ph => 6.70 mOhm/ph
MepEngineer? the above the answer is sufficient, IFF, it was the answer required on the PE Exam! However your sketch presents an incongruous situation! The cable shown, i.e., #1 AWG, having a length of 50-ft, will not yield the same reactance! Instead, the value will be as shown below:
Part II:
GIven: #1 AWG Cable, Length, L in ft, and XL, per NEC, in ?/k-ft
FInd: Cable Total Reactance, XT, in ? (Ignoring fact cable is 3 times more resistive!)
Equation: XT = (L/1,000)*XL
Solution: XT = (50/1,000)*0.046 = 0.0023 Ohm => 2.30 mOhm/ph
MepEngineer? are you sure about the parameters shown in your illustration?
Regards, Phil Corso