Here is an example with voltages and currents, note the neutral to ground voltage after the fault:
Primary impedance: 4607 amps 3-phase, 3990 amps 2-phase, 3455 amps 1-phase with X/R = 10
Secondary impedance: 1012 amps 3-phase, 876.3 amps 2-phase, 54 amps 1-phase with X/R = 0.17
30 kVA transformer (three 10 kVA units)
Load = 15 kW, 2 kvar at 480 volts wye connected at transformer terminals
PRE-FAULT:
At the source:
VAN = 7199<0
VBN = 7199<-120
VCN = 7199<120
VAB = 12470<29.99
VBC = 12470<-90.01
VCA = 12470<150.00
IA = 0.7009<171.2
IB = 0.7009<51.18
IC = 0.7009<-68.82
SA = 4987+j773.4
SB = 4987+j773.5
SC = 4987+j773.6
P = 14.96 kW
Q = 2.320 kvar
S = 15.14 kVA
pf = 98.82%
At the load:
Van = 275.7<-0.99
Vbn = 275.7<-121.00
Vcn = 275.7<119.00
Vab = 477.6<29.01
Vbc = 477.6<-90.99
Vca = 477.6<149.00
Ia = 18.11<-8.584
Ib = 18.11<-128.6
Ic = 18.11<111.4
Sa = 4949+j659.9
Sb = 4949+j659.9
Sc = 4949+j659.9
P = 14.85 kW
Q = 1.980 kvar
S = 14.98 kVA
pf = 99.12%
Now introduce a bolted secondary c-n fault: Ic = 54.03<109.2
At the source:
VAN = 12180<-31.2
VBN = 12550<-91.73
VCN = 381.2<-168.9
VAB = 12470<29.99
VBC = 12470<-90.02
VCA = 12470<150.00
VAG = 7198<-0.01
VBG = 7200<-120
VCG = 7199<120
VNG = 7085<117.1
IA = 1.186<140
IB = 1.222<79.45
IC = 2.080<-70.78
SA = 14280+j2215
SB = 15150+j2350
SC = 112.1+j784.8
P = 29.54 kW
Q = 5.349 kvar
S = 30.02 kVA
pf = 98.40%
At the load:
Van = 466.6<-32.18
Vbn = 480.6<-92.71
Vcn = 0<0
Vab = 477.6<29.01
Vbc = 480.6<-92.71
Vca = 466.6<147.8
Ia = 30.64<-39.77
Ib = 31.57<-100.3
Ic = 0<0
Sa = 14170+j1890
Sb = 15040+j2005
Sc = 0+j0
P = 29.21 kW
Q = 3.895 kvar
S = 29.47 kVA
pf = 99.12%