__Concept:__

In a single line to ground (LG) fault, the fault current is

\({I_n} = \frac{{3{E_a}}}{{{Z_1} + {Z_2} + {Z_0} + 3{Z_n}}}\)

Where,

Ea = terminal voltage

Z1 = positive sequence impedance

Z2 = negative sequence impedance

Z0 = zero sequence impedance

Zn = impedance in neutral grounding

The fault current can also be written as If = 3I_{a1}

__Calculation:__

Single line to ground fault, Fault current in \({I_f} = 3{I_{{a_1}}}\)

\({I_{{a_1}}} = \frac{{{E_a}}}{{{Z_1} + {Z_2} + {Z_o}}} = \frac{1}{{j0.2 + j0.2 + j0.05}}\)

= -j 2.223 Pu.

Fault current \(\left( {{I_f}} \right) = 3 \times {I_{{a_1}}}\)

= (3 × -j 2.222) = - j 6.666 pu

Base current \( = \frac{{100 \times {{10}^6}}}{{\sqrt 3 \times 25 \times {{10}^3}}} = 2309.4\ Pu\)

Fault current I_{f} = 15396 A.