CONCEPT:The following circuit shown is a system with line to ground fault.Here, Va = 0, Ib = 0, Ic = 0Now, \({I_{a1}} = \frac{1}{3}\LEFT( {{I_a} + k{I_b} + {k^2}{I_c}} \right)\)\({I_{A2}} = \frac{1}{3}\left( {{I_a} + {k^2}{I_b} + k{I_c}} \right)\)\({I_{a0}} = \frac{1}{3}\left( {{I_a} + {I_b} + {I_c}} \right)\)By substituting Ib and Ic values,\({I_{a1}} = {I_{a2}} = {I_{a0}} = \frac{{{I_a}}}{3}\)Hence in LG faults, all of the three COMPONENTS la0, la1 and la2 are equalApplication:IPositive = - j0.91 P.U.INegative = -j0.91 P.U.IZero = -j0.91 P.U.As all the sequence currents are equal, the fault is SINGLE line to ground fault.