Given that: $$\,{V_{s1}} = {V_{s2}} = 1 + j0\,\,p.u,\,\, + ve\,\,$$ sequence impedance are $$\,{Z_{s1}} = {Z_{s2}} = 0.001 + j0.01\,\,p.u\,\,$$ and $${Z_L} = 0.006 + j\,0.06\,\,p.u,\,\,3\phi .\,\,\,$$ Base $$MVA=100,$$ voltage base $$=400$$ $$kV(L-L).$$
Nominal system frequency $$= 50$$ $$Hz.$$ The reference voltage for phase $$'a'$$ is defined as $$\,\,V\left( t \right) = {V_m}\,\cos \left( {\omega t} \right).\,\,\,$$ A symmetrical $$3\phi $$ fault occurs at centre of the line, i.e., at point $$'F'$$ at time 'to' the $$+ve$$ sequence impedance from source $${S_1}$$ to point $$'F'$$ equals $$(0.004 + j \,\,0.04)$$ $$p.u.$$ The wave form corresponding to phase $$'a'$$ fault current from bus $$X$$ reveals that decaying $$d.c.$$ offset current is $$-ve$$ and in magnitude at its maximum initial value. Assume that the negative sequence are equal to $$+ve$$ sequence impedances and the zero sequence $$(Z)$$ are $$3$$ times $$+ve$$ sequence $$(Z).$$

The $$rms$$ value of the ac component of fault current $$\,\left( {{{\rm I}_x}} \right)$$ will be

Given that: $$\,{V_{s1}} = {V_{s2}} = 1 + j0\,\,p.u,\,\, + ve\,\,$$ sequence impedance are $$\,{Z_{s1}} = {Z_{s2}} = 0.001 + j0.01\,\,p.u\,\,$$ and $${Z_L} = 0.006 + j\,0.06\,\,p.u,\,\,3\phi .\,\,\,$$ Base $$MVA=100,$$ voltage base $$=400$$ $$kV(L-L).$$
Nominal system frequency $$= 50$$ $$Hz.$$ The reference voltage for phase $$'a'$$ is defined as $$\,\,V\left( t \right) = {V_m}\,\cos \left( {\omega t} \right).\,\,\,$$ A symmetrical $$3\phi $$ fault occurs at centre of the line, i.e., at point $$'F'$$ at time 'to' the $$+ve$$ sequence impedance from source $${S_1}$$ to point $$'F'$$ equals $$(0.004 + j \,\,0.04)$$ $$p.u.$$ The wave form corresponding to phase $$'a'$$ fault current from bus $$X$$ reveals that decaying $$d.c.$$ offset current is $$-ve$$ and in magnitude at its maximum initial value. Assume that the negative sequence are equal to $$+ve$$ sequence impedances and the zero sequence $$(Z)$$ are $$3$$ times $$+ve$$ sequence $$(Z).$$

The instant $$\,\left( {{t_0}} \right)\,\,$$ of the fault will be

Given that: $$\,{V_{s1}} = {V_{s2}} = 1 + j0\,\,p.u,\,\, + ve\,\,$$ sequence impedance are $$\,{Z_{s1}} = {Z_{s2}} = 0.001 + j0.01\,\,p.u\,\,$$ and $${Z_L} = 0.006 + j\,0.06\,\,p.u,\,\,3\phi .\,\,\,$$ Base $$MVA=100,$$ voltage base $$=400$$ $$kV(L-L).$$
Nominal system frequency $$= 50$$ $$Hz.$$ The reference voltage for phase $$'a'$$ is defined as $$\,\,V\left( t \right) = {V_m}\,\cos \left( {\omega t} \right).\,\,\,$$ A symmetrical $$3\phi $$ fault occurs at centre of the line, i.e., at point $$'F'$$ at time 'to' the $$+ve$$ sequence impedance from source $${S_1}$$ to point $$'F'$$ equals $$(0.004 + j \,\,0.04)$$ $$p.u.$$ The wave form corresponding to phase $$'a'$$ fault current from bus $$X$$ reveals that decaying $$d.c.$$ offset current is $$-ve$$ and in magnitude at its maximum initial value. Assume that the negative sequence are equal to $$+ve$$ sequence impedances and the zero sequence $$(Z)$$ are $$3$$ times $$+ve$$ sequence $$(Z).$$

Instead of the three phase fault, if a single line to ground fault occurs on phase $$' a '$$ at point $$' F '$$ with zero fault impedance, then the $$rms$$ of the ac component of fault current $$\left( {{{\rm I}_x}} \right)$$ for phase $$'a'$$ will be

Suppose we define a sequence transformation between ''a-b-c'' and ''p-n-0''' variables as follows:
$$\left[ {\matrix{ {{f_a}} \cr {{f_b}} \cr {{f_c}} \cr } } \right] = k\left[ {\matrix{ 1 & 1 & 1 \cr {{\alpha ^2}} & \alpha & 1 \cr \alpha & {{\alpha ^2}} & 1 \cr } } \right]\left[ {\matrix{ {{f_p}} \cr {{f_n}} \cr {{f_o}} \cr } } \right]$$ where $$\,\alpha = {e^{j{{2\pi } \over 3}}}\,\,$$ and $$k$$ is a constant
Now, if it is given that:
$$\left[ {\matrix{ {{V_p}} \cr {{V_n}} \cr {{V_o}} \cr } } \right] = k\left[ {\matrix{ {0.5} & 0 & 0 \cr 0 & {0.5} & 0 \cr 0 & 0 & {2.0} \cr } } \right]\left[ {\matrix{ {{i_p}} \cr {{I_n}} \cr {{i_o}} \cr } } \right]\,\,$$ and $$\left[ {\matrix{ {{V_a}} \cr {{V_b}} \cr {{V_c}} \cr } } \right] = z\left[ {\matrix{ {{i_a}} \cr {{I_b}} \cr {{i_c}} \cr } } \right]\,\,$$ then,