Voltage drop across the transmission line is given by the following equation:
$$$\left[ {\matrix{
{\Delta {V_a}} \cr
{\Delta {V_b}} \cr
{\Delta {V_c}} \cr
} } \right] = \left[ {\matrix{
{{Z_s}} & {{Z_m}} & {{Z_m}} \cr
{{Z_m}} & {{Z_s}} & {{Z_m}} \cr
{{Z_m}} & {{Z_m}} & {{Z_s}} \cr
} } \right]\left[ {\matrix{
{{i_a}} \cr
{{i_b}} \cr
{{i_c}} \cr
} } \right]$$$
Shunt capacitance of the line can be neglect. If the line has positive sequence impedance of $$15\,\,\Omega $$ and zero sequence in impedance of $$48\,\,\Omega ,$$ then the values of $${{Z_s}}$$ and $${{Z_m}}$$ will be
$${C_1}\left( {{P_{G1}}} \right) = {P_{G1}} + 0.055 \times P_{G1}^2$$
$${C_2}\left( {{P_{G2}}} \right) = 3{P_{G2}} + 0.03 \times P_{G2}^2$$
Where $${P_{G1}}$$ and $${P_{G2}}$$ are the MW injections from generator $${G_1}$$ and $${G_2}$$ respectively. Thus, the minimum cost dispatch will be
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
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