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When a 50 MVA, 11 kV, 3-phase generator is subjected to a 3-phase fault, the fault current is –j5 pu · When it is subjected to a line-to-line fault, the positive sequence current is j4 pu · The positive and negative sequence reactances are respectively
j0.2 and j0.05 pu
j0.2 and j0.25 pu
j0.25 and j0.25 pu
j0.05 and j0.05 pu
j0.2 and j0.05 pu
Quick Trick: For 3-phase fault, I = 1/X₁ · For line-to-line fault, Ipos = 1/(X₁+X₂) · Simply solve for X₁ then X₂ using the given fault currents.
For 3-phase fault, I = 1/X₁ · For line-to-line fault, Ipos = 1/(X₁+X₂) · Simply solve for X₁ then X₂ using the given fault currents.
Calculate Positive Sequence Reactance (X₁)
For a 3-phase fault, the fault current is If = 1/X₁ · Given If = 5 pu, 1/X₁ = 5, therefore X₁ = 1/5 = 0.2 pu.
Calculate Negative Sequence Reactance (X₂)
For a line-to-line fault, the positive sequence current is Ipos = 1/(X₁ + X₂) · Given Ipos = 4 pu, then 1/(0.2 + X₂) = 4 · This implies 0.2 + X₂ = 0.25, so X₂ = 0.05 pu.
B: X₂ = 0.25 leads to current = 1/(0.2+0.25) = 2.22, incorrect · C: X₁=0.25 contradicts 3-phase fault current of 5 · D: X₁=0.05 contradicts 3-phase fault current of 5.
A is correct because the 3-phase fault equation (I=1/X₁) gives X₁=0.2, and substituting this into the line-to-line fault equation (I=1/(X₁+X₂)) yields X₂=0.05.
Always verify if the fault current provided is total fault current or sequence component current, as formulas change accordingly.