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The following sequence currents were recorded in a power system under a fault condition IтВЪтВТтВЫс╡втВЬс╡вс╡етВС = j1.653 pu, IтВЩтВСgтВРтВЬс╡вс╡етВС = тАУ j 0.5 pu, IzтВСс╡гтВТ = тАУ j1.153 pu.The fault is
Line to ground
Three-phase
Line to line to ground
Line to line
Line to line to ground
Quick Trick: Check the magnitude equality: If I1, IтВВ, and I0 are all non-zero and unequal, it is an LLG fault; if they are equal, it is an LL fault; if IтВВ and I0 are zero, it is a 3-phase fault.
Check the magnitude equality: If I1, IтВВ, and I0 are all non-zero and unequal, it is an LLG fault; if they are equal, it is an LL fault; if IтВВ and I0 are zero, it is a 3-phase fault.
Evaluate current components
Observe the given sequence currents: I1 = j1.653, IтВВ = -j0.5, I0 = -j1.153. Note that I1 + IтВВ + I0 = j1.653 - j0.5 - j1.153 = 0.
Match with fault conditions
In an LLG (Line-to-Line-to-Ground) fault, all three sequence networks (positive, negative, and zero) are connected in parallel. Since all three values are non-zero and distinct, it matches the mathematical requirement for an LLG fault.
A: LG fault requires I1 = IтВВ = I0. B: 3-phase fault requires IтВВ = 0 and I0 = 0. D: LL fault requires I0 = 0 and I1 = -IтВВ.
C is correct because the existence of non-zero positive, negative, and zero sequence components that sum to zero is the unique signature of an unsymmetrical Line-to-Line-to-Ground (LLG) fault.
Always verify if the sum of sequence currents is zero for balanced phase faults or specific unsymmetrical connections to quickly rule out invalid fault types.