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The following sequence currents were recorded in a power system under a fault condition Iₚₒₛᵢₜᵢᵥₑ = j 1.753 pu, Iₙₑgₐₜᵢᵥₑ = – j 0.6 pu, Izₑᵣₒ = – j 1.153 pu.The fault is
Line to ground
Three-phase
Line to line to ground
Line to line
Line to line to ground
Given: Positive sequence current I1 = j1.753 pu, Negative sequence current I₂ = -j0.6 pu, Zero sequence current I0 = -j1.153 pu.
Positive sequence current I1 = j1.753 pu, Negative sequence current I₂ = -j0.6 pu, Zero sequence current I0 = -j1.153 pu.
I1=I2=I0⟹LG Fault,I1+I2+I0=0⟹LL Fault,I1+I2+I0=0⟹LLG Fault
Analyze the relationship between sequence currents
In an unbalanced fault analysis using symmetrical components, the relationship between sequence currents identifies the fault type. Here, we check the sum of sequence currents: Isum=Ipositive+Inegative+Izero.
Isum=j1.753−j0.6−j1.153
Calculate the sum
Adding the given values: Isum=j(1.753−0.6−1.153)=j(0)=0.
Isum=0
Determine fault type based on conditions
Since Ipositive+Inegative+Izero=0, this is a condition for a line-to-line fault. However, if any sequence current is non-zero and they are not equal, it indicates a double line-to-ground (LLG) fault. Specifically, in LLG faults, all three sequence networks are connected in parallel. Since all three are non-zero and their sum is zero, it represents a Line-to-Line-to-Ground fault.
I1+I2+I0=0 and I0=0
C is correct because the sum of the sequence currents is zero (I1+I2+I0=0) while I0 is non-zero, which confirms a line-to-line-to-ground fault.
This concept is fundamental for relay coordination and protection studies; remember that LLG faults involve the parallel combination of positive, negative, and zero sequence impedance networks.