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ElectricalPower System
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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

A

Line to ground

B

Three-phase

C

Line to line to ground

D

Line to line

Correct Answer

⚙️ TE • Technical Symmetrical Components AnalysisElectricalPower System
Option C

Line to line to ground

Quick Summary:

Given: Positive sequence current I1 = j1.753 pu, Negative sequence current I₂ = -j0.6 pu, Zero sequence current I0 = -j1.153 pu.

📐MAMath SolutionSymmetrical Components Analysis
📋 Given

Positive sequence current I1 = j1.753 pu, Negative sequence current I₂ = -j0.6 pu, Zero sequence current I0 = -j1.153 pu.

🔢 Formula Used

I1=I2=I0  ⟹  LG Fault,I1+I2+I0=0  ⟹  LL Fault,I1+I2+I0≠0  ⟹  LLG FaultI_1 = I_2 = I_0 \implies \text{LG Fault}, \quad I_1 + I_2 + I_0 = 0 \implies \text{LL Fault}, \quad I_1 + I_2 + I_0 \neq 0 \implies \text{LLG Fault}I1​=I2​=I0​⟹LG Fault,I1​+I2​+I0​=0⟹LL Fault,I1​+I2​+I0​=0⟹LLG Fault

🔢 Step-by-Step Solution
1

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+IzeroI_{sum} = I_{positive} + I_{negative} + I_{zero}Isum​=Ipositive​+Inegative​+Izero​.

Isum=j1.753−j0.6−j1.153I_{sum} = j1.753 - j0.6 - j1.153Isum​=j1.753−j0.6−j1.153

2

Calculate the sum

Adding the given values: Isum=j(1.753−0.6−1.153)=j(0)=0I_{sum} = j(1.753 - 0.6 - 1.153) = j(0) = 0Isum​=j(1.753−0.6−1.153)=j(0)=0.

Isum=0I_{sum} = 0Isum​=0

3

Determine fault type based on conditions

Since Ipositive+Inegative+Izero=0I_{positive} + I_{negative} + I_{zero} = 0Ipositive​+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≠0I_1 + I_2 + I_0 = 0 \text{ and } I_0 \neq 0I1​+I2​+I0​=0 and I0​=0

✅

C is correct because the sum of the sequence currents is zero (I1+I2+I0=0I_1 + I_2 + I_0 = 0I1​+I2​+I0​=0) while I0I_0I0​ is non-zero, which confirms a line-to-line-to-ground fault.

Core Concepts Used
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Symmetrical Components Power System Fault Analysis Sequence Networks
💡 EXAM TIP

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.

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