The severity of line to ground and three-phase faults at the terminals of an unloaded synchronous generator is to be same. If the terminal voltage is 1.0 pu, Z1 = Z2 = j0.1 pu and Z0 = j0.05 pu for the alternator, then the required inductive reactance for neutral grounding is

Option A: 0.0166 pu Explanation & Concepts: To equate the fault currents, the magnitude of the three-phase fault current ($I_{3\phi} = \frac{V}{Z_1}$) is set equal to the line-to-ground fault current ($I_{LG} = \frac{3V}{Z_1 + Z_2 + Z_0 + 3Z_n}$). By solving this equality with the given sequence impedances and $Z_n = jX_n$, we determine the required grounding reactance. Working Principle: In a symmetrical three-phase fault, only the positive sequence impedance limits the fault current. In a line-to-ground fault, the fault current depends on the series combination of positive, negative, and zero sequence impedances, where the neutral grounding impedance $Z_n$ is amplified by a factor of 3 in the zero-sequence network. Key Formulas: $I_{3\phi} = \frac{V}{Z_1}$ — Three-phase fault current; $I_{LG} = \frac{3V}{Z_1 + Z_2 + Z_0 + 3Z_n}$ — Line-to-ground fault current Key Points: For solid grounding, $Z_n = 0$.; Neutral reactance $X_n$ is used to limit the magnitude of LG faults to protect equipment.; Sequence networks are connected in parallel for LG faults.; Base values for per-unit system must be consistent across all impedances. Advantages: Limits fault currents to prevent equipment damage; Reduces mechanical stress on generator windings Disadvantages: Increases overvoltage during LG faults; Complicates relay coordination Applications: Synchronous generator protection; Power system stability control Additional Info: Calculation: $1/0.1 = 3 / (0.1 + 0.1 + 0.05 + 3X_n) \Rightarrow 10 = 3 / (0.25 + 3X_n) \Rightarrow 2.5 + 30X_n = 3 \Rightarrow 30X_n = 0.5 \Rightarrow X_n = 0.0166 pu$.. Option B (0.05) is the zero sequence reactance, not the neutral grounding reactance. Answer: A is correct — The required neutral grounding reactance is 0.0166 pu to ensure the LG fault current magnitude equals the three-phase fault current magnitude. Exam Tip: Always remember the factor of 3 for neutral impedance ($3Z_n$) in the zero-sequence network diagram for unbalanced faults.

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The severity of line to ground and three-phase faults at the terminals of an unloaded synchronous generator is to be same. If the terminal voltage is 1.0 pu, Z₁ = Z₂ = j0.1 pu and Z₀ = j0.05 pu for the alternator, then the required inductive reactance for neutral grounding is

0.0166 pu

0.05 pu

0.1 pu

0.15 pu

Correct Answer

⚙️ TE • Technical Concept & PrincipleElectricalPower System

Option A

0.0166 pu

Quick Summary:

To equate the fault currents, the magnitude of the three-phase fault current ( $I_{3\phi} = \frac{V}{Z_1}$ ) is set equal to the line-to-ground fault current ( $I_{LG} = \frac{3V}{Z_1 + Z_2 + Z_0 + 3Z_n}$ ). By solving this equality with the given sequence impedances and $Z_n = jX_n$ , we determine the required grounding reactance.

⚙️TETechnical SolutionConcept & Principle

💡 Explanation

🔢 Key Formulas

$I_{3\phi} = \frac{V}{Z_1}$ — Three-phase fault current

$I_{LG} = \frac{3V}{Z_1 + Z_2 + Z_0 + 3Z_n}$ — Line-to-ground fault current

⚙️ Working Principle

In a symmetrical three-phase fault, only the positive sequence impedance limits the fault current. In a line-to-ground fault, the fault current depends on the series combination of positive, negative, and zero sequence impedances, where the neutral grounding impedance $Z_n$ is amplified by a factor of 3 in the zero-sequence network.

📌 Key Points

▸
For solid grounding, $Z_n = 0$ .
▸
Neutral reactance $X_n$ is used to limit the magnitude of LG faults to protect equipment.
▸
Sequence networks are connected in parallel for LG faults.
▸
Base values for per-unit system must be consistent across all impedances.

✅ Advantages

▸
Limits fault currents to prevent equipment damage
▸
Reduces mechanical stress on generator windings

❌ Disadvantages / Limitations

▸
Increases overvoltage during LG faults
▸
Complicates relay coordination

🛠️ Applications / Uses

▸
Synchronous generator protection
▸
Power system stability control

📄 Additional Information

▸
Calculation: $1/0.1 = 3 / (0.1 + 0.1 + 0.05 + 3X_n) \Rightarrow 10 = 3 / (0.25 + 3X_n) \Rightarrow 2.5 + 30X_n = 3 \Rightarrow 30X_n = 0.5 \Rightarrow X_n = 0.0166 pu$ .
▸
Option B (0.05) is the zero sequence reactance, not the neutral grounding reactance.

📊 Diagram / Illustration

✅

A is correct — The required neutral grounding reactance is 0.0166 pu to ensure the LG fault current magnitude equals the three-phase fault current magnitude.

Core Concepts Used

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Symmetrical components Fault analysis Sequence networks

💡 EXAM TIP

Always remember the factor of 3 for neutral impedance ( $3Z_n$ ) in the zero-sequence network diagram for unbalanced faults.

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