---
title: "Sequence reactances from fault currents"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:sequence-reactances-from-fault-currents"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-reactances-from-fault-currents
site: BCT Engineering
---

# Sequence reactances from fault currents

Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems

- Asked in: 2082 Bhadra (new course), 7 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/sequence-reactances-from-fault-currents
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A 50 MVA, 12 kV, three-phase alternator were subjected to different types of faults. The fault currents were: (i) 1870 A for three-phase fault (ii) 2590 A for L-L fault (iii) 4130 A for L-G fault. The alternator neutral is solidly grounded. Find the three-sequence reactance of the alternator in per unit.

## Short answer

Base current is 2405.6 A. Fault currents in pu: 3-phase 0.7773, L-L 1.0766, L-G 1.7168. Using I_3ph = 1/X1, I_LL = √3/(X1+X2), I_LG = 3/(X1+X2+X0). X1 = 1.286 pu, X2 = 0.322 pu, X0 = 0.139 pu.

## Answer

**Assumptions:** the faults are at the terminals of the unloaded alternator, so the emf $E = 1$ pu (rated voltage); resistance neglected; neutral solidly grounded ($Z_n = 0$).

#### Base current
$$
I_{base} = \frac{50\times10^6}{\sqrt3 \times 12\times10^3} = 2405.6\ \text{A}
$$
| Fault | Current (A) | Current (pu) |
|---|---|---|
| 3-phase | 1870 | 0.7773 |
| L-L | 2590 | 1.0766 |
| L-G | 4130 | 1.7168 |

#### Fault current formulas
$$
I_{3\phi} = \frac{E}{X_1}, \qquad I_{LL} = \frac{\sqrt3 E}{X_1 + X_2}, \qquad I_{LG} = \frac{3E}{X_1 + X_2 + X_0}
$$

#### Positive sequence reactance
$$
X_1 = \frac{1}{0.7773} = 1.2864\ \text{pu}
$$

#### Negative sequence reactance
$$
\begin{aligned}
X_1 + X_2 &= \frac{\sqrt3}{1.0766} = 1.6088 \\
X_2 &= 1.6088 - 1.2864 = 0.3223\ \text{pu}
\end{aligned}
$$

#### Zero sequence reactance
$$
\begin{aligned}
X_1 + X_2 + X_0 &= \frac{3}{1.7168} = 1.7474 \\
X_0 &= 1.7474 - 1.6088 = 0.1387\ \text{pu}
\end{aligned}
$$

**Answer:** $X_1 = 1.286$ pu, $X_2 = 0.322$ pu, $X_0 = 0.139$ pu.

The large $X_1$ (it is the steady-state synchronous reactance here, since sustained currents were measured) and small $X_0$ explain why the L-G current is the highest and the 3-phase current the lowest.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which reactance does the 3-phase fault current give?
   A. X0 = 3E/I
   B. X1 = E/I
   C. X2 = E/I
   D. X1+X2 = E/I
   Answer: B. I(3 phase) = E/X1, so X1 = 1/0.7773 = 1.286 pu. (Key point: 3-phase fault gives X1)

2. How is X2 found from the L-L fault current?
   A. X2 = E/I
   B. X2 = sqrt3 E/I - X1
   C. X2 = 3E/I - X1
   D. X2 = sqrt3 E/I
   Answer: B. I(LL) = sqrt3 E/(X1+X2), so X1+X2 = 1.6088 and X2 = 0.322 pu. (Key point: X2 from L-L)

3. How is X0 found from the L-G current?
   A. X0 = 3E/I - (X1 + X2)
   B. X0 = E/I - X1
   C. X0 = sqrt3/I - X1
   D. X0 = 3E/I
   Answer: A. I(LG) = 3E/(X1+X2+X0), so X0 = 1.7474 - 1.6088 = 0.139 pu. (Key point: X0 from L-G)

## Related in Power System Analysis II

- [L-L fault line voltages, 25 MVA alternator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-25mva-line-voltages) — Power System Analysis II, chapter 5, asked 2×
- [Neutral reactance and L-G fault severity](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-neutral-reactance-lg-severity) — Power System Analysis II, chapter 5, asked 2×
- [Neutral reactance for 15 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-15mva) — Power System Analysis II, chapter 5
- [Line currents with isolated and grounded neutral](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-impedance-unbalanced-voltages-neutral) — Power System Analysis II, chapter 4
- [SLG fault on bus with three alternators](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-three-alternators-common-bus) — Power System Analysis II, chapter 5
- [Fault currents of isolated-neutral 380 V generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-faults-380v-isolated-neutral) — Power System Analysis II, chapter 5

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