---
title: "Fault currents of isolated-neutral 380 V generator"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:generator-faults-380v-isolated-neutral"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-faults-380v-isolated-neutral
site: BCT Engineering
---

# Fault currents of isolated-neutral 380 V generator

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

- Asked in: 2082 Baishakh, 6+2 marks
- Video: [LG fault on neutral isolated no loaded generator | BARC Actual interview question from Power system](https://www.youtube.com/watch?v=MVHbKloyOBs) — YourPedia Mock Interview GD GT
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/generator-faults-380v-isolated-neutral
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A 3-phase synchronous generator with isolated neutral at no-load is subjected to different types of faults at their terminal. Considering the line-to-line voltage rating of generator as 380V and positive, negative & zero sequence impedances of the generator are j2Ω, j0.5Ω & j0.25Ω respectively, determine the fault current for (i) L-G fault, (ii) L-L-G fault, and (iii) L-L-L-G fault. Comment on the results.

## Short answer

With isolated neutral, the zero-sequence network is open. (i) L-G fault: 0 A, no return path. (ii) L-L-G fault reduces to an L-L fault: |Ib|=|Ic|= sqrt(3)*219.39/(2+0.5)=152.0 A, ground current zero. (iii) Three-phase fault: 219.39/2=109.7 A. L-L current exceeds three-phase because Z2 < Z1.

## Answer

**Data:** $V_L = 380$ V, so phase emf at no load $E = 380/\sqrt3 = 219.39$ V. $Z_1 = j2\ \Omega$, $Z_2 = j0.5\ \Omega$, $Z_0 = j0.25\ \Omega$. Neutral is **isolated**, so the neutral impedance $Z_n = \infty$ and the zero-sequence network is open: no zero-sequence current can flow.

#### (i) L-G fault (phase a to ground)
$$
I_f = 3I_{a0} = \frac{3E}{Z_1 + Z_2 + Z_0 + 3Z_n}
$$
With $Z_n \to \infty$:
$$
I_f = 0\ \text{A}
$$
There is no return path for the current, so no fault current flows. Only the potential of the neutral shifts: the faulted phase goes to ground potential and the healthy phase voltages to ground rise to line voltage (380 V).

#### (ii) L-L-G fault (phases b, c to ground)
$I_{a0} = 0$ because the zero-sequence network is open. The fault behaves like an L-L fault:
$$
\begin{aligned}
I_{a1} &= \frac{E}{Z_1 + Z_2} = \frac{219.39}{j2.5} = -j87.76\ \text{A} \\
|I_b| = |I_c| &= \sqrt3\,|I_{a1}| = \frac{\sqrt3 \times 219.39}{2.5} = 152.0\ \text{A}
\end{aligned}
$$
Current to ground $= 3I_{a0} = 0$. The line currents are **152.0 A** each (flowing from b to c).

#### (iii) L-L-L-G fault
Balanced fault, only the positive sequence network acts:
$$
I_f = \frac{E}{|Z_1|} = \frac{219.39}{2} = 109.70\ \text{A}
$$

#### Results
| Fault | Fault current |
|---|---|
| L-G | 0 A |
| L-L-G | 152.0 A (ground current 0) |
| L-L-L-G | 109.7 A |

#### Comments
- With an isolated neutral, faults involving ground give no zero-sequence current; an L-G fault gives no fault current and an L-L-G fault reduces to an L-L fault.
- The L-L (and L-L-G) current, 152 A, is larger than the three-phase current, 109.7 A, because $Z_2$ (0.5 Ω) is much smaller than $Z_1$ (2 Ω): $I_{LL}/I_{3\phi} = \sqrt3 Z_1/(Z_1+Z_2) = 1.386$.
- If the neutral were solidly grounded, the L-G current would be $3E/(Z_1+Z_2+Z_0) = 3\times219.39/2.75 = 239.3$ A, the highest of all. Isolating the neutral therefore removes the severe earth fault current but lets healthy phases rise to line voltage, which stresses insulation.

## Check yourself

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

1. What is the per-phase EMF for a 380 V line-to-line generator?
   A. 380 V
   B. 219.4 V
   C. 658 V
   D. 190 V
   Answer: B. E = 380/sqrt3 = 219.39 V. (Key point: Phase EMF)

2. Why is the L-G fault current zero for the isolated neutral?
   A. Z1 is very large
   B. E is zero at no load
   C. The zero sequence network is open (Zn infinite), so no return path exists
   D. Z2 cancels Z1
   Answer: C. If = 3E/(Z1 + Z2 + Z0 + 3Zn) tends to 0 as Zn tends to infinity. (Key point: L-G fault with isolated neutral)

3. With zero sequence open, the L-L-G fault current in each faulted line is
   A. E/(Z1 + Z2) = 87.8 A
   B. 3E/(Z1 + Z2 + Z0) = 239 A
   C. sqrt3 E/(Z1 + Z2) = 152 A
   D. E/Z1 = 109.7 A
   Answer: C. Ia1 = 219.39/2.5 = 87.76 A, and |Ib| = sqrt3 x 87.76 = 152.0 A. (Key point: L-L-G fault reduces to L-L)

4. Why is the L-L current (152 A) larger than the 3-phase current (109.7 A)?
   A. Because Z2 (0.5 ohm) is much smaller than Z1 (2 ohm), giving ratio sqrt3 Z1/(Z1 + Z2) = 1.386
   B. Because zero sequence adds current
   C. Because the neutral is grounded
   D. Because the L-L fault involves three phases
   Answer: A. I_LL/I_3phase = sqrt3 Z1/(Z1 + Z2) = 1.386 for these values. (Key point: Comparison of fault currents)

## Related in Power System Analysis II

- [Equal positive and negative sequence impedances](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astatic-circuit-positive-negative-equal) — Power System Analysis II, chapter 4
- [No zero sequence in line-to-line fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-zero-sequence-absent) — Power System Analysis II, chapter 5, asked 2×
- [L-G fault through Zf: currents and voltages](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-with-zf-derivation) — 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
- [L-G fault of unloaded generator: equal currents](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-unloaded-generator-derivation) — Power System Analysis II, chapter 5
- [Generator and motor currents for terminal fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-current-generator-motor-2080) — Power System Analysis II, chapter 3

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