---
title: "SLG fault on 20 MVA generator"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:slg-fault-20mva-generator"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-20mva-generator
site: BCT Engineering
---

# SLG fault on 20 MVA generator

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

- Asked in: 2076 Asoj, 8 marks
- Video: [UNSYMMETRICAL FAULT CALCULATIONS - PART -17 - SINGLE LINE TO GROUND FAULT - PROBLEM](https://www.youtube.com/watch?v=u-dCxkvONTU) — zenmurali Sachithanandam
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/slg-fault-20mva-generator
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A single line to ground fault occur at generator terminal of 20 MVA, 13.8 KV and having Z1 = j0.20 pu, Z2 = j0.3 pu. Find the fault current and line to line voltage under fault condition.

## Short answer

Assuming Z0=j0.10 pu and solid grounding, fault current I_f = 3/(Z1+Z2+Z0) = 5 pu = 4.18 kA. Line-to-line voltages during fault: Vab = Vca = 8.29 kV, Vbc = 16.10 kV.

## Answer

**Data:** 20 MVA, 13.8 kV, $Z_1 = j0.20$ pu, $Z_2 = j0.30$ pu. The zero-sequence impedance is not given; **assume $Z_0 = j0.10$ pu** (typical value for such a generator, about half of $Z_1$) and a solidly grounded neutral. Unloaded, $E_a = 1\angle0^\circ$ pu.

Base current: $I_{base} = \dfrac{20\times10^6}{\sqrt3 \times 13.8\times10^3} = 836.7$ A; base phase voltage $= 13.8/\sqrt3 = 7.967$ kV.

#### Fault current (phase a to ground)
Sequence networks in series:
$$
\begin{aligned}
I_{a1} = I_{a2} = I_{a0} &= \frac{E_a}{Z_1+Z_2+Z_0} = \frac{1}{j0.6} = -j1.667\ \text{pu} \\
I_f &= 3I_{a1} = -j5.0\ \text{pu} \\
|I_f| &= 5 \times 836.7 = 4184\ \text{A}
\end{aligned}
$$

#### Sequence voltages
$$
\begin{aligned}
V_{a1} &= 1 - (j0.2)(-j1.667) = 0.6667 \\
V_{a2} &= -(j0.3)(-j1.667) = -0.5 \\
V_{a0} &= -(j0.1)(-j1.667) = -0.1667
\end{aligned}
$$

#### Phase voltages
$$
\begin{aligned}
V_a &= V_{a0}+V_{a1}+V_{a2} = 0 \\
V_b &= V_{a0} + a^2V_{a1} + aV_{a2} = 1.0408\angle{-103.90^\circ}\ \text{pu} \\
V_c &= V_{a0} + aV_{a1} + a^2V_{a2} = 1.0408\angle{103.90^\circ}\ \text{pu}
\end{aligned}
$$

#### Line-to-line voltages
$$
\begin{aligned}
V_{ab} &= V_a - V_b = 1.0408\angle{76.10^\circ}\ \text{pu} = 1.0408 \times 7.967 = 8.29\ \text{kV} \\
V_{bc} &= V_b - V_c = 2.0207\angle{-90^\circ}\ \text{pu} = 16.10\ \text{kV} \\
V_{ca} &= V_c - V_a = 1.0408\angle{103.90^\circ}\ \text{pu} = 8.29\ \text{kV}
\end{aligned}
$$

**Answer (with $Z_0 = j0.1$ pu):** $I_f = 5$ pu $= 4.18$ kA; $V_{ab} = V_{ca} = 8.29$ kV, $V_{bc} = 16.10$ kV.

General form for any $Z_0$: $I_f = \dfrac{3}{|Z_1+Z_2+Z_0|}$ pu, so with the data given the fault current is $3/(0.5 + X_0)$ pu. $V_{bc}$ rises above the rated 13.8 kV because $Z_2 > Z_1$.

## Check yourself

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

1. With Z1 = j0.2, Z2 = j0.3, Z0 = j0.1, what is the L-G fault current?
   A. 1.667 pu
   B. 5 pu = 4.18 kA
   C. 3.33 pu
   D. 10 pu
   Answer: B. Ia1 = 1/j0.6 = -j1.667; If = 3Ia1 = 5 pu; base current is 836.7 A so 4184 A. (Key point: Fault current)

2. What is Va1?
   A. -0.5
   B. 0.6667
   C. 1.0
   D. -0.1667
   Answer: B. Va1 = 1 - (j0.2)(-j1.667) = 0.6667 pu. (Key point: Sequence voltage of positive network)

3. Why does Vbc rise above rated voltage?
   A. Because the fault current is large
   B. Because Z2 > Z1, so the healthy-phase voltages exceed normal; Vbc = 2.02 pu = 16.1 kV
   C. Because Zf is high
   D. Because Z0 is zero
   Answer: B. The healthy phase voltages are about 1.04 pu with 76°/-90° geometry, giving Vbc = 2.0207 pu x 7.967 kV = 16.1 kV. (Key point: Line voltage during fault)

## Related in Power System Analysis II

- [DLG fault on 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-fault-30mva-generator) — Power System Analysis II, chapter 5
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