---
title: "Symmetrical fault at bus 3"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-bus3-delta-network"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-bus3-delta-network
site: BCT Engineering
---

# Symmetrical fault at bus 3

Power System Analysis II · Chapter 3: Power System Fault Calculation

- Asked in: 2072 Kartik, 4 marks
- Video: [symmetrical fault analysis in power system | symmetrical fault analysis solved problems | fault](https://www.youtube.com/watch?v=8HGInf9rYAU) — Mind of Engineer 
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-bus3-delta-network
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

A 3-phase symmetrical fault occurs at bus 3 in the given power system network. Determine the fault current. Per unit values of reactance are based on 100 MVA base. [Figure: generator j0.1 (Y grounded) through transformer j0.1 (Y grounded/Y grounded) to bus 1; generator j0.2 (Y grounded) through transformer j0.2 (Δ/Y grounded) to bus 2; line 1-2 j0.8; line 1-3 j0.4; line 2-3 j0.4]

## Short answer

Convert the line delta to star: X_th = 0.34 pu on 100 MVA base. Fault current = 1 / 0.34 = 2.94 pu, fault level = 294 MVA. Actual amperes depend on the bus voltage base.

## Answer

A three-phase fault is balanced, so only the positive-sequence network is used; the transformer connections (Y/Δ) and grounding do not affect it. Pre-fault voltage is 1.0 pu and load current is neglected.

#### Positive-sequence network
- Source 1 to bus 1: $0.1 + 0.1 = 0.2$ pu.
- Source 2 to bus 2: $0.2 + 0.2 = 0.4$ pu.
- Lines: $X_{12} = 0.8$, $X_{13} = 0.4$, $X_{23} = 0.4$ pu (a delta between buses 1, 2, 3).

#### Delta-star conversion of the line delta
$$
\begin{aligned}
X_1 &= \frac{X_{12}X_{13}}{X_{12} + X_{13} + X_{23}} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\
X_2 &= \frac{X_{12}X_{23}}{1.6} = \frac{0.8 \times 0.4}{1.6} = 0.20 \\
X_3 &= \frac{X_{13}X_{23}}{1.6} = \frac{0.4 \times 0.4}{1.6} = 0.10
\end{aligned}
$$

```text
 E=1-j0.2-bus1-j0.2-+
                    N--j0.1--bus3 (F)
 E=1-j0.4-bus2-j0.2-+
```

#### Thevenin reactance at bus 3
$$
\begin{aligned}
X_{path1} &= 0.2 + 0.20 = 0.40, \quad X_{path2} = 0.4 + 0.20 = 0.60 \\
X_{th} &= \frac{0.40 \times 0.60}{0.40 + 0.60} + 0.10 = 0.24 + 0.10 = 0.34\ \text{pu}
\end{aligned}
$$

(The same value is $Z_{33}$ of the bus impedance matrix.)

#### Fault current
$$
\begin{aligned}
I_f &= \frac{V_f}{jX_{th}} = \frac{1}{j0.34} = -j2.9412\ \text{pu} \\
\text{Fault MVA} &= \frac{100}{0.34} = 294.1\ \text{MVA}
\end{aligned}
$$

**Answer:** $I_f = 2.941$ pu (fault level 294.1 MVA on the 100 MVA base). The current in amperes is $I_f \times I_{base}$ once the bus voltage base is known.

## Check yourself

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

1. For a balanced 3-phase fault, which network is needed?
   A. Positive, negative and zero sequence in series
   B. Zero sequence only, because of the grounded transformers
   C. Positive sequence only; transformer connections do not matter
   D. Negative sequence only
   Answer: C. A balanced fault produces only positive-sequence current, so Y/Delta and grounding details do not affect the result. (Key point: Sequence network used)

2. Converting the line delta (0.8, 0.4, 0.4), the star arm to bus 3 is
   A. 0.8 x 0.4/1.6 = 0.20
   B. 0.4 x 0.4/1.6 = 0.10
   C. 0.8 + 0.4 + 0.4 = 1.6
   D. 0.4 x 0.4/0.8 = 0.20
   Answer: B. The arm at bus 3 is X13 x X23/(X12 + X13 + X23) = 0.16/1.6 = 0.10 pu. (Key point: Delta-star conversion)

3. With paths 0.40 and 0.60 in parallel and then 0.10 in series, the fault current is
   A. 1/0.24 = 4.17 pu
   B. 1/0.34 = 2.94 pu
   C. 1/1.1 = 0.91 pu
   D. 1/0.10 = 10 pu
   Answer: B. X_th = 0.24 + 0.10 = 0.34 pu, so I_f = 1/0.34 = 2.94 pu. (Key point: Thevenin and fault current)

## Related in Power System Analysis II

- [Fault through impedance at bus 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-impedance-bus2-actual-currents) — Power System Analysis II, chapter 3
- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [L-L fault current in kA at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-bus3-ka) — Power System Analysis II, chapter 5
- [L-G fault at bus 1 through Zf](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-bus1-parallel-lines) — Power System Analysis II, chapter 5
- [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5
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