---
title: "Fault MVA at bus 5 with inductive lines"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-mva-bus5-mh-lines"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-mva-bus5-mh-lines
site: BCT Engineering
---

# Fault MVA at bus 5 with inductive lines

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

- Asked in: 2081 Bhadra, 8 marks; 2083 Baishakh (new course), 4+3 marks (asked 2 times)
- 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-mva-bus5-mh-lines
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Figure below shows a power system network. Each alternator G1 and G2 are rated at 100 MVA, 11 kV and has a sub-transient reactance of 20%. Each of the transformer is rated at 100 MVA, 11/132 kV and has a leakage reactance of 5%. The line L1 of 1.2 mH/phase/km and a length of 100 km. Lines L2 and L3 have inductance of 1.0 mH/phase/km and has a length of 50 km each. Find the fault MVA and fault current for the fault at the bus 5 (point F). [Figure: G2 at bus 2 - T2 - bus 4; G1 at bus 1 - T1 - bus 3; L1 between buses 4 and 3; L2 from bus 4 to bus 5; L3 from bus 5 to bus 3; the fault F is at bus 5]

## Short answer

On 100 MVA base, each source branch 0.25 pu, L2 and L3 0.0902 pu; L1 carries no current. Thevenin reactance at bus 5 = 0.1701 pu. Fault MVA = 588.0 MVA, fault current = 2.572 kA at 132 kV.

## Answer

Base: 100 MVA, 11 kV (generator side), 132 kV (line side), $f = 50$ Hz. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation).

#### Per-unit reactances
- G1, G2: $X'' = 0.20$ pu each (already on 100 MVA, 11 kV).
- T1, T2: $X = 0.05$ pu each.
- So from each generator EMF to its 132 kV bus: $0.20 + 0.05 = 0.25$ pu.

Lines ($X = 2\pi fL$):

$$
\begin{aligned}
Z_{base} &= \frac{132^2}{100} = 174.24\ \Omega \\
X_{L1} &= 2\pi(50)(1.2 \times 10^{-3} \times 100) = 37.699\ \Omega = \frac{37.699}{174.24} = 0.2164\ \text{pu} \\
X_{L2} = X_{L3} &= 2\pi(50)(1.0 \times 10^{-3} \times 50) = 15.708\ \Omega = 0.0902\ \text{pu}
\end{aligned}
$$

#### Reactance diagram
```text
  E=1                         E=1
   |                           |
 j0.25 (G1+T1)               j0.25 (G2+T2)
   |                           |
 bus3 ------- L1 j0.216 ------- bus4
    \                         /
  L3 j0.090            L2 j0.090
       \                   /
            bus 5 (F)
```

#### Thevenin reactance at bus 5
Both generator EMFs are equal (1 pu), and the two halves of the network are identical (0.25 pu to each source, $X_{L2} = X_{L3}$). So buses 3 and 4 are always at the same potential and **line L1 carries no current**; it can be removed. The two paths (0.25 + 0.0902) are then in parallel:

$$
\begin{aligned}
X_{th} &= \frac{0.25 + 0.0902}{2} = 0.1701\ \text{pu}
\end{aligned}
$$

(The same value is obtained as $Z_{55}$ of the bus impedance matrix.)

#### Fault current and fault MVA
$$
\begin{aligned}
I_f &= \frac{1}{0.1701} = 5.8797\ \text{pu} \\
I_{base} &= \frac{100 \times 10^6}{\sqrt{3}(132 \times 10^3)} = 437.39\ \text{A} \\
I_f &= 5.8797 \times 437.39 = 2571.7\ \text{A} \\
\text{Fault MVA} &= \frac{100}{0.1701} = 587.97\ \text{MVA}
\end{aligned}
$$

**Answer:** fault MVA $\approx 588.0$ MVA and fault current $\approx 2.572$ kA at 132 kV.

## Check yourself

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

1. How is L2's per-unit reactance found (1 mH/km, 50 km, 50 Hz, 132 kV, 100 MVA)?
   A. 2 pi f L = 15.708 ohm, divided by Z_base 174.24 = 0.0902 pu
   B. 1 mH x 50 = 0.05 pu directly
   C. 15.708 / 100 = 0.157 pu
   D. 15.708 / 132 = 0.119 pu
   Answer: A. X = 2 pi (50)(0.05 H) = 15.708 ohm and Z_base = 174.24 ohm give 0.0902 pu. (Key point: Line reactance)

2. Why can line L1 be removed?
   A. Both sources are equal and the network is symmetrical, so buses 3 and 4 are at the same potential and L1 carries no current
   B. L1 is the longest line
   C. L1 is in series with the fault
   D. L1 has the highest reactance
   Answer: A. Equal EMFs and identical branches make V3 = V4, so no current flows in L1. (Key point: Symmetry)

3. What are X_th at bus 5 and the fault MVA?
   A. 0.1701 pu and about 588 MVA
   B. 0.3402 pu and about 294 MVA
   C. 0.25 pu and 400 MVA
   D. 0.0902 pu and 1109 MVA
   Answer: A. X_th = (0.25 + 0.0902)/2 = 0.1701 and MVA = 100/0.1701 = 588. (Key point: Thevenin and MVA)

## Related in Power System Analysis II

- [Fault at receiving end of 132 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-132kv-receiving-end-2071) — Power System Analysis II, chapter 3
- [L-G fault at bus 3, parallel lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-bus3-two-generators) — 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
- [Fault level at end of two 200 km lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-two-stations-132kv-200km) — Power System Analysis II, chapter 3, asked 2×
- [Fault at receiving end of 220 kV lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-220kv-two-generators) — 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

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