---
title: "Short circuit MVA at bus 2"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:short-circuit-mva-bus2-2076"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/short-circuit-mva-bus2-2076
site: BCT Engineering
---

# Short circuit MVA at bus 2

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

- Asked in: 2076 Chaitra, 6 marks
- Video: [SHORT CIRCUIT MVA / FAULT MVA WITH CURRENT LIMITING REACTANCE / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=SB1tqp0fGYY) — praveen raj
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/short-circuit-mva-bus2-2076
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

In the network given below, compute the short circuit MVA for a fault at bus 2. [Figure: generator j0.1 - transformer j0.12 - bus 1; line 1-2 j0.08; line 1-3 j0.06; line 2-3 j0.1; Sbase = 100 MVA]

## Short answer

Using Thevenin reactance at bus 2 on 100 MVA base, the short-circuit MVA is 365.9 MVA.

## Answer

Base: 100 MVA. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). Only one source (the generator) feeds the network.

```text
 E=1--j0.1--j0.12--bus1
                  /    \
             j0.08      j0.06
                /          \
            bus2 --j0.1-- bus3
             |
             F
```

#### Thevenin reactance at bus 2
- Source to bus 1: $0.1 + 0.12 = 0.22$ pu.
- Bus 3 has no source, so lines 1-3 and 3-2 are simply in series: $0.06 + 0.10 = 0.16$ pu.
- This path is in parallel with line 1-2 (0.08 pu):

$$
\begin{aligned}
X_{1-2} &= \frac{0.08 \times 0.16}{0.08 + 0.16} = 0.0533\ \text{pu} \\
X_{th} &= 0.22 + 0.0533 = 0.2733\ \text{pu}
\end{aligned}
$$

#### Short-circuit MVA
$$
\begin{aligned}
I_f &= \frac{1}{0.2733} = 3.6585\ \text{pu} \\
\text{SC MVA} &= \frac{S_{base}}{X_{th}} = \frac{100}{0.2733} = 365.85\ \text{MVA}
\end{aligned}
$$

**Answer:** short-circuit MVA at bus 2 $\approx 365.9$ MVA (fault current 3.659 pu).

## Check yourself

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

1. Bus 3 has no generator. How are lines 1-3 and 3-2 treated when finding the Thevenin reactance at bus 2?
   A. In parallel, giving 0.0375 pu
   B. Ignored completely
   C. In series, giving 0.06 + 0.10 = 0.16 pu
   D. As a delta that must be left unreduced
   Answer: C. With no source or load at bus 3, the same current flows through both lines, so they are simply in series. (Key point: Bus without source)

2. What is the reactance between bus 1 and bus 2 with the 0.16 pu path in parallel with line 1-2 (0.08 pu)?
   A. 0.24 pu
   B. 0.0533 pu
   C. 0.12 pu
   D. 0.08 pu
   Answer: B. 0.08 x 0.16/(0.08 + 0.16) = 0.0533 pu. (Key point: Parallel paths from bus 1 to bus 2)

3. With X_th = 0.1 + 0.12 + 0.0533 = 0.2733 pu on 100 MVA, the short-circuit MVA is
   A. 27.3 MVA
   B. 100 x 0.2733 MVA
   C. 365.9 MVA
   D. 1/0.2733 MVA
   Answer: C. SC MVA = S_base/X_th = 100/0.2733 = 365.9 MVA. (Key point: Short-circuit MVA)

## Related in Power System Analysis II

- [Fault MVA at bus 5 with inductive lines](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-mva-bus5-mh-lines) — Power System Analysis II, chapter 3, asked 2×
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