---
title: "HV bus fault and limiting reactor"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:hv-bus-fault-reactor-limit"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/hv-bus-fault-reactor-limit
site: BCT Engineering
---

# HV bus fault and limiting reactor

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

- Asked in: 2074 Chaitra, 14 marks
- Video: [Generator Reactors | Reactor Control of Short Circuit Currents | Electrical Engineering](https://www.youtube.com/watch?v=Z1qjopkJQmM) — Shoaib Ahmed Dayo
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/hv-bus-fault-reactor-limit
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Figure below shows a three-phase power system. (i) Calculate the short circuit MVA and the fault current when a 3-phase balanced short-circuit fault occurs at the High Voltage (HV) bus. (ii) Calculate the ohmic value of reactor 'X' to be placed on the secondary side of transformer 'T2' to limit the Fault Level to by 25%. Assume the system data as under: Generators: G1: 20 MVA, 11 kV, XG1 = 50%; G2: 30 MVA, 11 kV, XG2 = 50%. Transformers: T1: 20 MVA, 11/132 kV, XT1 = 5%; T2: 30 MVA, 11/132 kV, XT2 = 5%. [Figure: G1-T1 and G2-T2 both connected to the HV bus, which supplies a load; fault on the HV bus]

## Short answer

On 30 MVA base, Thevenin reactance at HV bus is 0.33 pu. Fault level is 90.9 MVA, fault current 397.6 A at 132 kV. To reduce fault level by 25% to 68.2 MVA, add a reactor of 0.393 pu (228 Ω/phase) in series with T2 secondary.

## Answer

Base: 30 MVA, 11 kV / 132 kV. Pre-fault voltage is taken as 1.0 pu and pre-fault load current is neglected (usual assumption for fault level calculation). The load on the HV bus is neglected.

#### Per-unit reactances (30 MVA base)
| Element | Calculation | pu |
|---|---|---|
| G1 | $0.50 \times 30/20$ | 0.75 |
| T1 | $0.05 \times 30/20$ | 0.075 |
| G2 | 0.50 (already on 30 MVA) | 0.50 |
| T2 | 0.05 | 0.05 |

```text
 G1--j0.75--T1 j0.075--+
                       +-- HV bus (F)
 G2--j0.50--T2 j0.05--[X]--+
```

#### (i) Fault at the HV bus
$$
\begin{aligned}
X_1 &= 0.75 + 0.075 = 0.825, \quad X_2 = 0.50 + 0.05 = 0.55 \\
X_{th} &= \frac{0.825 \times 0.55}{0.825 + 0.55} = 0.33\ \text{pu} \\
\text{SC MVA} &= \frac{30}{0.33} = 90.91\ \text{MVA} \\
I_{base} &= \frac{30 \times 10^6}{\sqrt{3} \times 132 \times 10^3} = 131.22\ \text{A} \\
I_f &= \frac{1}{0.33} \times 131.22 = 3.0303 \times 131.22 = 397.6\ \text{A}
\end{aligned}
$$

#### (ii) Reactor to reduce the fault level by 25%
"Limit the fault level by 25%" is read as reducing it by 25%, i.e. to 75% of the value above. (Reducing it *to* 25% is impossible here, because the G1 branch alone gives $30/0.825 = 36.4$ MVA, which is 40% of the present level.)

$$
\begin{aligned}
\text{New fault level} &= 0.75 \times 90.91 = 68.18\ \text{MVA} \\
X_{th,new} &= \frac{30}{68.18} = 0.44\ \text{pu}
\end{aligned}
$$

With reactor $X$ (pu) in series with T2 on its 132 kV (secondary) side:

$$
\begin{aligned}
\frac{1}{0.44} &= \frac{1}{0.825} + \frac{1}{0.55 + X} \\
\frac{1}{0.55 + X} &= 2.2727 - 1.2121 = 1.0606 \\
0.55 + X &= 0.9429 \\
X &= 0.3929\ \text{pu}
\end{aligned}
$$

In ohms on the 132 kV side, $Z_{base} = 132^2/30 = 580.8\ \Omega$:

$$
X = 0.3929 \times 580.8 = 228.2\ \Omega
$$

**Answer:** (i) SC MVA $= 90.91$ MVA, fault current $\approx 397.6$ A at 132 kV. (ii) A reactor of about 228 Ω per phase (0.393 pu on 30 MVA) on the 132 kV side of T2 reduces the fault level to 68.2 MVA.

## Check yourself

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

1. G1 (20 MVA, 50%) on the 30 MVA base has X equal to
   A. 0.50
   B. 0.333
   C. 0.75
   D. 0.25
   Answer: C. 0.50 x 30/20 = 0.75 pu. T1 similarly becomes 0.075 pu. (Key point: Per-unit conversion)

2. With branches 0.825 and 0.55 in parallel, SC MVA at the HV bus is
   A. 30/0.33 = 90.9 MVA
   B. 30 x 0.33 = 9.9 MVA
   C. 30/1.375 = 21.8 MVA
   D. 30/0.825 = 36.4 MVA
   Answer: A. X_th = 0.825 x 0.55/1.375 = 0.33 pu, so MVA = 30/0.33 = 90.9. (Key point: Fault MVA at HV bus)

3. The answer reads 'limit by 25%' as a reduction of the fault level by 25%. The new X_th is then
   A. 0.33/0.75 = 0.44 pu, since MVA falls to 75%
   B. 0.33 x 0.25 = 0.0825 pu
   C. 0.33 x 0.75 = 0.2475 pu
   D. 0.33 + 0.25 = 0.58 pu
   Answer: A. Fault level 0.75 x 90.91 = 68.18 MVA, so X_th = 30/68.18 = 0.44 pu. (Key point: Interpreting 25% limit and finding the new X_th)

4. Which equation gives the reactor X placed in series with T2?
   A. 1/0.44 = 1/0.825 + 1/(0.55 + X)
   B. 0.44 = 0.825 + 0.55 + X
   C. 1/0.44 = 1/(0.825 + X) + 1/0.55
   D. 0.44 = 0.33 + X
   Answer: A. The reactor adds to the G2-T2 branch (0.55 + X), which is in parallel with the G1 branch 0.825. Solving gives X = 0.393 pu, or 228 ohm on the 132 kV base (580.8 ohm). (Key point: Reactor value)

## Related in Power System Analysis II

- [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
- [Short circuit at load B (ohmic data)](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ashort-circuit-load-b-ohmic) — Power System Analysis II, chapter 3
- [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×
- [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
- [SLG fault on bus with three alternators](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-three-alternators-common-bus) — Power System Analysis II, chapter 5
- [DLG fault at bus 2, fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-fault-bus2-sequence-networks) — Power System Analysis II, chapter 5

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