---
title: "Fault at load end of overhead line"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-load-end-33kv-line"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-load-end-33kv-line
site: BCT Engineering
---

# Fault at load end of overhead line

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

- Asked in: 2073 Chaitra, 10 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-load-end-33kv-line
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

Figure below shows a three-phase system. Calculate the short circuit MVA and the fault current when 3-phase balanced short-circuit fault occurs at the load end of the transmission line. Also, determine the fault current supplied by each generator. [Figure: G1 10 MVA, 11 kV, X1'' = 10% and G2 5 MVA, 11 kV, X2'' = 7.5% on a common bus; transformer T1 15 MVA, 11/33 kV, XT = 6%; overhead line XTL = 20 Ω to the load, where the fault occurs]

## Short answer

On 15 MVA base, Thevenin reactance is 0.4255 pu. Fault level is 35.3 MVA, fault current 617 A at 33 kV. On the 11 kV side, total fault current is 1850 A; G1 supplies 1110 A, G2 supplies 740 A.

## Answer

Base: 15 MVA (transformer rating), 11 kV on the generator side and 33 kV on the line side. 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 (15 MVA)
| Element | Calculation | pu |
|---|---|---|
| G1 | $0.10 \times 15/10$ | 0.150 |
| G2 | $0.075 \times 15/5$ | 0.225 |
| T1 | 6% on 15 MVA | 0.06 |
| Line | $20/(33^2/15) = 20/72.6$ | 0.2755 |

```text
 G1 j0.150--+
               +--T1 j0.06--line j0.275--F (load end)
 G2 j0.225--+
```

#### Thevenin reactance
$$
\begin{aligned}
X_G &= \frac{0.150 \times 0.225}{0.150 + 0.225} = 0.090 \\
X_{th} &= 0.090 + 0.06 + 0.2755 = 0.4255\ \text{pu}
\end{aligned}
$$

#### Short-circuit MVA and fault current
$$
\begin{aligned}
\text{SC MVA} &= \frac{15}{0.4255} = 35.25\ \text{MVA} \\
I_f &= \frac{1}{0.4255} = 2.3503\ \text{pu} \\
I_{base,33} &= \frac{15 \times 10^6}{\sqrt{3} \times 33 \times 10^3} = 262.43\ \text{A} \Rightarrow I_f = 616.8\ \text{A at 33 kV}
\end{aligned}
$$

#### Current supplied by each generator
On the 11 kV side, $I_{base,11} = \dfrac{15 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 787.30$ A, so the fault current there is $2.3503 \times 787.30 = 1850.4$ A. It divides inversely as the generator reactances:

$$
\begin{aligned}
I_{G1} &= 1850.4 \times \frac{0.225}{0.150 + 0.225} = 1110.2\ \text{A} \\
I_{G2} &= 1850.4 \times \frac{0.150}{0.150 + 0.225} = 740.1\ \text{A}
\end{aligned}
$$

**Answer:** SC MVA $\approx 35.25$ MVA; fault current $\approx 617$ A at 33 kV; G1 supplies $\approx 1110$ A and G2 $\approx 740$ A (at 11 kV).

## Check yourself

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

1. The 20 ohm line at 33 kV on 15 MVA base is
   A. 20/(11^2/15) = 2.48 pu
   B. 20 x 72.6 pu
   C. 20/72.6 = 0.2755 pu
   D. 20/33 pu
   Answer: C. Z_base = 33^2/15 = 72.6 ohm, so X = 20/72.6 = 0.2755 pu. (Key point: Line per-unit)

2. X_th at the load end is
   A. 0.15 + 0.225 + 0.06 + 0.2755
   B. (0.15||0.225) + 0.06 + 0.2755 = 0.4255 pu
   C. 0.15||0.225||0.06||0.2755
   D. 0.15 + (0.225||0.06) + 0.2755
   Answer: B. The generators are in parallel (0.09 pu) and in series with T1 and the line, giving 0.4255 pu. (Key point: Thevenin reactance)

3. At the 11 kV bus the fault current is 1850 A. Why does G1 (X = 0.15) supply 1110 A and G2 (X = 0.225) supply 740 A?
   A. G1 has the larger MVA rating, so share is proportional to MVA only
   B. Current divides inversely as the generator reactances
   C. Current divides directly as reactances
   D. The two share equally
   Answer: B. I_G1 = I x 0.225/0.375 and I_G2 = I x 0.15/0.375, so the smaller reactance carries more. (Key point: Generator currents)

## Related in Power System Analysis II

- [Fault at far end of 33 kV line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-far-end-33kv-line) — Power System Analysis II, chapter 3
- [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
- [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
- [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-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
- [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×

## Related in other subjects

- [Per unit circuit of three-generator system, 30 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Apu-circuit-30mva-g1-g2-g3) — Power System Analysis I, chapter 3
- [Reactance diagram: 20 MVA generator, 15 kV/25 MVA base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-20mva-15kv-25mva) — Power System Analysis I, chapter 3
- [Fault current: two 50 MVA generators, feeder reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mva-feeder-reactor) — Power Plant Equipment, chapter 4, asked 3×
- [Differential protection: 11 kV, 100 MVA generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Agenerator-differential-11kv-6-ohm) — Switchgear and Protection, chapter 7