---
title: "Fault at generator terminal feeding motor"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-generator-terminal-motor-new"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-generator-terminal-motor-new
site: BCT Engineering
---

# Fault at generator terminal feeding motor

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

- Asked in: 2082 Bhadra (new course), 3+4 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-generator-terminal-motor-new
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

An alternator is connected to a synchronous motor through a transmission line and two transformers as shown in the figure below. The rated MVA, voltage rating and percentage reactance of the components are depicted in the figure. The transmission line reactance is also shown in the figure in Ω. Compute the fault current, if a 3-phase bolted fault occurs at the terminal of the generator. Consider the pre fault voltage as 1 p.u. Also determine the fault MVA at the generator terminal. [Figure: generator 50 MVA, 11 kV, 10% - T1 100 MVA, 11/132 kV, 5% - line j25 Ω - T2 50 MVA, 132/6.6 kV, 4% - motor 50 MVA, 6.6 kV, 6%]

## Short answer

Base 50 MVA, 11 kV. Motor-side series reactance = 0.1967 pu; in parallel with generator 0.1 pu gives Thevenin reactance 0.0663 pu. Fault current = 1/0.0663 = 15.08 pu = 39.6 kA. Fault MVA = 50/0.0663 = 754 MVA.

## Answer

For a bolted fault at the generator terminal, both the generator and the synchronous motor (acting as a source) feed the fault. $I_f = V_f / X_{th}$ with $V_f = 1$ pu.

#### Per unit values
Base: 50 MVA, 11 kV at the generator, so 132 kV on the line and 6.6 kV at the motor (voltage bases follow the transformer ratios, so no voltage correction is needed). $Z_{b,line} = 132^2/50 = 348.48\ \Omega$.

| Element | Calculation | X (pu) |
|---|---|---|
| Generator (50 MVA, 10%) | own base | 0.10 |
| T1 (100 MVA, 5%) | $0.05 \times 50/100$ | 0.025 |
| Line (25 Ω) | $25/348.48$ | 0.0717 |
| T2 (50 MVA, 4%) | own base | 0.04 |
| Motor (50 MVA, 6%) | own base | 0.06 |

```text
        F
 G      o   T1     Line     T2     M
(~)-0.1-+-0.025--0.0717--0.04--0.06-(~)
 |                                   |
 +====== reference (EMF = 1 pu) =====+
```

#### Fault current
Reactance on the motor side seen from F:
$$
X_m = 0.025 + 0.0717 + 0.04 + 0.06 = 0.1967\ \text{pu}
$$
$$
\begin{aligned}
X_{th} &= \frac{0.10 \times 0.1967}{0.10 + 0.1967} = 0.0663\ \text{pu} \\
I_f &= \frac{1}{0.0663} = 15.08\ \text{pu} \\
I_b &= \frac{50\times10^6}{\sqrt3 \times 11\times10^3} = 2624.3\ \text{A} \\
I_f &= 15.08 \times 2624.3 = 39\,582\ \text{A} \approx 39.6\ \text{kA}
\end{aligned}
$$
Contributions: generator $1/0.1 = 10$ pu (26.24 kA); motor side $1/0.1967 = 5.08$ pu (13.34 kA).

#### Fault MVA at the generator terminal
$$
\text{Fault MVA} = \frac{S_b}{X_{th}} = \frac{50}{0.0663} = 754.1\ \text{MVA}
$$
Check: $\sqrt3 \times 11 \times 39.58 = 754.1$ MVA.

**Answer: Fault current ≈ 15.08 pu ≈ 39.6 kA; fault MVA ≈ 754 MVA.**

## Check yourself

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

1. Why is the synchronous motor included as a source in the fault calculation?
   A. It acts as a generator and feeds current into the fault
   B. It has no effect on the fault
   C. It increases the line impedance
   D. It only absorbs current
   Answer: A. During a fault the motor's rotating field acts as an EMF behind its reactance and feeds the fault. (Key point: Motor as a source)

2. T1 (100 MVA, 5%) on 50 MVA base is
   A. 0.10 pu
   B. 0.05 pu
   C. 0.0025 pu
   D. 0.025 pu
   Answer: D. 0.05 x 50/100 = 0.025 pu. (Key point: Per-unit conversion)

3. The motor-side reactance 0.1967 pu is combined with the generator's 0.10 pu
   A. In series, giving 0.2967
   B. In parallel, giving 0.0663 pu
   C. By subtraction
   D. Only the larger is used
   Answer: B. Both sources feed the same fault point, so their branches are in parallel: 0.10 x 0.1967/0.2967 = 0.0663 pu. (Key point: Thevenin reactance at generator terminal)

4. With X_th = 0.0663 pu on 50 MVA, fault MVA is
   A. 50 x 0.0663
   B. 50/0.0663 = 754 MVA
   C. 1/0.0663 MVA
   D. 50/0.1 = 500 MVA
   Answer: B. Fault MVA = S_base/X_th = 754 MVA, which checks with sqrt(3) x 11 x 39.6 kA. (Key point: Fault MVA)

## Related in Power System Analysis II

- [Generator and motor currents for terminal fault](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-current-generator-motor-2080) — Power System Analysis II, chapter 3
- [L-L fault line voltages, 25 MVA alternator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-25mva-line-voltages) — Power System Analysis II, chapter 5, asked 2×
- [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
- [Neutral reactance for 15 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-15mva) — Power System Analysis II, chapter 5
- [SLG fault at mid-line through 4.033 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-mid-line-motor) — Power System Analysis II, chapter 5
- [L-L fault through Zf with two motors](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-motors-zf-new) — Power System Analysis II, chapter 5

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