---
title: "Subtransient currents for fault at motor terminals"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:subtransient-currents-motor-fault"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/subtransient-currents-motor-fault
site: BCT Engineering
---

# Subtransient currents for fault at motor terminals

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

- Asked in: 2076 Asoj, 8 marks
- Video: [Fault Analysis by Thevenin theorem | Numerical -1 | Power System Analysis | Solved Problem-1](https://www.youtube.com/watch?v=Z_aCanO-IOg) — Simple n Easy Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/subtransient-currents-motor-fault
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

A synchronous generator and motor are rated 30 MVA, 13.2 kV and both have subtransient reactances of 20%. The line connecting them has reactance of 8% on the base of the machine ratings. The motor is drawing 20 MW at 0.8 pf leading and terminal voltage of 12.8 kV when a symmetrical three phase fault occurs at the motor terminals. Find the subtransient currents in the generator, the motor and fault by using the internal voltages of machines.

## Short answer

With 30 MVA, 13.2 kV base and internal EMFs from pre-fault load, subtransient generator current is 3972 A, motor current is 7096 A, and fault current is 10906 A.

## Answer

Base: 30 MVA, 13.2 kV (the machine ratings). The motor terminal voltage is the reference. Internal EMFs are found from the pre-fault load, then each machine's contribution to the fault at the motor terminals is found separately.

#### Per-unit data
- $X''_g = X''_m = 0.20$ pu, line $X_L = 0.08$ pu.
- $V_f = \dfrac{12.8}{13.2} = 0.9697\angle 0^\circ$ pu.
- Motor load: $P = 20/30 = 0.6667$ pu at 0.8 pf leading.

```text
 Eg --j0.2--+--j0.08--+--j0.2-- Em
          (gen)       (Vf) motor terminals
                       |
                       F
```

#### Pre-fault current
$$
\begin{aligned}
|I_L| &= \frac{0.6667}{0.9697 \times 0.8} = 0.8594\ \text{pu} \\
I_L &= 0.8594\angle 36.87^\circ = 0.6875 + j0.5156\ \text{pu}
\end{aligned}
$$

#### Internal voltages
$$
\begin{aligned}
E''_g &= V_f + j(X_L + X''_g)I_L = 0.9697 + j0.28(0.6875 + j0.5156) = 0.8253 + j0.1925\ \text{pu} \\
E''_m &= V_f - jX''_mI_L = 0.9697 - j0.2(0.6875 + j0.5156) = 1.0728 - j0.1375\ \text{pu}
\end{aligned}
$$

#### Sub-transient currents with the fault at the motor terminals
$$
\begin{aligned}
I''_g &= \frac{E''_g}{j(X''_g + X_L)} = \frac{0.8253 + j0.1925}{j0.28} = 0.6875 - j2.9476\ \text{pu} \\
I''_m &= \frac{E''_m}{jX''_m} = \frac{1.0728 - j0.1375}{j0.2} = -0.6875 - j5.3641\ \text{pu} \\
I''_f &= I''_g + I''_m = -j8.3117\ \text{pu}
\end{aligned}
$$

Base current: $I_{base} = \dfrac{30 \times 10^6}{\sqrt{3} \times 13.2 \times 10^3} = 1312.2$ A.

| Current | pu | Amperes |
|---|---|---|
| Generator $I''_g$ | 3.0267 | 3972 |
| Motor $I''_m$ | 5.4080 | 7096 |
| Fault $I''_f$ | 8.3117 | 10906 |

Check (Thevenin): $I''_f = V_f\left(\frac{1}{j0.28} + \frac{1}{j0.2}\right) = -j8.3117$ pu, the same.

**Answer:** $I''_g \approx 3972$ A, $I''_m \approx 7096$ A, $I''_f \approx 10906$ A.

## Check yourself

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

1. With the fault at the motor terminals, which reactance limits the generator's current?
   A. X''g + X_L = 0.28 pu
   B. X''g = 0.20 pu
   C. X''m = 0.20 pu
   D. X_L = 0.08 pu
   Answer: A. The generator feeds the fault through its own reactance and the line, while the motor feeds directly through X''m. (Key point: Fault location effect)

2. How do the internal EMFs relate to the terminal voltage Vf and load current I_L?
   A. E''g = Vf + j(X_L + X''g) I_L; E''m = Vf - jX''m I_L
   B. E''g = Vf - j(X_L + X''g) I_L; E''m = Vf + jX''m I_L
   C. Both equal Vf
   D. Both equal 1 pu
   Answer: A. The generator is a source so its EMF is the terminal voltage plus the drop; the motor is a load so its EMF is the terminal voltage minus the drop. (Key point: Internal EMFs)

3. Which gives the total subtransient fault current in pu?
   A. I_f = I''g + I''m = 8.31 pu (10906 A)
   B. I_f = I''g only = 3.03 pu
   C. I_f = I''m - I''g
   D. I_f = 1/0.48 = 2.08 pu
   Answer: A. Contributions add: 3.03 + 5.41 (phasor sum -j8.3117 pu), times I_base 1312.2 A = 10906 A. (Key point: Totals)

## Related in Power System Analysis II

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- [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-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
- [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
- [DLG fault on 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Adlg-fault-30mva-generator) — Power System Analysis II, chapter 5
- [SLG fault on 20 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aslg-fault-20mva-generator) — Power System Analysis II, chapter 5

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