---
title: "Generator and motor currents for terminal fault"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:fault-current-generator-motor-2080"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/fault-current-generator-motor-2080
site: BCT Engineering
---

# Generator and motor currents for terminal fault

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

- Asked in: 2080 Bhadra, 8 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-current-generator-motor-2080
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

An 11 kV 100 MVA alternator having a sub-transient reactance of 0.25 pu is supplying a 50 MVA motor having a sub-transient reactance of 0.2 pu through a transmission line. The line reactance is 0.05 pu on a base of 100 MVA. The motor is drawing 40 MW at 0.8 p.f. leading with a terminal voltage of 10.95 kV when three phase fault occurs at the generator terminals. Calculate the total current in generator and motor under fault conditions.

## Short answer

Using sub-transient EMFs from pre-fault load on 100 MVA base, the generator fault current is 19.17 kA, motor fault current is 13.15 kA, and total fault current is 32.02 kA.

## Answer

Use the internal (sub-transient) EMFs of both machines found from the pre-fault load. Base: 100 MVA, 11 kV.

#### Per-unit values
- Generator: $X''_g = 0.25$ pu.
- Motor: $X''_m = 0.2 \times \dfrac{100}{50} = 0.4$ pu.
- Line: $X_L = 0.05$ pu.
- Motor terminal voltage: $V_m = \dfrac{10.95}{11} = 0.9955\angle 0^\circ$ pu (reference).
- Motor input: $P = 40/100 = 0.4$ pu at 0.8 pf leading.

```text
 Eg --j0.25--(Vt)--j0.05--(Vm)--j0.4-- Em
             |
             F (fault at generator terminals)
```

#### Pre-fault current
$$
\begin{aligned}
|I| &= \frac{P}{V_m\cos\phi} = \frac{0.4}{0.9955 \times 0.8} = 0.5023\ \text{pu} \\
I &= 0.5023\angle 36.87^\circ = 0.4018 + j0.3014\ \text{pu}\ \text{(leading)}
\end{aligned}
$$

#### Internal EMFs
$$
\begin{aligned}
V_t &= V_m + jX_LI = 0.9955 + j0.05(0.4018 + j0.3014) = 0.9804 + j0.0201\ \text{pu} \\
E''_g &= V_t + jX''_gI = 0.9804 + j0.0201 + j0.25(0.4018 + j0.3014) = 0.9050 + j0.1205 = 0.9130\angle 7.59^\circ\ \text{pu} \\
E''_m &= V_m - jX''_mI = 0.9955 - j0.4(0.4018 + j0.3014) = 1.1160 - j0.1607 = 1.1275\angle -8.20^\circ\ \text{pu}
\end{aligned}
$$

#### Currents during the fault
With the generator terminals shorted, each machine feeds the fault through its own reactance:

$$
\begin{aligned}
I''_g &= \frac{E''_g}{jX''_g} = \frac{0.9050 + j0.1205}{j0.25} = 0.4822 - j3.6202 = 3.6521\angle -82.41^\circ\ \text{pu} \\
I''_m &= \frac{E''_m}{j(X''_m + X_L)} = \frac{1.1160 - j0.1607}{j0.45} = -0.3572 - j2.4800 = 2.5056\angle -98.20^\circ\ \text{pu} \\
I_f &= I''_g + I''_m = 0.1250 - j6.1002 = 6.1015\angle -88.83^\circ\ \text{pu}
\end{aligned}
$$

Base current at 11 kV: $I_{base} = \dfrac{100 \times 10^6}{\sqrt{3} \times 11 \times 10^3} = 5248.6$ A.

| Current | pu | Amperes |
|---|---|---|
| Generator, $I''_g$ | 3.6521 | 19169 |
| Motor, $I''_m$ | 2.5056 | 13151 |
| Fault, $I_f$ | 6.1015 | 32024 |

Check: $I_f = V_t\left(\frac{1}{j0.25} + \frac{1}{j0.45}\right) = 0.1250 - j6.1002$ pu, the same.

**Answer:** generator current $\approx 19.17$ kA, motor current $\approx 13.15$ kA, total fault current $\approx 32.02$ kA.

## Check yourself

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

1. What is the motor's sub-transient reactance on the 100 MVA base?
   A. 0.4 pu
   B. 0.2 pu
   C. 0.1 pu
   D. 0.8 pu
   Answer: A. 0.2 x 100/50 = 0.4 pu. (Key point: Motor reactance on new base)

2. Why must internal EMFs of both machines be found from the pre-fault load?
   A. The motor also feeds the fault, and the load current makes the two EMFs differ from 1 pu and from each other
   B. To account for the transformer ratio
   C. Because the fault current is zero before the fault
   D. Because the line has resistance
   Answer: A. With prefault current flowing, E''g and E''m differ (0.9130 and 1.1275 pu), so each machine's contribution is computed from its own EMF. (Key point: Why internal EMFs)

3. What is the prefault current magnitude and its phase?
   A. 0.5023 pu at +36.87 deg (leading)
   B. 0.4 pu at 0 deg
   C. 0.5023 pu at -36.87 deg
   D. 0.8 pu at 36.87 deg
   Answer: A. I = P/(V cos phi) = 0.4/(0.9955 x 0.8) = 0.5023 pu, leading by acos(0.8) = 36.87 deg. (Key point: Prefault current)

4. With the fault at the generator terminals, which reactance limits the motor's contribution?
   A. X''m + X_L = 0.45 pu
   B. X''m = 0.4 pu only
   C. X''g = 0.25 pu
   D. X_L = 0.05 pu only
   Answer: A. The motor feeds the fault through its own reactance plus the line, while the generator feeds directly through X''g; the currents add to 6.10 pu = 32.0 kA. (Key point: Fault contributions)

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