---
title: "Initial symmetrical current, load short circuit"
subject: "Power System Analysis II"
chapter: 3
key: "semester_5/power_system_analysis_ii:initial-current-load-short-circuit"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/initial-current-load-short-circuit
site: BCT Engineering
---

# Initial symmetrical current, load short circuit

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

- Asked in: 2079 Bhadra, 4 marks
- Video: [Example Q1.1 Three Phase Short circuit current EMF Method](https://www.youtube.com/watch?v=QOXO_96Y-Co) — Francisco Gonzalez-Longatt
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/initial-current-load-short-circuit
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-3

## Question

A synchronous generator is rated 500 MVA, 20 kV, 60 Hz with having subtransient reactance (Xd'') = 0.20 per unit. It supplies a purely (resistive) load of 400 MW at 20 kV. The load is connected directly across the terminals of generator. If all the three phase of load are short circuited simultaneously, find the initial symmetrical rms current in the generator in per unit on a base of 500 MVA, 20 kV by using internal voltages of the machine.

## Short answer

Using internal voltage from pre-fault resistive load, the initial symmetrical rms current is 5.0636 pu (73.1 kA on 500 MVA, 20 kV base).

## Answer

Base: 500 MVA, 20 kV. The terminal voltage is $V_t = 1\angle 0^\circ$ pu.

#### Pre-fault load current
The load is purely resistive, so the current is in phase with $V_t$:

$$
I_L = \frac{P}{V_t} = \frac{400/500}{1} = 0.8\angle 0^\circ\ \text{pu}
$$

#### Internal (sub-transient) voltage
$$
\begin{aligned}
E'' &= V_t + jX''_dI_L = 1 + j0.2(0.8) \\
&= 1 + j0.16 = 1.0127\angle 9.09^\circ\ \text{pu}
\end{aligned}
$$

#### Fault current
When all three load phases are shorted, the terminals are at zero voltage and the generator sees only $X''_d$:

$$
\begin{aligned}
I'' &= \frac{E''}{jX''_d} = \frac{1 + j0.16}{j0.2} = 0.8000 - j5.0000\ \text{pu} \\
|I''| &= \sqrt{0.8^2 + 5^2} = 5.0636\ \text{pu}
\end{aligned}
$$

In amperes: $I_{base} = \dfrac{500 \times 10^6}{\sqrt{3} \times 20 \times 10^3} = 14433.8$ A, so $|I''| = 5.0636 \times 14433.8 \approx 73.09$ kA.

**Answer:** initial symmetrical rms current $= 5.0636$ pu (about 73.1 kA).

## Check yourself

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

1. What is the prefault current for a purely resistive 400 MW load on a 500 MVA base?
   A. 0.8 pu at 0 deg
   B. 0.8 pu lagging
   C. 1.25 pu
   D. 0.4 pu
   Answer: A. I = P/Vt = 0.8/1 = 0.8 pu, in phase with Vt for a resistive load. (Key point: Prefault load current)

2. What is the sub-transient internal voltage E''?
   A. 1 + j0.16 = 1.0127 at 9.09 deg
   B. 1 - j0.16
   C. 1.0 at 0 deg
   D. 1.16 at 0 deg
   Answer: A. E'' = Vt + jX''d I_L = 1 + j0.2 x 0.8. (Key point: Internal voltage)

3. With the load terminals shorted, what limits the generator current?
   A. Only X''d, so I'' = E''/jX''d = 5.06 pu
   B. The load resistance in series
   C. Vt divided by the load resistance
   D. X''d plus the load resistance
   Answer: A. Shorting the load makes terminal voltage zero, so the load drops out and I'' = (1 + j0.16)/j0.2, magnitude 5.0636 pu. (Key point: Fault current)

## Related in Power System Analysis II

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