---
title: "Fault current: 50 MW and 40 MW generators"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:fault-50mw-40mw-equal-current"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-50mw-40mw-equal-current
site: BCT Engineering
---

# Fault current: 50 MW and 40 MW generators

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2076 Asoj, 8 marks
- Video: [power plant design | calculation of fault current and fault MVA @smartstudy11](https://www.youtube.com/watch?v=mKqg6UYa6NI) — jitendra sah
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-50mw-40mw-equal-current
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

Below figure show the single line diagram of parallel operated generators. Their ratings are: G1: capacity = 50 MW, 11kV, X1 = 0.16 pu based on its rating; G2: capacity = 40 MW, 11kV, X2 = 0.12 pu based on its rating; XL = 2 ohms. [Figure: G1 (through X1) and G2 (through X2), both star-grounded, connected to a common busbar; the outgoing feeder leaves the busbar through XL] a) If a three-phase to ground fault occurs at outgoing feeder, calculate the fault current and fault power at the outgoing feeder and fault current supplied by each generator. b) Calculate the value of inductor (in mH) of the reactor to be connected in series with X2 so that both generators delivers equal amp of fault current during 3 phase to ground fault on the feeder.

## Short answer

Fault current ≈2903 A, fault power ≈55.32 MVA; G1 supplies ≈1405 A, G2 ≈1499 A. For equal currents, add a reactor of 0.01 pu (0.0242 Ω, ≈0.077 mH) in series with G2.

## Answer

Generator ratings are given in MW; they are taken as MVA (unity power factor assumed). Fault MVA $=\text{MVA}_b/X_{eq}$; ohms to pu: $X_{pu} = X_\Omega\,\text{MVA}_b/\text{kV}^2$.

**Base:** 50 MVA, 11 kV. $Z_{base} = 11^2/50 = 2.42\ \Omega$; $I_{base} = \dfrac{50\times10^6}{\sqrt3\times11\times10^3} = 2624.3\ \text{A}$.

```text
  G1 50MW,0.16    G2 40MW,0.12
      |               |
     X1              X2 (+ Xr in part b)
      |               |
  ====+=======+=======+==== 11 kV bus
              |
             XL = 2 ohm
              |
              F (3-ph to ground)
```

#### a) Fault current, fault power and generator currents

$$
\begin{aligned}
X_1 &= 0.16\ \text{pu},\quad X_2 = 0.12\times\frac{50}{40} = 0.15\ \text{pu} \\
X_1\parallel X_2 &= \frac{0.16\times0.15}{0.31} = 0.077419\ \text{pu} \\
X_L &= \frac{2}{2.42} = 0.82645\ \text{pu} \\
X_{eq} &= 0.077419 + 0.82645 = 0.90387\ \text{pu} \\
I_f &= \frac{2624.3}{0.90387} = 2903.4\ \text{A} \\
\text{Fault power} &= \frac{50}{0.90387} = 55.32\ \text{MVA}
\end{aligned}
$$

Current divides inversely as the branch reactances:

$$
\begin{aligned}
I_{G1} &= 2903.4\times\frac{0.15}{0.31} = 1404.9\ \text{A} \\
I_{G2} &= 2903.4\times\frac{0.16}{0.31} = 1498.5\ \text{A}
\end{aligned}
$$

#### b) Reactor in series with $X_2$ for equal currents

Both generators feed the same bus, so they deliver equal current only when their branch reactances are equal:

$$
\begin{aligned}
X_2 + X_r &= X_1 \;\Rightarrow\; X_r = 0.16 - 0.15 = 0.01\ \text{pu} \\
X_r(\Omega) &= 0.01\times2.42 = 0.0242\ \Omega \\
L &= \frac{X_r}{2\pi f} = \frac{0.0242}{2\pi\times50} = 7.70\times10^{-5}\ \text{H} = 0.0770\ \text{mH}
\end{aligned}
$$

Check: new $X_{eq} = 0.08 + 0.82645 = 0.90645$ pu, $I_f = 2895.2$ A, each generator supplies 1447.6 A.

**Answer:** (a) $I_f$ ≈ **2903 A**, fault power ≈ **55.32 MVA**, $I_{G1}$ ≈ 1405 A, $I_{G2}$ ≈ 1499 A. (b) $L$ ≈ **0.077 mH** (0.0242 Ω).

## Check yourself

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

1. On a 50 MVA base, what is G2 (40 MW, 0.12 pu)?
   A. 0.15 pu
   B. 0.096 pu
   C. 0.12 pu
   D. 0.375 pu
   Answer: A. 0.12 x 50/40 = 0.15 pu. (Key point: G2 base conversion)

2. What is the fault power with X_eq = 0.077 + 0.826 = 0.904 pu?
   A. 553 MVA
   B. 55.3 MVA
   C. 100 MVA
   D. 645 MVA
   Answer: B. 50/0.904 = 55.32 MVA. (Key point: Fault power)

3. Which generator supplies more current in part (a)?
   A. G1 because its rating is larger
   B. Both equal
   C. G2 (1499 A) because its branch reactance 0.15 is smaller than 0.16
   D. G1 because its reactance is larger
   Answer: C. Current divides inversely with branch reactance, so G2 takes 0.16/0.31 of the total. (Key point: Current sharing)

4. What gives equal currents from both, and what is L?
   A. Xr = X1 + X2 = 0.31 pu, L about 2.4 mH
   B. Xr = 0.0242 ohm, L about 7.7 mH
   C. Xr = 0.16 pu, L about 1.2 mH
   D. X2 + Xr = X1, so Xr = 0.01 pu = 0.0242 ohm, L about 0.077 mH
   Answer: D. Equal branch reactances give equal currents; L = X/(2 pi f) = 0.0242/314.2 = 0.077 mH. (Key point: Equal current condition)

## Related in Power Plant Equipment

- [Fault current: two 50 MW generators, reactor inductance](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mw-inductance) — Power Plant Equipment, chapter 4
- [Fault current: four 1600 kVA generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-1600kva-generators) — Power Plant Equipment, chapter 4
- [Fault level: two 25 MVA generators, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-25mva-50-percent) — Power Plant Equipment, chapter 4
- [Fault level: two 50 MVA generators, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-50mva-50-percent) — Power Plant Equipment, chapter 4
- [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×
- [Fault level: three generators, feeder reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-three-generators-feeder) — Power Plant Equipment, chapter 4

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