---
title: "Fault current: two 50 MVA generators, feeder reactor"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:fault-two-50mva-feeder-reactor"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-50mva-feeder-reactor
site: BCT Engineering
---

# Fault current: two 50 MVA generators, feeder reactor

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2082 Baisakh, 8 marks
- Video: [SHORT CIRCUIT MVA / FAULT MVA WITH CURRENT LIMITING REACTANCE / KTU/ POWER SYSTEM ANALYSIS](https://www.youtube.com/watch?v=SB1tqp0fGYY) — praveen raj
- 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-two-50mva-feeder-reactor
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

Two generators each of capacities 50 MVA, 11 kV having their sub transient reactance of 5% and 3% respectively operating in parallel and supplying to a common load through a feeder. If a 3 phase to ground fault occurs on the outgoing feeder, calculate the fault current supplied by each generator and fault power on the outgoing feeder. Also calculate the value of reactor to be connected in series with the outgoing feeder to reduce the fault level by 40 %.

## Short answer

On 50 MVA, 11 kV base with negligible feeder reactance, parallel reactance is 0.01875 pu. Fault current = 1/0.01875 = 53.33 pu = 139.96 kA (G1: 52.49 kA, G2: 87.48 kA), fault power 2666.7 MVA. To reduce by 40% (to 1600 MVA), add series reactor of 0.0303 Ω (0.0125 pu).

## Answer

The feeder reactance is not given, so it is taken as negligible: the fault is close to the bus. Fault MVA $= \text{MVA}_b/X_{pu}$.

**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 50MVA,5%   G2 50MVA,3%
     |             |
 ====+=====+=======+==== 11 kV bus
           |  outgoing feeder
           F
```

#### Fault current and fault power

$$
\begin{aligned}
X_{eq} &= \frac{0.05\times0.03}{0.05+0.03} = 0.01875\ \text{pu} \\
I_f &= \frac{1}{0.01875} = 53.33\ \text{pu} = 53.33\times2624.3 = 139{,}964\ \text{A} \approx 139.96\ \text{kA} \\
\text{Fault MVA} &= \frac{50}{0.01875} = 2666.7\ \text{MVA}
\end{aligned}
$$

#### Current from each generator

Current divides inversely as the reactances:

$$
\begin{aligned}
I_{G1} &= I_f\frac{X_{G2}}{X_{G1}+X_{G2}} = 139.96\times\frac{0.03}{0.08} = 52.49\ \text{kA} \\
I_{G2} &= I_f\frac{X_{G1}}{X_{G1}+X_{G2}} = 139.96\times\frac{0.05}{0.08} = 87.48\ \text{kA}
\end{aligned}
$$

#### Reactor for 40 % reduction

New fault MVA $= 0.6\times2666.7 = 1600$ MVA.

$$
\begin{aligned}
X_{eq,new} &= \frac{50}{1600} = 0.03125\ \text{pu} \\
X_{added} &= 0.03125 - 0.01875 = 0.0125\ \text{pu} \\
X_{added}(\Omega) &= 0.0125\times2.42 = 0.03025\ \Omega
\end{aligned}
$$

**Answer:** $I_f \approx 139.96$ kA (G1: 52.49 kA, G2: 87.48 kA), fault power = 2666.7 MVA; series reactor = **0.0125 pu ≈ 0.0303 Ω** (fault level becomes 1600 MVA).

## Check yourself

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

1. What is X_eq for G1 (0.05 pu) parallel G2 (0.03 pu), feeder reactance neglected?
   A. 0.01875 pu
   B. 0.08 pu
   C. 0.04 pu
   D. 0.0015 pu
   Answer: A. 0.05 x 0.03/0.08 = 0.01875 pu. (Key point: Equivalent reactance)

2. What is the fault power?
   A. 1250 MVA
   B. 2666.7 MVA
   C. 53.3 MVA
   D. 4000 MVA
   Answer: B. 50/0.01875 = 2666.7 MVA. (Key point: Fault MVA)

3. Which generator supplies more fault current and why?
   A. G1, because its reactance is higher
   B. Both equal because ratings are equal
   C. G2, because its reactance (0.03) is lower and current divides inversely with reactance
   D. G1, because current divides in proportion to reactance
   Answer: C. I_G2 = I_f x 0.05/0.08 = 87.48 kA versus 52.49 kA for G1. (Key point: Current sharing)

4. What series reactor reduces the fault level by 40%?
   A. 0.0075 pu
   B. 0.0125 ohm
   C. 0.03125 pu
   D. 0.0125 pu (about 0.0303 ohm)
   Answer: D. New X_eq = 50/1600 = 0.03125 pu; minus 0.01875 gives 0.0125 pu; times 2.42 gives 0.0303 ohm. (Key point: Reactor size)

## Close variants set in other exams

- [Fault level: two 40 MVA generators, 2 Ω feeder](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-two-40mva-2-ohm-feeder) — Power Plant Equipment, chapter 4, asked 5×

## Related in Power Plant Equipment

- [Feeder reactor: 40 and 30 MVA, 50% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-40-30mva-50-percent) — Power Plant Equipment, chapter 4, asked 4×
- [Feeder reactor: 40 and 30 MVA, 40% reduction](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-40-30mva-40-percent) — Power Plant Equipment, chapter 4, asked 4×
- [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
- [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 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 level: four 25 MVA generators, bus reactor](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Afault-four-25mva-bus-reactor) — Power Plant Equipment, chapter 4

## Related in other subjects

- [Reactance diagram: 50 MVA, 13.8 kV base, three generators](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-50mva-13-8kv) — Power System Analysis I, chapter 3
- [Reactance diagram: 300 MVA, 20 kV generator, two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-300mva-20kv) — Power System Analysis I, chapter 3
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