---
title: "Bus-bar reactor scheme: 30 MVA and 50 MVA"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:bus-bar-reactor-30-50-mva"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/bus-bar-reactor-30-50-mva
site: BCT Engineering
---

# Bus-bar reactor scheme: 30 MVA and 50 MVA

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2071 Shrawan, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_plant_equipment/bus-bar-reactor-30-50-mva
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

Figure below shows a bus-bar reactor scheme. If a 3 phase to ground fault occurs on the outgoing feeder, calculate the fault current (in kA) supplied by each generator and fault MVA required for the circuit breaker (CB) on the feeder. Rating of generators are given as follow: G1: 30MVA, 11kV, Xg1" = 0.1 pu; G2: 50MVA, 11kV, Xg2" = 0.1 pu. [Figure: G1 (0.1 pu) and G2 (0.1 pu) on two bus-bar sections joined by a bus-bar reactor X = 0.2 ohm; the feeder leaves the G1 section through the CB]

## Short answer

On 50 MVA base, G1 reactance is 0.16667 pu, G2 is 0.1 pu, bus reactor 0.08264 pu. Equivalent reactance is 0.087145 pu, so fault current is 30.11 kA and fault MVA is 573.76 MVA; G1 supplies 15.75 kA and G2 supplies 14.37 kA.

## Answer

G1 and G2 are on two bus sections joined by a bus-bar reactor. The feeder is on the G1 section; its reactance is not given, so the fault is taken just beyond the CB. G1 feeds directly; G2 feeds through the bus reactor. Fault MVA $=\text{MVA}_b/X_{eq}$.

**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 30MVA,0.1        G2 50MVA,0.1
      |                    |
 ===[Sec 1]====[X=0.2 ohm]====[Sec 2]===
      |
     CB
      |--- F
```

#### Reactances on 50 MVA base

$$
\begin{aligned}
X_{G1} &= 0.1\times\frac{50}{30} = 0.16667\ \text{pu} \\
X_{G2} &= 0.1\ \text{pu} \\
X_R &= \frac{0.2}{2.42} = 0.08264\ \text{pu} \\
X_{G2}+X_R &= 0.18264\ \text{pu}
\end{aligned}
$$

#### Fault current and fault MVA

$$
\begin{aligned}
X_{eq} &= \frac{0.16667\times0.18264}{0.16667+0.18264} = 0.087145\ \text{pu} \\
I_f &= \frac{2624.3}{0.087145} = 30{,}114\ \text{A} = 30.11\ \text{kA} \\
\text{Fault MVA} &= \frac{50}{0.087145} = 573.76\ \text{MVA}
\end{aligned}
$$

#### Current from each generator

$$
\begin{aligned}
I_{G1} &= 30.114\times\frac{0.18264}{0.34931} = 15.75\ \text{kA} \\
I_{G2} &= 30.114\times\frac{0.16667}{0.34931} = 14.37\ \text{kA}
\end{aligned}
$$

**Answer:** G1 supplies ≈ **15.75 kA**, G2 ≈ **14.37 kA** (total 30.11 kA); the feeder CB must be rated for about **574 MVA**.

## Check yourself

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

1. Why must the generator and reactor reactances be converted before combining?
   A. Per-unit values must share one MVA base (here 50 MVA)
   B. Ohms cannot be used in fault calculation
   C. The generators have different frequencies
   D. Only the larger generator reactance is needed
   Answer: A. Reactances are added only on a common base; here the 50 MVA base is used, with G1 = 0.1 x 50/30 = 0.1667 pu. (Key point: Common base)

2. What is the 0.2 ohm bus-bar reactor in per unit on the 50 MVA, 11 kV base?
   A. 0.2 x 2.42 = 0.484 pu
   B. 0.2/2.42 = 0.0826 pu
   C. 0.2/50 = 0.004 pu
   D. 0.2 x 50 = 10 pu
   Answer: B. X_pu = X(ohm)/Z_base with Z_base = 11^2/50 = 2.42 ohm, giving 0.0826 pu. (Key point: Reactor in pu)

3. How is X_eq at the feeder fault formed?
   A. G1 (0.1667) in parallel with G2 + reactor (0.1826) = 0.0871 pu
   B. G1, G2 and the reactor all in series = 0.349 pu
   C. G2 alone = 0.1 pu
   D. G1 + G2 in parallel = 0.0625 pu
   Answer: A. G1 feeds directly, while G2 reaches the fault only through the bus reactor, so the branches in parallel are 0.1667 and 0.1826 pu. (Key point: Network structure)

4. Which result is correct for the generators and the feeder CB?
   A. G1 supplies 14.37 kA, G2 15.75 kA, CB 574 MVA
   B. G1 supplies 15.75 kA, G2 14.37 kA, CB about 574 MVA
   C. Each supplies 15.06 kA, CB 287 MVA
   D. G1 supplies 30.11 kA, G2 zero, CB 301 MVA
   Answer: B. Total 30.11 kA divides inversely to branch reactance, so G1 (smaller reactance 0.1667 vs 0.1826) carries 15.75 kA; fault MVA = 50/0.0871 = 574. (Key point: Current division and CB rating)

## Close variants set in other exams

- [Bus-bar reactor scheme with four generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-four-generators) — Power Plant Equipment, chapter 4, asked 2×

## Related in Power Plant Equipment

- [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
- [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
- [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 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

## Related in other subjects

- [Short circuit MVA of feeder circuit breaker](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afeeder-cb-short-circuit-mva) — Power System Analysis II, chapter 3, asked 2×
- [Generator-reactor system fault MVA](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-reactor-fault-mva) — Power System Analysis II, chapter 3
- [L-L fault current in kA at bus 3](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3All-fault-bus3-ka) — Power System Analysis II, chapter 5
- [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