---
title: "Bus-bar reactor scheme with four generators"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:bus-bar-reactor-four-generators"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/bus-bar-reactor-four-generators
site: BCT Engineering
---

# Bus-bar reactor scheme with four generators

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2071 Chaitra, 5 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-four-generators
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

In figure below a bus-bar reactor scheme with four generators. 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/G2: [?]MVA, 11kV, Xg1" = 0.2 pu; G3/G4: [?]MVA, 11kV, Xg2" = 0.1 pu. [Figure: G1 (0.2 pu) and G2 (0.2 pu) on one bus-bar section, G3 (0.1 pu) and G4 (0.1 pu) on the other, the two sections joined by a bus-bar reactor X = 0.2 ohms; the feeders leave the G1/G2 section through the CB] (Generator MVA ratings are unreadable in the scan.)

## Short answer

Using a 50 MVA, 11 kV base and assuming G1/G2=30 MVA, G3/G4=50 MVA, the total fault current is 35.53 kA and CB fault MVA is 677 MVA. Each G1/G2 contributes 7.87 kA; each G3/G4 contributes 9.89 kA.

## Answer

The generator MVA ratings are not readable, so assume **G1 = G2 = 30 MVA** and **G3 = G4 = 50 MVA**, all 11 kV (the method is the same for any rating). The feeder reactance is not given, so the fault is taken just beyond the CB. G1 and G2 feed directly; G3 and G4 feed through the bus-bar reactor.

**Base:** 50 MVA, 11 kV; $Z_{base} = 2.42\ \Omega$; $I_{base} = 2624.3\ \text{A}$.

```text
  G1     G2            G3     G4
  0.2    0.2           0.1    0.1
  |      |             |      |
 ==[Section 1]==[X=0.2 ohm]==[Section 2]==
       |
      CB
       |--- F
```

#### Reactances on 50 MVA base

$$
\begin{aligned}
X_{G1} = X_{G2} &= 0.2\times\frac{50}{30} = 0.33333\ \text{pu} \;\Rightarrow\; X_{12} = 0.16667\ \text{pu} \\
X_{G3} = X_{G4} &= 0.1\ \text{pu} \;\Rightarrow\; X_{34} = 0.05\ \text{pu} \\
X_R &= \frac{0.2}{2.42} = 0.08264\ \text{pu} \\
X_{34} + X_R &= 0.13264\ \text{pu}
\end{aligned}
$$

#### Fault MVA and fault current

$$
\begin{aligned}
X_{eq} &= \frac{0.16667\times0.13264}{0.16667+0.13264} = 0.073861\ \text{pu} \\
\text{Fault MVA} &= \frac{50}{0.073861} = 676.95\ \text{MVA} \\
I_f &= \frac{2624.3}{0.073861} = 35{,}531\ \text{A} = 35.53\ \text{kA}
\end{aligned}
$$

#### Current from each generator

$$
\begin{aligned}
I_{G1+G2} &= 35.53\times\frac{0.13264}{0.29931} = 15.75\ \text{kA} \Rightarrow I_{G1} = I_{G2} = 7.87\ \text{kA} \\
I_{G3+G4} &= 35.53\times\frac{0.16667}{0.29931} = 19.78\ \text{kA} \Rightarrow I_{G3} = I_{G4} = 9.89\ \text{kA}
\end{aligned}
$$

**Answer (with the assumed ratings):** G1, G2 ≈ **7.87 kA** each; G3, G4 ≈ **9.89 kA** each; total 35.53 kA; CB fault rating ≈ **677 MVA**.

## Check yourself

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

1. With the feeder on the G1/G2 section, how do the generators feed the fault?
   A. All four feed directly
   B. G1 and G2 feed directly; G3 and G4 feed through the bus-bar reactor
   C. G3 and G4 feed directly; G1 and G2 feed through the reactor
   D. Only G1 feeds the fault
   Answer: B. The reactor sits between the sections, so only the far section (G3, G4) passes through it. (Key point: Which generators feed directly)

2. On the 50 MVA base, what are the two parallel branch reactances?
   A. 0.1667 pu (G1||G2) and 0.1326 pu (G3||G4 + reactor)
   B. 0.3333 pu and 0.2 pu
   C. 0.2 pu and 0.1 pu
   D. 0.0833 pu and 0.0826 pu
   Answer: A. G1||G2 = 0.3333/2 = 0.1667 pu; G3||G4 = 0.05 pu plus the reactor 0.0826 pu gives 0.1326 pu. (Key point: Equivalent branch reactances)

3. What fault MVA must the feeder CB handle?
   A. About 677 MVA (50/0.0739)
   B. About 300 MVA
   C. About 150 MVA
   D. About 1350 MVA
   Answer: A. X_eq = 0.1667 x 0.1326/(0.1667 + 0.1326) = 0.0739 pu, so fault MVA = 50/0.0739 = 677. (Key point: Fault MVA for the CB)

4. The generator MVA ratings were unreadable. How does the answer handle this?
   A. Assumes ratings (30 and 50 MVA) and notes the method is the same for any rating
   B. Ignores generator reactance altogether
   C. Declares the problem unsolvable
   D. Uses the reactor value as the generator rating
   Answer: A. Per-unit reactances must be converted to a common base, so ratings are assumed; the procedure is unchanged for other values. (Key point: Why ratings are assumed)

## Close variants set in other exams

- [Bus-bar reactor scheme: 30 MVA and 50 MVA](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-bar-reactor-30-50-mva) — 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: 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
- [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: 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 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×
- [Bus reactor between 100 MVA and 200 MVA generators](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_plant_equipment%3Abus-reactor-100-200-mva) — 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
- [HV bus fault and limiting reactor](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Ahv-bus-fault-reactor-limit) — Power System Analysis II, chapter 3