---
title: "Bus reactor between 100 MVA and 200 MVA generators"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:bus-reactor-100-200-mva"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/bus-reactor-100-200-mva
site: BCT Engineering
---

# Bus reactor between 100 MVA and 200 MVA generators

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2079 Baisakh, 6 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/bus-reactor-100-200-mva
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

A 100 MVA generator with 10% reactance and a 200 MVA generator with 8% reactance are connected to a common bus. The fault level on bus 1 is to be restricted to 1500 MVA. If a reactor is added in between these two generators in the bus bar, then on 100 MVA base Calculate the value of reactance added.

## Short answer

On 100 MVA base, generator reactances are 0.1 pu and 0.04 pu. To limit fault level to 1500 MVA, total equivalent reactance must be 0.0667 pu; solving the parallel paths yields a bus-bar reactor of 0.16 pu.

## Answer

The bus-bar is split into two sections joined by a reactor $X$: G1 on bus 1 and G2 on bus 2. For a fault on bus 1, G1 feeds directly and G2 feeds through $X$. Fault MVA $=\text{MVA}_b/X_{eq}$.

**Base:** 100 MVA.

$$
X_{G1} = 0.10\ \text{pu},\qquad X_{G2} = 0.08\times\frac{100}{200} = 0.04\ \text{pu}
$$

```text
  G1 100MVA,10%        G2 200MVA,8%
       |                     |
  ===[Bus 1]=====[ X ]=====[Bus 2]===
       |
       F (fault on bus 1)
```

**Without reactor:** $X_{eq} = 0.1\parallel0.04 = 0.02857$ pu, fault MVA $= 100/0.02857 = 3500$ MVA (too high).

**With reactor:** required

$$
X_{eq} = \frac{100}{1500} = 0.06667\ \text{pu}
$$

The two paths are in parallel:

$$
\begin{aligned}
\frac{1}{X_{eq}} &= \frac{1}{X_{G1}} + \frac{1}{X_{G2}+X} \\
15 &= 10 + \frac{1}{0.04+X} \\
0.04 + X &= \frac{1}{5} = 0.2 \\
X &= 0.16\ \text{pu}
\end{aligned}
$$

**Answer:** Bus-bar reactor $X$ = **0.16 pu (16 %) on 100 MVA base**. (If the bus voltage $V$ kV is known, $X_\Omega = 0.16\,V^2/100$; e.g. at 11 kV, 0.194 Ω.)

## Check yourself

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

1. For a fault on bus 1, how does G2 reach the fault?
   A. Through the bus-bar reactor X in series with its own reactance
   B. Directly with no reactance
   C. In parallel with the reactor
   D. It does not contribute
   Answer: A. G1 feeds directly, G2 feeds through X, and the two paths are in parallel. (Key point: Bus-bar reactor connection)

2. On the 100 MVA base, what is G2 (200 MVA, 8%)?
   A. 0.16 pu
   B. 0.04 pu
   C. 0.08 pu
   D. 0.4 pu
   Answer: B. 0.08 x 100/200 = 0.04 pu. (Key point: Base conversion)

3. For 1500 MVA on 100 MVA base, what X_eq is needed?
   A. 15 pu
   B. 0.15 pu
   C. 0.0667 pu
   D. 0.0286 pu
   Answer: C. X_eq = 100/1500 = 0.0667 pu, or 1/X_eq = 15. (Key point: Required reactance)

4. Solving 15 = 10 + 1/(0.04 + X), what is X?
   A. 0.2 pu
   B. 0.04 pu
   C. 0.24 pu
   D. 0.16 pu
   Answer: D. 1/(0.04 + X) = 5 so 0.04 + X = 0.2 and X = 0.16 pu. (Key point: Reactor value)

## Related in Power Plant Equipment

- [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 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×
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- [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
- [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 other subjects

- [Reactance diagram: 100 MVA base, two parallel paths](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-two-paths) — Power System Analysis I, chapter 3
- [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
- [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: three generators, 30 MVA 6.6 kV base](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-30mva-three-generators) — Power System Analysis I, chapter 3