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

# Fault level: two 50 MVA generators, 50% reduction

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2080 Bhadra, 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/fault-two-50mva-50-percent
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

For the power system network given below, calculate the fault level in MVA at the outgoing feeder for a 3 phase to ground fault on the feeder. Calculate the value of reactance to be connected in the feeder in order to reduce the fault level by 50%. Ratings of the generators are: G1 = 50 MVA, 11 kV, Xg1 = 0.15 p.u.; G2 = 50 MVA, 11 kV, Xg2 = 0.3 p.u. [Figure: G1 and G2 connected to a common busbar with a single outgoing feeder]

## Short answer

Fault level is 500 MVA. Halving it requires doubling the existing 0.1 pu reactance to 0.2 pu; thus add a feeder reactor of 0.1 pu (0.242 Ω), giving a new fault level of 250 MVA.

## Answer

Generators in parallel feed the fault through the bus. Fault MVA $=\text{MVA}_b/X_{eq}$. Feeder reactance (not given) is taken as zero before adding the reactor.

**Base:** 50 MVA, 11 kV; $Z_{base} = 11^2/50 = 2.42\ \Omega$. Both generators are already on 50 MVA.

```text
 G1 50MVA,0.15   G2 50MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |  feeder (+X)
              F
```

#### Fault level

$$
\begin{aligned}
X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\
\text{Fault MVA} &= \frac{50}{0.1} = 500\ \text{MVA} \\
I_f &= \frac{500\times10^6}{\sqrt3\times11\times10^3} = 26.24\ \text{kA}
\end{aligned}
$$

#### Reactance for 50 % reduction

New fault MVA = 250 MVA, so

$$
\begin{aligned}
X_{eq,new} &= \frac{50}{250} = 0.2\ \text{pu} \\
X &= 0.2 - 0.1 = 0.1\ \text{pu} \\
X(\Omega) &= 0.1\times2.42 = 0.242\ \Omega
\end{aligned}
$$

(Halving the fault level needs the total reactance to double, so the added reactance equals the existing $X_{eq}$.)

**Answer:** Fault level = **500 MVA**; feeder reactance = **0.1 pu = 0.242 Ω** (fault level becomes 250 MVA).

## Check yourself

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

1. What is X_eq for 0.15 pu and 0.3 pu in parallel?
   A. 0.1 pu
   B. 0.45 pu
   C. 0.225 pu
   D. 0.05 pu
   Answer: A. 0.15 x 0.3/0.45 = 0.1 pu. (Key point: Equivalent reactance)

2. What is the fault level?
   A. 250 MVA
   B. 500 MVA
   C. 5 MVA
   D. 1000 MVA
   Answer: B. 50/0.1 = 500 MVA. (Key point: Fault level)

3. To halve the fault level, what must happen to total reactance, and what reactor is needed?
   A. Total reactance halves, so add 0.05 pu
   B. Total reactance doubles, so add 0.2 pu
   C. Total reactance doubles, so add 0.1 pu = 0.242 ohm
   D. Total reactance stays; add 0.242 pu
   Answer: C. Fault MVA is inversely proportional to X; doubling X_eq (0.1 to 0.2) needs 0.1 pu x 2.42 = 0.242 ohm. (Key point: Halving the fault level)

## Related in Power Plant Equipment

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