---
title: "Fault level: two 25 MVA generators, 50% reduction"
subject: "Power Plant Equipment"
chapter: 4
key: "semester_7/power_plant_equipment:fault-two-25mva-50-percent"
source: https://www.bctengineer.com/q/semester_7/power_plant_equipment/fault-two-25mva-50-percent
site: BCT Engineering
---

# Fault level: two 25 MVA generators, 50% reduction

Power Plant Equipment · Chapter 4: Substation Equipments

- Asked in: 2079 Bhadra, 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/fault-two-25mva-50-percent
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_plant_equipment/chapter-4

## Question

For the system shown in figure below, calculate the fault level in MVA at out going feeder for a three phase to ground fault on this feeder. Calculate the value of reactance to be connected in the feeder in order to reduce the fault level by 50%. [Figure: G1 25 MVA, 11 kV, E = 1 pu, X1 = 0.15 pu and G2 25 MVA, 11 kV, E = 1 pu, X2 = 0.3 pu connected to a common busbar; outgoing feeder through a C.B. with fault point 'F' on the feeder]

## Short answer

Fault level is 250 MVA. To halve it, double the existing 0.1 pu reactance to 0.2 pu; add a feeder reactor of 0.1 pu (0.484 Ω), giving 125 MVA.

## Answer

Fault MVA $=\text{MVA}_b/X_{eq}$, with $E = 1$ pu. Feeder reactance (not given) is taken as zero before adding the reactor.

**Base:** 25 MVA, 11 kV; $Z_{base} = 11^2/25 = 4.84\ \Omega$.

```text
 G1 25MVA,0.15   G2 25MVA,0.3
      |               |
 =====+=======+=======+==== 11 kV
              |
             C.B.--[X]-- F
```

#### Fault level

$$
\begin{aligned}
X_{eq} &= \frac{0.15\times0.3}{0.15+0.3} = 0.1\ \text{pu} \\
\text{Fault MVA} &= \frac{25}{0.1} = 250\ \text{MVA} \\
I_f &= \frac{250\times10^6}{\sqrt3\times11\times10^3} = 13.12\ \text{kA}
\end{aligned}
$$

#### Reactance for 50 % reduction

New fault MVA = 125 MVA.

$$
\begin{aligned}
X_{eq,new} &= \frac{25}{125} = 0.2\ \text{pu} \\
X &= 0.2 - 0.1 = 0.1\ \text{pu} \\
X(\Omega) &= 0.1\times4.84 = 0.484\ \Omega
\end{aligned}
$$

**Answer:** Fault level = **250 MVA**; feeder reactance = **0.1 pu = 0.484 Ω** (fault level becomes 125 MVA).

## Check yourself

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

1. What is the equivalent reactance of 0.15 and 0.3 pu in parallel?
   A. 0.1 pu
   B. 0.45 pu
   C. 0.075 pu
   D. 0.2 pu
   Answer: A. 0.15 x 0.3/0.45 = 0.1 pu. (Key point: Equivalent reactance)

2. What is the fault level (25 MVA base)?
   A. 125 MVA
   B. 250 MVA
   C. 2.5 MVA
   D. 500 MVA
   Answer: B. 25/0.1 = 250 MVA. (Key point: Fault level)

3. What reactor reduces the level by 50%?
   A. 0.2 pu = 0.968 ohm
   B. 0.05 pu = 0.242 ohm
   C. 0.1 pu = 0.484 ohm
   D. 0.1 pu = 0.1 ohm
   Answer: C. X_eq must double to 0.2 pu, so add 0.1 pu; Z_base = 4.84 ohm gives 0.484 ohm. (Key point: Reactor)

## Related in Power Plant Equipment

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