---
title: "Neutral reactance for 15 MVA generator"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:neutral-reactance-15mva"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/neutral-reactance-15mva
site: BCT Engineering
---

# Neutral reactance for 15 MVA generator

Power System Analysis II · Chapter 5: Unsymmetrical Faults on Power Systems

- Asked in: 2073 Shrawan, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/neutral-reactance-15mva
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A 3-phase generator rated 15 MVA, 13.2 kV has a solidly grounded neutral. Its positive, negative and zero sequence reactance are 40%, 30% and 5% respectively. Find the value of reactance to be connected in neutral circuit so that fault current for a single line to ground fault (of negligible fault impedance) at No-load does not exceed line current.

## Short answer

Rated line current = 1 pu. Without reactance SLG fault current is 4 pu; with Xn, 3/(0.75+3Xn)=1 gives neutral reactance Xn=0.75 pu = 8.71 Ω.

## Answer

**Data:** 15 MVA, 13.2 kV, solidly grounded; $X_1 = 0.40$, $X_2 = 0.30$, $X_0 = 0.05$ pu. At no load $E = 1$ pu; rated line current $= 1$ pu.

#### Fault current without neutral reactance
$$
I_f = \frac{3E}{X_1+X_2+X_0} = \frac{3}{0.75} = 4\ \text{pu}
$$
This is four times rated current, so a neutral reactance $X_n$ is needed. $X_n$ carries $3I_{a0}$, so it appears as $3X_n$ in the zero-sequence network.

#### Required neutral reactance
$$
\begin{aligned}
I_f &= \frac{3E}{X_1+X_2+X_0+3X_n} \le 1 \\
0.40 + 0.30 + 0.05 + 3X_n &= 3 \\
X_n &= \frac{3 - 0.75}{3} = 0.75\ \text{pu}
\end{aligned}
$$

#### Ohmic value
$$
Z_{base} = \frac{(13.2)^2}{15} = 11.616\ \Omega
$$
$$
X_n = 0.75 \times 11.616 = 8.712\ \Omega
$$

**Answer: neutral reactance $X_n = 0.75$ pu $= 8.71\ \Omega$.**

Check: with $X_n$, $I_f = 3/(0.75 + 2.25) = 1$ pu = rated current $\left(\frac{15\times10^6}{\sqrt3\times13.2\times10^3} = 656\ \text{A}\right)$.

## Check yourself

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

1. With no neutral reactance, the L-G fault current is:
   A. 1.33 pu
   B. 4 pu, four times rated current
   C. 0.75 pu
   D. 3 pu
   Answer: B. I_f = 3/(0.40+0.30+0.05) = 4 pu. (Key point: Unrestricted fault current)

2. Why does Xn appear as 3Xn in the zero-sequence network?
   A. It is three-phase rated
   B. It is in parallel with three phases
   C. It carries 3Ia0, the neutral current
   D. Per-unit conversion
   Answer: C. The neutral current is 3Ia0, so the drop is 3Ia0 Xn, equal to Ia0 x 3Xn. (Key point: Why 3Xn)

3. Xn = 0.75 pu on Zbase = 13.2^2/15 gives:
   A. 8.71 ohm
   B. 11.6 ohm
   C. 4.36 ohm
   D. 0.75 ohm
   Answer: A. Zbase = 11.616 ohm, so Xn = 0.75 x 11.616 = 8.71 ohm. (Key point: Ohmic value)

## Related in Power System Analysis II

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