---
title: "Neutral reactance/resistance to limit L-G current"
subject: "Power System Analysis II"
chapter: 5
key: "semester_5/power_system_analysis_ii:neutral-reactance-resistance-limit-lg"
source: https://www.bctengineer.com/q/semester_5/power_system_analysis_ii/neutral-reactance-resistance-limit-lg
site: BCT Engineering
---

# Neutral reactance/resistance to limit L-G current

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

- Asked in: 2074 Asoj, 8 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-resistance-limit-lg
- Whole chapter: https://www.bctengineer.com/md/semester_5/power_system_analysis_ii/chapter-5

## Question

A 30 MVA, 13.2 kV synchronous generator has a solidly grounded neutral. Its positive, negative and zero sequence impedances are 0.30, 0.40 and 0.05 pu respectively. Determine the following: i) The value of reactance that must be placed in the generator neutral so that the fault current for a line-to-ground fault of zero fault impedance shall not exceed the rated line current. ii) The value of resistance to be placed in the neutral that will serve the same purpose.

## Short answer

Rated line current = 1 pu. To limit SLG fault current from 4 pu to 1 pu, set 3/|0.75+j3Xn|=1 → Xn=0.75 pu (4.36 Ω); for resistance, 3/√((3Rn)²+0.75²)=1 → Rn=0.9682 pu (5.62 Ω).

## Answer

**Data:** 30 MVA, 13.2 kV; $X_1 = 0.30$, $X_2 = 0.40$, $X_0 = 0.05$ pu. Rated line current $= 1$ pu. Pre-fault voltage $E = 1$ pu.
$$
Z_{base} = \frac{13.2^2}{30} = 5.808\ \Omega
$$
With a neutral impedance $Z_n$, the zero-sequence impedance becomes $Z_0 + 3Z_n$:
$$
I_f = \frac{3E}{|Z_1 + Z_2 + Z_0 + 3Z_n|}
$$
Without $Z_n$: $I_f = 3/0.75 = 4$ pu, i.e. four times rated current, so it must be limited to 1 pu.

#### (i) Neutral reactance $X_n$
$$
\begin{aligned}
\frac{3}{0.30 + 0.40 + 0.05 + 3X_n} &= 1 \\
0.75 + 3X_n &= 3 \\
X_n &= 0.75\ \text{pu} = 0.75 \times 5.808 = 4.356\ \Omega
\end{aligned}
$$

#### (ii) Neutral resistance $R_n$
The resistance is in quadrature with the reactances:
$$
\begin{aligned}
\frac{3}{\sqrt{(3R_n)^2 + 0.75^2}} &= 1 \\
(3R_n)^2 &= 9 - 0.5625 = 8.4375 \\
3R_n &= 2.9047, \quad R_n = 0.9682\ \text{pu} \\
R_n &= 0.9682 \times 5.808 = 5.624\ \Omega
\end{aligned}
$$

**Answer:** (i) neutral reactance $X_n = 0.75$ pu $= 4.36\ \Omega$; (ii) neutral resistance $R_n = 0.968$ pu $= 5.62\ \Omega$.

A larger ohmic value is needed with resistance because it adds to the network reactance in quadrature, not directly.

## Check yourself

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

1. A neutral impedance Zn appears in the zero-sequence network as:
   A. Zn
   B. Zn/3
   C. 3Zn
   D. Zn^2
   Answer: C. The neutral carries 3Ia0, so the zero network sees Z0 + 3Zn. (Key point: Neutral impedance in zero sequence)

2. To limit the L-G current to 1 pu, X1+X2+X0+3Xn must equal:
   A. 1 pu
   B. 4 pu
   C. 0.75 pu
   D. 3 pu
   Answer: D. I_f = 3/(sum) = 1 gives sum = 3, so 0.75 + 3Xn = 3 and Xn = 0.75 pu. (Key point: Reactance value)

3. Why is the required neutral resistance (0.968 pu) larger than the reactance (0.75 pu)?
   A. Resistance adds to the reactance in quadrature, not directly
   B. Resistance has a lower base
   C. Resistance reduces X0
   D. The fault current is higher for resistance
   Answer: A. |Z| = sqrt((3Rn)^2 + 0.75^2) = 3, so more ohmic value is needed than with a directly adding reactance. (Key point: Resistance in quadrature)

## Related in Power System Analysis II

- [Neutral reactance for 15 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-15mva) — Power System Analysis II, chapter 5
- [L-G fault and 3-phase fault, 30 MVA generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Alg-fault-30mva-voltages-3phase) — Power System Analysis II, chapter 5
- [Equal positive and negative sequence impedances](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astatic-circuit-positive-negative-equal) — Power System Analysis II, chapter 4
- [Zero sequence Thevenin impedance at bus 2](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Azero-sequence-thevenin-bus2) — Power System Analysis II, chapter 4
- [Fault currents of isolated-neutral 380 V generator](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Agenerator-faults-380v-isolated-neutral) — Power System Analysis II, chapter 5
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