---
title: "Earth fault neutralizer reactance"
subject: "Switchgear and Protection"
chapter: 5
key: "semester_6/switchgear_and_protection:earth-fault-neutralizer-reactance"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/earth-fault-neutralizer-reactance
site: BCT Engineering
---

# Earth fault neutralizer reactance

Switchgear and Protection · Chapter 5: System Earthing

- Asked in: 2079 Chaitra, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/earth-fault-neutralizer-reactance
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-5

## Question

In a 50 Hz overhead line, the capacitance of one line to earth was 1.5 μF. It was decided to use an earth fault neutralizer. Calculate the reactance to neutralize the capacitance of (i) 100% of length of the line (ii) 90% of length of line.

## Short answer

Reactance X_L = 1/(3ωC). For full 1.5 μF, X_L = 707.4 Ω. For 90% length (1.35 μF), X_L = 786.0 Ω. Reactance increases as capacitance decreases to maintain current balance during earth faults.

## Answer

An **earth fault neutraliser** (Peterson coil) is a reactor connected between the neutral and earth. During a line-to-earth fault its inductive current equals the capacitive current 3*V_ph*ω*C* of the healthy phases, so:

```text
V_ph / X_L = 3 V_ph ω C   →   X_L = 1 / (3 ω C)
```
For a fraction *k* of the line, *C* is replaced by *kC*.

#### Data
- *C* = 1.5 μF (line to earth, per phase), *f* = 50 Hz, ω = 314.16 rad/s

#### (i) 100% of the line
```text
X_L = 1 / (3 × 314.16 × 1.5×10⁻⁶)
    = 1 / 1.4137×10⁻³
    = 707.4 Ω
(L = X_L/ω = 2.252 H)
```
#### (ii) 90% of the line
```text
C' = 0.9 × 1.5 = 1.35 μF
X_L = 1 / (3 × 314.16 × 1.35×10⁻⁶)
    = 1 / 1.2723×10⁻³
    = 786.0 Ω
(L = 2.502 H)
```
**Answer: Reactance of the neutraliser = 707.4 Ω for 100% of the line and 786.0 Ω for 90% of the line.**

## Check yourself

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

1. What is the neutraliser reactance formula?
   A. X_L = 1/(w C)
   B. X_L = 1/(3 w C)
   C. X_L = 3/(w C)
   D. X_L = w C/3
   Answer: B. Equating V_ph/X_L = 3 V_ph w C gives X_L = 1/(3wC). (Key point: Reactance formula)

2. With C = 1.5 uF and f = 50 Hz, X_L for 100% is:
   A. 212 Ω
   B. 707.4 Ω
   C. 2122 Ω
   D. 786 Ω
   Answer: B. X_L = 1/(3 x 314.16 x 1.5e-6) = 707.4 Ω. (Key point: 100% of line)

3. Why is the reactance for 90% of the line (786 Ω) higher?
   A. C is reduced to 1.35 uF, so less capacitive current needs a larger reactance
   B. The reactance depends on length directly
   C. The frequency drops
   D. The fault current increases
   Answer: A. Smaller C gives X_L = 1/(3 w x 1.35e-6) = 786 Ω. (Key point: 90% of line)

## Related in Switchgear and Protection

- [Peterson coil for full and 90% length](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apeterson-coil-partial-length) — Switchgear and Protection, chapter 5
- [Arc suppression coil for 66 kV 90 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aarc-suppression-coil-66kv-90km) — Switchgear and Protection, chapter 5, asked 6×
- [Earth fault protection: 20 MVA alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aearth-fault-20mva-5-ohm) — Switchgear and Protection, chapter 7
- [Restricted earth fault for generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aref-generator-solidly-grounded) — Switchgear and Protection, chapter 7
- [Combined phase and earth fault protection of generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acombined-phase-earth-fault-generator) — Switchgear and Protection, chapter 7
- [Arc suppression coil for 0.0125 μF line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aarc-suppression-coil-0-0125) — Switchgear and Protection, chapter 5

## Related in other subjects

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- [Fault at load end of overhead line](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Afault-load-end-33kv-line) — Power System Analysis II, chapter 3
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [Reactance diagram: 90 MVA generator (15%), two motors](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areactance-diagram-90mva-generator-15) — Power System Analysis I, chapter 3, asked 2×