---
title: "Peterson coil for full and 90% length"
subject: "Switchgear and Protection"
chapter: 5
key: "semester_6/switchgear_and_protection:peterson-coil-partial-length"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/peterson-coil-partial-length
site: BCT Engineering
---

# Peterson coil for full and 90% length

Switchgear and Protection · Chapter 5: System Earthing

- Asked in: 2072 Magh, 4 marks
- Video: [Peterson coil](https://www.youtube.com/watch?v=PFyU59T_KOU) — Physics4U
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/peterson-coil-partial-length
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-5

## Question

A transmission line has a capacitance of 0.1 microfarad per phase. Determine the inductance of Peterson coil to neutralize the effect of capacitance of (i) complete length of (ii) 90% length of the line. f = 50 Hz.

## Short answer

For full line capacitance 0.1 μF, inductance L = 1/(3ω²C) = 33.77 H at 50 Hz. For 90% length, effective capacitance is 0.09 μF, giving L = 1/(3ω²×0.09 μF) = 37.53 H. Smaller capacitance requires larger inductance to maintain resonance.

## Answer

For a Peterson coil to neutralise the capacitive earth-fault current of a line of capacitance *C* per phase, ω*L* = 1/(3ω*C*), i.e. **L = 1/(3ω²C)**. If only a fraction *k* of the line length is to be neutralised, the capacitance to be tuned is *kC*, so *L* = 1/(3ω²*kC*).

#### Data
- *C* = 0.1 μF per phase (whole line), *f* = 50 Hz
- ω = 2π × 50 = 314.16 rad/s, ω² = 98696

#### (i) Complete length (k = 1)
```text
L = 1 / (3 × 98696 × 0.1×10⁻⁶)
  = 1 / 0.029609
  = 33.77 H
```
#### (ii) 90% of the length (k = 0.9)
```text
C' = 0.9 × 0.1 μF = 0.09 μF
L  = 1 / (3 × 98696 × 0.09×10⁻⁶)
   = 1 / 0.026648
   = 37.53 H
```
**Answer: (i) L = 33.77 H for the complete line; (ii) L = 37.53 H for 90% of the line.**

A smaller part of the line gives less capacitive current, so a larger inductance (smaller inductive current) is needed. In practice the coil has tappings so it can be retuned when the network length changes.

## Check yourself

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

1. If only a fraction k of the line is to be neutralised, which formula applies?
   A. L = k/(3 w^2 C)
   B. L = 1/(3 w^2 k C)
   C. L = 1/(3 w^2 C) regardless of k
   D. L = 3 w^2 kC
   Answer: B. The capacitance to tune is kC, so it replaces C in L = 1/(3w^2C). (Key point: Formula for fraction)

2. What is L for the complete line with C = 0.1 uF?
   A. 3.38 H
   B. 11.3 H
   C. 33.77 H
   D. 337.7 H
   Answer: C. L = 1/(3 x 98696 x 0.1e-6) = 33.77 H. (Key point: Complete length)

3. Why is the inductance for 90% of the line larger (37.53 H)?
   A. Less capacitance gives less capacitive current, so a smaller inductive current is needed
   B. More capacitance needs more inductance
   C. Frequency changes
   D. Voltage rises
   Answer: A. Less capacitive current means the coil must carry less current, which means a larger inductance. (Key point: 90% length)

## Related in Switchgear and Protection

- [Peterson coil for 33 kV 60 km line](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apeterson-coil-33kv-60km) — Switchgear and Protection, chapter 5, asked 6×
- [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×
- [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

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