---
title: "Peterson coil for 33 kV 60 km line"
subject: "Switchgear and Protection"
chapter: 5
key: "semester_6/switchgear_and_protection:peterson-coil-33kv-60km"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/peterson-coil-33kv-60km
site: BCT Engineering
---

# Peterson coil for 33 kV 60 km line

Switchgear and Protection · Chapter 5: System Earthing

- Asked in: 2073 Magh, 4 marks; 2074 Bhadra, 4 marks; 2075 Baisakh, 5 marks; 2075 Bhadra, 5 marks; 2080 Chaitra, 5 marks (asked 5 times)
- 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-33kv-60km
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-5

## Question

A 33 kV, 3 phase, 50 Hz overhead line 60 km long has a capacitance to ground of each line equal to 0.015 μF per km. Determine the inductance and kVA rating of the Peterson coil (arc suppression coil).

## Short answer

Tune the Peterson coil so inductive fault current cancels capacitive current: L = 1/(3ω²C). With total capacitance 0.9 μF, inductance is 3.753 H. The coil carries 16.16 A at 19.05 kV, giving a rating of about 307.9 kVA.

## Answer

A **Peterson coil** (arc suppression coil) is an iron-cored reactor connected between the system neutral and earth, tuned to the line-to-earth capacitance so that the earth-fault current is neutralised.

**Principle:** during an earth fault on one phase, the healthy phases send a capacitive current *I_C* = 3 *V_ph* ω*C* into the fault. The Peterson coil in the neutral sends an equal inductive current *I_L* = *V_ph* / (ω*L*) through the fault, which cancels *I_C*, so the arc goes out. For resonance:

```text
I_L = I_C
V_ph/(ωL) = 3 V_ph ωC
        L = 1 / (3 ω² C)
```
Rating of the coil = *V_ph* × *I_L* (the coil sees phase voltage during the fault).

#### Data
- Line voltage *V_L* = 33 kV, *f* = 50 Hz, ω = 2π × 50 = 314.16 rad/s
- Capacitance per phase *C* = 0.015 μF/km × 60 km = 0.9 μF = 0.9 × 10⁻⁶ F

#### Inductance of the coil
```text
L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 0.9 × 10⁻⁶)
  = 1 / 0.26648
  = 3.753 H
```
#### kVA rating
```text
V_ph = 33000/√3          = 19052.6 V
I_L  = I_C = 3 V_ph ω C
     = 3 × 19052.6 × 314.16 × 0.9×10⁻⁶
     = 16.16 A
Rating = V_ph × I_L
       = 19052.6 × 16.16
       = 307.9 kVA
```
**Answer: Inductance of Peterson coil = 3.753 H; kVA rating ≈ 307.9 kVA (coil current 16.16 A at 19.05 kV).**

## Check yourself

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

1. What condition determines the Peterson coil inductance?
   A. Inductive current equals the capacitive earth-fault current: L = 1/(3 w^2 C)
   B. L = 1/(w^2 C) with no factor 3
   C. Coil voltage equals line voltage
   D. L = 3 w^2 C
   Answer: A. The coil current V_ph/(wL) must equal 3 V_ph wC, giving L = 1/(3w^2C). (Key point: Resonance condition)

2. What is the capacitance per phase to use in the calculation?
   A. 0.015 uF
   B. 0.09 uF
   C. 0.9 uF (0.015 x 60)
   D. 54 uF
   Answer: C. 0.015 uF/km multiplied by 60 km gives 0.9 uF per phase. (Key point: Capacitance per phase)

3. What is the inductance of the coil?
   A. 1.88 H
   B. 3.753 H
   C. 33.8 H
   D. 0.375 H
   Answer: B. L = 1/(3 x 314.16^2 x 0.9e-6) = 3.753 H. (Key point: Inductance result)

4. How is the kVA rating of the coil found, and what is it?
   A. Line voltage x I_L, about 534 kVA
   B. Phase voltage x I_L = 19.05 kV x 16.16 A, about 308 kVA
   C. Phase voltage x I_L / 3, about 103 kVA
   D. Line voltage x I_L x sqrt(3), about 925 kVA
   Answer: B. The coil sees phase voltage during the fault, so rating = V_ph x I_L = 307.9 kVA. (Key point: kVA rating)

## Close variants set in other exams

- [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×

## Related in Switchgear and Protection

- [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
- [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

## Related in other subjects

- [280 km, 60 Hz line: nominal-π vs long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A280km-60hz-nominal-pi-long) — Power System Analysis I, chapter 5, asked 2×
- [8 km line receiving 15000 kVA at 33 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A8km-line-15000kva) — Power System Analysis I, chapter 6
- [400 km long line, 60 MVA at 220 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-400km-60mva) — Power System Analysis I, chapter 5
- [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6