---
title: "Arc suppression coil for 66 kV 90 km line"
subject: "Switchgear and Protection"
chapter: 5
key: "semester_6/switchgear_and_protection:arc-suppression-coil-66kv-90km"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/arc-suppression-coil-66kv-90km
site: BCT Engineering
---

# Arc suppression coil for 66 kV 90 km line

Switchgear and Protection · Chapter 5: System Earthing

- Asked in: 2077 Chaitra, 6 marks
- Video: [Arc Suppression Coil ⚡ Save Your System From DISASTER!](https://www.youtube.com/watch?v=TYaGkVqCBns) — Electrology
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/arc-suppression-coil-66kv-90km
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-5

## Question

A 90 km long 66 kV, 50 Hz, 3 phase overhead line has capacitance to earth equal to 0.02 μF per km. What should be the inductance and kVA rating of the arc suppression coil?

## Short answer

Total capacitance per phase is 1.8 μF (0.02 μF/km × 90 km). Inductance L = 1/(3ω²C) = 1.876 H at 50 Hz. Phase voltage 38.1 kV; inductive current I_L = 3V_phωC = 64.64 A, giving kVA rating = V_ph × I_L ≈ 2463 kVA (2.46 MVA).

## Answer

The arc suppression (Peterson) coil in the neutral must supply an inductive fault current equal to the capacitive current fed by the two healthy phases during a line-to-ground fault.

**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
- *V_L* = 66 kV, *f* = 50 Hz, ω = 314.16 rad/s
- *C* = 0.02 μF/km × 90 km = 1.8 μF per phase

#### Inductance
```text
L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 1.8 × 10⁻⁶)
  = 1.876 H
```
#### kVA rating
```text
V_ph = 66000/√3            = 38105.1 V
I_L  = 3 V_ph ω C
     = 3 × 38105.1 × 314.16 × 1.8×10⁻⁶
     = 64.64 A
Rating = V_ph × I_L = 38105.1 × 64.64
       = 2463.3 kVA ≈ 2.46 MVA
```
**Answer: Inductance = 1.876 H; kVA rating ≈ 2463 kVA.**

## Check yourself

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

1. What is the earth-fault capacitive current the coil must cancel?
   A. V_ph w C
   B. 3 V_ph w C, from the two healthy phases
   C. sqrt(3) V_L w C
   D. V_L / (w C)
   Answer: B. The two healthy phases together send 3 V_ph wC into the fault. (Key point: Capacitive fault current)

2. With C = 0.02 uF/km x 90 km = 1.8 uF, what is L?
   A. 0.625 H
   B. 1.876 H
   C. 5.63 H
   D. 18.8 H
   Answer: B. L = 1/(3 x 314.16^2 x 1.8e-6) = 1.876 H. (Key point: Inductance)

3. For the 66 kV line, which values are correct?
   A. V_ph = 66 kV, rating 4.27 MVA
   B. V_ph = 38.1 kV, I_L = 64.64 A, rating about 2463 kVA
   C. V_ph = 38.1 kV, rating about 1.4 MVA
   D. V_ph = 114 kV, rating 7.4 MVA
   Answer: B. V_ph = 66000/sqrt(3) = 38105 V and rating = 38105 x 64.64 = 2463 kVA. (Key point: Phase voltage and rating)

## Close variants set in other exams

- [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×

## 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

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