---
title: "Arc suppression coil for 0.0125 μF line"
subject: "Switchgear and Protection"
chapter: 5
key: "semester_6/switchgear_and_protection:arc-suppression-coil-0-0125"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/arc-suppression-coil-0-0125
site: BCT Engineering
---

# Arc suppression coil for 0.0125 μF line

Switchgear and Protection · Chapter 5: System Earthing

- Asked in: 2076 Bhadra, 5 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-0-0125
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-5

## Question

Calculate the inductance and kVA rating of the arc suppression coil of a 50 km long 33 kV, 50 Hz, 3 phase overhead transmission system having line to earth capacitance of each phase equal to 0.0125 μF.

## Short answer

Total capacitance per phase is 0.625 μF (0.0125 μF/km × 50 km). Using L = 1/(3ω²C) at 50 Hz gives inductance 5.404 H. Phase voltage is 19.05 kV; inductive current I_L = 3V_phωC = 11.22 A, so kVA rating = V_ph × I_L ≈ 213.8 kVA.

## Answer

An **arc suppression coil** (Peterson coil) is a reactor between neutral and earth whose inductive current cancels the capacitive earth-fault current, so the fault arc is self-extinguished.

**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 and assumption
The length (50 km) is given, so the capacitance 0.0125 μF is taken as **per km** of each phase:
- *C* = 0.0125 μF/km × 50 km = 0.625 μF per phase
- *V_L* = 33 kV, *f* = 50 Hz, ω = 314.16 rad/s

#### Inductance
```text
L = 1 / (3 ω² C)
  = 1 / (3 × 314.16² × 0.625 × 10⁻⁶)
  = 5.404 H
```
#### kVA rating
```text
V_ph = 33000/√3              = 19052.6 V
I_L  = 3 V_ph ω C
     = 3 × 19052.6 × 314.16 × 0.625×10⁻⁶
     = 11.22 A
Rating = V_ph × I_L = 19052.6 × 11.22
       = 213.8 kVA
```
**Answer: L = 5.404 H; kVA rating ≈ 213.8 kVA.**

(If 0.0125 μF were the total capacitance of the line, the same formula gives L ≈ 270 H and about 4.3 kVA, which is unrealistically small; the per-km reading is the intended one.)

## Check yourself

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

1. What is the coil inductance?
   A. 2.70 H
   B. 5.404 H
   C. 54.0 H
   D. 1.80 H
   Answer: B. L = 1/(3 x 314.16^2 x 0.625e-6) = 5.404 H. (Key point: Inductance)

2. What is the kVA rating?
   A. About 11.2 kVA
   B. About 213.8 kVA
   C. About 370 kVA
   D. About 4.3 kVA
   Answer: B. I_L = 3 V_ph wC = 11.22 A and V_ph = 19.05 kV, so rating = 213.8 kVA. (Key point: kVA rating)

## Related in Switchgear and Protection

- [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×
- [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×

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