---
title: "Earth fault protection: 20 MVA alternator"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:earth-fault-20mva-5-ohm"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/earth-fault-20mva-5-ohm
site: BCT Engineering
---

# Earth fault protection: 20 MVA alternator

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2072 Asoj, 8 marks
- Video: [Alternator Protection- Numerical on percentage of winding Protected|Switchgear & Protection.](https://www.youtube.com/watch?v=Fjt4NQX1ZvA) — Dhananjay Pawar Sir
- 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-20mva-5-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

The neutral point of a three-phase 20 MVA, 11 kV alternator is earthed through a resistance of 5 Ω, the relay is set to operate when there is an out of balance current of 1.5 A. The CTs have a ratio 1000/5. What percentage of winding is protected against an earth fault and what should be the minimum value of earthing resistance to protect 90% of the winding?

## Short answer

Primary operating current = 1.5×200 = 300 A; Vph = 6351 V. Unprotected = (300×5)/6351 = 0.236, so protected = 76.4%. For 90% protection, earthing resistance cannot exceed 0.10×6351/300 = 2.12 Ω.

## Answer

For an earth fault at fraction *x* of the winding from the neutral, the fault current is x·V_ph / R_n (winding impedance neglected). The relay operates when the corresponding CT secondary current reaches the setting.

**Data:** 20 MVA, 11 kV, star; R_n = 5 Ω; relay setting 1.5 A; CT 1000/5 (ratio 200).

#### Step 1: Primary operating current
```text
I_op (primary) = 1.5 × 200 = 300 A
V_ph = 11 000 / √3 = 6350.85 V
```

#### Step 2: Percentage of winding protected
```text
x × V_ph / R_n = I_op
x = 300 × 5 / 6350.85 = 0.2362   (unprotected)
Protected = 1 − 0.2362 = 0.7638 → 76.38%
```

#### Step 3: Earthing resistance to protect 90%
To protect 90%, only x = 0.10 may be unprotected:

```text
0.10 × V_ph / R = 300
R = 0.10 × 6350.85 / 300 = 2.117 Ω
```

Any resistance larger than this leaves more than 10% unprotected, so 2.117 Ω is the **limiting (maximum permissible)** value of earthing resistance; a lower resistance protects more winding but gives larger earth fault currents.

**Answer:** **76.38%** of the winding is protected; earthing resistance must be **2.117 Ω** (not more) to protect 90% of the winding.

## Check yourself

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

1. Relay setting 1.5 A with CT 1000/5 corresponds to a primary operating current of:
   A. 7.5 A
   B. 300 A
   C. 1500 A
   D. 30 A
   Answer: B. CT ratio is 200, so 1.5 x 200 = 300 A. (Key point: Primary operating current)

2. With R_n = 5 Ω and V_ph = 6350.85 V, what percentage of the winding is protected?
   A. 23.62%
   B. 90%
   C. 76.38%
   D. 50%
   Answer: C. Unprotected x = 300 x 5/6350.85 = 0.2362, so protected is 76.38%. (Key point: Percentage protected)

3. To protect 90% of the winding, the earthing resistance must be (at most):
   A. 2.12 Ω
   B. 21.2 Ω
   C. 5 Ω
   D. 0.21 Ω
   Answer: A. R = 0.10 x 6350.85 / 300 = 2.117 Ω; a larger resistance leaves more winding unprotected. (Key point: Resistance for 90% protection)

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