---
title: "Differential protection: 11 kV, 100 MVA generator"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:generator-differential-11kv-6-ohm"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/generator-differential-11kv-6-ohm
site: BCT Engineering
---

# Differential protection: 11 kV, 100 MVA generator

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2076 Bhadra, 6 marks
- Video: [11 Problems on generator differential protection percentage of generator winding to be protected](https://www.youtube.com/watch?v=0_3jyQFwuhQ) — Dr Padmini S Panneer Selvam
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/generator-differential-11kv-6-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

A 11 kV, 100 MVA generator is grounded through a resistance of 6 Ω. The CTs have a ratio of 1000/5. The relay is set to operate when there is an out of balance current of 1 A. What percentage of the generator winding will be protected by percentage differential scheme of protection?

## Short answer

Primary operating current = 1×200 = 200 A; Vph = 6351 V. Unprotected fraction = (200×6)/6351 = 0.189. Thus 81.1% of the winding is protected.

## Answer

In a differential scheme, an earth fault at fraction *x* of the winding from the neutral gives a fault current x·V_ph / R_n. The relay sees this as out-of-balance current; the part of the winding near the neutral, where the fault current is below the setting, is unprotected.

**Data:** 11 kV, 100 MVA; R_n = 6 Ω; CT 1000/5 (ratio 200); relay setting 1 A; winding impedance neglected.

```text
I_op (primary) = 1 × 200 = 200 A
V_ph = 11 000 / √3 = 6350.85 V

x = I_op × R_n / V_ph
  = 200 × 6 / 6350.85
  = 0.1890   (unprotected)

Protected = 1 − 0.1890 = 0.8110
```

```text
 Earth ──R_n=6Ω── N ──[ 18.9% ]──[   81.1%   ]── line
                      unprotected   protected
```

**Answer:** about **81.1%** of the generator winding is protected; the 18.9% nearest the neutral is unprotected.

## Check yourself

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

1. A 1 A relay setting on 1000/5 CTs equals what primary current?
   A. 200 A
   B. 1000 A
   C. 5 A
   D. 20 A
   Answer: A. CT ratio 200, so 1 x 200 = 200 A. (Key point: Primary operating current)

2. With R_n = 6 Ω and V_ph = 6350.85 V, the unprotected fraction x is about:
   A. 0.811
   B. 0.189
   C. 0.32
   D. 0.0189
   Answer: B. x = 200 x 6 / 6350.85 = 0.189. (Key point: Unprotected fraction)

3. What percentage of the winding is protected?
   A. 18.9%
   B. 100%
   C. 50%
   D. 81.1%
   Answer: D. Protected = 1 - 0.189 = 0.811, i.e. 81.1%; the portion nearest the neutral is unprotected. (Key point: Protected percentage)

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