---
title: "Merz-Price: protected winding (13.8 kV)"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:merz-price-13-8kv-protected"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/merz-price-13-8kv-protected
site: BCT Engineering
---

# Merz-Price: protected winding (13.8 kV)

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2071 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/merz-price-13-8kv-protected
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

A 13.8 kV, 125 MVA, star-connected alternator has a synchronous reactance of 1.4 per unit per phase and negligible resistance. It is protected by a Merz-Price balanced current system which operates when out-of-balance current exceeds 10% of the full load current. If the neutral point is earthed through a resistance of 2 Ω, determine what proportion of winding is protected against earth fault.

## Short answer

Protected winding = 1 − (I_op×Rn/Vph). I_op = 0.10×5229.6 = 523 A, Vph = 7967 V, Rn=2 Ω. x unprotected = 523×2/7967 = 0.1313, so protected = 86.9%.

## Answer

An earth fault at fraction *x* of the winding from the neutral drives a current x·V_ph through the neutral resistance (and the faulted part of the winding). The relay sees this current and operates only if it exceeds the setting.

**Data:** 13.8 kV, 125 MVA, star; X_s = 1.4 pu; R negligible; R_n = 2 Ω; setting = 10% of full-load current.

#### Step 1: Basic quantities
```text
V_ph = 13 800 / √3 = 7967.43 V
I_FL = 125 × 10⁶ / (√3 × 13 800) = 5229.62 A
I_op = 0.10 × 5229.62 = 522.96 A
Z_base = 13.8² / 125 = 1.5235 Ω
X_s = 1.4 × 1.5235 = 2.1329 Ω
```

#### Step 2: Unprotected fraction (neglecting winding reactance)
```text
x × V_ph / R_n = I_op
x = 522.96 × 2 / 7967.43 = 0.1313
Protected = 1 − 0.1313 = 0.8687 → 86.87%
```

#### Step 3: Including the reactance of the faulted part
The faulted part has reactance x·X_s (as an approximation, since synchronous reactance does not divide exactly in proportion to turns):

```text
x × 7967.43 = 522.96 × |2 + j x × 2.1329|
Solving numerically: x = 0.1326
Protected = 86.74%
```

The reactance changes the answer only slightly because x is small.

**Answer:** about **86.9%** of the winding is protected against earth faults (86.7% if winding reactance is included); the 13.1% nearest the neutral is unprotected.

## Check yourself

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

1. What is the full-load current of the 13.8 kV, 125 MVA alternator?
   A. 9056 A
   B. 5230 A
   C. 2615 A
   D. 523 A
   Answer: B. I_FL = 125e6/(sqrt3 x 13800) = 5229.6 A. (Key point: Full-load current)

2. With I_op = 10% of I_FL = 523 A, R_n = 2 Ω and V_ph = 7967 V, the unprotected fraction is:
   A. 0.131
   B. 0.869
   C. 0.0131
   D. 0.262
   Answer: A. x = 523 x 2 / 7967 = 0.1313. (Key point: Unprotected fraction)

3. Hence the proportion of winding protected against earth faults is about:
   A. 13.1%
   B. 10%
   C. 86.9%
   D. 98.7%
   Answer: C. Protected = 1 - 0.1313 = 0.8687, about 86.9%. (Key point: Protected proportion)

## Close variants set in other exams

- [Merz-Price: unprotected winding (5000 kVA)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Amerz-price-5000kva-unprotected) — Switchgear and Protection, chapter 7, asked 2×

## Related in Switchgear and Protection

- [Balanced circulating current: 10 kV alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acirculating-current-10kv-10-ohm) — Switchgear and Protection, chapter 7
- [Merz-Price CT ratio for 66000/6600 V](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Amerz-price-66kv-ct-ratio) — Switchgear and Protection, chapter 7
- [Earth fault protection: 20 MVA alternator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aearth-fault-20mva-5-ohm) — Switchgear and Protection, chapter 7
- [Percentage differential relay: 125 MVA transformer](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apercentage-differential-125mva) — Switchgear and Protection, chapter 7

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