---
title: "Merz-Price: unprotected winding (5000 kVA)"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:merz-price-5000kva-unprotected"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/merz-price-5000kva-unprotected
site: BCT Engineering
---

# Merz-Price: unprotected winding (5000 kVA)

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2070 Magh, 8 marks
- Video: [Merz Price Differential Protection Scheme | Power Systems | GATE (EE) | Ankit Goyal](https://www.youtube.com/watch?v=mjCTiUkOTcM) — Kreatryx GATE - EE, ECE & IN by Unacademy
- 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-5000kva-unprotected
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

A 5000 kVA, 6600 V, star-connected alternator has a synchronous reactance of 2 Ω per phase and 0.5 Ω resistance. It is protected by Merz-Price balanced current system which operates when out of balance current exceeds 30% of load current. Determine what proportion of the alternator winding is unprotected if the star is earthed through a resistance of 6.5 Ω.

## Short answer

Unprotected portion near neutral is found from x*Vph/Rn = I_op. Full-load current 437.4 A, tripping current 131.2 A, Vph 3810.5 V. x = (131.2×6.5)/3810.5 = 0.224, so about 22.4% of the winding is unprotected.

## Answer

For an earth fault at a point *x* (fraction of winding from the neutral), the voltage driving the fault current is *x* × phase voltage, and the current is limited mainly by the neutral earthing resistance. The relay operates only if this current exceeds the setting; faults close to the neutral give too little current, so that part of the winding is unprotected.

**Data:** 5000 kVA, 6600 V, star; R_n = 6.5 Ω; winding impedance 0.5 + j2 Ω per phase; relay operates at 30% of full-load current.

#### Step 1: Full-load and operating currents
```text
I_FL = 5000 × 10³ / (√3 × 6600) = 437.39 A
I_op = 0.30 × 437.39 = 131.22 A
V_ph = 6600 / √3 = 3810.51 V
```

#### Step 2: Unprotected portion (usual textbook method)
Neglecting the impedance of the short faulted part of the winding (small compared with 6.5 Ω):

```text
x × V_ph / R_n = I_op
x = I_op × R_n / V_ph
  = 131.22 × 6.5 / 3810.51
  = 0.2238
```

So **22.38%** of the winding (from the neutral end) is unprotected.

#### Step 3: Including the winding impedance (check)
The faulted part has impedance x(0.5 + j2) Ω, so:

```text
x × 3810.51 = 131.22 × |6.5 + x(0.5 + j2)|
Solving numerically: x = 0.2283
```

This gives 22.83% unprotected, only slightly more; the winding impedance has little effect because R_n dominates.

```text
 Neutral ─R_n(6.5Ω)─ N ──x──●F── (1−x) ── line
                             │
                            earth
```

**Answer:** about **22.4%** of the winding (22.8% if the winding impedance is included) near the neutral is unprotected; about 77.6% is protected.

## Check yourself

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

1. For the 5000 kVA, 6600 V alternator with relay setting 30% of full-load, the operating current is about:
   A. 437 A
   B. 131 A
   C. 1312 A
   D. 757 A
   Answer: B. I_FL = 5000k/(sqrt3 x 6600) = 437.4 A; 30% gives 131.2 A. (Key point: Operating current)

2. Which relation gives the unprotected fraction x of winding?
   A. x = I_op x R_n / V_ph
   B. x = V_ph / (I_op R_n)
   C. x = I_FL / I_op
   D. x = R_n / X_s
   Answer: A. Fault current x V_ph / R_n must equal I_op at the limit, so x = I_op R_n / V_ph. (Key point: Unprotected fraction formula)

3. With R_n = 6.5 Ω and V_ph = 3810.5 V, the unprotected proportion is about:
   A. 77.6%
   B. 50%
   C. 22.4%
   D. 6.5%
   Answer: C. x = 131.22 x 6.5 / 3810.5 = 0.2238, i.e. 22.4% near the neutral is unprotected. (Key point: Result)

## Close variants set in other exams

- [Merz-Price: protected winding (13.8 kV)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Amerz-price-13-8kv-protected) — Switchgear and Protection, chapter 7, asked 2×

## Related in Switchgear and Protection

- [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
- [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

## Related in other subjects

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- [Emf and regulation of 1500 kVA 13 kV alternator](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machines_ii%3Aemf-regulation-1500kva-13kv) — Electric Machines II, chapter 1
- [1500 kVA alternator: excitation emf, voltage at leading pf](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Aalt-1500kva-6600v-excitation-emf) — Electrical Machines, chapter 6
- [Design: 150 kVA 6600/400 V distribution transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-150kva-6600-400-regulation) — Electric Machine Design, chapter 3, asked 5×