---
title: "Balanced circulating current: 10 kV alternator"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:circulating-current-10kv-10-ohm"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/circulating-current-10kv-10-ohm
site: BCT Engineering
---

# Balanced circulating current: 10 kV alternator

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2073 Bhadra, 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/circulating-current-10kv-10-ohm
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

An alternator rated at 10 kV protected by the balanced circulating current system has its neutral grounded through a resistance of 10 ohms. The protective relay is set to operate when there is out of balance current of 1.8 A in the pilot wires, which are connected to the secondary windings of 1000/5 ratio current transformers. Determine: (i) the percentage winding which remain unprotected (ii) The minimum value of earthing resistance required to protect 80% of the winding.

## Short answer

Primary operating current = 1.8×200 = 360 A; Vph = 5774 V. Unprotected = (360×10)/5774 = 0.6235 (62.35%). To protect 80% of winding, maximum earthing resistance = 0.2×5774/360 = 3.21 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 CT secondary out-of-balance current reaches 1.8 A.

**Data:** 10 kV, star; R_n = 10 Ω; setting 1.8 A; CT 1000/5 (ratio 200).

#### Step 1: Primary operating current
```text
I_op = 1.8 × 200 = 360 A
V_ph = 10 000 / √3 = 5773.50 V
```

#### (i) Percentage unprotected
```text
x × V_ph / R_n = I_op
x = 360 × 10 / 5773.50 = 0.6235
```
So **62.35%** of the winding (from the neutral) is unprotected and only 37.65% is protected. The large earthing resistance limits the fault current too much.

#### (ii) Earthing resistance to protect 80%
Unprotected part allowed = 20% (x = 0.2):

```text
0.2 × V_ph / R = 360
R = 0.2 × 5773.50 / 360 = 3.208 Ω
```

**Answer:** (i) **62.35%** of the winding is unprotected; (ii) the earthing resistance must be reduced to **3.21 Ω** (or less) to protect 80% of the winding.

## Check yourself

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

1. Setting 1.8 A on 1000/5 CTs gives primary operating current of:
   A. 9 A
   B. 36 A
   C. 360 A
   D. 1800 A
   Answer: C. 1.8 x 200 = 360 A. (Key point: Primary operating current)

2. With R_n = 10 Ω and V_ph = 5773.5 V, the unprotected winding is about:
   A. 62.35%
   B. 37.65%
   C. 20%
   D. 6.2%
   Answer: A. x = 360 x 10 / 5773.5 = 0.6235 or 62.35%. (Key point: Unprotected percentage)

3. To protect 80% of the winding (x = 0.2), the earthing resistance should be:
   A. 32.1 Ω
   B. 3.21 Ω or less
   C. 10 Ω
   D. 0.32 Ω exactly
   Answer: B. R = 0.2 x 5773.5/360 = 3.208 Ω; reducing resistance increases fault current and protected fraction. (Key point: Resistance for 80% protection)

## Related in Switchgear and Protection

- [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×
- [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
- [Pilot current with tap change (10 MVA)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Apilot-current-10mva-tap) — Switchgear and Protection, chapter 7
- [Circulating current vs balanced voltage differential](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acirculating-vs-balanced-voltage) — Switchgear and Protection, chapter 7
- [Differential protection: 11 kV, 100 MVA generator](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Agenerator-differential-11kv-6-ohm) — Switchgear and Protection, chapter 7
- [Why and how the neutral is grounded](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aneutral-grounding-methods) — Switchgear and Protection, chapter 5, asked 4×

## Related in other subjects

- [Neutral reactance/resistance to limit L-G current](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Aneutral-reactance-resistance-limit-lg) — Power System Analysis II, chapter 5
- [Arrester current, resistance and rating on 132 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-132kv-line-3000kv) — High Voltage Engineering, chapter 4
- [Line currents with isolated and grounded neutral](https://www.bctengineer.com/question.md?key=semester_5%2Fpower_system_analysis_ii%3Astar-impedance-unbalanced-voltages-neutral) — Power System Analysis II, chapter 4
- [Arrester current and resistance for 750 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-750kv-z3000) — High Voltage Engineering, chapter 4, asked 2×