---
title: "Specifying breaker on 132/66 kV transformer"
subject: "Switchgear and Protection"
chapter: 6
key: "semester_6/switchgear_and_protection:breaker-x-specification"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/breaker-x-specification
site: BCT Engineering
---

# Specifying breaker on 132/66 kV transformer

Switchgear and Protection · Chapter 6: Circuit Breakers

- Asked in: 2070 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/breaker-x-specification
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-6

## Question

Figure shows a 132/66 kV power transformer is fed from an infinite bus. If the 66 kV bus is a load bus, specify for the breaker 'X': i) Rated continuous current ii) Breaking capacity in MVA.
[Figure: Infinite bus – breaker Y – 3-phase 132/66 kV, 50 MVA transformer with Z = 14% – breaker X – load bus.]

## Short answer

Rated continuous current is 437.4 A (standard 630 A) calculated as 50 MVA/(√3×66 kV). Breaking capacity is 357.14 MVA (3.12 kA) from fault MVA = 50 MVA/0.14.

## Answer

Breaker X is on the 66 kV side, between the transformer and the load bus. It must carry the full-load current of the transformer continuously, and it must interrupt the fault current that flows through it for a fault just beyond it (on the load bus).

**Assumptions:** the source is an infinite bus (zero impedance), the load bus has no other source, and the fault current is limited only by the transformer impedance (14% on a 50 MVA base).

#### i) Rated continuous current

```text
I_rated = S / (√3 × V)
        = 50 × 10⁶ / (√3 × 66 × 10³)
        = 437.39 A
```

So breaker X must have a continuous current rating of at least **437.4 A**; the next standard rating, **630 A**, is chosen (IEC R10 series).

#### ii) Breaking capacity

For a three-phase fault on the load bus (just after X), the only impedance is the transformer reactance:

```text
Fault MVA = Base MVA / Z(pu)
          = 50 / 0.14
          = 357.14 MVA

Fault current at 66 kV:
I_f = 357.14 × 10⁶ / (√3 × 66 × 10³)
    = 3124.2 A  ≈ 3.12 kA
```

A fault on the transformer side of X is fed directly from the infinite bus and is cleared by breaker Y, so X only has to break the current fed through the transformer.

**Making current** (peak, first cycle) = 2.55 × 3.124 = 7.97 kA.

**Answer:** rated continuous current = **437.4 A** (choose the 630 A standard); breaking capacity = **357.14 MVA** (about 3.12 kA at 66 kV). In practice a standard **72.5 kV, 630 A breaker with a rated short-circuit breaking current of 12.5 kA** (the lowest standard value commonly offered at this voltage, well above the required 3.12 kA) would be specified.

## Check yourself

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

1. What is the rated continuous current of breaker X on the 66 kV side of the 50 MVA transformer?
   A. 437.4 A
   B. 218.7 A
   C. 875 A
   D. 378.8 A
   Answer: A. I = 50x10^6/(sqrt(3) x 66x10^3) = 437.4 A; the next standard rating of 630 A is chosen. (Key point: Rated continuous current)

2. What is the breaking capacity of X for a fault just beyond it, with transformer Z = 14% on 50 MVA?
   A. 50 MVA
   B. 700 MVA
   C. 357.14 MVA
   D. 250 MVA
   Answer: C. Fault MVA = Base MVA / Z(pu) = 50/0.14 = 357.14 MVA, since the source is an infinite bus. (Key point: Breaking capacity)

3. Why is a fault on the transformer side of X not used to rate X?
   A. It is smaller than the load bus fault
   B. It is fed from the infinite bus and cleared by breaker Y
   C. It cannot occur
   D. It is limited by the load
   Answer: B. That fault current comes directly from the infinite bus through Y, so X only needs to break current fed through the transformer. (Key point: Fault location logic)

4. What is the fault current at 66 kV for 357.14 MVA?
   A. About 1.56 kA
   B. About 6.25 kA
   C. About 437 A
   D. About 3.12 kA
   Answer: D. I = 357.14x10^6/(sqrt(3) x 66x10^3) = 3124 A, about 3.12 kA. (Key point: Fault current)

## Related in Switchgear and Protection

- [Breaking MVA and making kA (800 A, 33 kV)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acb-breaking-making-800a) — Switchgear and Protection, chapter 6
- [Breaking, making, short-time ratings (500 MVA)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acb-ratings-500a-500mva) — Switchgear and Protection, chapter 6
- [IDMT relays across 132/33 kV transformer (2700 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-132-33kv-2700a-graph) — Switchgear and Protection, chapter 7
- [IDMT relays across 132/33 kV transformer (2000 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-132-33kv-2000a) — Switchgear and Protection, chapter 7
- [IDMT relays across 132/11 kV transformer (3500 A)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Aidmt-132-11kv-3500a) — Switchgear and Protection, chapter 7
- [CB ratings: 2500 A, 1500 MVA, 33 kV](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Acb-ratings-2500a-1500mva) — Switchgear and Protection, chapter 6

## Related in other subjects

- [Current chopping overvoltage in 132 kV breaker](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Achopping-overvoltage-10a) — High Voltage Engineering, chapter 3
- [Nominal-T 100 km line, 50 MVA at 132 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Anominal-t-100km-50mva) — Power System Analysis I, chapter 6
- [Design: 500 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-500kva-66-11kv) — Electric Machine Design, chapter 3, asked 6×
- [Design: 4000 kVA 66/11 kV power transformer](https://www.bctengineer.com/question.md?key=semester_5%2Felectric_machine_design%3Adesign-4000kva-66-11kv) — Electric Machine Design, chapter 3, asked 5×