---
title: "Fuse selection for 250 kVA transformer"
subject: "Switchgear and Protection"
chapter: 3
key: "semester_6/switchgear_and_protection:transformer-fuse-selection-250kva"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/transformer-fuse-selection-250kva
site: BCT Engineering
---

# Fuse selection for 250 kVA transformer

Switchgear and Protection · Chapter 3: Fuses

- Asked in: 2070 Magh, 6 marks
- Video: [Transformer Circuit breaker and Fuse size Calculation](https://www.youtube.com/watch?v=KkJhhgAZZkQ) — Rashid iqubal
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/transformer-fuse-selection-250kva
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-3

## Question

A transformer having capacity of 250 KVA, 11/0.415 KV, percentage impedance of 4.75% is required to be protected with fuses. Suggest guide line for selection of HV and LV side fuses. State any assumptions you make.

## Short answer

Full-load currents: HV =13.12A, LV =347.8A. LV short-circuit current =7.32kA (276A on HV side). Select HV fuse rated 25A HRC (1.5–2x full-load) and LV fuse rated 400A HRC, ensuring HV fuse withstands 157.5A inrush for 0.1s and discrimination: LV total I²t < HV pre-arcing I²t.

## Answer

**Given:** 250 kVA, 11/0.415 kV, 3-phase, impedance 4.75 %.

**Assumptions:** delta–star (Dyn11) transformer; infinite source on 11 kV side (gives the highest fault current); HRC (gG/back-up) fuses used on both sides; magnetising inrush ≈ 12 × full-load for 0.1 s.

#### Step 1: Full-load currents
```text
 I_FL(HV) = 250 000 / (√3 × 11 000) = 13.12 A
 I_FL(LV) = 250 000 / (√3 × 415)    = 347.80 A
```

#### Step 2: Fault current for a fault at LV terminals
```text
 I_sc(LV) = I_FL(LV) / Z_pu
          = 347.80 / 0.0475 = 7322 A ≈ 7.32 kA
 Seen on HV side = 13.12 / 0.0475 = 276.2 A
```

#### Step 3: Inrush current
```text
 I_inrush ≈ 12 × 13.12 = 157.5 A for about 0.1 s (HV side)
```

#### Guidelines for HV fuse
1. Rating must exceed full-load current with margin for permissible overload: about **1.5–2 × I_FL** → 19.7–26.2 A. Choose standard **25 A** HV HRC fuse.
2. It must **not melt on inrush**: the 0.1 s point of its T–I curve must be above 157.5 A (for a 25 A fuse this is normally satisfied; otherwise use 31.5 A).
3. It must clear a fault at the LV terminals (276 A on HV side, i.e. ≈ 11 × its rating) quickly, within the transformer short-circuit withstand time (2 s as per IEC 60076-5).
4. Breaking capacity must exceed the 11 kV system fault level.
5. It must **discriminate** with the LV fuse: for an LV feeder fault, the LV fuse must clear before the HV fuse melts (HV fuse curve, referred to LV via ratio 11/0.415 = 26.5, must lie above the LV fuse curve).

#### Guidelines for LV fuse
1. Rating ≥ full-load current: 347.8 A → choose standard **400 A** gG fuse (≈ 1.15 × I_FL), allowing normal overload.
2. Breaking capacity ≥ 7.32 kA (a standard 80 kA LV HRC fuse is ample).
3. It must clear an LV fault of 7.32 kA in a short time and protect the LV cables and busbars (its I²t below cable withstand).
4. It must be **selective** with downstream feeder fuses (rating ratio ≈ 1.6:1 or more) and with the HV fuse.

**Answer:** HV fuse ≈ **25 A** HRC (I_FL = 13.12 A); LV fuse ≈ **400 A** HRC (I_FL = 347.8 A); LV terminal fault current = **7.32 kA** (276 A on HV side). Final ratings are confirmed from manufacturer T–I curves.

## Check yourself

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

1. What is the HV full-load current of the 250 kVA, 11 kV transformer?
   A. 13.12 A
   B. 347.8 A
   C. 22.7 A
   D. 7.32 A
   Answer: A. I = 250000 / (sqrt(3) x 11000) = 13.12 A. (Key point: Full-load current)

2. With 4.75% impedance, what is the fault current at the LV terminals?
   A. About 276 A
   B. About 347.8 A
   C. About 7.32 kA
   D. About 73 kA
   Answer: C. I_sc = I_FL(LV)/Z_pu = 347.8/0.0475, about 7322 A. (Key point: LV fault current)

3. Why is the HV fuse chosen as about 1.5-2 x I_FL (25 A) rather than just above 13.12 A?
   A. To allow permissible overload and avoid melting on magnetising inrush
   B. To make it blow on every LV overload
   C. Because breaking capacity must equal full-load current
   D. To match the 11 kV system fault level
   Answer: A. The margin covers overload and the inrush of about 12 x I_FL for 0.1 s, so the fuse does not melt on energising. (Key point: HV fuse selection)

4. What must hold between the LV fuse and HV fuse for correct discrimination?
   A. The HV fuse must clear LV feeder faults first
   B. Both must have the same current rating
   C. The LV fuse must clear an LV feeder fault before the HV fuse melts
   D. The HV fuse must have a lower breaking capacity
   Answer: C. Referred to the same voltage, the HV fuse curve must lie above the LV fuse curve. (Key point: Discrimination)

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