---
title: "Percentage differential relay: 125 MVA transformer"
subject: "Switchgear and Protection"
chapter: 7
key: "semester_6/switchgear_and_protection:percentage-differential-125mva"
source: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/percentage-differential-125mva
site: BCT Engineering
---

# Percentage differential relay: 125 MVA transformer

Switchgear and Protection · Chapter 7: Protective Relays

- Asked in: 2075 Bhadra, 4 marks
- Video: [percentage differential relay | in hindi | percentage differential protection of transformer](https://www.youtube.com/watch?v=faMk98Gaptw) — Mind of Engineer 
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/switchgear_and_protection/percentage-differential-125mva
- Whole chapter: https://www.bctengineer.com/md/semester_6/switchgear_and_protection/chapter-7

## Question

A 125 MVA, 220/132 kV three phase power transformer is protected by percentage differential relays. The current transformers located on HV and LV sides of the power transformer are 400/5 A and 1200/5 respectively. If the HV side is delta connected and the LV side is star connected, determine: i) The output line currents of both CT at full load ii) The relay current at 15% overload iii) The minimum relay current setting to permit 25% overload.

## Short answer

At full load, HV star CT output is 328.04/80 = 4.10 A, LV delta CT line output is √3 x 546.73/240 = 3.95 A. Full-load unbalance is 0.1548 A; at 15% overload relay current is 0.178 A. Minimum setting for 25% overload is 1.25 x 0.1548 = 0.194 A.

## Answer

HV is delta connected, so its CTs are connected in **star**; LV is star connected, so its CTs are connected in **delta**. The relay current is the difference of the CT output line currents.

**Data:** 125 MVA, 220/132 kV; HV CTs 400/5 (ratio 80); LV CTs 1200/5 (ratio 240).

#### i) CT output line currents at full load
```text
I_HV = 125 × 10⁶ / (√3 × 220 × 10³) = 328.04 A
I_LV = 125 × 10⁶ / (√3 × 132 × 10³) = 546.73 A

HV CTs (star): output = 328.04 / 80 = 4.100 A
LV CTs (delta): CT secondary = 546.73 / 240 = 2.278 A
                output line = √3 × 2.278 = 3.946 A
```

#### ii) Relay current at 15% overload
```text
Relay current at full load = 4.100 − 3.946 = 0.1548 A
At 115% load = 1.15 × 0.1548 = 0.178 A
```

#### iii) Minimum relay setting for 25% overload
The relay must not operate up to 125% load:
```text
Setting ≥ 1.25 × 0.1548 = 0.1935 A
```

| Quantity | Value |
|---|---|
| HV CT output (FL) | 4.100 A |
| LV CT output (FL) | 3.946 A |
| Relay current at 115% | 0.178 A |
| Minimum setting (125%) | 0.194 A |

**Answer:** (i) **4.10 A** (HV) and **3.95 A** (LV); (ii) **0.178 A**; (iii) minimum relay setting **≈ 0.194 A** (about 3.9% of 5 A).

## Check yourself

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

1. HV is delta and LV is star. The CTs are connected:
   A. HV delta, LV star
   B. HV star, LV delta
   C. Both star
   D. Both delta
   Answer: B. Delta winding gets star CTs and star winding gets delta CTs. (Key point: CT connections)

2. LV CTs (1200/5, delta) carry 546.73 A. The CT output line current is:
   A. 2.278 A
   B. 3.946 A
   C. 4.100 A
   D. 1.315 A
   Answer: B. Secondary = 546.73/240 = 2.278 A; line output = sqrt(3) × 2.278 = 3.946 A. (Key point: LV CT output line current)

3. The relay current at full load is:
   A. 4.100 + 3.946 A
   B. 4.100 − 3.946 = 0.1548 A
   C. 0 A
   D. 3.946 − 2.278 A
   Answer: B. The relay carries the difference of CT output currents, 0.1548 A. (Key point: Relay current)

4. Minimum relay setting to permit 25% overload is:
   A. 0.1548 A
   B. 1.25 × 0.1548 = 0.194 A
   C. 0.8 × 0.1548 A
   D. 0.25 A
   Answer: B. The spill scales with load, so the setting must be at least 1.25 × full-load spill. (Key point: Setting for overload)

## Related in Switchgear and Protection

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