---
title: "500 kVA delta/star transformer efficiency"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:three-phase-500kva-33-11-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/three-phase-500kva-33-11-efficiency
site: BCT Engineering
---

# 500 kVA delta/star transformer efficiency

Electrical Machine · Chapter 2: Transformer

- Asked in: 2068 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/three-phase-500kva-33-11-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

A 500-KVA, 3-Ø, 50 Hz transformer has a voltage ratio (line voltage) of 33/11 kV and is delta/star connected. The resistances per phase are: High voltage 35 Ω, low voltage 0.876 Ω and the iron loss is 3050 W. Calculate the value of efficiency at full-load and one-half of full-load respectively at 0.8 p.f.

## Short answer

Full-load copper loss 4488 W (from phase currents), iron loss 3050 W. At 0.8 pf: η full load=98.15%, η half load=97.96%.

## Answer

The copper loss is found from the phase currents of the delta (HV) and star (LV) windings. The resistances are given per phase, so the loss is 3 × I_ph² × R. The load is taken at 0.8 pf lagging.

#### Rated phase currents
- HV (delta): phase voltage = 33 kV, so I_ph1 = 500 000 / (3 × 33 000) = 5.051 A
- LV (star): I_ph2 = I_L = 500 000 / (√3 × 11 000) = 26.24 A

#### Full-load copper loss
- HV: 3 × (5.051)² × 35 = 2678.3 W
- LV: 3 × (26.24)² × 0.876 = 1809.9 W
- P_cu(fl) = 4488.2 W
- Iron loss P_i = 3050 W

#### Full load at 0.8 pf
```text
Output = 500 000 x 0.8             = 400 000 W
Losses = 3050 + 4488.2             = 7538.2 W
η      = 400 000 / 407 538.2       = 0.9815
```
**η(full load) = 98.15 %**

#### Half load at 0.8 pf
P_cu = (1/2)² × 4488.2 = 1122.1 W
```text
Output = 200 000 W
Losses = 3050 + 1122.1             = 4172.1 W
η      = 200 000 / 204 172.1       = 0.9796
```
**η(half load) = 97.96 %**

| Load | Output (kW) | Cu loss (W) | Iron loss (W) | η (%) |
|---|---|---|---|---|
| Full | 400 | 4488.2 | 3050 | 98.15 |
| Half | 200 | 1122.1 | 3050 | 97.96 |

**Answer: 98.15 % at full load and 97.96 % at half load.**

## Check yourself

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

1. In the delta HV winding (33 kV), what is the rated phase current?
   A. 5.051 A
   B. 8.75 A
   C. 26.24 A
   D. 15.15 A
   Answer: A. I = 500000/(3 × 33000) = 5.051 A. (Key point: HV phase current)

2. Why is each winding's copper loss multiplied by 3?
   A. Resistances are given per phase and there are three phases
   B. Because of delta connection only
   C. Because of 3rd harmonic
   D. To include iron loss
   Answer: A. Loss = 3 × Iph² × R. (Key point: 3 phases)

3. What is the copper loss at half load?
   A. 4488 W
   B. 2244 W
   C. 1122 W
   D. 561 W
   Answer: C. (0.5)² × 4488.2 = 1122.1 W. (Key point: Half-load Cu loss)

4. What is the efficiency at full load, 0.8 pf?
   A. 97.96 %
   B. 99.0 %
   C. 96.5 %
   D. 98.15 %
   Answer: D. 400000/(400000 + 3050 + 4488.2) = 98.15 %. (Key point: Efficiency full load)

## Related in Electrical Machine

- [500 kVA 3-phase transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-33-11kv-efficiency) — Electrical Machines, chapter 2
- [Currents and efficiency of 132/11 kV transformer at 5 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-5mw-currents) — Electrical Machines I, chapter 2, asked 3×
- [Delta/star 120 kVA transformer values](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Adelta-star-120kva-line-phase-values) — Electrical Machine, chapter 2
- [Currents and efficiency of 132/11 kV transformer at 4 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-4mw-currents) — Electrical Machines I, chapter 2, asked 3×
- [Primary current and efficiency of 11 kV/380 V Δ/Y transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Adelta-star-11kv-380v-40kw-efficiency) — Electrical Machines I, chapter 2, asked 3×
- [3-phase 500 kVA transformer primary current](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-500kva-11kv-400v-primary-current) — Electrical Machines, chapter 2

## Related in other subjects

- [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×
- [CT ratio for 33/6.6 kV star/delta (400/5)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Adifferential-ct-33-6-6kv-400) — Switchgear and Protection, chapter 7
- [CT ratio for 33/6.6 kV star/delta (300:5)](https://www.bctengineer.com/question.md?key=semester_6%2Fswitchgear_and_protection%3Adifferential-ct-33-6-6kv-300) — Switchgear and Protection, chapter 7
- [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×