---
title: "500 kVA 3-phase transformer efficiency"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-500kva-33-11kv-efficiency"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-33-11kv-efficiency
site: BCT Engineering
---

# 500 kVA 3-phase transformer efficiency

Electrical Machines · Chapter 2: Transformer

- Asked in: 2068 Magh, 8 marks
- Video: [Transformer efficiency: at Full and Half Load problem.](https://www.youtube.com/watch?v=ek2AQfUmAKo) — Dr Bharat Singh Choudhary
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-500kva-33-11kv-efficiency
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 500-kVA, 3-phase, 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 $35\ \Omega$, low voltage $0.876\ \Omega$ 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 is 4488.2 W, iron loss 3050 W. Efficiency at full load is 98.15% and at half load is 97.96% at 0.8 power factor.

## Answer

Copper loss is found from the phase currents and phase resistances of both windings. The HV side is delta (phase current = line current $/\sqrt{3}$) and the LV side is star (phase current = line current).

#### Phase currents at full load

HV (delta), line voltage = phase voltage = 33 kV:

$$
I_{ph1} = \frac{500000}{3 \times 33000} = 5.051\ \text{A}
$$

LV (star), line current = phase current:

$$
I_{ph2} = \frac{500000}{\sqrt{3}\times 11000} = 26.243\ \text{A}
$$

#### Full-load copper loss

$$
\begin{aligned}
P_{cu} &= 3\left(I_{ph1}^2R_1 + I_{ph2}^2R_2\right) \\
&= 3\left[(5.051)^2(35) + (26.243)^2(0.876)\right] \\
&= 3\,[892.8 + 603.3] = 4488.2\ \text{W}
\end{aligned}
$$

Iron loss $P_i = 3050$ W.

#### Efficiency at full load (0.8 p.f.)

$$
\eta = \frac{500000\times 0.8}{500000\times 0.8 + 3050 + 4488.2}\times 100 = \frac{400000}{407538.2}\times 100 = 98.15\%
$$

#### Efficiency at half load (0.8 p.f.)

Copper loss $= (0.5)^2 \times 4488.2 = 1122.1$ W.

$$
\eta = \frac{200000}{200000 + 3050 + 1122.1}\times 100 = \frac{200000}{204172.1}\times 100 = 97.96\%
$$

**Answer:** $\eta_{\text{full load}} = 98.15\%$ and $\eta_{\text{half load}} = 97.96\%$ at 0.8 p.f.

## Check yourself

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

1. For the delta-connected HV winding (500 kVA, 33 kV), the phase current is
   A. 500000/(sqrt(3) x 33000) = 8.75 A
   B. 500000/(3 x 33000) = 5.051 A
   C. 500000/33000 = 15.15 A
   D. 5.051 x sqrt(3) = 8.75 A
   Answer: B. In delta, phase voltage = line voltage, so Iph = S/(3 VL) = 5.051 A (line current is 8.75 A). (Key point: Phase current in a delta winding)

2. For the star-connected LV winding (11 kV line), the phase current is
   A. Line current / sqrt(3)
   B. Line current x sqrt(3)
   C. Equal to line current, 26.24 A
   D. 500000/(3 x 11000) = 15.15 A
   Answer: C. In star, line current = phase current = S/(sqrt(3) VL) = 26.243 A. (Key point: Phase current in a star winding)

3. Full-load copper loss for the transformer is
   A. I1^2 R1 + I2^2 R2 for one phase = 1496 W
   B. 3(Iph1^2 R1 + Iph2^2 R2) = 4488 W
   C. Iph1^2 R1 only
   D. Equal to the iron loss
   Answer: B. Per phase loss 892.8 + 603.3 = 1496.1 W, multiplied by 3 phases = 4488.2 W. (Key point: Three-phase copper loss)

4. At half load, 0.8 pf, the efficiency is
   A. 98.15%
   B. 97.96%
   C. 96.5%
   D. 98.9%
   Answer: B. Output 200 kW; losses 3050 + 0.25 x 4488.2 = 4172 W; eta = 200000/204172 = 97.96%. (Key point: Half-load efficiency)

## Related in Electrical Machines

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## Related in other subjects

- [500 kVA delta/star transformer efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Athree-phase-500kva-33-11-efficiency) — Electrical Machine, chapter 2
- [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