---
title: "250/500 V transformer: max-efficiency output"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:transformer-250-500-max-efficiency-output"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-250-500-max-efficiency-output
site: BCT Engineering
---

# 250/500 V transformer: max-efficiency output

Electrical Machine · Chapter 2: Transformer

- Asked in: 2068 Magh, 8 marks
- Video: [43. Transformer Efficiency, Load KVA, Maximum Efficiency Condition, Maximum Efficiency  Calculation.](https://www.youtube.com/watch?v=EsDMulvvKHo) — Electrical Tutorial
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-250-500-max-efficiency-output
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

Test data on a 1-Ø, 250/500 V, 50 Hz transformer are:
O.C. Test: 250 V, 1 A, 80 W (carried on L.V. side)
S.C. Test: 20 V, 12 A, 100 W (carried on H.V. side)
Then draw the equivalent circuit referred to primary side and find out the output power to obtain maximum efficiency at 0.9 lag p.f.

## Short answer

Shunt branch: Ro=781.3 Ω, Xo=263.9 Ω; series referred to primary: R01=0.174 Ω, X01=0.379 Ω. For maximum efficiency at 0.9 pf lag, output power is 4.83 kW (efficiency ≈ 96.8%).

## Answer

**Given:** 1-φ, 250/500 V, 50 Hz. OC (LV): 250 V, 1 A, 80 W. SC (HV): 20 V, 12 A, 100 W. Primary = 250 V side; K = 2.

**Assumption:** the rated kVA is not given. The SC test current of 12 A on the HV side is taken as rated, so S = 500 × 12 = 6 kVA.

#### Shunt branch (primary side)
```text
cos φ0 = 80/(250 × 1) = 0.32
Iw = 0.32 A ; Iμ = √(1 − 0.32²) = 0.947 A
Ro = 250/0.32 = 781.3 Ω ; Xo = 250/0.947 = 263.9 Ω
```

#### Series branch
On the HV side: Z02 = 20/12 = 1.667 Ω; R02 = 100/12² = 0.694 Ω; X02 = √(1.667² − 0.694²) = 1.515 Ω. Referred to the primary (÷ 4):
```text
Z01 = 0.417 Ω ; R01 = 0.1736 Ω ; X01 = 0.379 Ω
```

#### Equivalent circuit (primary side)
```text
        R01=0.174Ω  X01=0.379Ω
 o──┬────┬───/\/\/───────(((────────o
    │    │                          
 781.3Ω 263.9Ω                  load
    │    │                       V2'
 o──┴────┴───────────────────────────o
```

#### Output for maximum efficiency at 0.9 lagging
```text
Iron loss Pi = 80 W ; full-load Cu loss Pcu = 100 W
x = √(Pi/Pcu) = √(80/100) = 0.894
Load kVA at max η = 0.894 × 6 = 5.37 kVA
Output power = 5.37 × 0.9 = 4.83 kW
η_max = 4830/(4830 + 2 × 80) = 96.8 %
```
**Answer: output power ≈ 4.83 kW (0.894 of full load) for maximum efficiency; η_max ≈ 96.8 %.**

## Check yourself

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

1. How is the rated kVA found since it is not given?
   A. Using OC current: 250 × 1
   B. Using SC current as rated HV current: 500 × 12 = 6 kVA
   C. Using the SC voltage
   D. It is assumed 10 kVA
   Answer: B. The 12 A SC current on the HV side is taken as rated. (Key point: Assumed rating)

2. What fraction of full load gives maximum efficiency?
   A. 0.894
   B. 0.8
   C. 1.25
   D. 0.64
   Answer: A. x = sqrt(80/100) = 0.894. (Key point: Load for maximum efficiency)

3. What is the output power at maximum efficiency and 0.9 pf?
   A. 5.37 kW
   B. 4.83 kW
   C. 6.0 kW
   D. 3.86 kW
   Answer: B. Load = 0.894 × 6 = 5.37 kVA, so P = 5.37 × 0.9 = 4.83 kW. (Key point: Output power)

4. How is η_max computed?
   A. Output/(Output + Pcu)
   B. Output/(Output + Pi)
   C. Output/(Output + 2 × 80 W)
   D. Output/(Output + 100 W)
   Answer: C. At max η, total loss = 2 Pi = 160 W. (Key point: Max efficiency formula)

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