---
title: "8 kVA 400/120 V tests, half-load efficiency"
subject: "Electrical Machine"
chapter: 2
key: "semester_4/electrical_machine:transformer-8kva-400-120-hv-side"
source: https://www.bctengineer.com/q/semester_4/electrical_machine/transformer-8kva-400-120-hv-side
site: BCT Engineering
---

# 8 kVA 400/120 V tests, half-load efficiency

Electrical Machine · Chapter 2: Transformer

- Asked in: 2071 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_machine/transformer-8kva-400-120-hv-side
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machine/chapter-2

## Question

The following test results were obtained for open circuit and short circuit tests on a 8 kVA, 400/120 V, 50 Hz transformer:
Open-circuit Test (LV Side): 120 V, 4 A, 75 W
Short-circuit Test (HV Side): 9.5 V, 20 A, 110 W
Calculate the equivalent circuit parameters referred to high voltage side. Also calculate the efficiency at half full load and 0.8 power factor lagging load.

## Short answer

Referred to HV side: Ro=2133 Ω, Xo=337.5 Ω, R01=0.275 Ω, X01=0.387 Ω. At half load and 0.8 pf lagging, efficiency = 96.9%.

## Answer

**Given:** 8 kVA, 400/120 V, 50 Hz. OC test (LV): 120 V, 4 A, 75 W. SC test (HV): 9.5 V, 20 A, 110 W. HV side is 400 V; K = 120/400 = 0.3.

#### Shunt branch
On the LV side:
```text
cos φ0 = 75/(120 × 4) = 0.15625
Iw = 4 × 0.15625 = 0.625 A ;  Iμ = √(4² − 0.625²) = 3.951 A
Ro = 120/0.625 = 192 Ω ;  Xo = 120/3.951 = 30.37 Ω
```
Refer to the HV side (× 1/K² = 11.11):
```text
Ro(HV) = 192 × 11.11 = 2133 Ω
Xo(HV) = 30.37 × 11.11 = 337.5 Ω
```

#### Series branch (SC test is on the HV side)
```text
Z01 = 9.5/20 = 0.475 Ω
R01 = 110/20² = 0.275 Ω
X01 = √(0.475² − 0.275²) = 0.387 Ω
```

#### Equivalent circuit (HV side)
```text
        R=0.275Ω    X=0.387Ω
 o──┬────┬───/\/\/───────(((────────o
    │    │                          
 2133Ω  337.5Ω                  load
    │    │                       V2'
 o──┴────┴───────────────────────────o
```

#### Efficiency at half load, 0.8 pf lagging
Rated HV current = 8000/400 = 20 A, the same as the SC test current, so the full-load copper loss is 110 W.
```text
Output   = 0.5 × 8000 × 0.8 = 3200 W
Iron loss = 75 W
Cu loss  = 0.5² × 110 = 27.5 W
η = 3200/(3200 + 75 + 27.5) = 96.9 %
```
**Answer: Ro = 2133 Ω, Xo = 337.5 Ω, R01 = 0.275 Ω, X01 = 0.387 Ω (HV side); η at half load, 0.8 pf = 96.9 %.**

## Check yourself

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

1. OC test on LV (120 V). How are Ro = 192 Ω and Xo = 30.37 Ω referred to the HV side?
   A. Multiplied by 1/K² = 11.11
   B. Divided by 11.11
   C. Multiplied by 3.33
   D. Unchanged
   Answer: A. HV has the higher voltage, so impedances increase by (400/120)² = 11.11. (Key point: Shunt referral to HV)

2. SC on HV side gives which R01?
   A. 0.55 Ω
   B. 0.275 Ω
   C. 5.5 Ω
   D. 0.0248 Ω
   Answer: B. R01 = 110/20² = 0.275 Ω. (Key point: Series branch)

3. What is the copper loss at half load?
   A. 55 W
   B. 110 W
   C. 27.5 W
   D. 37.5 W
   Answer: C. 0.5² × 110 = 27.5 W. (Key point: Half-load Cu loss)

4. What is the efficiency at half load, 0.8 pf?
   A. 96.9 %
   B. 93.4 %
   C. 99.0 %
   D. 98.1 %
   Answer: A. 3200/(3200 + 75 + 27.5) = 96.9 %. (Key point: Efficiency)

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