---
title: "Efficiency and regulation of 5 kVA transformer"
subject: "Electrical Machines I"
chapter: 2
key: "semester_4/electrical_machines_i:sc-5kva-220-400-efficiency-regulation"
source: https://www.bctengineer.com/q/semester_4/electrical_machines_i/sc-5kva-220-400-efficiency-regulation
site: BCT Engineering
---

# Efficiency and regulation of 5 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2074 Bhadra, 8 marks
- Video: [Transformer OC test and SC test problems | effeciency, voltage regulation calculations](https://www.youtube.com/watch?v=aB2j_Fqd4zo) — The PiSquare Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/electrical_machines_i/sc-5kva-220-400-efficiency-regulation
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

Open circuit and Short circuit test on 5 kVA, 220/400 V, 50 Hz, single phase transformer gave the following results. Short circuit test (on H.V. side): 40 V, 11.4 A, 200 watts. Determine the efficiency and the voltage regulation of the transformer at full load at 0.9 pf lagging.

## Short answer

Assuming iron loss 100 W, full-load copper loss is scaled from SC test: 200×(12.5/11.4)^2 = 240.5 W. Output 4500 W at 0.9 pf. Efficiency = 92.97%. Voltage regulation using R02=1.539 Ω, X02=3.153 Ω and 0.9 pf lagging = 8.62%.

## Answer

Only the short-circuit data is given; the open-circuit result is needed for iron loss. **Assumption:** the usual OC data for this textbook problem, **220 V, 2 A, 100 W on the LV side**, i.e. iron loss $P_i = 100$ W. (If a different OC wattmeter reading is given, replace 100 W.)

Rated HV current: $I_2 = \frac{5000}{400} = 12.5$ A. The SC test was done at 11.4 A, so its copper loss must be scaled to full load.

#### Parameters from SC test (HV side)
$$
\begin{aligned}
Z_{02} &= \frac{40}{11.4} = 3.509\ \Omega \\
R_{02} &= \frac{200}{11.4^2} = 1.539\ \Omega \\
X_{02} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega
\end{aligned}
$$

#### Full-load copper loss
$$
P_{cu,FL} = 200 \times \left(\frac{12.5}{11.4}\right)^2 = I_2^2R_{02} = 12.5^2 \times 1.539 = 240.5\ \text{W}
$$

#### Efficiency at full load, 0.9 pf lagging
$$
\begin{aligned}
\text{Output} &= 5000 \times 0.9 = 4500\ \text{W} \\
\text{Losses} &= P_i + P_{cu} = 100 + 240.5 = 340.5\ \text{W} \\
\eta &= \frac{4500}{4500 + 340.5} \times 100 = 92.97\%
\end{aligned}
$$

#### Voltage regulation at full load, 0.9 pf lagging
$\cos\phi = 0.9$, $\sin\phi = 0.4359$.
$$
\begin{aligned}
\%\text{Reg} &= \frac{I_2(R_{02}\cos\phi + X_{02}\sin\phi)}{V_2}\times 100 \\
&= \frac{12.5\,(1.539 \times 0.9 + 3.153 \times 0.4359)}{400}\times 100 \\
&= \frac{12.5 \times (1.385 + 1.375)}{400}\times 100 = 8.62\%
\end{aligned}
$$
The regulation does not depend on the OC test, so 8.62% holds whatever the iron loss is.

**Answer:** Full-load efficiency at 0.9 pf lagging $\approx$ **92.97%** (with $P_i$ = 100 W); voltage regulation = **8.62%**.

## Check yourself

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

1. Rated HV current of the 5 kVA, 400 V side is:
   A. 11.4 A
   B. 12.5 A
   C. 22.7 A
   D. 9.1 A
   Answer: B. I2 = 5000/400 = 12.5 A, which differs from the SC test current of 11.4 A. (Key point: Full-load current)

2. The SC test was at 11.4 A with 200 W. Full-load copper loss is found by:
   A. Using 200 W directly
   B. Multiplying 200 W by 12.5/11.4
   C. Multiplying 200 W by (12.5/11.4)^2
   D. Dividing 200 W by (12.5/11.4)^2
   Answer: C. Copper loss varies as current squared, so Pcu = 200 x (12.5/11.4)^2 = 240.5 W. (Key point: Scaling copper loss)

3. With iron loss 100 W, Pcu = 240.5 W and output 4500 W, the full-load efficiency is about:
   A. 92.97%
   B. 96.3%
   C. 89.5%
   D. 98.2%
   Answer: A. Efficiency = 4500/(4500+340.5) = 92.97%. (Key point: Efficiency)

4. For 0.9 pf lagging, regulation is computed from:
   A. I2(R02 cos(phi) - X02 sin(phi))/V2
   B. I2(R02 cos(phi) + X02 sin(phi))/V2
   C. I2 R02 /V2 only
   D. I2 X02 /V2 only
   Answer: B. For lagging load the reactive term adds. The result is 8.62%, independent of the OC test. (Key point: Regulation formula)

## Related in Electrical Machines I

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- [Equivalent circuit of 5 kVA 200/400 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-5kva-200-400-primary) — Electrical Machines I, chapter 2, asked 2×

## Related in other subjects

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