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

# Equivalent circuit, efficiency and regulation of 5 kVA transformer

Electrical Machines I · Chapter 2: Transformer

- Asked in: 2079 Chaitra, 6 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/oc-sc-5kva-220-400-lv
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines_i/chapter-2

## Question

Open-circuit and short-circuit tests on a 5 kVA, 220/400 V, 50 Hz single phase transformer gave the following results:
O.C. test: 220 V, 2 A, 100 W (carried on LV side)
S.C. test: 40 V, 11.4 A, 200 W (carried on HV side)
Determine: i) The equivalent circuit parameters referred to LV side. ii) The efficiency and approximate regulation of the transformer at full load 0.9 pf lagging.

## Short answer

From OC test shunt parameters referred to LV are R0 = 484 ohm and X0 = 112.96 ohm; from SC test referred to LV series parameters are Req = 0.466 ohm and Xeq = 0.954 ohm. Full-load 0.9 pf lagging efficiency is 92.97% and regulation is 8.62%.

## Answer

OC test on LV gives the shunt branch directly on LV. SC test on HV gives series values on HV, which are referred to LV by $a^2 = (220/400)^2 = 0.3025$. Rated HV current $= 5000/400 = 12.5$ A, so the SC reading at 11.4 A must be scaled to full load.

#### i) Equivalent circuit parameters referred to LV

OC test:

$$
\begin{aligned}
\cos\phi_0 &= \frac{100}{220\times2} = 0.2273 \\
I_c &= 2\times0.2273 = 0.4545\ \text{A}, \quad I_m = 2\times0.9738 = 1.9477\ \text{A} \\
R_0 &= \frac{220}{0.4545} = 484.0\ \Omega, \quad X_0 = \frac{220}{1.9477} = 112.96\ \Omega
\end{aligned}
$$

SC test (HV side):

$$
\begin{aligned}
R_{eq,HV} &= \frac{200}{11.4^2} = 1.539\ \Omega, \quad Z_{eq,HV} = \frac{40}{11.4} = 3.509\ \Omega \\
X_{eq,HV} &= \sqrt{3.509^2 - 1.539^2} = 3.153\ \Omega
\end{aligned}
$$

Referred to LV:

$$
R_{eq,LV} = 1.539\times0.3025 = 0.4655\ \Omega, \quad X_{eq,LV} = 3.153\times0.3025 = 0.9539\ \Omega
$$

#### ii) Efficiency at full load, 0.9 pf lag

$$
\begin{aligned}
P_{cu,fl} &= 200\left(\frac{12.5}{11.4}\right)^2 = 240.46\ \text{W} \\
P_{out} &= 5000\times0.9 = 4500\ \text{W} \\
\eta &= \frac{4500}{4500 + 100 + 240.46} = 92.97\%
\end{aligned}
$$

#### Approximate regulation (on HV side, $\sin\phi = 0.4359$)

$$
\begin{aligned}
\%\text{Reg} &= \frac{I(R\cos\phi + X\sin\phi)}{V}\times100 \\
&= \frac{12.5(1.539\times0.9 + 3.153\times0.4359)}{400}\times100 \\
&= \frac{17.31 + 17.18}{400}\times100 = 8.62\%
\end{aligned}
$$

**Answer:** $R_0 = 484\ \Omega$, $X_0 = 112.96\ \Omega$, $R_{eq} = 0.466\ \Omega$, $X_{eq} = 0.954\ \Omega$ (all LV side); efficiency = **92.97%**; regulation ≈ **8.62%**.

## Check yourself

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

1. In this problem, which parameters does each test give?
   A. OC test gives shunt R0, X0; SC test gives series Req, Xeq
   B. OC gives Xeq only
   C. OC gives series branch; SC gives shunt branch
   D. Both tests give only the shunt branch
   Answer: A. OC test at rated voltage with open secondary gives the shunt branch; SC at reduced voltage gives series impedance. (Key point: Which test gives which branch)

2. How are the SC (HV side) resistance and reactance referred to the LV side?
   A. Divide by 0.3025
   B. Multiply by (400/220)^2
   C. Multiply by (220/400)^2 = 0.3025
   D. Multiply by 220/400
   Answer: C. LV-referred values equal HV values times a^2 = (220/400)^2 = 0.3025. (Key point: Referring HV values to LV)

3. The SC test was at 11.4 A but rated HV current is 12.5 A. How is full-load copper loss found?
   A. 200 x (11.4/12.5)^2 = 166 W
   B. 200 x (12.5/11.4)^2 = 240.46 W
   C. 200 W directly
   D. 200 x (12.5/11.4) = 219 W
   Answer: B. Copper loss varies as the square of current, so scale 200 W by (12.5/11.4)^2. (Key point: Scaling copper loss)

4. What are the output and total loss used for efficiency at full load, 0.9 pf?
   A. Output 5000 W; loss 100 W only
   B. Output 4500 W; loss 100 + 240.46 W
   C. Output 4500 W; loss 200 W only
   D. Output 5000 W; loss 100 + 240.46 W
   Answer: B. Output = 5000 x 0.9 = 4500 W, losses = iron 100 W + copper 240.46 W, giving 92.97%. (Key point: Efficiency)

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