---
title: "25 kVA 250/2500 V tests, equivalent circuit"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-25kva-250-2500-parameters"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-25kva-250-2500-parameters
site: BCT Engineering
---

# 25 kVA 250/2500 V tests, equivalent circuit

Electrical Machines · Chapter 2: Transformer

- Asked in: 2071 Magh (old course), 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/tf-25kva-250-2500-parameters
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 25 kVA, 250 V/2500 V, 50 Hz single phase transformer gave the following test results:
No load test (on L.V. side): 250 V, 1.4 A, 105 watts
Short circuit test (on H.V. side): 120 V, 8 A, 320 watts
Calculate the equivalent circuit parameters referred to primary side and draw the equivalent circuit.

## Short answer

Referred to primary: R0=595.2 Ω, X0=187.2 Ω, R01=0.05 Ω, X01=0.1414 Ω.

## Answer

**Given:** 25 kVA, 250 V/2500 V. Primary = LV (250 V), secondary = HV (2500 V), $K = 10$, $K^2 = 100$.

#### Open-circuit test (on the LV side)

The no-load power is almost entirely core loss, so the shunt branch is found from it.

$$
\begin{aligned}
\cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{105}{250\times1.4} = 0.3 \\
I_w &= I_0\cos\phi_0 = 0.42\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 1.336\ \text{A} \\
R_0 &= \frac{V_0}{I_w} = 595.2\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 187.2\ \Omega\quad(\text{on the LV side})
\end{aligned}
$$

#### Short-circuit test (on the HV side)

At the reduced test voltage the core loss is negligible, so the wattmeter reading is the copper loss.

$$
\begin{aligned}
Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{120}{8} = 15\ \Omega \\
R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{320}{8^2} = 5\ \Omega \\
X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 14.14\ \Omega\quad(\text{on the HV side})
\end{aligned}
$$

#### Referring to the LV side

The turns ratio is $K = 2500/250 = 10$, so $K^2 = 100$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$
\begin{aligned}
R_{eq} &= 5 \div 100 = 0.05\ \Omega \\ X_{eq} &= 14.14 \div 100 = 0.1414\ \Omega
\end{aligned}
$$

#### Equivalent circuit referred to the LV side

```text
   Req=0.05 ohm   Xeq=0.141 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=595.2] [X0=187.2]   Load
    |      (in parallel)
 o--+---------------------------o
```

$R_{eq} = 0.05\ \Omega$, $X_{eq} = 0.1414\ \Omega$ ($Z_{eq} = 0.15\ \Omega$), shunt branch $R_0 = 595.2\ \Omega$ in parallel with $X_0 = 187.2\ \Omega$.

**Answer (referred to primary):** $R_0 = 595.2\ \Omega$, $X_0 = 187.2\ \Omega$, $R_{01} = 0.05\ \Omega$, $X_{01} = 0.1414\ \Omega$.

**Note:** the 25 kVA rating is not needed for the parameters. The SC test current of 8 A is the rated HV current only for a 20 kVA unit (a 25 kVA unit has 10 A on the HV side), so the test is below full load. The equivalent circuit parameters are the same, because they do not depend on the test current if the transformer is linear. The copper loss at the true full load would be $320\times(10/8)^2 = 500$ W.

## Check yourself

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

1. Does the 25 kVA rating affect the equivalent circuit parameters?
   A. Yes, they scale with kVA
   B. Yes, but only R0
   C. No, they come from test readings and do not depend on the test current if linear
   D. Only the series branch changes
   Answer: C. The note says parameters are the same irrespective of test current for a linear transformer. (Key point: Rating relevance)

2. Is the 8 A SC test at full load for this 25 kVA unit?
   A. Yes, rated HV current is 8 A
   B. No, rated HV current is 10 A
   C. Yes, rated is 80 A
   D. No, rated is 2.5 A
   Answer: B. 25000/2500 = 10 A. (Key point: Test level)

3. What are R_eq and X_eq referred to the primary?
   A. 0.05 ohm and 0.1414 ohm
   B. 5 ohm and 14.14 ohm
   C. 0.5 ohm and 1.414 ohm
   D. 0.05 ohm and 0.5 ohm
   Answer: A. Zsc = 15 ohm, Rsc = 5 ohm, Xsc = 14.14 ohm on HV; divide by 100. (Key point: Series branch)

## Close variants set in other exams

- [20 kVA 250/2500 V tests, equivalent circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-20kva-250-2500-parameters) — Electrical Machines, chapter 2, asked 2×

## Related in Electrical Machines

- [Equivalent circuit of 20 kVA 250/2500 V transformer from tests](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-20kva-250-2500) — Electrical Machines I, chapter 2, asked 3×
- [20 kVA 250/2500 V tests, regulation, efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-20kva-250-2500-regulation-efficiency) — Electrical Machines, chapter 2
- [Equivalent circuit of 20 kVA 220/2200 V transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Aoc-sc-20kva-220-2200-primary) — Electrical Machines I, chapter 2
- [8 kVA 400/120 V tests, parameters on HV side](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-8kva-400-120-efficiency-hv) — Electrical Machines, chapter 2
- [Equivalent circuit and half-load efficiency of 50 kVA transformer](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3A50kva-2500-250-test-secondary-half) — Electrical Machines I, chapter 2
- [10 kVA 250/1000 V tests, equivalent circuit](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atf-10kva-250-1000-parameters) — Electrical Machines, chapter 2

## Related in other subjects

- [20 kVA 250/2500 V test parameters](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-20kva-250-2500-primary) — Electrical Machine, chapter 2, asked 2×
- [8 kVA 400/120 V tests, half-load efficiency](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-8kva-400-120-hv-side) — Electrical Machine, chapter 2
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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×