---
title: "1000 VA 230/115 V transformer equivalent circuit"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-1000va-230-115-equivalent-circuit"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-1000va-230-115-equivalent-circuit
site: BCT Engineering
---

# 1000 VA 230/115 V transformer equivalent circuit

Electrical Machines · Chapter 2: Transformer

- Asked in: 2078 Chaitra, 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-1000va-230-115-equivalent-circuit
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

A 1000-VA 230/115-V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below.

| Open-circuit test (on secondary side) | Short-circuit test (on primary side) |
|---|---|
| $V_{oc}$ = 115 V | $V_{sc}$ = 17.1 V |
| $I_{oc}$ = 0.11 A | $I_{sc}$ = 8.7 A |
| $P_{oc}$ = 3.9 W | $P_{sc}$ = 38.1 W |

(i) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer.
(ii) Determine the transformer's efficiency at rated conditions and 0.8 PF lagging.

## Short answer

From OC test referred LV, shunt Rc=3391 Ω, Xm=1099 Ω; from SC test referred LV, Req=0.126 Ω, Xeq=0.475 Ω. At rated 115 V, 8.696 A and 0.8 pf lagging, efficiency is 98.35%.

## Answer

**Given:** 1000 VA, 230/115 V. HV = 230 V, LV = 115 V, so $K = 2$ and $K^2 = 4$.

The OC test was done on the LV side at rated voltage, so the shunt branch is already on the LV side. The SC test was done on the HV side, so the series branch must be divided by $K^2 = 4$ to refer it to the LV side.

#### (i) Equivalent circuit referred to the LV side

**Open-circuit test (LV side):** $V_{oc} = 115$ V, $I_{oc} = 0.11$ A, $P_{oc} = 3.9$ W.

$$
\begin{aligned}
\cos\phi_0 &= \frac{3.9}{115\times0.11} = 0.3083 \\
I_w &= 0.11\times0.3083 = 0.03391\ \text{A},\quad I_\mu = \sqrt{0.11^2 - 0.03391^2} = 0.1046\ \text{A} \\
R_c &= \frac{115}{0.03391} = 3391\ \Omega,\qquad X_m = \frac{115}{0.1046} = 1099\ \Omega
\end{aligned}
$$

**Short-circuit test (HV side):** $V_{sc} = 17.1$ V, $I_{sc} = 8.7$ A, $P_{sc} = 38.1$ W.

$$
\begin{aligned}
Z_{eq,H} &= \frac{17.1}{8.7} = 1.966\ \Omega,\qquad R_{eq,H} = \frac{38.1}{8.7^2} = 0.5034\ \Omega \\
X_{eq,H} &= \sqrt{1.966^2 - 0.5034^2} = 1.900\ \Omega
\end{aligned}
$$

**Referred to the LV side:**

$$
R_{eq,L} = \frac{0.5034}{4} = 0.1258\ \Omega,\qquad X_{eq,L} = \frac{1.900}{4} = 0.4750\ \Omega
$$

```text
   Req=0.1258 ohm   Xeq=0.475 ohm
 o--+-----/\/\/\-----UUUU------o
    |
   [Rc=3391] [Xm=1099]    Load (LV side, 115 V)
    |
 o--+---------------------------o
```

#### (ii) Efficiency at rated load, 0.8 pf lagging

Rated LV current: $I_2 = \dfrac{1000}{115} = 8.696$ A.

$$
\begin{aligned}
P_{cu} &= I_2^2R_{eq,L} = 8.696^2\times0.1258 = 9.515\ \text{W} \\
P_{core} &= 3.9\ \text{W (OC test at rated voltage)} \\
P_{out} &= 1000\times0.8 = 800\ \text{W} \\
\eta &= \frac{800}{800 + 9.515 + 3.9}\times100 = 98.35\%
\end{aligned}
$$

(The same copper loss follows from the HV side: rated HV current $4.348$ A, $4.348^2\times0.5034 = 9.515$ W.)

**Answer: $R_c = 3391\ \Omega$, $X_m = 1099\ \Omega$, $R_{eq} = 0.126\ \Omega$, $X_{eq} = 0.475\ \Omega$ (LV side); $\eta = 98.35\%$**

## Check yourself

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

1. How is the SC result measured on the HV side moved to the LV side?
   A. Multiply by 4
   B. Divide by 2
   C. Multiply by 2
   D. Divide by K^2 = 4
   Answer: D. HV-to-LV impedance conversion divides by K^2. (Key point: Referring SC to LV)

2. Why need the shunt branch not be referred?
   A. The OC test was already done on the LV side at rated voltage
   B. Shunt elements are independent of the side
   C. The SC test gives the shunt branch
   D. The shunt branch is zero
   Answer: A. OC data on the LV side directly give Rc and Xm on the LV side. (Key point: Shunt branch)

3. What is the efficiency at rated load, 0.8 pf?
   A. 99.2%
   B. 98.35%
   C. 97.0%
   D. 98.9%
   Answer: B. Output 800 W; losses 9.515 + 3.9 W; eta = 800/813.4 = 98.35%. (Key point: Efficiency)

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