---
title: "250/500 V test; equivalent circuit, max efficiency output"
subject: "Electrical Machines"
chapter: 2
key: "semester_4/electrical_machines:tf-250-500-max-efficiency-output"
source: https://www.bctengineer.com/q/semester_4/electrical_machines/tf-250-500-max-efficiency-output
site: BCT Engineering
---

# 250/500 V test; equivalent circuit, max efficiency output

Electrical Machines · Chapter 2: Transformer

- Asked in: 2068 Magh, 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-250-500-max-efficiency-output
- Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2

## Question

Test data on a 1-phase, 250/500 V, 50 Hz transformer are:
O.C. Test: 250 V, 1 A, 80 W (carried on L.V. side)
S.C. Test: 20 V, 12 A, 100 W (carried on H.V. side)
Then draw the equivalent circuit referred to primary side and find out the output power to obtain maximum efficiency at 0.9 lag p.f.

## Short answer

Equivalent circuit referred to 250 V primary: R0=781 Ω, X0=264 Ω, Req=0.174 Ω, Xeq=0.379 Ω. Maximum efficiency occurs at x=√(80/100)=0.8944 of rated kVA, i.e. 5.37 kVA; at 0.9 pf lagging output is 4.83 kW. Efficiency there is 96.8%.

## Answer

**Given:** 250/500 V, 50 Hz. Primary = 250 V (LV), $K = 2$. The kVA rating is not given, so the SC-test current (12 A on the HV side) is taken as the rated current, giving $S = 500\times12 = 6$ kVA. (The output at maximum efficiency found below does not actually depend on this assumption.)

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

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

$$
\begin{aligned}
\cos\phi_0 &= \frac{P_0}{V_0I_0} = \frac{80}{250\times1} = 0.32 \\
I_w &= I_0\cos\phi_0 = 0.32\ \text{A},\qquad I_\mu = \sqrt{I_0^2 - I_w^2} = 0.9474\ \text{A} \\
R_0 &= \frac{V_0}{I_w} = 781.25\ \Omega,\qquad X_0 = \frac{V_0}{I_\mu} = 263.9\ \Omega\quad(\text{on the LV side})
\end{aligned}
$$

#### 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.

$$
\begin{aligned}
Z_{sc} &= \frac{V_{sc}}{I_{sc}} = \frac{20}{12} = 1.667\ \Omega \\
R_{sc} &= \frac{P_{sc}}{I_{sc}^2} = \frac{100}{12^2} = 0.6944\ \Omega \\
X_{sc} &= \sqrt{Z_{sc}^2 - R_{sc}^2} = 1.515\ \Omega\quad(\text{on the HV side})
\end{aligned}
$$

#### Referring to the LV side

The turns ratio is $K = 500/250 = 2$, so $K^2 = 4$. Impedances are multiplied by $K^2$ when moved from LV to HV, and divided by $K^2$ when moved from HV to LV.

$$
\begin{aligned}
R_{eq} &= 0.6944 \div 4 = 0.1736\ \Omega \\ X_{eq} &= 1.515 \div 4 = 0.3788\ \Omega
\end{aligned}
$$

#### Equivalent circuit referred to the LV side

```text
   Req=0.174 ohm   Xeq=0.379 ohm
 o--+-----/\/\/\-----UUUU------o
    |                          
   [R0=781.25] [X0=263.9]   Load
    |      (in parallel)
 o--+---------------------------o
```

$R_{eq} = 0.1736\ \Omega$, $X_{eq} = 0.3788\ \Omega$ ($Z_{eq} = 0.4167\ \Omega$), shunt branch $R_0 = 781.25\ \Omega$ in parallel with $X_0 = 263.9\ \Omega$.

#### Output power for maximum efficiency at 0.9 lag

The core loss is the OC wattmeter reading, $P_i = 80$ W. The copper loss at the SC-test current is the SC wattmeter reading, 100 W, which is taken as the full-load copper loss $P_{cu,fl}$.

Maximum efficiency occurs when copper loss equals iron loss:

$$
\begin{aligned}
x^2P_{cu,fl} &= P_i \Rightarrow x = \sqrt{\frac{80}{100}} = 0.8944 \\
\text{kVA at } \eta_{max} &= 0.8944\times6 = 5.367\ \text{kVA} \\
P_{out} &= 5.367\times0.9 = 4.830\ \text{kW}
\end{aligned}
$$

Without assuming a rating: at maximum efficiency the copper loss must equal 80 W, so the HV current is $12\sqrt{0.8} = 10.73$ A and $P_{out} = 500\times10.73\times0.9 = 4.83$ kW, the same result.

Check of efficiency: $\eta_{max} = \dfrac{4830}{4830 + 2\times80}\times100 = 96.8\%$.

**Answer: output power at maximum efficiency $\approx 4.83$ kW (5.37 kVA at 0.9 pf lagging); equivalent circuit parameters are as above**

## Check yourself

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

1. What is the condition for maximum efficiency?
   A. Copper loss is twice iron loss
   B. Load is at full rating
   C. Copper loss equals iron loss
   D. Iron loss is zero
   Answer: C. x^2 Pcu,fl = Pi. (Key point: Max efficiency condition)

2. What load fraction x gives maximum efficiency here (Pi = 80 W, Pcu,fl = 100 W)?
   A. 0.894
   B. 0.8
   C. 1.25
   D. 0.64
   Answer: A. x = sqrt(80/100) = 0.8944. (Key point: Load fraction)

3. What is the output at maximum efficiency and 0.9 pf lagging (rating 6 kVA)?
   A. 5.37 kW
   B. 4.83 kW
   C. 5.4 kW
   D. 4.3 kW
   Answer: B. kVA = 0.8944 x 6 = 5.367; P = 5.367 x 0.9 = 4.83 kW. (Key point: Output power)

## Related in Electrical Machines

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