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Transformer Efficiency

Electronics & PhysicsPower and Energy
What you know

From the open-circuit test; roughly constant at any load.

I²R loss in the windings; scales with the square of load.

Efficiency —
Input power (kW)
—
Total losses (kW)
—
Losses as share of input
—
Maximum efficiency
—

Transformer efficiency is output power divided by input power, where input equals output plus core (iron) loss plus copper loss. A transformer delivering 80 kW with 1 kW of core loss and 1.5 kW of copper loss draws 82.5 kW, so its efficiency is 80 ÷ 82.5 = 96.97%.

About this tool

Electrical students, test engineers and anyone sizing a distribution transformer need the same figure: how much of the power going in comes out. Enter the output power and either the two loss components from the open-circuit and short-circuit tests, or the input power read on the primary side. The calculator returns the efficiency, the total losses in kilowatts and as a share of input, and, when you give the loss split, the load fraction at which efficiency peaks (where copper loss equals core loss). It treats power as real power in kW, so convert kVA to kW with the power factor first. It does not model all-day efficiency across a varying load profile; for that, work out energy losses per hour of the day separately.

How to use it

  1. Pick what you know

    Choose output and losses for test-sheet data, or input and output for two wattmeter readings.

  2. Enter output power

    Real power delivered by the secondary in kW, i.e. kVA times power factor.

  3. Enter losses or input

    Core loss and copper loss at the same load, or the measured primary input power.

  4. Read the result

    Efficiency, total losses and, in loss mode, where peak efficiency falls relative to this load.

Examples

100 kVA at 0.8 power factor

What you know
Output and losses
Output power
80
Core (iron) loss
1
Copper loss at this load
1.5

Result Efficiency: 96.97%
Input power (kW): 82.5
Total losses (kW): 2.5
Losses as share of input: 3.03%
Maximum efficiency: Peak efficiency near 81.6% of this load

  1. Input = 80 + 1 + 1.5 = 82.5 kW
  2. Efficiency = 80 ÷ 82.5 × 100 = 96.97 %
  3. Peak efficiency where copper loss = core loss: load fraction = √(1 ÷ 1.5) = 0.816

A 100 kVA unit delivering 80 kW with 1 kW iron loss and 1.5 kW copper loss at full load.

Bench test, 50 kW in, 48 kW out

What you know
Input and output
Output power
48
Input power
50

Result Efficiency: 96%
Input power (kW): 50
Total losses (kW): 2
Losses as share of input: 4%
Maximum efficiency: Needs the loss split

  1. Losses = 50 − 48 = 2 kW
  2. Efficiency = 48 ÷ 50 × 100 = 96 %

Two wattmeters read 50 kW on the primary and 48 kW on the secondary.

Losses equal: peak efficiency

What you know
Output and losses
Output power
200
Core (iron) loss
0.5
Copper loss at this load
0.5

Result Efficiency: 99.5%
Input power (kW): 201
Total losses (kW): 1
Losses as share of input: 0.5%
Maximum efficiency: Peak efficiency near 100% of this load

  1. Input = 200 + 0.5 + 0.5 = 201 kW
  2. Efficiency = 200 ÷ 201 × 100 = 99.5 %
  3. Peak efficiency where copper loss = core loss: load fraction = √(0.5 ÷ 0.5) = 1

When copper loss equals core loss the transformer is already at its peak efficiency point.

How it is calculated

η = Pout ÷ (Pout + Pcore + Pcu) × 100

η
efficiency in percent
Pout
real output power, kW
Pcore
core (iron) loss from hysteresis and eddy currents, nearly constant with load
Pcu
copper (I²R) loss in the windings at the given load

Input power is output plus every loss, so efficiency is output over that sum. Copper loss varies with the square of load: at half load it is a quarter of the full-load value. Efficiency is highest where copper loss equals core loss, which happens at a load fraction of √(Pcore ÷ Pcu) relative to the load you entered.

When not to use it

  • Do not use it for all-day (energy) efficiency over a changing daily load.
  • Do not enter apparent power in kVA as output without applying the power factor.
  • It does not cover switch-mode power supplies, whose losses follow different physics.

Common mistakes

  • Using full-load copper loss when the transformer runs at part load.
  • Mixing watts and kilowatts between the output and loss fields.
  • Entering kVA rating as output power at a power factor below 1.

Frequently asked questions

Why are transformer efficiencies so high?

A transformer has no moving parts, so there is no friction or windage loss. Only core loss and winding resistance loss remain, and large units routinely reach 98–99.5%.

What is the difference between core loss and copper loss?

Core loss comes from hysteresis and eddy currents in the laminated steel and depends on voltage and frequency, so it is nearly constant whenever the unit is energised. Copper loss is I²R heating in the windings and grows with the square of load current.

How do I find copper loss at part load?

Multiply the full-load copper loss by the square of the load fraction. At 50% load, a 1.5 kW full-load copper loss becomes 1.5 × 0.25 = 0.375 kW.

At what load is a transformer most efficient?

Efficiency peaks when copper loss equals core loss. With 1 kW core loss and 1.5 kW full-load copper loss, that is √(1 ÷ 1.5) ≈ 81.6% of full load. Distribution transformers are often designed to peak below full load because they spend most hours lightly loaded.

Does power factor change efficiency?

Yes. Losses depend on current, but output in kW depends on power factor. At the same kVA and the same losses, a lower power factor delivers fewer kW, so the efficiency figure drops.

Where do the loss figures come from?

Core loss is measured in the open-circuit test at rated voltage, and full-load copper loss in the short-circuit test at rated current. Manufacturers list both on the test certificate.

Cite this tool

Using this tool in an assignment, article or report? Copy a citation, or a link for your website.

APA

Tools Directory Editorial Team. (2026, October 12). Transformer Efficiency. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/electronics-physics/power-and-energy/transformer-efficiency/

MLA

Tools Directory Editorial Team. “Transformer Efficiency.” Tools Directory, 12 Oct. 2026, https://tools.milunir.com/electronics-physics/power-and-energy/transformer-efficiency/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “Transformer Efficiency.” Tools Directory. Last modified October 12, 2026. Accessed October 27, 2026. https://tools.milunir.com/electronics-physics/power-and-energy/transformer-efficiency/.

Link<a href="https://tools.milunir.com/electronics-physics/power-and-energy/transformer-efficiency/">Transformer Efficiency – Tools Directory</a>