Calculate transformer losses from no-load (core) and full-load (copper) losses at any load, or from measured input and output power, voltage and current.

Loss = no-load (core) loss + full-load (copper) loss × load². Rating and power factor are only needed for efficiency.

100 = full rated load. Change it for part-load losses.

Use loss data at the same reference conditions; the model scales this term with load².

Efficiency, annual energy and cost (optional)

Leave blank to omit annual estimates. The entered hours all use this constant operating point.

Leave blank for energy only. A price requires annual hours.

Transformer Loss Formula

Measured transformer loss is real input power minus real output power. The operating model instead uses constant no-load loss plus a full-load loss term scaled by load fraction squared, at fixed voltage/frequency and reference loss conditions. This is an AC transformer estimate, not a claim that an ordinary transformer operates on steady DC. See the Eaton design guide for the loss model.

Modeled loss = P₀ + Pₖ × x²
Measured loss = Pᵢₙ − Pₒᵤₜ

Where:

  • Pâ‚€ = no-load/core loss.
  • Pâ‚– = incremental full-load winding/load-loss term, not total full-load loss.
  • x = load fraction, entered as percent of rated apparent power divided by 100.
  • Pᵢₙ and Pₒᵤₜ = simultaneous total real input and output powers.
  • Use consistent power units. The calculator converts W, kW and available MW inputs and reports loss in kW.

Operating mode requires an explicit load power factor from 0 to 1. In balance mode, rating and PF may both be omitted; they are needed only for the optional real-output/efficiency calculation. The positive equal-loss point x = √(P₀/Pₖ) is an unconstrained efficiency optimum only for constant positive PF and positive loss terms. A balance above 100% rating is not overload authorization; within 0–100%, the fixed-positive-PF optimum is capped at 100%. Zero loss terms can make the balance nonexistent or nonunique, and zero PF gives no unique efficiency optimum.

Input and Output Power Relationships

The measured loss equation subtracts total real output from total real input. The voltage/current/PF expressions below are background for obtaining single-phase AC real powers in watts; they are not separate calculator input fields. Use directly measured real powers when available.

Pᵢₙ = Iᵢ × Vᵢ × PFᵢ
Pₒᵤₜ = Iₒ × Vₒ × PFₒ
TL = Pᵢₙ - Pₒᵤₜ

A positive measured difference represents real power not delivered at the output measurement boundary. A small difference can be dominated by instrument uncertainty. Output greater than input is rejected under this passive operating-point model; verify measurements, units, boundaries and timing.

How to Use the Transformer Loss Calculator

  1. For operating loss, enter percent load (100 for full load), no-load loss and incremental full-load loss term using matching reference conditions. Rating and load PF are an optional pair for efficiency/output.
  2. For the core/load-loss balance, enter the two positive loss terms. Rating and PF are an optional pair for efficiency/output.
  3. For measured loss, enter simultaneous total real input and output powers and select their units, or use the voltage, current and power factor mode for single-phase readings.
  4. Optionally enter annual energized hours at the same operating point. Add USD/kWh price only when hours are supplied. Leave both blank to omit annual outputs.
  5. Select Calculate to obtain loss in kW or the balance load in percent. Calculation details explain the model and assumptions.

For the most meaningful result, use measurements taken at the same time and under the same load condition.

What Transformer Loss Represents

Measured mode gives net real-power difference; modeled modes estimate no-load loss and a load-squared term. Actual losses can include the following components:

  • Core loss — caused by hysteresis and eddy currents in the magnetic core; often present whenever the transformer is energized.
  • Copper loss — caused by winding resistance and approximately proportional to current squared at a fixed resistance/reference condition.
  • Stray loss — caused by leakage flux inducing currents in structural parts and conductors.
  • Dielectric and miscellaneous losses — generally smaller, but still part of total inefficiency.

The model displays the supplied no-load term and the scaled load-loss term separately. Measured mode reports only their net real-power difference and does not identify physical loss components. Temperature, harmonics and other conditions can change actual losses; manufacturer data must match the reference conditions.

Efficiency and Transformer Loss

Transformer loss and efficiency are directly related. Once input and output power are known, efficiency can be calculated as:

eta = (Pₒᵤₜ) / (Pᵢₙ) × 100

As loss increases, efficiency decreases. Even highly efficient transformers still have some loss, especially at higher current levels where winding heating becomes more significant.

Example

As a single-phase AC background example, suppose simultaneous measurements give the following values. Convert them to total real powers before using measured mode:

  • Input current = 120 A
  • Input voltage = 80 V
  • Input power factor = 1
  • Output current = 100 A
  • Output voltage = 90 V
  • Output power factor = 1

First calculate input power:

Pᵢₙ = 120 × 80 × 1 = 9600

Then calculate output power:

Pₒᵤₜ = 100 × 90 × 1 = 9000

Now subtract output power from input power:

TL = 9600 - 9000 = 600

The difference is 600 W, or 0.6 kW, with efficiency 9000/9600 × 100 = 93.75%. In measured mode, enter 9.6 kW input and 9.0 kW output, or select W and enter 9600 and 9000.

Power Factor Notes

The operating model uses the explicitly entered load PF to convert apparent output to real output. Measured mode already takes real power, so it has no PF entry. RMS voltage/current alone do not determine AC real power without the appropriate power-factor information.

  • Ordinary AC power transformers must not be supplied with steady DC; see Schneider’s transformer guidance.
  • Use measured PF values for more accurate AC results.
  • If only apparent power is known, convert to real power before interpreting the loss value.

Common Reasons for Unexpected Results

  • Output exceeding input — rejected for this passive model; verify consistent real-power measurements, boundaries and units.
  • Very large loss — check W versus kW, MVA versus kVA, percent load, and whether the rated load-loss term excludes the no-load term.
  • Very small measured difference — instrument uncertainty may be large relative to the inferred loss; equal readings do not prove zero physical loss.

Practical Measurement Tips

  • Use consistent total real-power readings for measured mode; RMS voltage/current products alone are not sufficient without PF and phase information.
  • Measure input and output under the same operating condition.
  • Confirm unit selections before calculating.
  • For lightly loaded transformers, fixed core losses may dominate.
  • For heavily loaded transformers, copper losses usually become a larger share of the total.

Single-Phase vs. Three-Phase Use

Measured mode accepts total real input and output powers for either single- or three-phase AC systems. Model mode uses total rated kVA/MVA, load fraction and PF, so total three-phase rating must be entered once. The model is not a thermal or harmonic simulation. Annual energy equals loss kW × the entered energized hours at the same load; an energized unloaded interval still incurs its no-load term. A varying duty profile needs separate intervals.

Transformer Loss FAQ

Is transformer loss always present?

Real energized transformers generally have no-load loss, with additional load loss as current rises. Zero terms in this model are idealized inputs, not proof that a physical transformer is lossless.

Why can the output voltage be higher than expected while power is lower?

A transformer can step voltage up or down while still losing some total power internally. Power depends on both voltage and current, and for AC systems also on power factor.

Can this calculator be used for efficiency checks?

Yes, the calculator reports a modeled or measured real-power efficiency ratio when the needed inputs are supplied. A zero-input ratio is undefined. The result alone does not establish acceptable temperature, normal condition, suitable loading or equipment capability.