Electrical · magnetics

Transformer Calculator

Drag the turns. Watch the coils — and the physics — change together.
V₁/V₂ = N₁/N₂ = I₂/I₁
10 : 2 turns · 230V in

Windings

modify live

Supply & load

Common ratios

same behaviour

Core & windings

Efficiency vs. load

Primary & secondary, live

primary voltage secondary voltage both windings share one AC cycle — no phase shift in an ideal transformer
Field notes

Why turns ratio is the whole story — until losses show up

Limits

An ideal transformer with the losses bolted back on

The turns ratio is exact and always will be — it is a property of the winding, not of the load. Everything else here is a first approximation with the two biggest losses included and the rest left out.

The load is treated as resistive

Enter watts and the tool assumes those watts arrive at unity power factor, so current and voltage stay in step. Most real loads are not like that. A motor at 0.8 power factor draws 25% more current for the same useful watts, and a transformer is sized in kVA rather than kW precisely because the copper has to carry that current whether it does useful work or not. If you are sizing something real, work in kVA.

The secondary voltage does not droop

Secondary voltage is held at V₁/a whatever the load. A real transformer sags a few per cent between no load and full load as current passes through the winding resistance and leakage reactance — that figure is its regulation, and it is why a supply measured off-load reads higher than one under load.

Nothing here happens at switch-on

Energising an unloaded transformer can draw a magnetising inrush of several times rated current for the first few cycles, as the core saturates before the flux settles into its steady swing. It is the reason transformer circuits use time-delay fuses and why a breaker sized on steady-state current alone will nuisance-trip.

What the efficiency curve is, and is not

The curve is built from two components: a fixed core loss that is present whenever the transformer is energised, and a copper loss rising with the square of load. Efficiency peaks where the two are equal, which is the textbook result and is why the peak sits below full load rather than at it.

The three classes are representative figures, not a specific product. And a real winding warms as it works, so its resistance climbs and copper loss with it — there is no temperature in this model at all.

One thing this page used to get wrong. The primary current was computed as load / V₁ — the ideal, lossless answer — while the panel beside it drew an efficiency curve reading 94%. The two could not both be true. It now divides by the efficiency at the load you have chosen, which raises the primary current by up to 12% on a small transformer at light load. That is the figure you would size a primary fuse from, so it is worth having right. Verified in transformer_model.py.

How it works

Turns ratio, then reality

An ideal transformer is defined entirely by its turns ratio: V₁/V₂ = N₁/N₂. Because power in equals power out for an ideal device, current runs the other way — I₂/I₁ = N₁/N₂ — so a step-down winding that halves the voltage doubles the current. Drag either turns slider here and every other number updates instantly, because they're all derived from that one ratio.

Worked example

230V into a 1000:100 turns transformer (a 10:1 ratio) gives 23V out. A 100W load at 23V draws 4.35A on the secondary and 0.43A on the primary — checked against ampere-turns balance, 1000 × 0.43A matches 100 × 4.35A exactly.

Why isn't a real transformer 100% efficient?

Two loss mechanisms eat into it: core (iron) losses from magnetising the core, which are present constantly whenever the transformer is energised regardless of load, and copper losses from winding resistance, which scale with the square of the load current. Distribution transformers typically run 97–99% efficient at full load; large power transformers can exceed 99.5%.

Why does the efficiency chart peak partway up, not at full load?

Efficiency is maximised exactly where copper loss equals core loss — since core loss is constant and copper loss grows with current squared, that crossover typically happens around 40–70% of rated load for distribution transformers, which is also why they're deliberately sized a little larger than their average expected load.

Why does the visual only show a handful of turns for a real 20:1 transformer?

A real mains transformer might use hundreds or thousands of turns per winding — this tool draws the turns ratio at a legible scale (the presets pick small whole numbers with the same ratio) so the physics stays identical while the diagram stays readable.

Does turns ratio alone determine everything about a transformer?

Only for the ideal case. Real transformers also have leakage inductance, magnetising current, and winding resistance that this tool's core equations don't model — turns ratio gets you the voltage and current relationship exactly right, which is most of what matters for a first-pass calculation.

Why are the two waveforms perfectly in phase, never shifted?

Because they're driven by the same changing magnetic flux in the same core at the same instant — there's no physical mechanism for an ideal transformer to delay one winding relative to the other. A phase shift would only show up from added leakage inductance or a reactive load, neither of which this ideal-transformer model includes.

Results are for reference only. These are ideal-transformer relationships — real units have winding resistance, leakage reactance, and core losses. Do not size or connect equipment from these figures alone.
Version history · 1 release
  1. v3.842026-08-21The transformer drew a 94% efficiency curve beside a primary current that assumed 100%

Releases in which this page changed, newest last. Derived from the archived copy of every release, not from notes written afterwards — so it reflects what actually shipped. Site-wide passes are left out; they are in the full changelog.