Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Ideal Transformer Calculator — Turns Ratio, Voltage, Current and Impedance

Solve a lossless transformer from its turns or voltages: secondary voltage and current, turns ratio, winding count, volt-amps and reflected impedance.

Ideal Transformer Calculator

Solve for
Turns on the winding fed by the source. Only the ratio matters to the voltages, but the absolute count is needed to report the secondary turns.
Turns on the winding feeding the load. Ignored when you are solving for the turns ratio or for the secondary turns themselves.
RMS magnitude applied to the primary. This page has no sign or phase convention — enter magnitudes only.
V
RMS magnitude at the secondary terminals. Used when you are solving for the primary voltage, the turns ratio or the secondary turns.
V
The current drawn on the primary side. It sets the secondary current, the apparent power and both impedance figures, and it must be greater than zero — a transformer with no current has no load impedance to report.
A
Secondary voltage
240
RMS voltage at the secondary terminals in volts, V_P × N_S ÷ N_P; in a real transformer this is the no-load value.
Primary voltage
12,000 V
Turns ratio N_P : N_S
50
Secondary turns
100
Secondary current
100 A
Apparent power through the transformer
24,000 VA
Load impedance on the secondary
2.4 Ω
Impedance seen at the primary
6,000 Ω
Reading the result
Step-down transformer: 50 primary turns for every secondary turn, taking 12000 V down to 240 V and pushing the current up from 2 A to 100 A. The ratio is quoted here as N_P:N_S = 50:1; written the other way round, as the N_S/N_P that the source equations use, it is 0.02. Check which convention a nameplate uses before comparing. Both sides carry the same 24000 VA, because an ideal transformer transfers power without losing any of it. A 2.4 Ω load on the secondary therefore looks like 6000 Ω at the primary — impedance is scaled by the square of the turns ratio, not by the ratio itself. Ideal means no winding resistance, no leakage flux, no core loss and no magnetising current: efficiency is exactly 100 % here and voltage regulation exactly 0 %. A real unit is neither — a good transformer still loses about 1 % of the transmitted power, and its secondary voltage sags under load. These are RMS magnitudes with no phase angle, so nothing on this page is a phasor calculation.

Background.

Wind two coils on one iron core, drive one with alternating current, and every turn on both coils picks up the same voltage from the shared magnetic flux. That single fact is the transformer. Put 12,000 V across a primary of 5,000 turns and each turn carries 2.4 V; a secondary of 100 turns collects 100 × 2.4 = 240 V. Change the secondary turn count and the output voltage changes in proportion.

Current moves the other way. A lossless transformer cannot create power, so when the voltage falls fiftyfold the current must rise fiftyfold to keep volts times amps equal on both sides: 2 A drawn from a 12 kV line becomes 100 A at 240 V. That is why the low-voltage cables leaving a distribution transformer are so much thicker than the high-voltage line feeding it.

Solve the proportion for whichever of four values is missing — either voltage, the turns ratio, or the secondary turns for a target voltage — and you also get the secondary current, the volt-amps, and the load impedance seen from each side. Impedance is where intuition fails: it transforms with the square of the turns ratio, so a 2.4 Ω load behind a 50:1 transformer looks like 6,000 Ω from the primary.

These are the ideal relations, with no winding resistance, stray flux or core loss. Good real transformers come close, losing as little as 1% of the power they carry according to OpenStax, but their output voltage drops as load increases, which is what the voltage regulation calculator estimates.

What is ideal transformer calculator?

An ideal transformer is a model of two windings coupled perfectly through a lossless core: all the flux from one winding links the other, nothing is dissipated, and no current is needed just to magnetise the core. Under those assumptions the voltages are in the ratio of the turns and the currents in the inverse ratio. More secondary turns than primary turns makes a step-up transformer; fewer makes step-down; equal turns make a 1:1 isolation transformer. The ratio is reported here as a = N_P ÷ N_S, so a value above 1 means step-down, while OpenStax and many textbooks write the reciprocal N_S ÷ N_P.

How to use this calculator.

  1. Under "Solve for" choose the unknown: secondary voltage, primary voltage, turns ratio, or secondary turns. Fields that mode does not need are ignored.
  2. Enter Primary turns N_P, and Secondary turns N_S if the mode asks for it. Only the ratio sets the voltages, so 50 and 1 give the same voltages as 5,000 and 100.
  3. Enter the voltages as RMS values, the figures a multimeter or nameplate gives, without signs or phase angles.
  4. Enter Primary current I_P, the current drawn from the supply. It must be above zero because the current, power and impedance results all scale from it.

The formula.

V_S ⁄ V_P = N_S ⁄ N_P I_S ⁄ I_P = N_P ⁄ N_S V_P·I_P = V_S·I_S Z_primary = (N_P⁄N_S)² · Z_load

Because the same changing flux links every turn, volts per turn are equal on both windings: V_P ÷ N_P = V_S ÷ N_S, which rearranges to V_S = V_P × N_S ÷ N_P. Solving for the ratio or the turns from two voltages uses the same equation backwards: a = V_P ÷ V_S and N_S = N_P × V_S ÷ V_P.

Current follows from conservation of power rather than from the winding geometry. With no losses, V_P × I_P = V_S × I_S, so I_S = I_P × a: the current is multiplied by exactly the factor the voltage is divided by. The common product is the apparent power S = V_P × I_P, in volt-amps.

Impedance is voltage over current. On the secondary the load is Z_load = V_S ÷ I_S; at the primary the source sees V_P ÷ I_P. Voltage has been multiplied by a while current has been divided by it, so the ratio between the two impedances is a², not a: Z_primary = a² × Z_load.

Everything is a magnitude. The model assumes a passive load and carries no phase angle, so it cannot split volt-amps into real and reactive power, and it says nothing about frequency, core saturation or the inrush current when a transformer is switched on.

A worked example.

Example

OpenStax's University Physics works a distribution transformer that takes 12 kV down to 240 V, drawing 2.0 A on the primary. Model it with 5,000 primary turns and 100 secondary turns. The turns ratio is a = 5,000 ÷ 100 = 50. The secondary voltage is 12,000 ÷ 50 = 240 V, and the secondary current is 2 × 50 = 100 A, matching the textbook. Apparent power is 12,000 × 2 = 24,000 VA on the primary and 240 × 100 = 24,000 VA on the secondary, as it must be with no losses. The load on the secondary is 240 ÷ 100 = 2.4 Ω, while the supply sees 12,000 ÷ 2 = 6,000 Ω. Dividing, 6,000 ÷ 2.4 = 2,500, which is 50². In the textbook's own convention the same transformer has N_S ÷ N_P = 1/50 = 0.02, so a nameplate reading 1:50 and a calculation giving 50 describe the same hardware.

secondary Turns100
secondary Voltage240
primary Turns5,000
primary Current2
solve ForsecondaryVoltage
primary Voltage12,000

Frequently asked questions.

Should a turns ratio be written N_P to N_S or the other way round?
Both are in use. Here it is primary turns per secondary turn, so 50 means a step-down unit; OpenStax writes N_S ÷ N_P and would call the same transformer 1/50. The result summary prints both numbers, but check the convention of any nameplate or datasheet before comparing.
How many secondary turns do I need for a particular output voltage?
Choose Secondary turns and enter the primary turns and both voltages. A 230 V primary of 1,150 turns needs 1,150 × 12 ÷ 230 = 60 turns for 12 V. That is the no-load figure; a real winding sags under load, so it needs a few more turns, roughly in proportion to its voltage regulation.
Why does impedance scale with the square of the turns ratio?
Impedance is voltage divided by current, and a transformer multiplies one by the ratio while dividing the other by it, so the ratio enters twice. That is how transformers match loads: making an 8 Ω loudspeaker look like 5,000 Ω takes a ratio of √(5,000 ÷ 8) = 25, not 625.
Why is the power given in volt-amps instead of watts?
Watts depend on the load's power factor, which the ideal relations do not include. For a resistive load the two are equal. For a motor load at a power factor of 0.8, 24,000 VA delivers 19,200 W of real power and 14,400 VAr of reactive power. Transformers are rated in kVA because winding heating depends on current regardless of phase.
What is a 1:1 isolation transformer for if the voltage stays the same?
It removes the direct electrical connection between supply and load. Voltage and current pass through unchanged, but the secondary is no longer tied to the supply's neutral or earth, which breaks ground loops in audio and test equipment and reduces shock risk from touching one conductor. Here it shows a ratio of exactly 1.
Will a transformer work on a DC supply?
No. A transformer needs a changing magnetic flux to induce voltage, and steady direct current produces a constant flux, so nothing appears on the secondary. Worse, the only thing limiting primary current on DC is the winding's small resistance, so connecting a mains transformer to a DC source can overheat it.
How far is a real transformer from these ideal numbers?
Close, but not equal. OpenStax gives losses as low as about 1% of transmitted power for a good transformer, and Kuphaldt's benchmark for a good power transformer is voltage regulation under 3% on a resistive load. A real unit also draws magnetising current with nothing connected, and small transformers lose a larger share of their power.

References& sources.

  1. [1]OpenStax (Rice University), University Physics Volume 2, §15.6 'Transformers'. Primary source for every relation implemented: the voltage relation 'v_S(t) = (N_S/N_P)v_P(t)' with the abbreviated form V_S/V_P = N_S/N_P; the current relation 'i_S(t) = (N_P/N_S)i_P(t)'; power conservation 'i_P(t)v_P(t) = i_S(t)v_S(t)'; the definitions that a step-up transformer 'increases voltage and decreases current' while a step-down transformer 'decreases voltage and increases current'; and the real-world figure that 'a good transformer can have losses as low as 1% of the transmitted power'. Example 15.6, used as this page's worked example, steps 12 kV down to 240 V with N_S/N_P = 1/50 and an input current of 2.0 A, giving an output current of 100 A. Retrieved 2026-07-29.
  2. [2]Tony R. Kuphaldt, 'Electric Circuits II — Alternating Current' (Lessons in Electric Circuits, Volume II), Workforce LibreTexts §10.2 'Step-up and Step-down Transformers'. Independent second authority, consulted to check this page rather than to derive it. States that 'a transformer that increases voltage from primary to secondary (more secondary winding turns than primary winding turns) is called a step-up transformer', that 'a transformer designed to reduce voltage from primary to secondary is called a step-down transformer', and that 'the voltage and current transformation ratio is equal to the ratio of winding turns between primary and secondary'. Its simulated 10:1 example gives a secondary of 0.9962 V and 0.9962 mA from a primary of 10 V and 0.09975 mA — against the exactly 1.0000 V and 0.9975 mA this ideal model returns, a 0.4 % gap that is precisely the winding loss the ideal model omits. Retrieved 2026-07-29.
  3. [3]Tony R. Kuphaldt, 'Electric Circuits II — Alternating Current', Workforce LibreTexts §10.6 'Voltage Regulation'. Cited for the boundary of this page's model: an ideal transformer has zero voltage regulation by construction, while a real one does not, and the source's benchmark is that 'a good power transformer should exhibit a regulation percentage of less than 3%' on a resistive load. This is the quantity the secondary-voltage figure on this page does not include, and the reason the two pages cross-link. Retrieved 2026-07-29.

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V_S ⁄ V_P = N_S ⁄ N_P I_S ⁄ I_P = N_P ⁄ N_S V_P·I_P = V_S·I_S Z_primary = (N_P⁄N_S)² · Z_load
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