Physics · Electrodynamics

Transformer: ideal conversion and real limits

In an ideal transformer RMS voltage follows the turns ratio and load current scales inversely. Polarity, power factor and real losses are treated separately.

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Formula

Calculate

\frac{U _{1}}{U _{2}} = \frac{n _{1}}{n _{2}} = \frac{I _{2}}{I _{1}}

U₁, U₂ in V · n₁, n₂ dimensionless · I₁, I₂ in A

Variables & units

Variables & units
Symbol	Meaning	Unit
U₁, U₂	Nonnegative RMS voltage across primary and secondary windings	V
n₁, n₂	Positive turn counts	dimensionless
I₁, I₂	Nonnegative RMS values of ideally transferred load currents	A

LaTeX

\frac{U_1}{U_2} = \frac{n_1}{n_2} = \frac{I_2}{I_1}

QUANTA STUDYquanta-study.de

Original transformer sketch with shared core and dotted terminals. Values are RMS magnitudes. The secondary circuit closes through a 12 Ω load; voltages scale 20:1 and currents 1:20. The few drawn loops do not represent 800 or 40 actual turns. — Ideal loaded 20:1 model: primary 240 V and 50 mA, secondary 12 V and 1 A. Both sides transfer 12 W to the 12 Ω load. Positive current enters the primary and leaves the matching dotted secondary terminal; windings are schematic.

Transformer calculator

Find the voltage or current of an ideal transformer from the numbers of turns, for alternating current and without losses.

Calculation model

Find

Primary rms voltage U₁

Rms voltage across the input coil, e.g. 230 V from a European mains socket.

Primary turns N₁

Number of turns on the input coil as a whole number, e.g. 1150.

Secondary turns N₂

Number of turns on the output coil as a whole number, e.g. 60.

Filled with example values. Type your own and the result updates straight away.

Result: Secondary rms voltage U₂

U₂ = 12 V

With U₁ = 230 V, N₁ = 1150 and N₂ = 60, U₂ = 12 V.

Working · rearrange for U₂, substitute, evaluate

1. Rearrange for U₂
$U_2=U_1\frac{N_2}{N_1}$
2. Substitute values (in SI units)
$U_2=230\cdot \frac{60}{1150}=12$
3. Result
U₂ = 12 V

Quick tips

Voltages are in the ratio of the turns, currents in the inverse ratio: U₁ : U₂ = N₁ : N₂, but I₁ : I₂ = N₂ : N₁. Where the voltage is high, the current is small.
Check with power: for an ideal transformer U₁ · I₁ = U₂ · I₂, e.g. 230 V · 0.12 A = 12 V · 2.3 A = 27.6 W.

Assumptions and limits

Ideal transformer sharing time-varying flux, with positive integer turn counts. Rms voltage magnitudes satisfy U₂/U₁ = N₂/N₁. No steady DC operation, losses, saturation or real power limits; polarity is not determined.

Derivation & background

Shared changing flux induces the same voltage per turn. Ideal complete coupling gives the voltage relation even under load. Current arrows and winding polarity determine signs; magnitude conversion alone does not describe them. Constant DC voltage cannot sustain indefinitely bounded steady core flux. Real losses and magnetising current need an extended model.

Exam blueprint

Validity range

Ideal transformer with complete coupling, no winding/core losses or magnetising current; consistent RMS magnitudes in steady changing operation. Ideal voltage conversion also holds under load. Current quotients require nonzero load currents; signs need terminal and current references. Real sinusoidal power calculations use $P=UI\cos\varphi$ , not universally $P=UI$ .

Derivation steps

A shared changing flux determines both induced voltages; ideal coupling refers the load without power gain.

1Define primary/secondary sides and matching voltage/current measures. Let U and I be nonnegative RMS values.
2Equally oriented turns link the same flux change: $\mathcal E_j=-n_j\frac{\mathrm d\Phi}{\mathrm dt}$ .
3The common factor gives $U_2=U_1\frac{n_2}{n_1}$ in the ideal winding model, including ideal loaded operation.
4Choose current entering the primary and leaving the corresponding secondary terminal; two inward arrows produce a minus sign.
5Without losses or magnetising current $i_1=\frac{n_2}{n_1}i_2$ , hence $I_1=\frac{n_2}{n_1}I_2$ .
6The load determines absolute currents. Check sinusoidal real power with its power factor; ideal no-load operation leaves both load currents zero.

Rearrangements

Secondary voltage

U_2 = U_1 \cdot \frac{n_2}{n_1}

More secondary turns mean higher voltage (stepping up).

Number of turns

n_2 = n_1 \cdot \frac{U_2}{U_1}

Sizing a secondary winding for a target voltage.

Primary current

I_1 = I_2 \cdot \frac{n_2}{n_1}

At equal transferred power and equal power factor, higher voltage means lower current and hence lower resistive line loss.

Task variant

An ideal transformer should supply $U_2=5\,\mathrm V$ AC RMS; primary values are $n_1=920$ and $U_1=230\,\mathrm V$ RMS. How many secondary turns are needed? Charger rectification is outside this calculation.

$n_2=n_1\frac{U_2}{U_1}=920\frac{5\,\mathrm V}{230\,\mathrm V}=20$ Turns in the ideal AC model; additional charger rectification is not specified.

The secondary carries 4 A at 11.5 V. What current flows in the 230 V primary?

$I_1=I_2\frac{U_2}{U_1}=(4\,\mathrm A)\frac{11.5\,\mathrm V}{230\,\mathrm V}=0.20\,\mathrm A$ RMS values for the ideal transformer.
For a resistive load, both sides have 46 W real power.

Common mistakes

Scaling the currents like the voltages.

Currents scale inversely: $\frac{I_2}{I_1}=\frac{n_1}{n_2}$ , otherwise energy would be created.

Calculating a transformer on DC.

Bounded steady core flux requires balanced voltage-time area. AC or suitable pulse sequences can achieve this; constant voltage drives continuing flux change in an ideal resistance-free model. Real DC switch-on needs resistance and a core model.

Setting up the ratio the wrong way round.

Mnemonic: more turns, more voltage; U and n belong to the same side.

Exam context

Tasks combine the turns ratio, power conservation and the rationale of high-voltage transmission (line loss $P_{\mathrm{loss}}=I^2R$ ).

Formula cluster

Induction and applications

A direct application of the induction law to coupled coils.

ℰ = −N dΦB/dt

Faraday’s law of induction

U_ind = −L·(dI/dt)

Self-Induction of a Coil

Worked example

Ideal: $n_1=1000$ , $n_2=50$ , $U_1=230\,\mathrm V$ RMS. Then $U_2=(230\,\mathrm V)\frac{50}{1000}=11.5\,\mathrm V$ RMS and under load $I_2=20I_1$ . An absolute current requires the load.

Applications

Ideal AC voltage conversion, load matching and explaining lower line currents at higher transmission voltage; concrete design needs a real model.

How do you calculate with Transformer turns ratio?

Here is how to work through a typical Transformer turns ratio (U₁/U₂ = n₁/n₂ = I₂/I₁) task step by step:

1
Task
An ideal transformer should supply $U_2=5\,\mathrm V$ AC RMS; primary values are $n_1=920$ and $U_1=230\,\mathrm V$ RMS. How many secondary turns are needed? Charger rectification is outside this calculation.
Solution path
$n_2=n_1\frac{U_2}{U_1}=920\frac{5\,\mathrm V}{230\,\mathrm V}=20$ Turns in the ideal AC model; additional charger rectification is not specified.
2
Task
The secondary carries 4 A at 11.5 V. What current flows in the 230 V primary?
Solution path
1. $I_1=I_2\frac{U_2}{U_1}=(4\,\mathrm A)\frac{11.5\,\mathrm V}{230\,\mathrm V}=0.20\,\mathrm A$ RMS values for the ideal transformer.
2. For a resistive load, both sides have 46 W real power.

Understand and apply the equation

A transformer changes voltages and currents through a shared changing magnetic flux. Its turns ratio describes ideal conversion, not an extra energy source. A reliable calculation specifies voltage measures, current directions and the loss model.

Convert voltage and transfer power

Two windings connect electrical circuits through a shared changing magnetic flux. The secondary receives a different voltage, but its usable power comes from the primary source. Lower voltage can mean higher current with a suitable load; without a load it produces no arbitrary current. This distinction takes us from turns ratio to power balance.[1][3]

We start with the ideal model and matching voltage references. RMS values are compared only with RMS values. Real losses and a specified power factor enter a separate later budget; they are not silently mixed into ideal current conversion.[1][3]

How do you calculate secondary voltage from turn counts?

Subscripts 1 and 2 identify primary and secondary windings. The primary receives power and the secondary supplies a load. Capital voltage and current symbols represent nonnegative RMS values in steady operation. Turn counts are positive integers. Both voltages must use the same convention: RMS with RMS, not a mains RMS value with a secondary peak value. The rms value of sinusoidal quantities connects the peak value to the voltage measure used here.[1]

\begin{aligned}\frac{U_1}{U_2}&=\frac{n_1}{n_2}\\\frac{I_2}{I_1}&=\frac{n_1}{n_2}\end{aligned}

Ideal, fully coupled, without magnetising current or losses; current quotients require nonzero currents.[1]

Fewer secondary turns give lower voltage. The connected load then determines the actual current. Conversion therefore does not promise a large current in an open secondary circuit. Both load currents vanish in ideal no-load operation; their quotient is undefined while the voltage relation still holds.[1]

How does Faraday’s law give the voltage ratio?

Faraday’s law of induction describes induced electromotive force. For equally oriented turns linked by the same flux per turn, it scales with the turn count. Terminal polarity must also be specified: reversing the voltage reference reflects the signed instantaneous waveform without changing the turns ratio or RMS magnitude. In the ideal model no additional voltage drops occur across winding resistance or leakage inductance.[2]

\begin{aligned}\mathcal E_1&=-n_1\frac{\mathrm d\Phi}{\mathrm dt}\\\mathcal E_2&=-n_2\frac{\mathrm d\Phi}{\mathrm dt}\\\frac{\mathcal E_2}{\mathcal E_1}&=\frac{n_2}{n_1}\end{aligned}

The identical flux derivatives cancel when nonzero; the linear relation remains valid at zero crossings.[2]

For matching terminal orientations the relation also holds for time-dependent voltages. Its constant scale factor carries over to RMS values. A sinusoidal waveform additionally permits the familiar peak-to-RMS conversion; that conversion comes from waveform shape, not from the transformer.[1]

Why does current scale inversely with the turns ratio?

\begin{aligned}u_2(t)&=\frac{n_2}{n_1}u_1(t)\\i_1(t)&=\frac{n_2}{n_1}i_2(t)\\u_1(t)i_1(t)&=u_2(t)i_2(t)\end{aligned}

Here i₁ enters the positive primary terminal and i₂ leaves the corresponding positive secondary terminal.[1]

If both current arrows instead point into the dotted terminals, the current relation acquires a minus sign. The RMS magnitudes above need no such sign. Voltage and current are in phase for a purely resistive load. An ideal transformer also transfers the phase relationship of a reactive load to the primary. Power conservation then means equality of real powers; the RMS product alone is apparent power. Electrical power therefore distinguishes real and apparent power within the matching AC model.[3]

\begin{aligned}P&=UI\cos\varphi\\S&=UI\\U_{\mathrm{peak}}&=\sqrt2\,U_{\mathrm{rms}}\end{aligned}

These power-factor and peak formulas apply to sinusoidal voltages and currents.[3]

Calculating turn count, current and referred load resistance

\begin{aligned}U_2&=U_1\frac{n_2}{n_1}\\n_2&=n_1\frac{U_2}{U_1}\\I_1&=I_2\frac{n_2}{n_1}\\R_{\mathrm{in}}&=\left(\frac{n_1}{n_2}\right)^2R_{\mathrm{load}}\end{aligned}

The input resistance applies to an ideal transformer with a purely resistive secondary load.[1]

The resistance relation follows directly from the primary voltage-to-current ratio: voltage gains one factor while current gains its reciprocal. Resistance therefore transfers with the square of the turns ratio. This is an input equivalent, not an additional physical resistor. Volt and ampere remain SI units; turn counts and their ratio are dimensionless.[1]

View diagram: load current and conversion

The load sets actual current

QUANTA STUDYquanta-study.de

For the first example’s ideal 20:1 transformer at 240 V primary,

U_2=12\,\mathrm V

. Secondary conductance G₂ sets

I_2=U_2G_2

; I₁ is one twentieth. At

G_2=0

both load currents vanish. Different current magnitudes create no extra power.[1]

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Calculating resistive load, efficiency and power factor

A 20:1 transformer with a specified load

Given: Ideal: $n_1=800$ , $n_2=40$ , sinusoidal $U_1=240\,\mathrm V$ RMS; secondary load $R=12\,\Omega$ .
Find: Find secondary voltage, both load currents and the input equivalent resistance.

Approach: Apply ideal voltage conversion, then Ohm’s law to the known load and inverse current conversion.
$\begin{aligned}U_2&=U_1\frac{n_2}{n_1}\\I_2&=\frac{U_2}{R}\\I_1&=I_2\frac{n_2}{n_1}\end{aligned}$
Substitution: Both sides use RMS values of the same sinusoidal waveform.
$\begin{aligned}U_2&=(240\,\mathrm V)\frac{40}{800}=12\,\mathrm V\\I_2&=\frac{12\,\mathrm V}{12\,\Omega}=1\,\mathrm A\\I_1&=(1\,\mathrm A)\frac{40}{800}=0.050\,\mathrm A\end{aligned}$
Evaluate equivalent resistance with the squared ratio.
$\begin{aligned}R_{\mathrm{in}}&=20^2(12\,\Omega)=4800\,\Omega\\P_2&=(12\,\mathrm V)(1\,\mathrm A)=12\,\mathrm W\end{aligned}$

Result: Secondary: 12 V and 1 A; primary: 50 mA. The source ideally sees 4.8 kΩ.

Check: Check: $U_1I_1=12\,\mathrm W=U_2I_2$ ; independently $\frac{240\,\mathrm V}{0.050\,\mathrm A}=4800\,\Omega$ . Lower voltage accompanies larger load current.

Account separately for losses and power factor

Given: Measured secondary values: $U_2=24\,\mathrm V$ , $I_2=3\,\mathrm A$ , resistive load. At this operating point $\eta=0.90$ ; primary $U_1=230\,\mathrm V$ , sinusoidal current with $\cos\varphi_1=0.80$ . All values are RMS.
Find: Calculate real powers, losses and primary current.

Approach: Efficiency refers to real power. Find input current using the specified power factor.
$\begin{aligned}P_2&=U_2I_2\\P_1&=\frac{P_2}{\eta}\\P_{\mathrm{loss}}&=P_1-P_2\\I_1&=\frac{P_1}{U_1\cos\varphi_1}\end{aligned}$
Substitution: Do not additionally impose the ideal current relation on this real budget.
$\begin{aligned}P_2&=(24\,\mathrm V)(3\,\mathrm A)=72\,\mathrm W\\P_1&=\frac{72\,\mathrm W}{0.90}=80\,\mathrm W\\P_{\mathrm{loss}}&=8\,\mathrm W\\I_1&=\frac{80\,\mathrm W}{(230\,\mathrm V)(0.80)}\\&\approx0.435\,\mathrm A\end{aligned}$

Result: The transformer takes 80 W, supplies 72 W and loses 8 W. Primary RMS current is about 0.435 A.

Check: Check: $230\cdot0.4347826\cdot0.80=80$ in watts and $\frac{72}{80}=0.90$ . Apparent power is 100 VA, distinct from 80 W real power.

Why is constant DC voltage not steady transformer operation?

The statement “only AC” needs a model boundary. Bounded steady core flux requires balanced voltage-time area over a period. Pulsed operation can also satisfy this condition. A constant voltage applied indefinitely to an ideal resistance-free winding instead drives continuing flux change, not immediately constant flux. With a finite core and real losses, this idealised state cannot persist without limit.[2]

\begin{aligned}\Delta\Phi&=\frac{1}{n_1}\int_{t_0}^{t_1}u_1(t)\,\mathrm dt\\\int_0^{T_{\mathrm{drive}}}u_1(t)\,\mathrm dt&=0\quad\text{(periodic flux)}\end{aligned}

Voltage orientation chosen so that $u_1=n_1\frac{\mathrm d\Phi}{\mathrm dt}$ in the ideal winding model.[2]

For a real primary winding connected to DC, switch-on, saturation and winding resistance may matter. Any eventual steady current produces no continuing changing flux and no persistent secondary voltage. Whether and how quickly overheating occurs does not follow from the turns ratio alone. Claiming every transformer immediately burns out would be as inaccurate as applying the ideal AC formula to an arbitrary DC state.[2]

Why does high voltage help transmit energy?

For the same real power and power factor, line current decreases when transmission voltage increases. At fixed line resistance, Joule loss decreases with current squared. This explains the benefit of high voltage; it is not a complete grid calculation because voltage drop, reactive power and insulation requirements are not represented.[1][3]

A transformer alone does not turn sinusoidal AC into smooth DC. A DC-device power supply requires additional stages. An ideal doorbell transformer with 1000 to 50 turns converts 230 V primary RMS to 11.5 V secondary RMS. A load specification must be added before an absolute secondary current can be found. A calculated turn count likewise does not replace thermal, magnetic or electrical design.[1]

How do ideal and real transformers differ?

The ideal model ignores winding resistance, leakage flux, core losses and input current needed for magnetisation. Real voltage deviation under load therefore depends on operating point and load behaviour. A single efficiency value does not generally determine a new voltage ratio. Measured secondary voltage, load current and input power factor are separate quantities; the second example explicitly specified them.[1][3]

RMS values remain defined for nonsinusoidal waveforms, but a single phase angle generally does not describe the whole power budget. Time averaging the voltage-current product remains the reliable definition. Near an open load, real magnetising current becomes important relative to load current; the ideal inverse-current rule is particularly easy to overinterpret there.[3]

Explore further: pulses and flux drift

Equal voltage-time areas keep flux periodic

Given: An ideal winding model has primary $n_1=100$ and secondary $n_2=25$ . In each period primary voltage is $+12.0\,\mathrm V$ for $1.00\,\mathrm{ms}$ , then $-6.00\,\mathrm V$ for $2.00\,\mathrm{ms}$ . Choose polarity so $u_1=n_1\dot\Phi$ . Assume complete coupling with negligible losses and magnetising current; initial flux is unspecified.
Find: Find both flux changes, secondary pulse levels and both RMS voltages.

Approach: integrate voltage over each constant interval. RMS follows from duration-weighted squares, not the sine formula.
$\begin{aligned}\Delta\Phi_j&=\frac{u_{1,j}\Delta t_j}{n_1}\\u_{2,j}&=\frac{n_2}{n_1}u_{1,j}\\U_{1,\mathrm{rms}}^2&=\frac{u_a^2t_a+u_b^2t_b}{t_a+t_b}\end{aligned}$
Substitution: the smaller negative level lasts twice as long and balances the positive pulse.
$\begin{aligned}\Delta\Phi_a&=+0.120\,\mathrm{mWb}\\\Delta\Phi_b&=-0.120\,\mathrm{mWb}\\u_{2,a}&=+3.00\,\mathrm V\\u_{2,b}&=-1.50\,\mathrm V\\U_{1,\mathrm{rms}}&=\sqrt{\frac{12.0^2(1.00)+(-6.00)^2(2.00)}{3.00}}\,\mathrm V\\&=\sqrt{72}\,\mathrm V\approx8.48528\,\mathrm V\\U_{2,\mathrm{rms}}&=\frac14U_{1,\mathrm{rms}}\approx2.12132\,\mathrm V\end{aligned}$

Result: Flux returns to its initial value after each period; its absolute level remains unknown without an initial value. Secondary levels are +3.00 V and −1.50 V; RMS is about 2.121 V.

Check: Both signed voltage means are zero. Squaring secondary levels directly gives $U_{2,\mathrm{rms}}^2=4.50\,\mathrm{V^2}$ . Balanced area prevents flux drift, but without initial flux, core area and material limits it does not prove freedom from saturation.

Five exercises on calculation, models and errors

Calculate · Foundation

Ideal: 600 primary turns, 75 secondary turns, 120 V RMS primary. Find secondary voltage.

Show hint

Voltage and turns belong on the same side of the ratio.

Learning goal: Apply voltage conversion with explicit RMS values.

Show solution and calculation

Approach: Shared changing flux gives the ideal voltage relation.
Substitution yields 15 V RMS.

Check: Eight times fewer turns give eight times less voltage.

U_2=U_1\frac{n_2}{n_1}=(120\,\mathrm V)\frac{75}{600}=15\,\mathrm V

[1]

Rearrange · Application

Ideally obtain 18 V RMS from 180 V RMS; $n_1=900$ . A resistive secondary load draws 2 A. Find n₂ and I₁.

Show hint

Turns follow voltage; primary current uses the inverse relation.

Learning goal: Translate a target voltage into turns and a load-current budget.

Show solution and calculation

Approach: First find n₂ from target voltage, then refer the specified load current.
The answers are 90 turns and 0.20 A primary.

Check: Both sides transfer 36 W.

\begin{aligned}n_2&=900\frac{18}{180}=90\\I_1&=(2\,\mathrm A)\frac{90}{900}=0.20\,\mathrm A\end{aligned}

[1]

Interpret · Application

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u₂(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Show hint

An open load circuit draws no load current; the sign depends on the measurement arrow.

Learning goal: Distinguish magnitude conversion from polarity and no-load behaviour.

Show solution and calculation

The turns ratio and RMS voltage magnitude remain unchanged; the signed oscilloscope waveform u₂(t) is reflected across the time axis.
A normal AC RMS meter still displays the same nonnegative RMS value.
Ideal secondary load current remains zero, as does ideal primary load current.
A quotient of two zero currents is undefined.
Real magnetising current would be an additional variable excluded here.

\begin{aligned}U_2&=\frac{U_1}{10}\\I_1&=I_2=0\end{aligned}

[1]

Error analysis · Application

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims $I_1=10\,\mathrm A$ because primary voltage is five times larger. Correct it.

Show hint

Compare real power on both sides first.

Learning goal: Detect incorrect current conversion using the power budget.

Show solution and calculation

The error is scaling current like voltage rather than inversely.
Secondary power is 40 W, so ideal primary current must be 0.40 A.
The incorrect 10 A would imply 1000 W input with unexplained 960 W losses, contradicting the specified ideal model.

I_1=\frac{U_2I_2}{U_1}=\frac{(20\,\mathrm V)(2\,\mathrm A)}{100\,\mathrm V}=0.40\,\mathrm A

[1]

Transfer · Application

Two idealised transmissions deliver the same real power at the same power factor through the same line resistance. Case B has four times the voltage. How do current and Joule loss change?

Show hint

Find current from the equal power requirement before squaring it.

Learning goal: Transfer voltage stepping to relative line losses.

Show solution and calculation

Approach: Equal power at equal power factor requires one quarter of the current.
Loss becomes one sixteenth.
This is a relative comparison under the specified conditions, not a complete real-line design.

\begin{aligned}\frac{I_B}{I_A}&=\frac{U_A}{U_B}=\frac14\\\frac{P_{\mathrm{loss},B}}{P_{\mathrm{loss},A}}&=\left(\frac{I_B}{I_A}\right)^2=\frac1{16}\end{aligned}

[1][3]

In brief

Shared changing flux sets ideal voltage conversion. Load sets current; power and reference directions explain inverse current scaling. Losses and power factor remain separate quantities. Balanced voltage-time area permits periodic flux but does not replace real core design.

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Made turn/voltage/current ratios, referred load, power factor, DC and the existing pulsed-flux balance accessible through learning questions; added matching induction, rms and electrical-power links.
October 1, 2026Added an original DE/EN explanation with two checked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

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Quanta. Transformer turns ratio. Updated October 1, 2026. https://quanta-study.de/en/formula-sheet/transformator

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Frequently asked questions about Transformer turns ratio

How do you calculate the secondary voltage of a transformer?+

First specify matching voltage measures, such as RMS values. Ideally $U_2=U_1\frac{n_2}{n_1}$ , because both windings link the same changing flux. This relation holds for an ideal transformer both at no load and under load. With 1000 primary turns, 50 secondary turns and 230 V primary, the secondary has 11.5 V. This is AC rather than DC smoothed by the transformer. Absolute current requires a load specification. Winding resistance, leakage flux and magnetisation can introduce deviations in a real device; their magnitude does not follow from the turns ratio alone. Winding orientation and measurement arrows determine sign, which a positive magnitude calculation cannot establish.

Why do the currents scale inversely to the voltages?+

The ideal model refers the load without losses or additional magnetising current. With matching current references, $I_1=I_2\frac{n_2}{n_1}$ . For a resistive load you can also check real power directly with $P=UI$ : secondary values of 20 V and 2 A give 40 W, requiring an ideal primary current of 0.40 A at 100 V. For sinusoidal reactive loads real power includes power factor; an ideal transformer transfers the phase relationship as well. With an open ideal secondary both load currents vanish, making their quotient undefined. Real input current contains additional components, so a generic efficiency percentage cannot replace a complete current calculation for an unspecified load.

Why does a transformer not work with DC voltage?+

Bounded steady core flux requires balanced voltage-time area. Constant voltage across an ideal resistance-free winding would change flux continuously; claiming that flux immediately becomes constant is incorrect in that model. A real core cannot sustain this behaviour indefinitely. Saturation, winding resistance and switch-on behaviour matter. If a real current eventually becomes constant, its flux stops changing and no persistent secondary voltage is induced. Alternating operation or suitable pulse sequences permit recurring flux changes. Actual current and heating cannot be predicted from the turns ratio alone. A switching power supply therefore requires an appropriate switching and flux-reset arrangement rather than simply applying the ideal AC ratio to DC.

Why is electricity transmitted across country at high voltage?+

For equal transferred real power and equal power factor, higher voltage requires lower current. Joule loss in a specified line resistance is $P_{\mathrm{loss}}=I^2R$ . Four times the voltage therefore means one quarter of the current and one sixteenth of the losses. This relative calculation explains the benefit of high transmission voltage but does not describe a complete grid. Voltage drop, load behaviour, reactive power and insulation must also be considered. If a simplified calculation yields losses larger than the supposedly delivered useful power, that chiefly signals a transmission scenario that cannot operate self-consistently as assumed. A transformer creates no extra energy and does not remove line resistance.

What does the ideal transformer mean and how realistic is it?+

The ideal model neglects winding resistance, leakage flux, core losses and required magnetising current. Voltage and load-current relations are then unambiguous and transferred real power is conserved. This is a calculation reference, not a guarantee about every real device. Efficiency and voltage deviation depend on design, frequency and operating point. An unsupported generic percentage does not reliably capture those differences. For example, supplying 72 W at a specified operating-point efficiency of 90% requires 80 W input and gives 8 W losses. Primary-current calculation still needs input voltage and, for sinusoidal AC, input power factor. These are separate specifications rather than consequences of turn count.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Vol. 2, §15.6 Transformers

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Transformer: ideal conversion and real limits

In an ideal transformer RMS voltage follows the turns ratio and load current scales inversely. Polarity, power factor and real losses are treated separately.

BasicExam-relevantUpper secondaryUniversity/collegeBachelor foundations

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Symbol

Meaning

Unit

U₁, U₂

Nonnegative RMS voltage across primary and secondary windings

n₁, n₂

Positive turn counts

dimensionless

I₁, I₂

Nonnegative RMS values of ideally transferred load currents

Derivation & background

Exam blueprint

Validity range

Derivation steps

A shared changing flux determines both induced voltages; ideal coupling refers the load without power gain.

1Define primary/secondary sides and matching voltage/current measures. Let U and I be nonnegative RMS values.
2Equally oriented turns link the same flux change: $\mathcal E_j=-n_j\frac{\mathrm d\Phi}{\mathrm dt}$ .
3The common factor gives $U_2=U_1\frac{n_2}{n_1}$ in the ideal winding model, including ideal loaded operation.
4Choose current entering the primary and leaving the corresponding secondary terminal; two inward arrows produce a minus sign.
5Without losses or magnetising current $i_1=\frac{n_2}{n_1}i_2$ , hence $I_1=\frac{n_2}{n_1}I_2$ .
6The load determines absolute currents. Check sinusoidal real power with its power factor; ideal no-load operation leaves both load currents zero.

How do you calculate with Transformer turns ratio?

Here is how to work through a typical Transformer turns ratio (U₁/U₂ = n₁/n₂ = I₂/I₁) task step by step:

Task

Solution path

$n_2=n_1\frac{U_2}{U_1}=920\frac{5\,\mathrm V}{230\,\mathrm V}=20$ Turns in the ideal AC model; additional charger rectification is not specified.

Task

The secondary carries 4 A at 11.5 V. What current flows in the 230 V primary?

Solution path

$I_1=I_2\frac{U_2}{U_1}=(4\,\mathrm A)\frac{11.5\,\mathrm V}{230\,\mathrm V}=0.20\,\mathrm A$ RMS values for the ideal transformer.
For a resistive load, both sides have 46 W real power.

Convert voltage and transfer power

How do you calculate secondary voltage from turn counts?

\begin{aligned}\frac{U_1}{U_2}&=\frac{n_1}{n_2}\\\frac{I_2}{I_1}&=\frac{n_1}{n_2}\end{aligned}

Ideal, fully coupled, without magnetising current or losses; current quotients require nonzero currents.[1]

How does Faraday’s law give the voltage ratio?

\begin{aligned}\mathcal E_1&=-n_1\frac{\mathrm d\Phi}{\mathrm dt}\\\mathcal E_2&=-n_2\frac{\mathrm d\Phi}{\mathrm dt}\\\frac{\mathcal E_2}{\mathcal E_1}&=\frac{n_2}{n_1}\end{aligned}

The identical flux derivatives cancel when nonzero; the linear relation remains valid at zero crossings.[2]

Why does current scale inversely with the turns ratio?

\begin{aligned}u_2(t)&=\frac{n_2}{n_1}u_1(t)\\i_1(t)&=\frac{n_2}{n_1}i_2(t)\\u_1(t)i_1(t)&=u_2(t)i_2(t)\end{aligned}

Here i₁ enters the positive primary terminal and i₂ leaves the corresponding positive secondary terminal.[1]

\begin{aligned}P&=UI\cos\varphi\\S&=UI\\U_{\mathrm{peak}}&=\sqrt2\,U_{\mathrm{rms}}\end{aligned}

These power-factor and peak formulas apply to sinusoidal voltages and currents.[3]

Calculating turn count, current and referred load resistance

\begin{aligned}U_2&=U_1\frac{n_2}{n_1}\\n_2&=n_1\frac{U_2}{U_1}\\I_1&=I_2\frac{n_2}{n_1}\\R_{\mathrm{in}}&=\left(\frac{n_1}{n_2}\right)^2R_{\mathrm{load}}\end{aligned}

The input resistance applies to an ideal transformer with a purely resistive secondary load.[1]

View diagram: load current and conversion

The load sets actual current

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For the first example’s ideal 20:1 transformer at 240 V primary,

U_2=12\,\mathrm V

. Secondary conductance G₂ sets

I_2=U_2G_2

; I₁ is one twentieth. At

G_2=0

both load currents vanish. Different current magnitudes create no extra power.[1]

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Calculating resistive load, efficiency and power factor

A 20:1 transformer with a specified load

Given: Ideal: $n_1=800$ , $n_2=40$ , sinusoidal $U_1=240\,\mathrm V$ RMS; secondary load $R=12\,\Omega$ .
Find: Find secondary voltage, both load currents and the input equivalent resistance.

Approach: Apply ideal voltage conversion, then Ohm’s law to the known load and inverse current conversion.
$\begin{aligned}U_2&=U_1\frac{n_2}{n_1}\\I_2&=\frac{U_2}{R}\\I_1&=I_2\frac{n_2}{n_1}\end{aligned}$
Substitution: Both sides use RMS values of the same sinusoidal waveform.
$\begin{aligned}U_2&=(240\,\mathrm V)\frac{40}{800}=12\,\mathrm V\\I_2&=\frac{12\,\mathrm V}{12\,\Omega}=1\,\mathrm A\\I_1&=(1\,\mathrm A)\frac{40}{800}=0.050\,\mathrm A\end{aligned}$
Evaluate equivalent resistance with the squared ratio.
$\begin{aligned}R_{\mathrm{in}}&=20^2(12\,\Omega)=4800\,\Omega\\P_2&=(12\,\mathrm V)(1\,\mathrm A)=12\,\mathrm W\end{aligned}$

Result: Secondary: 12 V and 1 A; primary: 50 mA. The source ideally sees 4.8 kΩ.

Check: Check: $U_1I_1=12\,\mathrm W=U_2I_2$ ; independently $\frac{240\,\mathrm V}{0.050\,\mathrm A}=4800\,\Omega$ . Lower voltage accompanies larger load current.

Account separately for losses and power factor

Given: Measured secondary values: $U_2=24\,\mathrm V$ , $I_2=3\,\mathrm A$ , resistive load. At this operating point $\eta=0.90$ ; primary $U_1=230\,\mathrm V$ , sinusoidal current with $\cos\varphi_1=0.80$ . All values are RMS.
Find: Calculate real powers, losses and primary current.

Approach: Efficiency refers to real power. Find input current using the specified power factor.
$\begin{aligned}P_2&=U_2I_2\\P_1&=\frac{P_2}{\eta}\\P_{\mathrm{loss}}&=P_1-P_2\\I_1&=\frac{P_1}{U_1\cos\varphi_1}\end{aligned}$
Substitution: Do not additionally impose the ideal current relation on this real budget.
$\begin{aligned}P_2&=(24\,\mathrm V)(3\,\mathrm A)=72\,\mathrm W\\P_1&=\frac{72\,\mathrm W}{0.90}=80\,\mathrm W\\P_{\mathrm{loss}}&=8\,\mathrm W\\I_1&=\frac{80\,\mathrm W}{(230\,\mathrm V)(0.80)}\\&\approx0.435\,\mathrm A\end{aligned}$

Result: The transformer takes 80 W, supplies 72 W and loses 8 W. Primary RMS current is about 0.435 A.

Check: Check: $230\cdot0.4347826\cdot0.80=80$ in watts and $\frac{72}{80}=0.90$ . Apparent power is 100 VA, distinct from 80 W real power.

Why is constant DC voltage not steady transformer operation?

\begin{aligned}\Delta\Phi&=\frac{1}{n_1}\int_{t_0}^{t_1}u_1(t)\,\mathrm dt\\\int_0^{T_{\mathrm{drive}}}u_1(t)\,\mathrm dt&=0\quad\text{(periodic flux)}\end{aligned}

Voltage orientation chosen so that $u_1=n_1\frac{\mathrm d\Phi}{\mathrm dt}$ in the ideal winding model.[2]

Why does high voltage help transmit energy?

How do ideal and real transformers differ?

Explore further: pulses and flux drift

Equal voltage-time areas keep flux periodic

Given: An ideal winding model has primary $n_1=100$ and secondary $n_2=25$ . In each period primary voltage is $+12.0\,\mathrm V$ for $1.00\,\mathrm{ms}$ , then $-6.00\,\mathrm V$ for $2.00\,\mathrm{ms}$ . Choose polarity so $u_1=n_1\dot\Phi$ . Assume complete coupling with negligible losses and magnetising current; initial flux is unspecified.
Find: Find both flux changes, secondary pulse levels and both RMS voltages.

Approach: integrate voltage over each constant interval. RMS follows from duration-weighted squares, not the sine formula.
$\begin{aligned}\Delta\Phi_j&=\frac{u_{1,j}\Delta t_j}{n_1}\\u_{2,j}&=\frac{n_2}{n_1}u_{1,j}\\U_{1,\mathrm{rms}}^2&=\frac{u_a^2t_a+u_b^2t_b}{t_a+t_b}\end{aligned}$
Substitution: the smaller negative level lasts twice as long and balances the positive pulse.
$\begin{aligned}\Delta\Phi_a&=+0.120\,\mathrm{mWb}\\\Delta\Phi_b&=-0.120\,\mathrm{mWb}\\u_{2,a}&=+3.00\,\mathrm V\\u_{2,b}&=-1.50\,\mathrm V\\U_{1,\mathrm{rms}}&=\sqrt{\frac{12.0^2(1.00)+(-6.00)^2(2.00)}{3.00}}\,\mathrm V\\&=\sqrt{72}\,\mathrm V\approx8.48528\,\mathrm V\\U_{2,\mathrm{rms}}&=\frac14U_{1,\mathrm{rms}}\approx2.12132\,\mathrm V\end{aligned}$

Result: Flux returns to its initial value after each period; its absolute level remains unknown without an initial value. Secondary levels are +3.00 V and −1.50 V; RMS is about 2.121 V.

Five exercises on calculation, models and errors

Calculate · Foundation

Ideal: 600 primary turns, 75 secondary turns, 120 V RMS primary. Find secondary voltage.

Show hint

Voltage and turns belong on the same side of the ratio.

Learning goal: Apply voltage conversion with explicit RMS values.

Show solution and calculation

Approach: Shared changing flux gives the ideal voltage relation.
Substitution yields 15 V RMS.

Check: Eight times fewer turns give eight times less voltage.

U_2=U_1\frac{n_2}{n_1}=(120\,\mathrm V)\frac{75}{600}=15\,\mathrm V

[1]

Rearrange · Application

Ideally obtain 18 V RMS from 180 V RMS; $n_1=900$ . A resistive secondary load draws 2 A. Find n₂ and I₁.

Show hint

Turns follow voltage; primary current uses the inverse relation.

Learning goal: Translate a target voltage into turns and a load-current budget.

Show solution and calculation

Approach: First find n₂ from target voltage, then refer the specified load current.
The answers are 90 turns and 0.20 A primary.

Check: Both sides transfer 36 W.

\begin{aligned}n_2&=900\frac{18}{180}=90\\I_1&=(2\,\mathrm A)\frac{90}{900}=0.20\,\mathrm A\end{aligned}

[1]

Interpret · Application

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u₂(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Show hint

An open load circuit draws no load current; the sign depends on the measurement arrow.

Learning goal: Distinguish magnitude conversion from polarity and no-load behaviour.

Show solution and calculation

The turns ratio and RMS voltage magnitude remain unchanged; the signed oscilloscope waveform u₂(t) is reflected across the time axis.
A normal AC RMS meter still displays the same nonnegative RMS value.
Ideal secondary load current remains zero, as does ideal primary load current.
A quotient of two zero currents is undefined.
Real magnetising current would be an additional variable excluded here.

\begin{aligned}U_2&=\frac{U_1}{10}\\I_1&=I_2=0\end{aligned}

[1]

Error analysis · Application

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims $I_1=10\,\mathrm A$ because primary voltage is five times larger. Correct it.

Show hint

Compare real power on both sides first.

Learning goal: Detect incorrect current conversion using the power budget.

Show solution and calculation

The error is scaling current like voltage rather than inversely.
Secondary power is 40 W, so ideal primary current must be 0.40 A.
The incorrect 10 A would imply 1000 W input with unexplained 960 W losses, contradicting the specified ideal model.

I_1=\frac{U_2I_2}{U_1}=\frac{(20\,\mathrm V)(2\,\mathrm A)}{100\,\mathrm V}=0.40\,\mathrm A

[1]

Transfer · Application

Two idealised transmissions deliver the same real power at the same power factor through the same line resistance. Case B has four times the voltage. How do current and Joule loss change?

Show hint

Find current from the equal power requirement before squaring it.

Learning goal: Transfer voltage stepping to relative line losses.

Show solution and calculation

Approach: Equal power at equal power factor requires one quarter of the current.
Loss becomes one sixteenth.
This is a relative comparison under the specified conditions, not a complete real-line design.

\begin{aligned}\frac{I_B}{I_A}&=\frac{U_A}{U_B}=\frac14\\\frac{P_{\mathrm{loss},B}}{P_{\mathrm{loss},A}}&=\left(\frac{I_B}{I_A}\right)^2=\frac1{16}\end{aligned}

[1][3]

In brief

Sources for this explanation

Quanta · Updated October 1, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Made turn/voltage/current ratios, referred load, power factor, DC and the existing pulsed-flux balance accessible through learning questions; added matching induction, rms and electrical-power links.
October 1, 2026Added an original DE/EN explanation with two checked examples and five distinct solved exercises.
October 1, 2026Expanded the learning sequence and model reasoning, added a third worked example and an original quantitative diagram, and refined mathematical typography and the concluding summary.

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Frequently asked questions about Transformer turns ratio

How do you calculate the secondary voltage of a transformer?+

Why do the currents scale inversely to the voltages?+

Why does a transformer not work with DC voltage?+

Why is electricity transmitted across country at high voltage?+

What does the ideal transformer mean and how realistic is it?+

Retain Transformer turns ratio for exams

Create a curated FSRS exam set for U₁/U₂ = n₁/n₂ = I₂/I₁: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential

Transformer: ideal conversion and real limits

Variables & units

Transformer calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction and applications

Worked example

Applications

How do you calculate with Transformer turns ratio?

Convert voltage and transfer power

How do you calculate secondary voltage from turn counts?

How does Faraday’s law give the voltage ratio?

Why does current scale inversely with the turns ratio?

Calculating turn count, current and referred load resistance

The load sets actual current

Calculating resistive load, efficiency and power factor

A 20:1 transformer with a specified load

Account separately for losses and power factor

Why is constant DC voltage not steady transformer operation?

Why does high voltage help transmit energy?

How do ideal and real transformers differ?

Explore further: pulses and flux drift

Equal voltage-time areas keep flux periodic

Five exercises on calculation, models and errors

Ideal: 600 primary turns, 75 secondary turns, 120 V RMS primary. Find secondary voltage.

Ideally obtain 18 V RMS from 180 V RMS; n1=900n_1=900n1​=900. A resistive secondary load draws 2 A. Find n2 and I1.

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u2(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims I1=10 AI_1=10\,\mathrm AI1​=10A because primary voltage is five times larger. Correct it.

Two idealised transmissions deliver the same real power at the same power factor through the same line resistance. Case B has four times the voltage. How do current and Joule loss change?

In brief

Sources for this explanation

Frequently asked questions about Transformer turns ratio

Subject references and further reading

Related formulas

More formulas: Electricity and magnetism

Retain Transformer turns ratio for exams

Transformer: ideal conversion and real limits

Variables & units

Transformer calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Induction and applications

Worked example

Applications

How do you calculate with Transformer turns ratio?

Convert voltage and transfer power

How do you calculate secondary voltage from turn counts?

How does Faraday’s law give the voltage ratio?

Why does current scale inversely with the turns ratio?

Calculating turn count, current and referred load resistance

The load sets actual current

Calculating resistive load, efficiency and power factor

A 20:1 transformer with a specified load

Account separately for losses and power factor

Why is constant DC voltage not steady transformer operation?

Why does high voltage help transmit energy?

How do ideal and real transformers differ?

Explore further: pulses and flux drift

Equal voltage-time areas keep flux periodic

Five exercises on calculation, models and errors

Ideal: 600 primary turns, 75 secondary turns, 120 V RMS primary. Find secondary voltage.

Ideally obtain 18 V RMS from 180 V RMS; n1=900n_1=900n1​=900. A resistive secondary load draws 2 A. Find n2 and I1.

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u2(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims I1=10 AI_1=10\,\mathrm AI1​=10A because primary voltage is five times larger. Correct it.

Two idealised transmissions deliver the same real power at the same power factor through the same line resistance. Case B has four times the voltage. How do current and Joule loss change?

In brief

Ideally obtain 18 V RMS from 180 V RMS; $n_1=900$ . A resistive secondary load draws 2 A. Find n₂ and I₁.

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u₂(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims $I_1=10\,\mathrm A$ because primary voltage is five times larger. Correct it.

Ideally obtain 18 V RMS from 180 V RMS; $n_1=900$ . A resistive secondary load draws 2 A. Find n₂ and I₁.

An ideal 10:1 transformer has an open secondary. An oscilloscope records signed instantaneous voltage u₂(t); the secondary measurement orientation is reversed. Do the turns ratio, RMS voltage magnitude or load current change?

Ideal: 100 V primary, 20 V secondary and 2 A secondary into a resistor. A solution claims $I_1=10\,\mathrm A$ because primary voltage is five times larger. Correct it.