---
title: Transformers
module: Alternating Current
moduleNumber: 9
lessonNumber: 5
order: 905
summary: >
  Two coils sharing an iron core exchange no charge, yet a changing current in one
  drives a voltage in the other, and the ratio of their turns sets how voltage and
  current trade off between the windings. That lets a transformer step a voltage up or
  down, isolate two circuits, and make a load look larger or smaller to the source by
  the square of the turns ratio. We build the ideal ratio element from Faraday's law and
  the dot convention, derive the reflected-impedance rule, then add the winding
  resistance, leakage, magnetizing current, and core loss that turn the ideal ratios
  into real regulation, efficiency, and a bounded voltage-frequency range.
topics: [Alternating Current]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 29 — Alternating-Current Circuits; §29-3 The Transformer"
---

## Coupled coils and mutual inductance

A transformer contains two electrically separate windings linked by a shared magnetic
path. The primary winding connects to the source. The secondary winding connects to a
load, an instrument, or another circuit. Conductive isolation is retained between the
windings; energy crosses the insulation through the time-varying magnetic field in the
core. The arrangement requires a changing current or changing magnetic flux. A steady
DC current establishes a static field after its switching transient and induces no
continuing secondary voltage.

Each winding has self-inductance. A changing current in winding 1 produces a voltage
in that same winding and a voltage in winding 2. With selected terminal references,
the coupled-winding equations are

$$
v_1=L_1\frac{\d i_1}{\d t}+M\frac{\d i_2}{\d t},
\qquad
v_2=M\frac{\d i_1}{\d t}+L_2\frac{\d i_2}{\d t}.
$$

The self terms $L_1$ and $L_2$ depend on turn count, geometry, and core permeability.
The mutual inductance $M$ measures the flux linkage shared by the two windings. Its
algebraic sign follows the chosen terminal references. Geometry alone does not give a
signed $M$ until the reference directions have been stated. The dot convention records
that sign on the circuit drawing and prevents a common error in coupled-circuit work:
combining induced voltages with the wrong polarity.

$$
% caption: Two windings share a high-permeability core. A changing primary current drives circulating core flux $\Phi$ that links both windings, while the copper circuits stay electrically isolated.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,very thick] (1.55,0.45) rectangle (4.95,3.05);
\draw[black,very thick] (2.25,1.15) rectangle (4.25,2.35);
\draw[acc,thick,->] (2.60,2.72) arc (150:35:0.95 and 0.28);
\draw[acc,thick,->] (4.30,0.78) arc (-30:-150:0.95 and 0.28);
\draw[acc,very thick] (0.95,0.95) arc (-90:90:0.21) arc (-90:90:0.21) arc (-90:90:0.21) arc (-90:90:0.21);
\draw[acc,very thick] (5.55,0.95) arc (-90:-270:0.21) arc (-90:-270:0.21) arc (-90:-270:0.21) arc (-90:-270:0.21);
\draw[black,thick] (0.95,2.63)--(0.45,2.63);
\draw[black,thick] (0.95,0.95)--(0.45,0.95);
\draw[black,thick] (5.55,2.63)--(6.05,2.63);
\draw[black,thick] (5.55,0.95)--(6.05,0.95);
\node[below] at (0.82,0.72) {primary};
\node[below] at (5.68,0.72) {secondary};
\node at (3.25,1.75) {core};
\node[above,acc] at (3.25,3.05) {$B$};
\end{tikzpicture}
$$

Faraday's law gives a direct physical definition. Let $\Phi_{21}$ be the part of the
magnetic flux produced by winding 1 that links winding 2, and let $N_2$ be the number
of secondary turns. The induced voltage due to the primary current is

$$
v_{2\leftarrow1}=\frac{\d(N_2\Phi_{21})}{\d t}=M\frac{\d i_1}{\d t},
$$

where the sign is included in the oriented flux linkage. A larger turn count raises
both flux linkage and induced voltage. Tighter core coupling raises the fraction of
the primary-produced flux that reaches the secondary. Air gaps, separated coils,
fringing, and a core operated near saturation all reduce the ideal shared-path model.

Mutual inductance has a magnitude bound:

$$
|M|\leq\sqrt{L_1L_2}.
$$

The dimensionless coupling coefficient is conventionally written as

$$
k=\frac{|M|}{\sqrt{L_1L_2}},
\qquad 0\leq k\leq1.
$$

An ideal common-core transformer is modeled with $k$ very close to one. The equality
case says that all magnetic energy associated with either winding shares the same
mode. It does not imply zero winding resistance, zero core loss, or unlimited power.
Those requirements enter the transformer model separately. Small signal transformers,
power transformers, RF transformers, and current transformers can all have a large
$k$ while differing strongly in loss, voltage rating, and frequency range.

### Reference directions and dots

Place a dot at one end of each winding. If a current enters the dotted terminal of
winding 1 and rises, the induced voltage makes the dotted terminal of winding 2
positive. Equivalently, a rising current entering one dotted terminal produces a
positive mutual-voltage term when the voltage reference on the other winding is from
its dotted terminal to its undotted terminal. In that reference system the mutual
terms in the two equations are $+M\,\d i/\d t$.

$$
% caption: Dot convention. With both voltage references running from the dotted to the undotted terminal, a rising current $i_1$ entering the primary dot makes the secondary dot positive, so both mutual terms are $+M\,\d i/\d t$.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[acc,very thick] (1.70,0.70) arc (-90:90:0.24) arc (-90:90:0.24) arc (-90:90:0.24) arc (-90:90:0.24);
\draw[acc,very thick] (4.60,0.70) arc (-90:90:0.24) arc (-90:90:0.24) arc (-90:90:0.24) arc (-90:90:0.24);
\draw[black,thick] (1.70,2.62)--(1.70,3.10);
\draw[black,thick] (1.70,0.70)--(1.70,0.30);
\draw[black,thick] (4.60,2.62)--(4.60,3.10);
\draw[black,thick] (4.60,0.70)--(4.60,0.30);
\filldraw[draw=acc,fill=acc!12] (1.70,2.74) circle (2.3pt);
\filldraw[draw=acc,fill=acc!12] (4.60,2.74) circle (2.3pt);
\node[right] at (1.80,2.78) {dot};
\node[right] at (4.70,2.78) {dot};
\draw[->,black,thick] (1.70,3.10)--(1.70,2.92) node[midway,left] {$i_1$};
\draw[<->,black] (1.05,0.30)--(1.05,3.05) node[midway,left] {$v_1$};
\draw[<->,black] (5.25,0.30)--(5.25,3.05) node[midway,right] {$v_2$};
\end{tikzpicture}
$$

Changing one reference reverses one sign. A current arrow entering an undotted
terminal is equivalent to a current leaving the dotted terminal. A voltage measured
from undotted to dotted is the negative of the dot-to-undotted voltage. Neither
choice is wrong. The error appears when a diagram uses one choice and the equations
silently use another. Mark current arrows, voltage polarities, and dots before using
mesh equations or phasors.

For sinusoidal steady state, derivatives become multiplication by $j\omega$ in a
phasor description. A coupled pair can therefore be written

$$
\widetilde V_1=j\omega L_1\widetilde I_1+j\omega M\widetilde I_2,
\qquad
\widetilde V_2=j\omega M\widetilde I_1+j\omega L_2\widetilde I_2,
$$

when both current arrows enter dots and both voltages are dot-to-undotted. If one
terminal convention is reversed, replace the relevant mutual sign. This compact
form is valuable for finite-coupling circuits, tuned transformers, and leakage-aware
models. The ideal-transformer relations developed below impose stronger assumptions
than these general coupled-inductor equations.

## Ideal transformer relations

The ideal transformer model separates a high-permeability shared core from all loss
and leakage mechanisms. It assumes that every turn on either winding links the same
core flux $\Phi(t)$, that winding voltage drops are negligible, and that the core
requires negligible magnetizing current. These assumptions give a ratio element. They
do not state that the secondary voltage exists independently of the primary source or
that the output can supply arbitrary current.

A winding with $N$ turns has terminal-voltage magnitude and sign set by Faraday's law
through the oriented flux linkage:

$$
v=N\frac{\d\Phi}{\d t}.
$$

The sign convention used here treats the winding terminal voltage as the applied
voltage balancing the induced back emf. Applying the same orientation on both
windings gives

$$
\frac{v_2(t)}{v_1(t)}=\frac{N_2}{N_1}=a,
\qquad
\frac{V_{2,\mathrm{rms}}}{V_{1,\mathrm{rms}}}=a.
$$

The turns ratio $a=N_2/N_1$ sets voltage scaling. A step-up transformer has $a>1$.
A step-down transformer has $a<1$. The waveform shape transfers with the ratio while
the core remains in its linear operating region. A sinusoidal primary produces a
sinusoidal secondary at the same frequency. A distorted source drives a proportionally
distorted secondary voltage in the ideal model.

$$
% caption: Ideal voltage scaling. Both windings link the same core flux $\Phi(t)$, so each terminal voltage is proportional to its turn count: $V_2/V_1 = N_2/N_1$. Here the secondary has more turns, so the connection steps the voltage up.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,very thick] (1.55,0.45) rectangle (4.95,3.05);
\draw[black,very thick] (2.25,1.10) rectangle (4.25,2.40);
\draw[acc,thick,->] (2.60,2.74) arc (150:35:0.95 and 0.26);
\draw[acc,very thick] (1.05,1.10) arc (-90:90:0.24) arc (-90:90:0.24) arc (-90:90:0.24);
\draw[acc,very thick] (5.45,0.60) arc (-90:-270:0.22) arc (-90:-270:0.22) arc (-90:-270:0.22) arc (-90:-270:0.22) arc (-90:-270:0.22);
\draw[black,thick] (1.05,2.54)--(0.55,2.54);
\draw[black,thick] (1.05,1.10)--(0.55,1.10);
\draw[black,thick] (5.45,2.80)--(6.05,2.80);
\draw[black,thick] (5.45,0.60)--(6.05,0.60);
\draw[<->,black] (0.40,1.10)--(0.40,2.54) node[midway,left] {$V_1$};
\draw[<->,black] (6.15,0.60)--(6.15,2.80) node[midway,right] {$V_2$};
\node[below] at (1.05,0.86) {$N_1$};
\node[below] at (5.45,0.44) {$N_2$};
\node[above,acc] at (3.25,3.05) {$B$};
\end{tikzpicture}
$$

The instantaneous relation is stronger than an RMS ratio. If $N_2/N_1=0.10$ and
the primary is a 120 V RMS sine wave, the unloaded secondary is 12 V RMS in the
ideal model. Its peak voltage is also one-tenth of the primary peak voltage. Confusing
an RMS rating with a peak value changes a turns calculation by a factor of $\sqrt2$.
Use one amplitude convention throughout a calculation. Voltage ratings on ordinary
power transformers are almost always RMS values at a stated frequency and load.

With an open secondary, an actual primary still draws a small current. This is the
magnetizing current $I_m$. It establishes the alternating core flux required by the
applied voltage. In a good power transformer it is generally much smaller than the
rated load current, though it can be large enough to matter in a lightly loaded
system. The ideal-ratio model sets it to zero in the load-current balance but does
not deny that a real core needs magnetic field strength.

Under sinusoidal voltage, the idealized magnetizing branch is inductive. Its current
lags the applied voltage by nearly a quarter cycle. Core hysteresis and eddy losses add
an in-phase component, so a real no-load current is neither zero nor perfectly
quadrature. A wattmeter reading on an unloaded transformer therefore measures core
loss plus a small winding loss, even when the secondary terminals are unconnected.

Loading the secondary changes the primary current. A load current $I_2$ produces a
magnetomotive force $N_2I_2$ whose core effect opposes the original source-driven
flux. The source voltage constrains the required rate of change of core flux. The
primary responds by drawing an additional load component $I_1'$ such that, under the
ideal approximation,

$$
N_1\widetilde I_1'+N_2\widetilde I_2=0.
$$

Thus the current ratio is

$$
\frac{\widetilde I_2}{\widetilde I_1'}=-\frac{N_1}{N_2}=-\frac{1}{a}.
$$

The minus sign is the winding-reference sign. It records opposing ampere-turns, not
a claim that input and output powers have opposite physical directions. Under passive
load references, the source delivers positive average power to the primary terminals
and the secondary load absorbs positive average power.

$$
% caption: Ampere-turn balance under load. Secondary current $I_2$ sets up an opposing magnetomotive force $N_2 I_2$; the primary draws an added load component (shown here as $I_1$) so that $N_1 I_1$ balances $N_2 I_2$ and the required core flux is restored.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,very thick] (1.55,0.45) rectangle (4.95,3.05);
\draw[black,very thick] (2.25,1.15) rectangle (4.25,2.35);
\draw[acc,thick,->] (2.60,2.70) arc (150:35:0.95 and 0.28);
\draw[acc,thick,->] (4.30,0.80) arc (-30:-150:0.95 and 0.28);
\draw[acc,very thick] (1.05,0.95) arc (-90:90:0.21) arc (-90:90:0.21) arc (-90:90:0.21) arc (-90:90:0.21);
\draw[acc,very thick] (5.45,0.95) arc (-90:-270:0.21) arc (-90:-270:0.21) arc (-90:-270:0.21) arc (-90:-270:0.21);
\draw[black,thick] (1.05,2.63)--(0.55,2.63)--(0.55,3.05);
\draw[black,thick] (1.05,0.95)--(0.55,0.95)--(0.55,0.55);
\draw[black,thick] (5.45,2.63)--(5.95,2.63)--(5.95,3.05);
\draw[black,thick] (5.45,0.95)--(5.95,0.95)--(5.95,0.55);
\draw[->,black,thick] (0.55,3.05)--(0.55,2.80) node[midway,left] {$I_1$};
\draw[->,black,thick] (5.95,2.80)--(5.95,3.05) node[midway,right] {$I_2$};
\node[below] at (0.90,0.55) {$N_1$};
\node[below] at (5.60,0.55) {$N_2$};
\end{tikzpicture}
$$

The complete primary current is the phasor sum of its magnetizing part and its reflected
load part. At rated load, the latter is commonly much larger. A current probe on the
primary will therefore show an increased current when a secondary load is connected,
even though no conductive path joins the two circuits. This observation is a direct
test of energy transfer by mutual induction and of the ampere-turn balance.

## Reflected impedance and source loading

The transformer ratio converts a secondary load into an input load seen by the source.
This result is often more practical than the separate voltage and current ratios. It
allows a primary-side circuit to be analyzed without redrawing the secondary network.
The reflected impedance depends on the square of the turns ratio, a consequence of one
factor from the voltage ratio and another from the current ratio.

Use two different secondary-current symbols while deriving the result. Let $I_2$ enter
the dotted secondary terminal, as required by the coupled-winding sign convention. Let
$I_L$ leave that terminal into the external load. They are related by

$$
\widetilde I_2=-\widetilde I_L.
$$

In an ideal transformer, the ampere-turn equation becomes

$$
N_1\widetilde I_1-N_2\widetilde I_L=0,
\qquad
\widetilde I_1=a\widetilde I_L,
$$

where $a=N_2/N_1$. The voltage relation is $\widetilde V_2=a\widetilde V_1$. If the
secondary load has impedance $Z_L=\widetilde V_2/\widetilde I_L$, then

$$
Z_{\mathrm{in}}
=\frac{\widetilde V_1}{\widetilde I_1}
=\frac{\widetilde V_2/a}{a\widetilde I_L}
=\left(\frac{N_1}{N_2}\right)^2Z_L
=\frac{Z_L}{a^2}.
$$

The equation applies to resistance, inductive impedance, capacitive impedance, or a
network containing all three. The ideal transformer preserves the impedance angle while
changing its magnitude by a positive real factor. If a secondary resistor is connected,
the input is a resistor of the reflected value. If a secondary network has a positive
imaginary part, the reflected network also has a positive imaginary part. The transformer
does not reverse capacitive or inductive character.

$$
% caption: Reflected impedance. A secondary load $Z_L$ appears at the source terminals as $Z_L/a^2$ with $a=N_2/N_1$, so a primary-side analysis can use the single reflected element without redrawing the secondary network.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[thick] (0.55,0.85)--(0.55,2.35);
\draw[thick] (0.55,1.60) circle (0.30);
\node[left] at (0.20,1.60) {source};
\draw[thick] (0.55,2.35)--(2.45,2.35)--(2.45,2.29);
\draw[thick] (0.55,0.85)--(2.45,0.85)--(2.45,1.15);
\draw[acc,very thick] (2.45,1.15) arc (-90:90:0.19) arc (-90:90:0.19) arc (-90:90:0.19);
\draw[black,thick] (2.78,0.95)--(2.78,2.55);
\draw[black,thick] (2.92,0.95)--(2.92,2.55);
\draw[acc,very thick] (3.25,1.15) arc (-90:-270:0.19) arc (-90:-270:0.19) arc (-90:-270:0.19);
\draw[thick] (3.25,2.29)--(3.25,2.35)--(4.55,2.35);
\draw[thick] (3.25,1.15)--(3.25,0.85)--(4.55,0.85);
\draw[fill=acc!10,draw=acc,thick] (4.55,1.05) rectangle (5.65,2.15);
\node at (5.10,1.60) {$Z_L$};
\draw[thick] (4.55,2.35)--(4.55,2.15);
\draw[thick] (4.55,0.85)--(4.55,1.05);
\node[below] at (2.60,0.85) {$N_1$};
\node[below] at (3.10,0.85) {$N_2$};
\node[below] at (0.55,0.55) {$\frac{Z_L}{a^2}$};
\end{tikzpicture}
$$

The reflected-impedance rule should be used on a fully specified two-port drawing.
Voltage polarity, the turns ratio direction, and which side is being reflected must be
clear. Defining $a=N_2/N_1$ gives $Z_{\mathrm{in}}=Z_L/a^2$ when the load lies on side 2
and the input lies on side 1. Reflecting a source-side impedance to side 2 multiplies it
by $a^2$. Reversing the naming of the windings without changing the formula is a common
source of factors inverted by four or five orders of magnitude.

### Power relation at an ideal ratio element

The ideal transformer has no independent source of energy. In phasor form, with
passive terminal conventions selected on the load side, the input and output complex
powers have equal magnitudes and opposite port signs. For real power delivered from
the primary source to a secondary load,

$$
P_{\mathrm{in}}=V_{1,\mathrm{rms}}I_{1,\mathrm{rms}}\cos\phi_1
=V_{2,\mathrm{rms}}I_{L,\mathrm{rms}}\cos\phi_L
=P_L
$$

in the lossless approximation. The primary load-current component inherits the load
power factor when it is reflected through an ideal transformer. The magnetizing
component adds a largely reactive no-load current in a physical transformer. Its
presence changes the actual primary current without changing the current-ratio rule
for the reflected load component.

A resistive secondary load permits a power relation without a phase angle. Combining
$V_2=aV_1$ with $I_1=aI_L$ gives

$$
V_1I_1=V_1(aI_L)=(aV_1)I_L=V_2I_L.
$$

A step-down transformer decreases voltage and increases available secondary current
by reciprocal factors. A step-up transformer increases voltage and decreases current.
The phrase "increases current" needs the load condition. An open secondary has no
load current, regardless of its turns ratio. A transformer cannot force a stated
current into an arbitrary load because its winding resistance, leakage reactance,
core limits, and source impedance constrain the operating point.

An ideal transformer can isolate a load from DC continuity, but it cannot transfer a
DC power waveform through steady mutual induction. A rectangular drive transfers
energy during its rising and falling intervals. The core flux changes during each
nonzero applied-voltage interval, then remains at its latest value while the voltage
is zero. Repeated unipolar pulses accumulate core flux and eventually drive a finite
core into saturation. Power-conversion circuits use alternating polarity, controlled
reset intervals, or a gapped magnetic design to keep the core within an allowed flux
range.

### Impedance matching

Many low-frequency power transformers are chosen for voltage conversion, whereas
small-signal transformers are often chosen to place a source and load at compatible
impedance levels. A source with a significant internal resistance delivers its maximum
available sinusoidal power to a purely resistive reflected load equal to that resistance.
If a load resistor $R_L$ is connected to secondary turns $N_2$, select primary turns
$N_1$ so that

$$
R_s=\left(\frac{N_1}{N_2}\right)^2R_L,
\qquad
\frac{N_1}{N_2}=\sqrt{\frac{R_s}{R_L}}.
$$

This condition describes a particular source model and a particular objective. It
does not maximize efficiency when source resistance represents an unwanted loss. In a
power supply, a low source resistance and a load that meets its voltage requirement are
usually preferred. In a measurement system, input impedance, bandwidth, noise, phase
error, and isolation may take priority over maximum transferred power.

$$
% caption: Impedance matching. Choosing the turns ratio so that $R_s = (N_1/N_2)^2 R_L$ makes the reflected load equal the source resistance, delivering the source's maximum available sinusoidal power without a direct conductive connection.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[thick] (0.55,0.85)--(0.55,2.35);
\draw[thick] (0.55,1.60) circle (0.30);
\node[left] at (0.20,1.60) {source};
\draw[thick] (0.55,2.35)--(1.15,2.35);
\draw[fill=black!08,draw=black,thick] (1.15,2.10) rectangle (2.05,2.60);
\node at (1.60,2.35) {$R_s$};
\draw[thick] (2.05,2.35)--(2.55,2.35)--(2.55,2.29);
\draw[thick] (0.55,0.85)--(2.55,0.85)--(2.55,1.15);
\draw[acc,very thick] (2.55,1.15) arc (-90:90:0.19) arc (-90:90:0.19) arc (-90:90:0.19);
\draw[black,thick] (2.88,0.95)--(2.88,2.55);
\draw[black,thick] (3.02,0.95)--(3.02,2.55);
\draw[acc,very thick] (3.35,1.15) arc (-90:-270:0.19) arc (-90:-270:0.19) arc (-90:-270:0.19);
\draw[thick] (3.35,2.29)--(3.35,2.35)--(4.75,2.35);
\draw[thick] (3.35,1.15)--(3.35,0.85)--(4.75,0.85);
\draw[fill=acc!10,draw=acc,thick] (4.75,1.05) rectangle (5.85,2.15);
\node at (5.30,1.60) {$R_L$};
\draw[thick] (4.75,2.35)--(4.75,2.15);
\draw[thick] (4.75,0.85)--(4.75,1.05);
\node[below] at (2.95,0.85) {$N_1:N_2$};
\end{tikzpicture}
$$

When an impedance has resistance and reactance, transform the entire complex value:

$$
Z_L=R_L+jX_L
\quad\Longrightarrow\quad
Z_{\mathrm{in}}=\frac{R_L}{a^2}+j\frac{X_L}{a^2}.
$$

The ratio therefore scales both real and imaginary components equally. A transformer
does not tune a reactive load to resonance. It can place the required component values
or source impedance in a more convenient range, but any cancellation of reactance
must come from the surrounding circuit and its frequency dependence.

Polarity matters when transformer windings are connected in series. Two equal
secondary windings can produce a doubled voltage when their marked polarities are
series-aiding. Reversing one connection produces series opposition and a near-zero
terminal voltage for equal voltages. Manufacturers mark a start lead, dot, or lead
number to preserve this information. Verify it at low voltage before connecting
windings in a higher-energy circuit.

## Nonideal transformer behavior

The ideal model is a reference point, not a complete device model. A physical
transformer contains resistance in both windings, finite magnetic permeability, flux
that links only one winding, and a core with loss. At higher frequency it also contains
distributed capacitance between turns, windings, and core. These effects determine the
voltage regulation, no-load current, heat generation, frequency response, and maximum
safe operating voltage.

A standard low-frequency equivalent circuit places primary winding resistance $R_1$
and primary leakage reactance $X_{\sigma1}$ in series with the source. The core is
represented by a shunt magnetizing reactance $X_m$ in parallel with a resistance $R_c$.
The resistance $R_c$ accounts for core-loss power at the applied voltage. Secondary
winding resistance $R_2$ and secondary leakage reactance $X_{\sigma2}$ lie in series
with the load. Values on one side can be reflected to the other side using the square
turns rule, reducing the model to a single circuit with no ideal-ratio symbol.

$$
% caption: Low-frequency equivalent circuit referred to the primary. Series $R_1$ and leakage reactance $X_{\sigma 1}$ carry the load current; the shunt branch (core-loss resistance $R_c$ in parallel with magnetizing reactance $X_m$) carries the no-load current, ahead of the ideal turns ratio $a$.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[thick] (0.40,2.05)--(0.85,2.05);
\draw[fill=acc!10,draw=acc,thick] (0.85,1.80) rectangle (1.65,2.30);
\node at (1.25,2.05) {$R_1$};
\draw[thick] (1.65,2.05)--(2.05,2.05);
\draw[fill=acc!10,draw=acc,thick] (2.05,1.80) rectangle (2.95,2.30);
\node at (2.50,2.05) {$X_{s1}$};
\draw[thick] (2.95,2.05)--(5.10,2.05);
\draw[thick] (0.40,0.45)--(5.10,0.45);
\draw[thick] (0.40,2.05)--(0.40,0.45);
\node[left] at (0.30,1.25) {$V_1$};
\draw[thick] (3.35,2.05)--(3.35,1.55);
\draw[fill=acc!10,draw=acc,thick] (3.01,1.05) rectangle (3.69,1.55);
\node at (3.35,1.30) {$R_c$};
\draw[thick] (3.35,1.05)--(3.35,0.45);
\draw[thick] (4.25,2.05)--(4.25,1.55);
\draw[fill=acc!10,draw=acc,thick] (3.91,1.05) rectangle (4.59,1.55);
\node at (4.25,1.30) {$X_m$};
\draw[thick] (4.25,1.05)--(4.25,0.45);
\draw[acc,very thick] (5.10,0.75) arc (-90:90:0.18) arc (-90:90:0.18) arc (-90:90:0.18);
\draw[thick] (5.10,2.05)--(5.10,1.83);
\draw[thick] (5.10,0.75)--(5.10,0.45);
\draw[black,thick] (5.42,0.70)--(5.42,1.90);
\draw[black,thick] (5.54,0.70)--(5.54,1.90);
\draw[acc,very thick] (5.86,0.75) arc (-90:-270:0.18) arc (-90:-270:0.18) arc (-90:-270:0.18);
\draw[thick] (5.86,1.83)--(5.86,2.05)--(6.30,2.05);
\draw[thick] (5.86,0.75)--(5.86,0.45)--(6.30,0.45);
\node[below] at (5.48,0.55) {$a$};
\node[right] at (6.35,1.25) {$V_2$};
\end{tikzpicture}
$$

The circuit is an approximation with a frequency range and operating range. It assumes
sinusoidal operation near the chosen test frequency and a core below severe saturation.
At audio, switching, or radio frequencies, extra distributed elements may be needed.
At high loading, winding temperature changes the resistances and the operating voltage
drop. Engineering data sheets commonly report a regulation percentage, no-load current,
short-circuit impedance, loss at a stated temperature, and a rated VA value instead of
every lumped model parameter.

### Winding resistance and copper loss

Each turn has resistance. Current produces an ohmic voltage drop and a heat rate

$$
P_{\mathrm{cu},1}=I_{1,\mathrm{rms}}^2R_1,
\qquad
P_{\mathrm{cu},2}=I_{2,\mathrm{rms}}^2R_2.
$$

At a fixed load current, this loss rises with the square of current. It therefore grows
quickly as the transformer approaches overload. Resistance also reduces the terminal
voltage under load. The induced secondary emf exists inside the winding, while the load
sees that emf after subtracting the drops across secondary resistance and leakage
reactance. The primary series terms similarly require a larger source voltage to hold
the intended core flux.

Copper resistance rises with temperature. For ordinary copper near room temperature,
the fractional resistance rise is about $0.0039$ per degree Celsius. A transformer
tested while cool will therefore have a smaller winding loss than the same transformer
after a long rated-load run. Ratings use a permitted temperature rise, insulation class,
ambient temperature, enclosure, and cooling arrangement. A winding current below its
headline rating can still be unsafe when airflow is blocked or ambient temperature is
high.

$$
% caption: Phasor construction of the terminal voltage. The induced secondary emf $E_2$ loses the resistive drop $I_2 R$ and the leakage-reactance drop $I_2 X_s$ (a quadrature phasor), leaving a terminal voltage $V_2$ below $E_2$ for this lagging load.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0.72,0.62)--(5.96,0.62) node[right] {in phase};
\draw[->,black] (0.82,0.48)--(0.82,2.60) node[above] {lag};
\draw[->,acc,very thick] (0.82,0.62)--(5.10,0.62);
\node[below,acc] at (4.20,0.60) {$E_2$};
\draw[->,black,very thick] (5.10,0.62)--(4.20,1.42);
\node[right,black] at (4.78,1.18) {$I_2X_s$};
\draw[->,black,very thick] (4.20,1.42)--(3.28,1.42);
\node[above,black] at (3.74,1.48) {$I_2R$};
\draw[->,black,very thick] (0.82,0.62)--(3.28,1.42);
\node[above] at (1.90,1.10) {$V_2$};
\draw[dashed,black] (3.28,0.62)--(3.28,1.42);
\end{tikzpicture}
$$

High-frequency transformers face an additional copper effect. Alternating current
crowds toward the surface of a conductor and toward regions influenced by nearby
conductors. The resulting AC resistance can exceed the DC resistance. Foil windings,
litz wire, interleaved layers, and conductor geometry are selected to limit those
effects within a design band. A DC resistance measurement alone cannot predict full-load
loss at a much higher operating frequency.

### Leakage inductance

Some magnetic field closes through air and links one winding more strongly than the
other. This portion is represented as leakage inductance. Its reactance increases with
frequency and appears in series with the winding. It limits short-circuit current and
causes load-dependent voltage drop. In a line-frequency power transformer, a modest
leakage value can be desirable because it limits fault current. In a wideband signal
transformer, leakage inductance restricts high-frequency response and is usually
reduced by close winding placement and interleaving.

Leakage voltage drop is a phasor quantity. A resistive load current aligns closely with
secondary voltage, so the resistive and leakage contributions form different directions
in the voltage diagram. A lagging load can produce a larger positive regulation drop.
A leading load can partly offset the series leakage contribution and can yield a
terminal voltage above its no-load value. The sign of voltage regulation is therefore
load-power-factor dependent, even when the transformer itself has only positive
resistance and inductance.

Winding placement changes both leakage and capacitance. Placing primary and secondary
layers adjacent reduces mean magnetic path length between them. Interleaving layers
reduces leakage further, but increases interwinding capacitance and can transmit more
common-mode noise. A design specifies an acceptable compromise according to insulation
requirements, required isolation, source impedance, load impedance, and frequency band.

### Core magnetization and saturation

An applied sinusoidal winding voltage with negligible series drop gives the familiar
volts-per-turn relation on integration of Faraday's law

$$
V_{\mathrm{rms}}=4.44\,fN\Phi_{\mathrm{pk}}.
$$

Here $f$ is frequency, $N$ is the driven winding's turn count, and $\Phi_{\mathrm{pk}}$
is peak core flux. The factor 4.44 belongs to a sine wave. It changes for a square wave,
trapezoid, or any other waveform because the relevant constraint is the time integral
of voltage. Frequency and applied voltage must therefore be specified together.
Reducing frequency while holding RMS voltage constant increases peak flux.
Increasing applied voltage at fixed frequency has the same effect.

The core's $B$-$H$ curve has a high-slope region followed by a knee and a lower-slope
saturated region. Below the knee, a small magnetizing current produces the required
flux density. Near saturation, a much larger current produces little additional flux
change. Primary current then becomes sharply non-sinusoidal and can exceed the winding
or source rating even with an open secondary. The secondary voltage may distort at the
same time.

$$
% caption: The core magnetization curve. Below the knee a small magnetizing field $H$ produces the required flux density $B$; beyond it, much larger $H$ yields little further $B$, so the magnetizing current turns sharply non-sinusoidal.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0.72,0.58)--(5.94,0.58) node[right] {$H$};
\draw[->,black] (0.82,0.45)--(0.82,3.00) node[above] {$B$};
\draw[acc,very thick] (0.92,0.64) .. controls (1.55,1.75) and (2.10,2.45) .. (3.05,2.66)
  .. controls (4.05,2.80) and (4.90,2.85) .. (5.68,2.88);
\draw[dashed,black] (3.05,0.58)--(3.05,2.66);
\node[below] at (3.05,0.56) {knee};
\node[acc] at (1.62,2.52) {linear};
\node[below,acc] at (5.05,2.62) {sat};
\end{tikzpicture}
$$

Core saturation is controlled by several related quantities:

- **Applied volt-seconds.** The integral $\int v\,\d t$ determines flux change. A DC
  offset in the voltage waveform displaces the core operating cycle toward one limit.
- **Frequency.** Lower frequency gives each half cycle more time to accumulate flux at
  a fixed voltage amplitude.
- **Turns and core area.** More turns or larger effective core area reduce flux density
  for a stated voltage and frequency.
- **Temperature and material.** Permeability, loss, and saturation flux density vary
  with core material and temperature.

An inrush transient occurs when energizing a transformer at an unfavorable source phase
and with residual core flux. The new flux transient can add to the retained flux and
cross the saturation knee for several cycles. Inrush current is limited by source
impedance, winding resistance, and leakage. It is not reliably predicted by dividing
rated voltage by a single magnetizing reactance measured under small-signal conditions.

### Core loss and frequency range

Magnetic materials dissipate energy during each reversal of magnetization. Hysteresis
loss corresponds to the area inside the material's $B$-$H$ loop. Eddy currents are
circulating currents induced inside conductive core material; their heating depends on
changing flux and available current paths. Laminating a power-frequency core into thin
insulated sheets interrupts large eddy-current loops. Ferrite cores have high electrical
resistivity and are common at higher switching frequencies.

At a fixed sinusoidal voltage, increasing frequency lowers the required flux amplitude
and often reduces the low-frequency saturation risk. Core loss may still rise because
the material is cycled more times per second. Transformer core selection therefore
requires a material loss curve at the intended frequency and flux density, not a simple
rule that higher frequency is always safer. The winding AC resistance and parasitic
capacitance also become more important as frequency rises.

Distributed capacitance provides a high-frequency path between windings and from a
winding to a grounded core or shield. It can couple common-mode voltage even when the
windings are galvanically isolated. Electrostatic shields reduce that coupling when
properly connected, but a continuous turn around the core would behave as a shorted
winding. Shields are constructed with a break so they do not form a closed conductive
loop.

## Regulation, efficiency, and ratings

Transformer nameplate voltage is normally a full-load RMS rating at a stated frequency.
The open-circuit secondary voltage is often higher because almost no secondary copper
or leakage drop occurs. Connecting a load increases current and produces a terminal
drop that depends on the series impedance and on the load phase. Voltage regulation is
reported as

$$
\mathrm{VR}=\frac{V_{\mathrm{nl}}-V_{\mathrm{fl}}}{V_{\mathrm{fl}}}\times100\%,
$$

where $V_{\mathrm{nl}}$ is the no-load secondary voltage and $V_{\mathrm{fl}}$ is the
secondary voltage at specified full load. A positive value describes a falling terminal
voltage for the stated load. The calculation must report the load power factor because
leakage reactance changes the result for leading and lagging loads.

$$
% caption: Secondary terminal voltage against load current. The voltage is highest at no load ($V_{nl}$) and falls as winding and leakage drops grow with current; the fractional drop to full load ($V_{fl}$) is the voltage regulation.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0.72,0.58)--(6.04,0.58) node[right] {load current};
\draw[->,black] (0.82,0.42)--(0.82,3.00) node[above] {$V_2$};
\draw[acc,very thick] (0.96,2.62) .. controls (2.20,2.52) and (3.90,2.20) .. (5.55,1.52);
\draw[dashed,black] (0.82,2.62)--(0.96,2.62);
\draw[dashed,black] (5.55,0.58)--(5.55,1.52);
\node[left] at (0.80,2.62) {$V_{nl}$};
\node[right] at (5.62,1.52) {$V_{fl}$};
\end{tikzpicture}
$$

The same transformer can have different regulation values for different loads. A
resistive load gives a drop associated with both winding resistance and leakage
reactance. A lagging current gives a leakage-reactance contribution in a direction that
usually increases the drop. A leading current can make the phasor subtraction smaller;
some transformers show negative regulation at a sufficiently leading power factor. The
result means the terminal voltage rises under the stated load condition. It does not
indicate power gain.

Use a defined testing order. Measure the primary voltage at the transformer terminals,
not at a distant source outlet. Measure secondary voltage at the load terminals, not at
the ends of a separate lead pair. Lead resistance and measurement-point differences can
otherwise be recorded as transformer regulation. Hold frequency and primary voltage
constant, allow the unit to reach an appropriate temperature state, and specify the
load current and power factor with the result.

Efficiency compares real output power with real input power:

$$
\eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}
=\frac{P_{\mathrm{out}}}{P_{\mathrm{out}}+P_{\mathrm{core}}+P_{\mathrm{cu}}+P_{\mathrm{stray}}}.
$$

The core-loss term is approximately constant over a modest range of load currents when
primary voltage and frequency are held fixed. Copper loss grows approximately as the
square of load current. Stray loss includes eddy-current loss caused by leakage fields
in conductors, clamps, and nearby metallic structure. It often increases more rapidly
than the simplest $I^2R$ estimate. At small loads, fixed core loss dominates. At high
loads, copper and stray loss dominate.

$$
% caption: Loss against load current. Core loss stays nearly constant at fixed voltage and frequency; copper loss grows as the square of current. Their sum sets the thermal limit, and the efficiency peak falls where the two losses are equal.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0.72,0.58)--(6.02,0.58) node[right] {load current};
\draw[->,black] (0.82,0.42)--(0.82,3.00) node[above] {loss};
\draw[black,thick] (0.96,1.28)--(5.35,1.28) node[right] {core};
\draw[black,thick] (0.96,0.64) .. controls (2.40,0.72) and (4.30,1.30) .. (5.35,2.48) node[right] {copper};
\draw[acc,very thick] (0.96,1.32) .. controls (2.50,1.44) and (4.30,1.95) .. (5.35,2.92) node[right] {sum};
\draw[dashed,black] (3.55,0.58)--(3.55,1.28);
\node[below] at (3.55,0.56) {equal};
\end{tikzpicture}
$$

With core loss treated as constant and copper loss proportional to $I^2$, maximum
efficiency occurs near the load current for which copper loss equals core loss. That
condition is a design result, not a recommendation to operate every transformer at the
maximum-efficiency point. Voltage regulation, temperature rise, overload margin,
required output voltage, and duty cycle may select a different operating current.
Efficiency also changes with load power factor because a lower-power-factor load can
draw the same RMS current while delivering less real output power.

The apparent-power rating is

$$
S_{\mathrm{rated}}=V_{\mathrm{rated}}I_{\mathrm{rated}}.
$$

It establishes a voltage-current thermal envelope for the windings and insulation.
It is not automatically a real-power rating in watts. A 100 VA transformer may supply
100 W to a unity-power-factor resistive load at rated voltage and current, but only
80 W to a 0.80 power-factor load at the same rated RMS quantities. The winding heating
is largely tied to RMS current, so the VA rating remains relevant for both loads.

Nameplate information also has a time basis. A continuous rating assumes a thermal
equilibrium condition. A short-duration overload may be permissible under a published
duty cycle because the winding and core have finite heat capacity. Repeated overloads
can accumulate temperature rise if the cooling interval is too short. A small sealed
transformer, an open-frame transformer, and a fan-cooled unit can have the same nominal
VA value under different ambient-temperature and installation conditions.

Short-circuit impedance is another rating-related property. With the secondary
shorted under a controlled test, only a small fraction of rated primary voltage may be
needed to circulate rated current. The ratio of that applied voltage to rated voltage,
often stated as a percentage, summarizes the transformer series impedance. It limits
fault current and contributes to regulation. A transformer with lower short-circuit
impedance has lower load drop but can deliver substantially larger fault current from a
stiff source.

## Measurement and validation

Transformer measurements begin with terminals, references, and an energy limit. Record
the nameplate voltage, frequency, VA rating, winding identifiers, insulation class, and
any stated tap connections. Apply only the voltage and frequency intended for the test
configuration. A practical lab uses an isolated low-voltage AC source, current limiting,
fused leads, and appropriately rated probes. Mains-connected windings, high-energy
power transformers, and unknown equipment require a supervised setup and a documented
isolation plan.

The turns ratio can be measured with a low-amplitude sinusoidal test applied to one
winding while every other winding remains open. Measure the RMS terminal voltage on the
driven winding and on each open winding. The observed ratio gives

$$
\frac{N_2}{N_1}\approx\frac{V_2}{V_1}.
$$

The approximation includes small winding and magnetizing drops. Repeating the test in
both drive directions checks that the selected terminal pairs have been identified
correctly. A result far from the expected ratio can indicate a wrong tap, a series
connection error, a meter bandwidth limitation, or an input waveform that differs from
the stated amplitude convention.

The polarity test identifies winding starts and ends. Energize one winding at a safe
test level. Connect one lead from it to one lead from a second winding, then measure
the voltage between the two remaining leads. The reading is close to the sum of the two
winding voltages for a series-aiding connection and close to the difference for a
series-opposing connection. Label the dots or leads after the measurement. This result
is more reliable than inferring polarity from physical lead position on an unknown
component.

The open-circuit test estimates the shunt branch. The secondary is left open, the
primary receives rated-frequency voltage at its rated terminal value, and the test
records $V_1$, no-load current $I_0$, and input real power $P_0$. Because the no-load
current is usually small, primary copper loss is small. The measured real power then
approximates core loss at that voltage and frequency. The measured current can be
resolved into an in-phase loss component and a quadrature magnetizing component after
the measurement references and meter bandwidth are checked.

The short-circuit test estimates the series branch. The secondary is shorted with an
approved low-resistance path. A reduced primary voltage is increased only enough to
produce the stated test current. Record the applied voltage, current, and real power.
The applied voltage is small compared with rated voltage, so core flux and core loss
are small; the measurement is dominated by winding resistance and leakage impedance.
This test is part of a controlled lab or manufacturer procedure. A direct secondary
short on an energized mains transformer without current limiting can create destructive
current and is not a diagnostic method.

Load testing checks the values that matter to the intended application. Record primary
terminal voltage, secondary terminal voltage, primary current, secondary current,
input real power, output real power, waveform frequency, and load type. Increase the
load in planned steps up to the permitted rating. Compare the measured full-load voltage
with the regulation prediction and compare total measured loss with the temperature
rise. A discrepancy can indicate an unmodeled lead drop, an incorrect turns ratio, meter
phase error, saturation, a load whose waveform is not sinusoidal, or a transformer
outside the model's frequency range.

Measurements on high-impedance transformer circuits require attention to instrument
loading. A meter with finite input capacitance or resistance changes the secondary load,
especially in small-signal and high-frequency work. A current probe has its own phase
and amplitude error. A two-channel oscilloscope with a shared earth-referenced ground
can short two circuit nodes through its probe grounds if the test arrangement is not
isolated. Use differential probes or a documented isolated measurement arrangement
where the circuit requires it, and state probe scale factors in the measurement record.

## Worked calculations and design checks

> **Worked example (step-down turns, currents, and reflected load).** A 60 VA
> transformer has a 120 V RMS primary of $N_1=1200$ turns and a 12 V RMS secondary.
> Find the secondary turns, the rated primary and secondary currents, and the primary
> resistance seen when the secondary drives a $2.4\ \Omega$ resistor.
>
> The turns follow the voltage ratio:
> $$
> N_2=N_1\frac{V_2}{V_1}=1200\cdot\frac{12}{120}=120\ \text{turns}.
> $$
> The rated currents follow from the VA rating at each winding's voltage:
> $$
> I_{2,\mathrm{rated}}=\frac{60\ \mathrm{VA}}{12\ \mathrm V}=5.0\ \mathrm A,
> \qquad
> I_{1,\mathrm{rated}}=\frac{60\ \mathrm{VA}}{120\ \mathrm V}=0.50\ \mathrm A.
> $$
> A $2.4\ \Omega$ resistor draws $12/2.4=5.0\ \mathrm A$ at 12 V, so it is the rated
> load. With $a=N_2/N_1=0.10$, its reflected resistance is
> $$
> R_{\mathrm{in}}=\frac{R_L}{a^2}=\frac{2.4\ \Omega}{(0.10)^2}=240\ \Omega,
> $$
> and the source-side current $120/240=0.50\ \mathrm A$ matches the VA calculation. A
> real unit draws slightly more primary current because core and winding losses add to
> the input power.

> **Worked example (voltage regulation and efficiency).** A 24 V secondary reads 25.2 V
> at no load and 24.0 V while delivering 5.0 A to a resistive load, and a simultaneous
> input wattmeter reads 130 W. Find the voltage regulation and the efficiency.
>
> Regulation compares the no-load and full-load secondary voltages:
> $$
> \mathrm{VR}=\frac{V_{\mathrm{nl}}-V_{\mathrm{fl}}}{V_{\mathrm{fl}}}\times100\%
> =\frac{25.2-24.0}{24.0}\times100\%=5.0\%.
> $$
> The real output power is $P_{\mathrm{out}}=(24.0\ \mathrm V)(5.0\ \mathrm A)=120\ \mathrm W$,
> so with $130\ \mathrm W$ in, the loss is $10\ \mathrm W$ and
> $$
> \eta=\frac{P_{\mathrm{out}}}{P_{\mathrm{in}}}=\frac{120\ \mathrm W}{130\ \mathrm W}
> =0.923=92.3\%.
> $$
> Efficiency needs the real input power from the wattmeter, not the VA nameplate. If the
> same 5.0 A load were at 0.80 power factor with the secondary held at 24.0 V, the output
> real power would be $96\ \mathrm W$, while copper heating stayed near its 5.0 A value,
> so efficiency would fall.

> **Worked example (peak flux density and a frequency change).** A 120 V RMS, 60 Hz
> winding has 900 turns and an effective core area of
> $12\ \mathrm{cm^2}=1.20\times10^{-3}\ \mathrm{m^2}$. Neglecting the series drop, find
> the peak core flux and flux density, then the flux density if the same winding runs at
> 50 Hz and 120 V.
>
> The volts-per-turn relation $V_{\mathrm{rms}}=4.44fN\Phi_{\mathrm{pk}}$ gives
> $$
> \Phi_{\mathrm{pk}}=\frac{120}{(4.44)(60)(900)}=5.00\times10^{-4}\ \mathrm{Wb},
> \qquad
> B_{\mathrm{pk}}=\frac{\Phi_{\mathrm{pk}}}{A}
> =\frac{5.00\times10^{-4}}{1.20\times10^{-3}}=0.417\ \mathrm T.
> $$
> At fixed voltage, $\Phi_{\mathrm{pk}}\propto1/f$, so lowering the frequency to 50 Hz
> raises the flux density by the frequency ratio:
> $$
> B_{\mathrm{pk},50}=0.417\cdot\frac{60}{50}=0.500\ \mathrm T.
> $$
> Compare this against the core material's allowable flux density and loss data at the
> operating temperature and waveform: reducing frequency at the same voltage has driven
> the flux density up by 20 percent.

> **Worked example (reflecting a complex load).** An $8+j6\ \Omega$ secondary load
> connects to a transformer with $N_1/N_2=5$. Find the impedance the primary source
> sees.
>
> Reflection from secondary to primary multiplies the whole impedance by $(N_1/N_2)^2=25$:
> $$
> Z_{\mathrm{in}}=25(8+j6)\ \Omega=200+j150\ \Omega.
> $$
> Both rectangular parts scale by the same factor, so the angle is preserved: the source
> sees a magnitude $|Z_{\mathrm{in}}|=\sqrt{200^2+150^2}=250\ \Omega$ at
> $\arctan(150/200)=36.9^\circ$. Source current, primary series-drop, and input power
> factor follow from this reflected circuit; add the transformer's own series and shunt
> branches when the accuracy needed exceeds the ideal model.

> **Worked example (impedance-matching turns ratio).** A source of $50\ \Omega$ internal
> resistance drives an $800\ \Omega$ resistive load through an ideal transformer. Find
> the turns ratio that makes the reflected load equal the source resistance.
>
> The load reflects to the source side as $(N_1/N_2)^2R_L$; set it equal to $R_s$:
> $$
> 50=\left(\frac{N_1}{N_2}\right)^2(800),
> \qquad
> \frac{N_1}{N_2}=\sqrt{\frac{50}{800}}=\frac14.
> $$
> The secondary needs four times the primary turns, so the secondary voltage is four
> times the primary voltage while the source-side current is four times the load current.
> The match holds only in the ideal model; winding loss, bandwidth, insulation, and the
> secondary voltage and power ratings remain separate checks.

## Model selection and calculation checks

Transformer calculations require a model whose assumptions match the quantity being
estimated. A turns ratio is sufficient for an unloaded secondary-voltage estimate at
the intended frequency. It is insufficient for a full-load voltage estimate, a thermal
rating calculation, an inrush prediction, or a high-frequency isolation measurement.
Writing the active assumptions beside the calculation prevents a correct formula from
being applied to the wrong operating condition.

- **Ideal ratio model.** Use $V_2/V_1=N_2/N_1$ and the reciprocal current ratio for
  rated-voltage conversions, first-pass reflected-load calculations, and conceptual
  power flow. State that winding drops, magnetizing current, core loss, and leakage are
  being neglected.
- **Referred series model.** Add reflected winding resistance and leakage reactance
  when computing load voltage, short-circuit current, or regulation. The load phase
  must accompany this model because the series drop is a phasor difference.
- **Shunt-core model.** Include magnetizing reactance and core-loss resistance for
  no-load current, no-load real-power, and source-loading estimates. The values are
  valid near the voltage, frequency, and flux density at which they were measured.
- **Frequency-aware model.** Add distributed capacitance, AC winding resistance, and
  material loss data when operating outside the low-frequency range. A 60 Hz ratio
  measurement cannot establish a transformer’s bandwidth or common-mode behavior.

Units provide a quick consistency check. Turns ratios and regulation are dimensionless.
Impedance reflection multiplies ohms by a dimensionless squared ratio, leaving ohms.
The product of RMS volts and RMS amperes has units of VA. Real power requires the
relevant power factor or a direct time-domain wattmeter measurement. Flux is measured
in webers, flux density in teslas, and the core cross-sectional area in square metres.
A missing area conversion from square centimetres to square metres introduces a factor
of $10^4$ in a flux-density calculation.

Use the following record for a calculation or test report:

- terminal labels, dot locations, voltage polarity, and current-arrow directions;
- RMS, peak, or instantaneous amplitude convention for every stated voltage and
  current;
- operating frequency, source impedance, load impedance, and load power factor;
- temperature state and allowed VA or current rating;
- model parameters that were measured rather than assumed;
- the location of voltage probes and current sensors.

Several recurring errors can be found from these entries. A voltage ratio greater than
one paired with a secondary current ratio greater than one signals a failed reciprocal
check. A computed reflected resistance that moves in the wrong direction often means
that $N_1/N_2$ and $N_2/N_1$ were interchanged. A predicted secondary voltage above its
no-load value under a resistive load calls for a review of voltage references and series
drop signs. A current calculation above the VA rating requires a thermal and protective
review even if the turns ratio remains correct.

The equivalent circuit is also an experimental claim. Compare its predicted no-load
current with an open-circuit record, its predicted short-circuit voltage with the
series test, and its predicted loaded secondary voltage with a load test. Agreement at
one point does not validate the model at every voltage, frequency, or temperature.
Saturation, nonlinear core loss, lead resistance, and waveform distortion can alter the
parameters outside the test condition. Report the range where agreement was checked.