====== Block 09/10 — Transformers and Magnetic Coupling ======

===== Learning objectives =====
<callout>
After this 90-minute block, you can

  * explain how two coils can exchange energy by a common magnetic flux \(\Phi\).
  * use the ideal transformer equations
\[
\begin{align*}
\frac{\underline{U}_1^\phantom{O}}{\underline{U}_2^\phantom{O}}=\frac{N_1}{N_2}=n,
\qquad
\frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}
\end{align*}
\]
with a clear sign convention.
  * explain mutual inductance \(M\) using flux linkage and magnetic reluctance \(R_{\rm m}\).
  * distinguish **main flux**, **leakage flux**, **copper losses**, and **iron losses** in a real transformer.
  * refer secondary-side quantities to the primary side using \( \underline{U}'_2=n\underline{U}_2\), \( \underline{I}'_2=\frac{1}{n}\underline{I}_2\), \(R'_2=n^2R_2\), and \(X'_{2\sigma}=n^2X_{2\sigma}\).
  * interpret the no-load test and short-circuit test using the reduced equivalent circuit.
  * calculate short-circuit voltage \(\rm u_k\), continuous short-circuit current \(I_{\rm 1k}\), and an estimated initial peak short-circuit current.
  * connect transformer parameters to engineering applications in mechatronics and robotics, such as isolated power supplies, motor current measurement, welding transformers, and safety transformers.
</callout>

===== Preparation at Home =====

Well, again

  * read through the present chapter and write down anything you did not understand.
  * Repeat the EEE1 ideas of [[:electrical_engineering_and_electronics_1:block18|magnetic flux and induction]], [[:electrical_engineering_and_electronics_1:block19|magnetic circuits]], and [[:electrical_engineering_and_electronics_1:block20|inductance and magnetic energy]].
  * Repeat from EEE2 the use of [[block03|sinusoidal quantities]], [[block04|complex calculation]], and [[block06|complex power]].

For checking your understanding please do the quick checks in the exercise section.

===== 90-minute plan =====

  * **Warm-up (10 min):**
    * Where do transformers occur in robots and automation systems?
    * Recall: Faraday induction from EEE1 — a changing magnetic flux induces a voltage.
    * Recall: in AC analysis we use RMS phasors \(\underline{U}\), \(\underline{I}\), and impedances \(j\omega L\).

  * **Core concepts and derivations (55 min):**
    * Ideal transformer: common flux, voltage ratio, current ratio, power balance.
    * Mutual inductance: how flux from one coil links another coil.
    * Magnetic coupling with reluctance \(R_{\rm m}\).
    * Real transformer: winding resistances, leakage inductances, iron-loss resistance.
    * Reduced equivalent circuit: refer secondary quantities to the primary side.
    * No-load and short-circuit operation: what can be measured, what can be neglected.

  * **Practice (20 min):**
    * Quick ratio calculations for step-up and step-down transformers.
    * Short-circuit current calculation for a transformer used in an actuator supply.

  * **Wrap-up (5 min):**
    * Summary box: ideal transformer, mutual inductance, real transformer, reduced circuit, short-circuit parameters.
    * Common pitfalls checklist.

===== Conceptual overview =====
<callout icon="fa fa-lightbulb-o" color="blue">
  * A transformer is **not** a DC component. It needs a changing magnetic flux. In normal operation this is usually a sinusoidal flux created by AC voltage.
  * The transformer does not “create power”. Ideally, it trades voltage for current:
\[
\begin{align*}
\text{higher voltage} \quad \Longleftrightarrow \quad \text{lower current}
\end{align*}
\]
  * The link between the two windings is the magnetic field in the iron core. This continues directly from EEE1:
    * [[:electrical_engineering_and_electronics_1:block18|induction]] explains why a changing flux induces voltage.
    * [[:electrical_engineering_and_electronics_1:block19|magnetic circuits]] explains why the iron core guides the flux.
    * [[:electrical_engineering_and_electronics_1:block20|inductance]] explains how flux linkage and current are connected.
  * Mutual inductance \(M\) measures how strongly one coil “notices” the changing current in another coil.
  * A real transformer is almost ideal, but not quite:
    * \(R_1, R_2\): copper losses in the windings.
    * \(L_{1\sigma}, L_{2\sigma}\): leakage flux that does not couple both windings.
    * \(R_{\rm Fe}\): iron losses in the core.
    * \(L_{\rm H}\): main magnetizing inductance needed to create the main flux.
  * In engineering, transformer data such as \(\rm u_k\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
</callout>

~~PAGEBREAK~~ ~~CLEARFIX~~

===== Core content =====

<panel type="info" title="Short Review of the Flux">
In EEE1 we considered magnetic flux \(\Phi\), flux linkage / linked flux \(\Psi\), and induction.  \\
For one coil with \(N\) turns the flux linkage is

\[
\begin{align*}
\Psi = N\Phi .
\end{align*}
\]

Faraday's law gives

\[
\begin{align*}
u(t)=\frac{{\rm d}\Psi}{{\rm d}t}=N\frac{{\rm d}\Phi}{{\rm d}t}.
\end{align*}
\]

(Be aware of Lenz law: Here u(t) is the terminal voltage according to the chosen voltage reference arrow. The induced voltage $u_{\rm ind}$ according to Faraday–Lenz would have the opposite sign) \\
In sinusoidal steady state this becomes the phasor equation

\[
\begin{align*}
\underline{U}=j\omega\underline{\Psi}=j\omega N\underline{\Phi}.
\end{align*}
\]

This is the starting point for the transformer.
</panel>

==== Polarity and the dot convention ====

Before we start with the transformer, we have to look at a common convention for the orientation of two coils to each other. \\

<WRAP>
<panel type="default">
<imgcaption fig_direction_of_coupling|Dot convention: the dots indicate corresponding winding ends.></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:directionofcoupling.svg}}
</panel>
</WRAP>

<callout>
**Rule of thumb**

  * If both currents enter dotted terminals, the fluxes support each other.
  * If one current enters a dotted terminal and the other current leaves a dotted terminal, the fluxes oppose each other.
</callout>


==== Ideal single-phase transformer ====

{{drawio>electrical_engineering_and_electronics_2:ideal_transformer.svg}}

For an ideal transformer we assume:

  * both windings are linked by the same magnetic flux \( \underline{\Phi} \),
  * there is no leakage flux,
  * there are no winding resistances,
  * there are no iron losses,
  * the transformer stores no net energy over one period.

Let \(N_1\) be the number of turns of the primary winding and \(N_2\) the number of turns of the secondary winding.

\[
\begin{align*}
\underline{\Psi}_1 &= N_1\underline{\Phi},
&
\underline{U}_1 &= j\omega\underline{\Psi}_1
= j\omega N_1\underline{\Phi},
\\
\underline{\Psi}_2 &= N_2\underline{\Phi},
&
\underline{U}_2 &= j\omega\underline{\Psi}_2
= j\omega N_2\underline{\Phi}.
\end{align*}
\]

Dividing the two voltage equations gives the **turns ratio**

\[
\begin{align*}
\boxed{
\frac{\underline{U}_1^\phantom{O}}{\underline{U}_2^\phantom{O}}
=
\frac{N_1}{N_2}
=
n
}
\end{align*}
\]

with

\[
\begin{align*}
n=\frac{N_1}{N_2}.
\end{align*}
\]

<panel type="danger" title="Remember: ideal transformer ratios">
With the indicated reference arrows and a lossless transformer, the resulting complex power as a sum of input and output power $\underline{S}$ must be zero:

\[
\begin{align*}
\underline{S}_1+\underline{S}_2= \underline{U}_1\underline{I}^*_1+\underline{U}_2\underline{I}^*_2=0
\end{align*}
\]

and therefore

\[
\begin{align*}
\boxed{
\frac{\underline{I}_1}{\underline{I}_2}
=
-\frac{\underline{U}_2^\phantom{O}}{\underline{U}_1^\phantom{O}}
=
-\frac{N_2}{N_1}
=
-\frac{1}{n}
}
\end{align*}
\]

The minus sign is not a “loss”. It is caused by the chosen current arrows. \\ 
The primary side **absorbs** power while the secondary side **delivers** power to the load.
</panel>


<panel type="info" title="Analogy: gearbox for voltage and current">
An ideal transformer behaves like a lossless gearbox:

  * a gearbox can trade speed for torque,
  * a transformer can trade voltage for current.

For a step-down transformer:

\[
\begin{align*}
\text{lower voltage} \quad \Longleftrightarrow \quad \text{higher current}.
\end{align*}
\]

The power is ideally conserved, just as mechanical power is ideally conserved in a lossless gearbox.
</panel>

<panel type="info" title="Physical interpretation">
  * If \(N_2<N_1\), the transformer steps the voltage down: \(U_2<U_1\).
  * At the same time, the secondary current can be higher: \(I_2>I_1\).
  * This is useful in robotics power supplies: a mains-side transformer or isolated converter stage may reduce voltage while increasing available current for actuators.
</panel>


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  step-down transformer for a robot controller
#@TaskText_HTML@#

A transformer has \(N_1=800\) turns and \(N_2=80\) turns.  
The primary RMS voltage is \(U_1=230~{\rm V}\).

\[
\begin{align*}
n &= \frac{N_1}{N_2}
= \frac{800}{80}
=10,
\\
U_2 &= \frac{U_1}{n}
= \frac{230~{\rm V}}{10}
=23~{\rm V}.
\end{align*}
\]

If the secondary side supplies \(I_2=4~{\rm A}\), the ideal primary current magnitude is

\[
\begin{align*}
I_1 &= \frac{I_2}{n}
= \frac{4~{\rm A}}{10}
=0.4~{\rm A}.
\end{align*}
\]

The apparent power is equal on both sides:

\[
\begin{align*}
S_1 &= U_1I_1=230~{\rm V}\cdot 0.4~{\rm A}=92~{\rm VA},
\\
S_2 &= U_2I_2=23~{\rm V}\cdot 4~{\rm A}=92~{\rm VA}.
\end{align*}
\]

#@TaskEnd_HTML@#


~~PAGEBREAK~~ ~~CLEARFIX~~
==== Linked fluxes and mutual inductance ====

For a single coil we already know that its flux linkage $\Psi = N\Phi$ is proportional to the current $i$ through the coil

\[
\begin{align*}
\Psi=L i
\end{align*}
\]

\[
\begin{align*}
\underline{\Psi}=L \underline{I} .
\end{align*}
\]

For two coupled coils $1$ and $2$, each flux linkage $\underline{\Psi}_1 = N\underline{\Phi}_1$ and $\underline{\Psi}_2 = N\underline{\Phi}_2$ depend on both currents $\underline{I}_1$ and $\underline{I}_2$. \\

{{drawio>electrical_engineering_and_electronics_2:ideal_transformer01.svg}}
\\ \\
Not only the current through the coil generates a part of the flux linkage, but also the other coil provides a part for the flux linkage.

\[
\begin{align*}
\underline{\Psi}_1
&=
\underbrace{{\color{green}{L_{11}\underline{I}_1}}}_{\text{self-linkage of coil 1}}
+
\underbrace{{\color{blue}{M_{12}\underline{I}_2}}}_{\text{mutual linkage from coil 2}},
\\[4pt]
\underline{\Psi}_2
&=
\underbrace{{\color{blue}{M_{21}\underline{I}_1}}}_{\text{mutual linkage from coil 1}}
+
\underbrace{{\color{green}{L_{22}\underline{I}_2}}}_{\text{self-linkage of coil 2}}.
\end{align*}
\]

For most transformer calculations we use the symmetric case of the mutual inductances. \\
(this is true for passive, stationary, and reciprocal situations, like transformers, but not necessarily for motors or complex setups)

\[
\begin{align*}
{\color{blue}{M_{12}}}={\color{blue}{M_{21}}}={\color{blue}{M}}.
\end{align*}
\]

Often, the self-inductances are abbreviated:
\[
\begin{align*}
\color{green}{L_{11}} \rightarrow \color{green}{L_{1}} \\
\color{green}{L_{22}} \rightarrow \color{green}{L_{2}} \\
\end{align*}
\]

Then

\[
\begin{align*}
\boxed{
\begin{pmatrix}
\underline{\Psi}_1\\
\underline{\Psi}_2
\end{pmatrix}
=
\begin{pmatrix}
{\color{green}{L_{1}}} & {\color{blue}{M}}\\
{\color{blue}{M}} & {\color{green}{L_{2}}}
\end{pmatrix}
\begin{pmatrix}
\underline{I}_1\\
\underline{I}_2
\end{pmatrix}
}
\end{align*}
\]

and

\[
\begin{align*}
M=k\sqrt{L_{1}L_{2}}.
\end{align*}
\]

Here \(k\) is the coupling coefficient. In the shown transformer \(k\) is 1 since all flux from $1$ flows through $2$ and vice versa. \\
In reality that is not the case as explained in the next chapters.

<tabcaption tab_coupling_coefficient|Meaning of the coupling coefficient \(k\)>

^ Coupling coefficient ^ Interpretation ^ Typical example ^
| \(k=0\)         | no useful flux from one coil links the other coil  | coils far apart                                |
| \(0<k<1\)       | partial coupling                                   | wireless charger with air gap or misalignment  |
| \(k\approx 1\)  | almost all useful flux links both coils            | transformer with iron core                     |
</tabcaption>

\\ \\

The mutual inductance \(M\) answers the question:

> How much flux linkage appears in coil \(2\) when the current in coil \(1\) changes?

  * A large \(M\) means strong interaction.  
  * A small \(M\) means weak interaction.

\\

<panel type="info" title="Engineering example: wireless charging">
In wireless charging, the transmitter coil and receiver coil are separated by an air gap.  
The coupling coefficient \(k\) is much smaller than in a transformer with an iron core.

If the receiver is misaligned, less flux from the transmitter passes through it.  
Then \(M\) decreases, the induced voltage decreases, and the transmitted power decreases.
</panel>

<panel type="info" title="Analogy: shared bed sheet">
<WRAP right>{{electrical_engineering_and_electronics_2:20077953-370d-452d-acc9-7756a95ba26e.png?400}}</WRAP>
Imagine two people lying on a bed holding the same bed sheet at different positions.

  * \(L_{1}\): \\ how strongly winding 1 couples its current into the shared magnetic path, \\ like person 1 moving the sheet at their hand.
  * \(L_{2}\): \\ how strongly winding 2 couples its current into the shared magnetic path, \\ like person 2 moving the sheet at their hand.
  * \(M\): \\ how strongly the motion from one hand is felt at the other hand through the same sheet.

In transformer language, \(L_{1}\) and \(L_{2}\) describe each winding's connection to the shared main flux path. \\
The mutual inductance \(M\) describes the transfer between the two windings through this shared path.
</panel>


==== voltages by mutual inductances and resistances ====

For positive coupling, we get the following complex representation (since $u(t) = L\frac{{\rm d}i}{{\rm d}t} \: \longrightarrow  \:\underline{U}=j\omega L \underline{I}$):
\\ \\
\[
\begin{align*}
\underline{U}_1 &= R_1 \underline{I}_1 + {\color{green} {j\omega L_{1} \underline{I}_1 }} + {\color{blue}{j\omega M       \underline{I}_2 }},
\\[4pt]
\underline{U}_2 &= R_2 \underline{I}_2 + {\color{blue}  {j\omega M      \underline{I}_1 }} + {\color{green}{j\omega L_{2} \underline{I}_2 }}.
\end{align*}
\]
\\
For negative coupling, the sign of the \(M\)-term changes in the chosen equation system, see <imgref fig_positive_coupling> and <imgref fig_negative_coupling>.

<WRAP group>
<WRAP column half>
<panel type="default">
<imgcaption fig_positive_coupling|Positive coupling: currents enter corresponding dotted terminals.></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:poscoupling.svg}}
</panel>
</WRAP>

<WRAP column half>
<panel type="default">
<imgcaption fig_negative_coupling|Negative coupling: only one current enters a dotted terminal.></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:negcoupling.svg}}
</panel>
</WRAP>
</WRAP>

<panel type="info" title="Tunnel Analogy for AC circuits">
The dots are like matching openings for magnetic action. \\
A positive reference current (e.g. $i_1$) entering the dotted terminal of one winding produces a positive induced voltage (e.g. aligned with $u_2$) at the dotted terminal of the other winding. \\
With only a load $R_2$ connected to the secondary side, this voltage tends to drive current out of the dotted terminal into the load. \\ 
</panel>

~~PAGEBREAK~~ ~~CLEARFIX~~
==== Stray and Leakage ====

The image of the ideal transformer shall be consecutively developed to a more realistic transformer model. \\
To do so, we look at the situation of two coils near each other and expand this formula for the induced voltage. 

<WRAP>
<panel type="default">
<imgcaption fig_mutual_induction_two_coils|Mutual induction of two coils: only part of the flux created by coil \(1\) links coil \(2\).></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:mutualinductiontwocoils1.svg}}
</panel>
</WRAP>

The flux created by coil \(1\) can be split into

\[
\begin{align*}
\Phi_{11}
=
{\color{blue}{\Phi_{21}}}
+
{\color{orange}{\Phi_{1 \rm \sigma}}}.
\end{align*}
\]

  * \(\Phi_{11}\): total flux created by coil \(1\). 
  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\). 
  * \({\color{orange}{\Phi_{1 \rm \sigma}}}\): stray or leakage flux that does **not** link coil \(2\). The Greek letter sigma $\sigma$ is used to denote leakage or "stray" quantities.

For an example, we will have a look at the instantaneous voltage induced in coil \(2\):

\[
\begin{align*}
u_{2,\rm ind}(t)
=
\frac{{\rm d}{\color{blue}{\Psi_{21}}}}{{\rm d}t}
=
N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}.
\end{align*}
\]

The complex voltage induced in coil \(2\) is

\[
\begin{align*}
\underline{U}_{2}
=
j\omega {\color{blue}{ \underline{\Psi}_{21} }}
=
j\omega N_2 {\color{blue}{ \underline{\Phi}_{21} }}
.
\end{align*}
\]

\\ \\

<panel type="info" title="Analogies">

<fs large>**Analogy 1: two pendulums connected by a spring**</fs> \\
Imagine two pendulums connected by a weak spring.

  * If pendulum \(1\) moves, the spring can make pendulum \(2\) move as well.
  * A strong spring transfers the motion strongly.
  * A weak spring transfers the motion only weakly.
  * If the spring is missing, pendulum \(2\) does not react.

For coupled coils:

  * the changing motion corresponds to changing current,
  * the spring corresponds to the magnetic coupling,
  * the motion transferred to the second pendulum corresponds to the induced voltage,
  * weak coupling means that only a small part of the magnetic flux links both coils.

<fs large>**Analogy 2: a leaky magnetic pipe**</fs> \\
The magnetic core can be imagined as a pipe guiding magnetic flux.

  * A good iron core is like a wide, low-resistance pipe: most flux reaches the second coil.
  * A large air gap is like a narrow, difficult path: less flux reaches the second coil.
  * Leakage flux is like flow escaping through side paths: it belongs to the first coil but does not help the second coil.

This image is helpful for transformers, wireless charging coils, and current sensors.
</panel>

<panel type="info" title="Engineering examples">
  * **Transformer:** very strong coupling because the iron core guides most of the flux through both windings.
  * **Wireless charger:** weaker coupling because the flux must cross an air gap and the coils may be misaligned.
  * **Current transformer:** the measured conductor acts like a one-turn primary winding; the secondary winding detects the changing magnetic field.
  * **Relay coil near signal wiring:** unwanted coupling can induce noise voltages in nearby loops.
</panel>

==== Real transformer: leakage and losses ====

In a real transformer, not all flux links both windings.

  * The **main flux** \(\Phi_{\rm H}\) links primary and secondary winding. The 'H' denotes the German word "Haupt" (sometimes also given as 'm' for "main").
  * The **primary leakage flux** \(\Phi_{1\sigma}\) mainly links only the primary winding.
  * The **secondary leakage flux** \(\Phi_{2\sigma}\) mainly links only the secondary winding.

The leakage flux can be interpreted as an additional virtual branch in the mechanical setup, see <imgref fig_main_and_leakage_flux>.

<WRAP>
<panel type="default">
<imgcaption fig_main_and_leakage_flux|Main flux and leakage fluxes in a real transformer.></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:real_transformer.svg}}
</panel>
</WRAP>

The total main flux linking is given by (see <imgref fig_main_and_leakage_flux>)
\[
\begin{align*}
\underline{\Phi}_{1 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H} 
\\[4pt]
\underline{\Phi}_{2 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H} 
\end{align*}
\]

For the flux linkage also the leakage flux has to be considered:
\[
\begin{align*}
\underline{\Psi}_{1} =N_1 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{1\sigma})
\\[4pt]
\underline{\Psi}_{2} =N_2 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{2\sigma})
\end{align*}
\]

The real flux linkage equations become

\[
\begin{align*}
\underline{\Psi}_1 
&= \underbrace{{\color{blue  }{L_{1{\rm H}}\underline{I}_1 + M\underline{I}_2}}  }_{\text{main magnetic path}}
+  \underbrace{{\color{orange}{L_{1\sigma }\underline{I}_1}}                     }_{\text{primary leakage}   },
\\[4pt]
\underline{\Psi}_2 
&= \underbrace{{\color{blue  }{M\underline{I}_1 + L_{2{\rm H}}\underline{I}_2 }} }_{\text{main magnetic path}}
+  \underbrace{{\color{orange}{L_{2\sigma }\underline{I}_2}}                     }_{\text{secondary leakage} }.
\end{align*}
\]

Equivalently,

\[
\begin{align*}
\underline{\Psi}_1
&=
L_1\underline{I}_1+M\underline{I}_2,
&
L_1&=L_{1{\rm H}}+L_{1\sigma},
\\
\underline{\Psi}_2
&=
L_2\underline{I}_2+M\underline{I}_1,
&
L_2&=L_{2{\rm H}}+L_{2\sigma}.
\end{align*}
\]

The winding resistances \(R_1\) and \(R_2\) cause copper losses:

\[
\begin{align*}
P_{\rm Cu,1}=R_1I_1^2,
\qquad
P_{\rm Cu,2}=R_2I_2^2.
\end{align*}
\]

\[
\begin{align*}
\underline{U}_1
&=
\underbrace{{\color{red}{R_1\underline{I}_1}}}_{\text{primary copper drop}}
+
\underbrace{{\color{orange}{j\omega L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage drop}}
+
\underbrace{{\color{blue}{j\omega L_{1{\rm H}}\underline{I}_1+j\omega M\underline{I}_2}}}_{\text{main magnetic coupling}},
\\[6pt]
\underline{U}_2
&=
\underbrace{{\color{red}{R_2\underline{I}_2}}}_{\text{secondary copper drop}}
+
\underbrace{{\color{orange}{j\omega L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage drop}}
+
\underbrace{{\color{blue}{j\omega L_{2{\rm H}}\underline{I}_2+j\omega M\underline{I}_1}}}_{\text{main magnetic coupling}}.
\end{align*}
\]

Based, on the [[electrical_engineering_and_electronics_1:block19]] and [[electrical_engineering_and_electronics_1:block20]] of last semester, the inductances can be calculated by the reluctance $R_{\rm mFe}$ of the iron core and the number of turns $N_1$, $N_2$:

\[
\begin{align*}
\boxed{
L_{1 \rm H} = \frac{N_1^2}{R_{\rm mFe} }  \\
L_{2 \rm H} = \frac{N_2^2}{R_{\rm mFe} }  \\
M = \frac{N_1 N_2}{R_{\rm mFe} }  \\
}
\end{align*}
\]

<panel type="info" title="Color scheme for the equivalent equations">
In the previous formulas:

  * <fc #FF0000>red terms: \\ winding resistance and copper loss, which convert electrical energy into heat.</fc>
  * <fc #FFA500>orange terms: \\ leakage flux, which is unwanted but unavoidable, because magnetic field lines can also close through the surrounding air.</fc>
  * <fc #0000FF>blue terms: \\ useful main magnetic coupling, which is responsible for transformer action.</fc>
</panel>


<panel type="info" title="Analogy: useful road and side roads">
Think of the main flux as traffic on the useful road between two cities.  
Traffic on side roads still exists, but it does not help transport goods between the two cities.

  * main flux: useful road between primary and secondary winding,
  * leakage flux: side roads that return locally,
  * winding resistance: friction that turns useful energy into heat.
</panel>

~~PAGEBREAK~~ ~~CLEARFIX~~

==== Reduced equivalent circuit referred to the primary side ====

Since we know, that we can transform the current and voltage by the transformer, we can use this also to simplify the circuit.

<imgcaption fig_red_eq_circ01|equivalent circuit of a real transformer></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:reduced_eq_circuit.svg}}

For calculations it is convenient to move all secondary-side quantities to the primary side. \\
This is called **referring** or **transforming** the secondary side to the primary side.

\[
\begin{align*}
\boxed{
\underline{U}'_2=n\underline{U}_2
}
\qquad
\boxed{
\underline{I}'_2=\frac{1}{n}\underline{I}_2
}
\end{align*}
\]

The secondary resistance and leakage reactance are transformed by \(n^2\) since $R = U / I$:

\[
\begin{align*}
\boxed{
R'_2=n^2R_2
}
\qquad
\boxed{
X'_{2\sigma}=n^2X_{2\sigma}
}
\end{align*}
\]

<imgcaption fig_red_eq_circ02|reduced equivalent circuit of a real transformer></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:reduced_eq_circuit02.svg}}

In the reduced equivalent circuit:

  * \(R_1\) and \(R'_2\) model copper losses.
  * \(jX_{1\sigma}\) and \(jX'_{2\sigma}\) model leakage flux.
  * \(jX_{1{\rm H}}\) models the magnetizing branch.
  * \(R_{\rm Fe}\) is placed parallel to \(jX_{1{\rm H}}\) to model iron losses.
  * \(R_{\rm L}\) is a load resistor for the phasor diagram.

The phasor diagram is shown in <imgref fig_phasor_eq_circ01>. \\ 
  * The upper one only shows the main voltages and currents. \\ For an ohmic load current is parallel to $\underline{U}'_2$. However, $\underline{I}'_2$ is drawn as a current entering the transformer secondary port. Since the transformer delivers power to the load, $\underline{I}'_2$ is antiparallel to the load current and therefore antiparallel $\underline{U}'_2$.
  * The lower one shows all voltages and currents. 


<imgcaption fig_phasor_eq_circ01|phasor diagram of a real transformer></imgcaption>
{{drawio>electrical_engineering_and_electronics_2:phasor_real_trafo01.svg}}


<panel type="info" title="Why the reduced circuit is useful">
Once all quantities are referred to one side, the transformer can be calculated like an AC network with impedances.  \\
This uses the same method as [[block04|complex network calculation]]: replace components by impedances and apply Kirchhoff's laws.
</panel>

==== No-load operation of the real transformer ====

No-load operation means that the secondary side is open:

\[
\begin{align*}
\underline{I}_2=0.
\end{align*}
\]

The primary side still draws a small no-load current \(\underline{I}_{10}\). This current has two parts:

\[
\begin{align*}
\underline{I}_{10}
=
\underline{I}_{\rm Fe}
+
\underline{I}_{\rm m}.
\end{align*}
\]

  * \(\underline{I}_{\rm Fe}\): current through \(R_{\rm Fe}\), in phase with voltage, represents iron losses.
  * \(\underline{I}_{\rm m}\): magnetizing current through \(jX_{1{\rm H}}\), approximately \(90^\circ\) lagging.

<WRAP>
<panel type="default">
<imgcaption fig_no_load_phasor|No-load phasor diagram: the no-load current is the sum of iron-loss current and magnetizing current.></imgcaption>
{{drawio>block09_no_load_phasor_diagram.svg}}
</panel>
</WRAP>

The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{\rm u}\):

\[
\boxed{
\begin{align*}
\ddot{\rm u}
=
\frac{\text{higher voltage}}{\text{lower voltage}}
\bigg|_{\rm no~load}.
\end{align*}
}
\]

For a step-down transformer:

\[
\begin{align*}
\ddot{\rm u}
=
\frac{U_{1{\rm N}}}{U_{20}}.
\end{align*}
\]

Here \(U_{1{\rm N}}\) is the rated primary voltage and \(U_{20}\) is the open-circuit secondary voltage.

<callout>
Because of real voltage drops and magnetizing effects,

\[
\begin{align*}
\ddot{\rm u}\neq n,
\end{align*}
\]

but for many practical transformers

\[
\begin{align*}
\ddot{\rm u}\approx n.
\end{align*}
\]
</callout>

==== Short-circuit operation of the real transformer ====

In the short-circuit test, the secondary side is shorted, so $\underline{U}_2=0$. \\
The primary voltage $\underline{U}_1$ is increased only until rated current flows. \\ 
Because the short-circuit impedance is small, this requires only a small fraction of the rated primary voltage. \\
Therefore the main flux and the magnetizing current are small, so the magnetizing branch can usually be neglected.


\[
\begin{align*}
\Big|\; jX_{1{\rm H}} \; || \;R_{\rm Fe} \;\Big|
\;\; \gg \;\;
\Big|\; jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 \;\Big| .
\end{align*}
\]

<WRAP>
<panel type="default">
<imgcaption fig_short_circuit_equivalent|Short-circuit equivalent circuit of a real transformer.></imgcaption>
{{drawio>block09_short_circuit_equivalent_circuit.svg}}
</panel>
</WRAP>

This gives the short-circuit equivalent circuit with

\[
\begin{align*}
\boxed{
R_{\rm k}=R_1+R'_2
}
\qquad
\boxed{
X_{\rm k}=X_{1\sigma}+X'_{2\sigma}
}
\end{align*}
\]

and

\[
\begin{align*}
\underline{Z}_{\rm k}
=
R_{\rm k}+jX_{\rm k}.
\end{align*}
\]



<panel type="info" title="Definition: rated short-circuit voltage">
The **rated short-circuit voltage** \(U_{1{\rm k}}\) is the primary voltage that must be applied while the secondary side is shorted so that rated primary current \(I_{1{\rm N}}\) flows.

As a relative value:

\[
\begin{align*}
\boxed{
{\rm u_k}
=
\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
}
\end{align*}
\]

  * Small \(\rm u_k\) means: small internal impedance. \\ When a short-circuit fault on the secondary side happens, the input current can get very high.  
  * Large \(\rm u_k\) means: high internal impedance. \\ Stronger current limitation, but also larger voltage drop under load.
</panel>

The continuous short-circuit current for rated primary voltage is

\[
\begin{align*}
\boxed{
I_{1{\rm k}}
=
\frac{U_{1{\rm N}}}{U_{1{\rm k}}}\cdot I_{1{\rm N}}
=
I_{1{\rm N}}\cdot \frac{100~\%}{{\rm u_k}}
}
\end{align*}
\]

where \(\rm u_k\) is inserted as a percentage value.

<WRAP column 100%>
<panel type="info" title="Definition: approximation for the first peak current">
For a first approximation of the maximum instantaneous current the following formula can be used:

\[
\begin{align*}
\boxed{
i_{{\rm peak}}
=
2.54 \cdot I_{1 \rm k}
}
\end{align*}
\]

</panel>
</WRAP>


==== Real transformer under load ====

Under load, the short-circuit equivalent circuit is often sufficient for engineering estimates.

\[
\begin{align*}
\underline{U}_{\rm k}
=
\left(R_{\rm k}+jX_{\rm k}\right)\underline{I}_1.
\end{align*}
\]

<WRAP>
<panel type="default">
<imgcaption fig_kapp_triangle|Kapp triangle: approximate voltage drop under load using R_k*I and X_k*I.></imgcaption>
{{drawio>block09_kapp_triangle.svg}}
</panel>
</WRAP>

This voltage drop is subtracted vectorially from the primary-side voltage relation. \\ 

\[
\begin{align*}
\boxed{
\underline{U}_1
-
\underline{U}_{\rm k}
=
\underline{U}'_2
=
n \underline{U}_2
}
\end{align*}
\]

If the magnetizing branch is neglected for the load calculation, then the series current relation is approximately $\underline{I}'_2 \approx -\underline{I}_1$. Therefore the ideal current transformation can still be used as an approximation.

\[
\begin{align*}
\boxed{
\frac{\underline{I}_1}{\underline{I}_2}
=
-\frac{N_2}{N_1}
=
-\frac{1}{n}
}
\end{align*}
\]


The Kapp triangle (see yellow triangle in <imgref fig_kapp_triangle>) represents the formula for $\underline{U}_{\rm k}$. 

<panel>
<imgcaption fig_kapp_triangle02 | animation of Kapp triangle>
</imgcaption> \\
<collapse id="geogebra_kapp" collapsed="true"><well>
{{url>https://www.geogebra.org/material/iframe/id/nswtdkpk/width/800/height/800/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/true/ld/false/sri/true/at/preferhtml5 800,800 noborder}}

</well></collapse>

<collapse id="geogebra_kapp" collapsed="false">
<button type="warning" collapse="geogebra_kapp">press here for the animation</button>
</collapse>
</panel>


<panel type="info" title="Engineering use: voltage regulation">
In a robot with motors, the supply transformer may show a lower output voltage during high acceleration because the motor currents increase.  
The Kapp triangle helps estimate this voltage drop. This is important for:

  * selecting transformer size,
  * checking whether the DC link after a rectifier remains high enough,
  * designing fuses and protective devices,
  * avoiding undervoltage resets in control electronics.
</panel>

==== Construction types and cooling ====

Transformer behavior is influenced by construction.

<WRAP>
<panel type="default">
<imgcaption fig_transformer_types|Transformer types></imgcaption>
{{drawio>block09_core_type_transformer.svg}}
</panel>
</WRAP>

Cooling types:

  * **Dry-type transformer:** air cooling, often used inside machines or buildings at lower and medium power.
  * **Oil transformer:** oil provides insulation and heat transfer, typical for higher power.
\\ 
<panel type="info" title="Mechatronics examples">
  * **Isolating transformer:** safe diagnostic supply for laboratory setups.
  * **Control transformer:** supplies \(24~{\rm V}\) or similar low-voltage control circuits.
  * **Current transformer:** measures large motor currents with galvanic isolation.
  * **Welding transformer:** intentionally high short-circuit voltage and current limitation for welding processes.
</panel>

==== Typical technical transformer data ====

<tabcaption tab_transformer_types|Typical transformer types, short-circuit voltage, and secondary voltage>

^ Name / use ^ Typical \({\rm u_k}\) ^ Secondary voltage \(U_2\) ^ Important note ^
| Power transformer      | \(4\ldots 12~\%\)   | application-dependent                    | low voltage drop, high fault currents possible            |
| Isolating transformer  | \(\approx 10~\%\)   | max. \(250~{\rm V}\)                     | galvanic isolation for safety and measurement             |
| Toy transformer        | \(\approx 20~\%\)   | max. \(24~{\rm V}\)                      | current limitation is desired                             |
| Doorbell transformer   | \(\approx 40~\%\)   | max. \(12~{\rm V}\), often several taps  | simple robust low-voltage supply                          |
| Ignition transformer   | \(\approx 100~\%\)  | \(\leq 14~{\rm kV}\)                     | high voltage, limited current                             |
| Welding transformer    | \(\approx 100~\%\)  | max. \(70~{\rm V}\)                      | large current, strong current limitation                  |
| Voltage transformer    | \(<1~\%\)           | \(100~{\rm V}\)                          | operate with high load resistance, approximately no-load  |
| Current transformer    | \(100~\%\)          | \(0~{\rm V}\) ideal secondary voltage    | operate with low burden, approximately short-circuit      |
</tabcaption>

\\ \\

<WRAP column 100%>
<panel type="danger" title="Important safety note: current transformers">
A current transformer secondary must not be opened while primary current flows.  \\
If the secondary circuit is open, the transformer tries to maintain the magnetic balance and can generate dangerous high voltages.
</panel>
</WRAP>

===== Exercises =====

#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: ideal transformer voltage and current ratio
#@TaskText_HTML@#

A transformer has $N_1=1200$ turns and $N_2=300$ turns.  \\
The primary RMS voltage is $U_1=230~{\rm V}$.  \\
The secondary side supplies a load current $I_2=2.0~{\rm A}$.

1. Calculate the turns ratio $n$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2001~@#
<WRAP leftalign>
The turns ratio of an ideal transformer is defined as:
\begin{align*}
n=\frac{N_1}{N_2}
\end{align*}

Insert the given values:
\begin{align*}
n
&=
\frac{1200}{300}
\\
&=
4
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2001~@#
\begin{align*}
n=4
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate the ideal secondary voltage $U_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2002~@#
<WRAP leftalign>
For an ideal transformer, the voltage ratio follows the turns ratio:
\begin{align*}
n=\frac{U_1}{U_2}
\end{align*}

Therefore:
\begin{align*}
U_2
&=
\frac{U_1}{n}
\\
&=
\frac{230~{\rm V}}{4}
\\
&=
57.5~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2002~@#
\begin{align*}
U_2=57.5~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate the magnitude of the ideal primary current $I_1$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2003~@#
<WRAP leftalign>
For the ideal transformer, the current ratio is inverse to the voltage ratio:
\begin{align*}
I_1=\frac{I_2}{n}
\end{align*}

Insert the values:
\begin{align*}
I_1
&=
\frac{2.0~{\rm A}}{4}
\\
&=
0.50~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2003~@#
\begin{align*}
I_1=0.50~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. State whether this is a step-up or step-down transformer.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2004~@#
<WRAP leftalign>
Compare the primary and secondary voltages:
\begin{align*}
U_1 &= 230~{\rm V}
\\
U_2 &= 57.5~{\rm V}
\end{align*}

Since
\begin{align*}
U_2<U_1
\end{align*}

the transformer reduces the voltage.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2004~@#
The transformer is a step-down transformer.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: mutual inductance from reluctance
#@TaskText_HTML@#

Two coils are wound on the same ideal magnetic core.  
The main magnetic reluctance is

\begin{align*}
R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}.
\end{align*}

The number of turns is $N_1=500$ and $N_2=100$.

1. Calculate $L_{1{\rm H}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2005~@#
<WRAP leftalign>
The main-flux inductance of coil 1 is:
\begin{align*}
L_{1{\rm H}}
=
\frac{N_1^2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
L_{1{\rm H}}
&=
\frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
\\
&=
0.125~{\rm H}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2005~@#
\begin{align*}
L_{1{\rm H}}=0.125~{\rm H}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate $L_{2{\rm H}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2006~@#
<WRAP leftalign>
The main-flux inductance of coil 2 is:
\begin{align*}
L_{2{\rm H}}
=
\frac{N_2^2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
L_{2{\rm H}}
&=
\frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
\\
&=
0.0050~{\rm H}
\\
&=
5.0~{\rm mH}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2006~@#
\begin{align*}
L_{2{\rm H}}
=
0.0050~{\rm H}
=
5.0~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate $M$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2007~@#
<WRAP leftalign>
The mutual inductance is:
\begin{align*}
M
=
\frac{N_1N_2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
M
&=
\frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
\\
&=
0.025~{\rm H}
\\
&=
25~{\rm mH}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2007~@#
\begin{align*}
M
=
0.025~{\rm H}
=
25~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Check whether the units are correct.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2008~@#
<WRAP leftalign>
The reluctance is given in:
\begin{align*}
[R_{\rm mH}]=\frac{1}{\rm H}
\end{align*}

The number of turns is dimensionless. Therefore:
\begin{align*}
\left[
\frac{N^2}{R_{\rm mH}}
\right]
=
\frac{1}{1/{\rm H}}
=
{\rm H}
\end{align*}

The same argument applies to the mutual inductance:
\begin{align*}
\left[
\frac{N_1N_2}{R_{\rm mH}}
\right]
=
{\rm H}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2008~@#
The unit is correct because
\begin{align*}
\frac{1}{1/{\rm H}}={\rm H}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Mutual inductance and leakage from a magnetic path
#@TaskText_HTML@#

Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance

\begin{align*}
R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}.
\end{align*}

The numbers of turns are

\begin{align*}
N_1=400,
\qquad
N_2=100.
\end{align*}

The leakage inductances are

\begin{align*}
L_{1\sigma}=4.0~{\rm mH},
\qquad
L_{2\sigma}=0.30~{\rm mH}.
\end{align*}

1. Calculate $L_{1{\rm H}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2009~@#
<WRAP leftalign>
The main-flux self-inductance of coil 1 is:
\begin{align*}
L_{1{\rm H}}
=
\frac{N_1^2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
L_{1{\rm H}}
&=
\frac{400^2}{1.6\cdot 10^6~1/{\rm H}}
\\
&=
0.100~{\rm H}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2009~@#
\begin{align*}
L_{1{\rm H}}=0.100~{\rm H}=100~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate $L_{2{\rm H}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2010~@#
<WRAP leftalign>
The main-flux self-inductance of coil 2 is:
\begin{align*}
L_{2{\rm H}}
=
\frac{N_2^2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
L_{2{\rm H}}
&=
\frac{100^2}{1.6\cdot 10^6~1/{\rm H}}
\\
&=
0.00625~{\rm H}
\\
&=
6.25~{\rm mH}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2010~@#
\begin{align*}
L_{2{\rm H}}
=
0.00625~{\rm H}
=
6.25~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate $M$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2011~@#
<WRAP leftalign>
The mutual inductance is:
\begin{align*}
M
=
\frac{N_1N_2}{R_{\rm mH}}
\end{align*}

Insert the values:
\begin{align*}
M
&=
\frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}}
\\
&=
0.025~{\rm H}
\\
&=
25~{\rm mH}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2011~@#
\begin{align*}
M
=
0.025~{\rm H}
=
25~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Calculate the total self-inductances $L_1$ and $L_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2012~@#
<WRAP leftalign>
The total self-inductance is the sum of main-flux inductance and leakage inductance.

For coil 1:
\begin{align*}
L_1
&=
L_{1{\rm H}}+L_{1\sigma}
\\
&=
100~{\rm mH}+4.0~{\rm mH}
\\
&=
104~{\rm mH}
\end{align*}

For coil 2:
\begin{align*}
L_2
&=
L_{2{\rm H}}+L_{2\sigma}
\\
&=
6.25~{\rm mH}+0.30~{\rm mH}
\\
&=
6.55~{\rm mH}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2012~@#
\begin{align*}
L_1 &= 104~{\rm mH}
\\
L_2 &= 6.55~{\rm mH}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

5. Calculate the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2013~@#
<WRAP leftalign>
The coupling coefficient is:
\begin{align*}
k
=
\frac{M}{\sqrt{L_1L_2}}
\end{align*}

Insert the values in henry:
\begin{align*}
k
&=
\frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}}
\\
&\approx
0.96
\end{align*}

The coupling is strong, but not ideal, because leakage inductances are present.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2013~@#
\begin{align*}
k\approx 0.96
\end{align*}

The coupling is strong, but not ideal.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: referring secondary quantities to the primary side
#@TaskText_HTML@#

A transformer has $n=5$.  
The secondary winding resistance is $R_2=0.20~\Omega$ and the secondary leakage reactance is $X_{2\sigma}=0.35~\Omega$.

Calculate the values $R'_2$ and $X'_{2\sigma}$ referred to the primary side.

1. Calculate $R'_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2014~@#
<WRAP leftalign>
When a resistance is referred from the secondary side to the primary side, it is multiplied by $n^2$:
\begin{align*}
R'_2
=
n^2R_2
\end{align*}

Insert the values:
\begin{align*}
R'_2
&=
5^2\cdot 0.20~\Omega
\\
&=
25\cdot 0.20~\Omega
\\
&=
5.0~\Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2014~@#
\begin{align*}
R'_2=5.0~\Omega
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate $X'_{2\sigma}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2015~@#
<WRAP leftalign>
The secondary leakage reactance is also referred to the primary side by multiplying with $n^2$:
\begin{align*}
X'_{2\sigma}
=
n^2X_{2\sigma}
\end{align*}

Insert the values:
\begin{align*}
X'_{2\sigma}
&=
5^2\cdot 0.35~\Omega
\\
&=
25\cdot 0.35~\Omega
\\
&=
8.75~\Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2015~@#
\begin{align*}
X'_{2\sigma}=8.75~\Omega
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Check the unit.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2016~@#
<WRAP leftalign>
The turns ratio $n$ is dimensionless:
\begin{align*}
[n]=1
\end{align*}

Therefore, multiplying by $n^2$ does not change the unit:
\begin{align*}
[R'_2] &= \Omega
\\
[X'_{2\sigma}] &= \Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2016~@#
The unit remains $\Omega$, because $n$ is dimensionless.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: short-circuit voltage and fault current
#@TaskText_HTML@#

A transformer has a rated primary current $I_{1{\rm N}}=10~{\rm A}$ and a short-circuit voltage ${\rm u_k}=5~\%$.

1. Calculate the continuous short-circuit current $I_{1{\rm k}}$ when rated primary voltage is applied.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2017~@#
<WRAP leftalign>
The short-circuit current can be estimated from the rated current and the relative short-circuit voltage:
\begin{align*}
I_{1{\rm k}}
=
I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
\end{align*}

Insert the values:
\begin{align*}
I_{1{\rm k}}
&=
10~{\rm A}\cdot \frac{100~\%}{5~\%}
\\
&=
200~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2017~@#
\begin{align*}
I_{1{\rm k}}=200~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Estimate the initial peak short-circuit current $i_{\rm p}$ using $i_{\rm p}\approx 2.54 I_{1{\rm k}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2018~@#
<WRAP leftalign>
The initial peak current is estimated by:
\begin{align*}
i_{\rm p}
\approx
2.54\cdot I_{1{\rm k}}
\end{align*}

Insert the continuous short-circuit current:
\begin{align*}
i_{\rm p}
&\approx
2.54\cdot 200~{\rm A}
\\
&=
508~{\rm A}
\end{align*}

The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2018~@#
\begin{align*}
i_{\rm p}\approx 508~{\rm A}
\end{align*}

Protection devices must be selected accordingly.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Longer exercise: transformer equivalent circuit for an actuator supply
#@TaskText_HTML@#

A single-phase transformer supplies an actuator driver.  
Rated data and equivalent circuit data are:

\begin{align*}
U_{1{\rm N}}&=230~{\rm V},
&
U_{2{\rm N}}&=23~{\rm V},
&
I_{2{\rm N}}&=5.0~{\rm A},
\\
R_1&=1.2~\Omega,
&
X_{1\sigma}&=1.8~\Omega,
\\
R_2&=0.012~\Omega,
&
X_{2\sigma}&=0.018~\Omega.
\end{align*}

Assume $n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}$. The magnetizing branch is neglected for the loaded operating point.

1. Calculate $n$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2019~@#
<WRAP leftalign>
The turns ratio is estimated from the rated voltages:
\begin{align*}
n
=
\frac{U_{1{\rm N}}}{U_{2{\rm N}}}
\end{align*}

Insert the values:
\begin{align*}
n
&=
\frac{230~{\rm V}}{23~{\rm V}}
\\
&=
10
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2019~@#
\begin{align*}
n=10
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Refer $R_2$ and $X_{2\sigma}$ to the primary side.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2020~@#
<WRAP leftalign>
Secondary quantities are referred to the primary side by multiplying them with $n^2$:
\begin{align*}
R'_2 &= n^2R_2
\\
X'_{2\sigma} &= n^2X_{2\sigma}
\end{align*}

With $n=10$:
\begin{align*}
R'_2
&=
10^2\cdot 0.012~\Omega
\\
&=
1.2~\Omega
\\[4pt]
X'_{2\sigma}
&=
10^2\cdot 0.018~\Omega
\\
&=
1.8~\Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2020~@#
\begin{align*}
R'_2 &= 1.2~\Omega
\\
X'_{2\sigma} &= 1.8~\Omega
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate $R_{\rm k}$ and $X_{\rm k}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2021~@#
<WRAP leftalign>
The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities:
\begin{align*}
R_{\rm k}
&=
R_1+R'_2
\\
X_{\rm k}
&=
X_{1\sigma}+X'_{2\sigma}
\end{align*}

Insert the values:
\begin{align*}
R_{\rm k}
&=
1.2~\Omega+1.2~\Omega
\\
&=
2.4~\Omega
\\[4pt]
X_{\rm k}
&=
1.8~\Omega+1.8~\Omega
\\
&=
3.6~\Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2021~@#
\begin{align*}
R_{\rm k} &= 2.4~\Omega
\\
X_{\rm k} &= 3.6~\Omega
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2022~@#
<WRAP leftalign>
For an ideal transformer, the primary current magnitude is:
\begin{align*}
I_{1{\rm N}}
=
\frac{I_{2{\rm N}}}{n}
\end{align*}

Insert the values:
\begin{align*}
I_{1{\rm N}}
&=
\frac{5.0~{\rm A}}{10}
\\
&=
0.50~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2022~@#
\begin{align*}
I_{1{\rm N}}=0.50~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

5. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2023~@#
<WRAP leftalign>
First calculate the magnitude of the short-circuit impedance:
\begin{align*}
|\underline{Z}_{\rm k}|
=
\sqrt{R_{\rm k}^2+X_{\rm k}^2}
\end{align*}

Insert the values:
\begin{align*}
|\underline{Z}_{\rm k}|
&=
\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2}
\\
&=
4.33~\Omega
\end{align*}

Now calculate the internal voltage drop:
\begin{align*}
U_{\rm k}
&\approx
|\underline{Z}_{\rm k}|I_{1{\rm N}}
\\
&=
4.33~\Omega\cdot 0.50~{\rm A}
\\
&=
2.17~{\rm V}
\end{align*}

This is a primary-side voltage drop. On the secondary side:
\begin{align*}
\frac{2.17~{\rm V}}{10}
=
0.217~{\rm V}
\end{align*}

For a $23~{\rm V}$ actuator supply this is small but not zero.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2023~@#
\begin{align*}
|\underline{Z}_{\rm k}| &= 4.33~\Omega
\\
U_{\rm k} &\approx 2.17~{\rm V}
\end{align*}

Secondary-side equivalent:
\begin{align*}
U_{\rm k,2}\approx 0.217~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Short-circuit voltage from the transformer impedance
#@TaskText_HTML@#

A transformer has the rated primary data

\begin{align*}
U_{1{\rm N}}=230~{\rm V},
\qquad
I_{1{\rm N}}=2.0~{\rm A}.
\end{align*}

The short-circuit equivalent impedance referred to the primary side is

\begin{align*}
R_{\rm k}=1.5~\Omega,
\qquad
X_{\rm k}=4.0~\Omega.
\end{align*}


1. Calculate $|\underline{Z}_{\rm k}|$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2024~@#
<WRAP leftalign>
The short-circuit impedance magnitude is:
\begin{align*}
|\underline{Z}_{\rm k}|
=
\sqrt{R_{\rm k}^2+X_{\rm k}^2}
\end{align*}

Insert the values:
\begin{align*}
|\underline{Z}_{\rm k}|
&=
\sqrt{(1.5~\Omega)^2+(4.0~\Omega)^2}
\\
&=
4.27~\Omega
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2024~@#
\begin{align*}
|\underline{Z}_{\rm k}|=4.27~\Omega
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate the rated short-circuit voltage $U_{1{\rm k}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2025~@#
<WRAP leftalign>
The primary voltage required to drive rated current through the short-circuited transformer is:
\begin{align*}
U_{1{\rm k}}
=
|\underline{Z}_{\rm k}|I_{1{\rm N}}
\end{align*}

Insert the values:
\begin{align*}
U_{1{\rm k}}
&=
4.27~\Omega\cdot 2.0~{\rm A}
\\
&=
8.54~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2025~@#
\begin{align*}
U_{1{\rm k}}=8.54~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate the relative short-circuit voltage ${\rm u_k}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2026~@#
<WRAP leftalign>
The relative short-circuit voltage is:
\begin{align*}
{\rm u_k}
=
\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
\end{align*}

Insert the values:
\begin{align*}
{\rm u_k}
&=
\frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\%
\\
&=
3.71~\%
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2026~@#
\begin{align*}
{\rm u_k}=3.71~\%
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Calculate the prospective continuous short-circuit current $I_{1{\rm k}}$ for rated primary voltage.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2027~@#
<WRAP leftalign>
The prospective continuous short-circuit current is:
\begin{align*}
I_{1{\rm k}}
=
I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
\end{align*}

Insert the values:
\begin{align*}
I_{1{\rm k}}
&=
2.0~{\rm A}\cdot \frac{100}{3.71}
\\
&=
53.9~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2027~@#
\begin{align*}
I_{1{\rm k}}=53.9~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

5. Calculate the approximate first peak current $i_{\rm peak}\approx 2.54 I_{1{\rm k}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2028~@#
<WRAP leftalign>
The approximate first peak current is:
\begin{align*}
i_{\rm peak}
\approx
2.54 I_{1{\rm k}}
\end{align*}

Insert the short-circuit current:
\begin{align*}
i_{\rm peak}
&\approx
2.54\cdot 53.9~{\rm A}
\\
&=
137~{\rm A}
\end{align*}

Even though the rated current is only $2.0~{\rm A}$, a short-circuit fault could lead to a much larger current until protection reacts.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2028~@#
\begin{align*}
i_{\rm peak}\approx 137~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Voltage drop under load using the Kapp approximation
#@TaskText_HTML@#

A transformer has the turns ratio $n=10$.

The short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.4~\Omega$, $X_{\rm k}=3.6~\Omega$. \\
The secondary load current is $I_2=4.0~{\rm A}$. The load has the power factor $\cos\varphi=0.8$ and is inductive.

Estimate the voltage drop on the secondary side using

\begin{align*}
\Delta U_1
\approx
I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
\end{align*}

and

\begin{align*}
\Delta U_2\approx \frac{\Delta U_1}{n}.
\end{align*}

1. Calculate the primary current magnitude $I_1$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2029~@#
<WRAP leftalign>
The primary current magnitude is approximately:
\begin{align*}
I_1
=
\frac{I_2}{n}
\end{align*}

Insert the values:
\begin{align*}
I_1
&=
\frac{4.0~{\rm A}}{10}
\\
&=
0.40~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2029~@#
\begin{align*}
I_1=0.40~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Determine $\sin\varphi$ for the inductive load.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2030~@#
<WRAP leftalign>
For an inductive load with $\cos\varphi=0.8$:
\begin{align*}
\sin\varphi
=
\sqrt{1-\cos^2\varphi}
\end{align*}

Insert the value:
\begin{align*}
\sin\varphi
&=
\sqrt{1-0.8^2}
\\
&=
0.6
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2030~@#
\begin{align*}
\sin\varphi=0.6
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Estimate the primary-side voltage drop $\Delta U_1$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2031~@#
<WRAP leftalign>
Use the Kapp approximation:
\begin{align*}
\Delta U_1
&\approx
I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
\end{align*}

Insert the values:
\begin{align*}
\Delta U_1
&\approx
0.40~{\rm A}
\left(
2.4~\Omega\cdot 0.8
+
3.6~\Omega\cdot 0.6
\right)
\\
&=
0.40~{\rm A}
\left(
1.92~\Omega+2.16~\Omega
\right)
\\
&=
1.63~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2031~@#
\begin{align*}
\Delta U_1\approx 1.63~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Estimate the secondary-side voltage drop $\Delta U_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2032~@#
<WRAP leftalign>
The secondary-side voltage drop is:
\begin{align*}
\Delta U_2
\approx
\frac{\Delta U_1}{n}
\end{align*}

Insert the values:
\begin{align*}
\Delta U_2
&\approx
\frac{1.63~{\rm V}}{10}
\\
&=
0.163~{\rm V}
\end{align*}

The secondary voltage decreases by approximately $0.16~{\rm V}$ for this operating point.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2032~@#
\begin{align*}
\Delta U_2\approx 0.163~{\rm V}
\end{align*}

The secondary voltage decreases by approximately $0.16~{\rm V}$.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Why the magnetizing branch can be neglected in the short-circuit test
#@TaskText_HTML@#

A transformer has the rated primary voltage $U_{1{\rm N}}=230~{\rm V}$.
Its rated primary current is $I_{1{\rm N}}=3.0~{\rm A}$.

At rated voltage and no-load operation, the magnetizing current is approximately $I_{\rm m,N}=0.12~{\rm A}$.
The short-circuit voltage is ${\rm u_k}=6.0~\%$.

Assume that the magnetizing current is approximately proportional to the applied voltage.
\\ \\ \\
1. Calculate the short-circuit test voltage $U_{1{\rm k}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2033~@#
<WRAP leftalign>
The short-circuit test voltage is:
\begin{align*}
U_{1{\rm k}}
=
\frac{\rm u_k}{100~\%}\cdot U_{1{\rm N}}
\end{align*}

Insert the values:
\begin{align*}
U_{1{\rm k}}
&=
0.06\cdot 230~{\rm V}
\\
&=
13.8~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2033~@#
\begin{align*}
U_{1{\rm k}}=13.8~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Estimate the magnetizing current $I_{\rm m,k}$ during the short-circuit test.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2034~@#
<WRAP leftalign>
The magnetizing current is assumed to be proportional to the voltage:
\begin{align*}
I_{\rm m,k}
=
I_{\rm m,N}\cdot \frac{U_{1{\rm k}}}{U_{1{\rm N}}}
\end{align*}

Insert the values:
\begin{align*}
I_{\rm m,k}
&=
0.12~{\rm A}\cdot \frac{13.8~{\rm V}}{230~{\rm V}}
\\
&=
0.0072~{\rm A}
\\
&=
7.2~{\rm mA}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2034~@#
\begin{align*}
I_{\rm m,k}
=
0.0072~{\rm A}
=
7.2~{\rm mA}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Compare $I_{\rm m,k}$ with $I_{1{\rm N}}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2035~@#
<WRAP leftalign>
Compare the short-circuit magnetizing current with the rated current:
\begin{align*}
\frac{I_{\rm m,k}}{I_{1{\rm N}}}
=
\frac{0.0072~{\rm A}}{3.0~{\rm A}}
\end{align*}

Calculate the ratio:
\begin{align*}
\frac{I_{\rm m,k}}{I_{1{\rm N}}}
&=
0.0024
\\
&=
0.24~\%
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2035~@#
\begin{align*}
\frac{I_{\rm m,k}}{I_{1{\rm N}}}
=
0.24~\%
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Explain why the magnetizing branch can be neglected.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2036~@#
<WRAP leftalign>
During the short-circuit test, the applied voltage is much smaller than the rated voltage:
\begin{align*}
U_{1{\rm k}} \ll U_{1{\rm N}}
\end{align*}

Since the magnetizing current is assumed to be approximately proportional to the applied voltage, the magnetizing current is also very small:
\begin{align*}
I_{\rm m,k}=7.2~{\rm mA}
\end{align*}

Compared with the rated current:
\begin{align*}
I_{1{\rm N}}=3.0~{\rm A}
\end{align*}

the magnetizing current is only $0.24~\%$.

So during the short-circuit test, almost all current flows through the short-circuit path consisting of $R_{\rm k}$ and $X_{\rm k}$.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2036~@#
The magnetizing branch $R_{\rm Fe}\parallel jX_{1{\rm H}}$ can usually be neglected in the short-circuit test.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Why the short-circuit equivalent circuit is often sufficient under load
#@TaskText_HTML@#

A transformer has the turns ratio $n=10$. The short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.0~\Omega$, $X_{\rm k}=4.0~\Omega$.

The secondary load current is $I_2=5.0~{\rm A}$. The load is ohmic-inductive with $\cos\varphi=0.8$.

The no-load current on the primary side is approximately $I_{10}=0.03~{\rm A}$.
\\ \\ \\
1. Calculate the load-related primary current $I'_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2037~@#
<WRAP leftalign>
The load-related primary current is:
\begin{align*}
I'_2
=
\frac{I_2}{n}
\end{align*}

Insert the values:
\begin{align*}
I'_2
&=
\frac{5.0~{\rm A}}{10}
\\
&=
0.50~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2037~@#
\begin{align*}
I'_2=0.50~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Estimate the primary-side voltage drop.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2038~@#
<WRAP leftalign>
The voltage drop is estimated with:
\begin{align*}
\Delta U_1
\approx
I'_2
\left(
R_{\rm k}\cos\varphi
+
X_{\rm k}\sin\varphi
\right)
\end{align*}

For $\cos\varphi=0.8$:
\begin{align*}
\sin\varphi
&=
\sqrt{1-\cos^2\varphi}
\\
&=
\sqrt{1-0.8^2}
\\
&=
0.6
\end{align*}

Insert the values:
\begin{align*}
\Delta U_1
&\approx
0.50~{\rm A}
\left(
2.0~\Omega\cdot 0.8
+
4.0~\Omega\cdot 0.6
\right)
\\
&=
0.50~{\rm A}
\left(
1.6~\Omega+2.4~\Omega
\right)
\\
&=
2.0~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2038~@#
\begin{align*}
\Delta U_1\approx 2.0~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Calculate the secondary-side voltage drop $\Delta U_2\approx \frac{\Delta U_1}{n}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2039~@#
<WRAP leftalign>
The secondary-side voltage drop is:
\begin{align*}
\Delta U_2
\approx
\frac{\Delta U_1}{n}
\end{align*}

Insert the values:
\begin{align*}
\Delta U_2
&\approx
\frac{2.0~{\rm V}}{10}
\\
&=
0.20~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2039~@#
\begin{align*}
\Delta U_2\approx 0.20~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Estimate an upper bound for the neglected voltage drop caused by $I_{10}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2040~@#
<WRAP leftalign>
First calculate the magnitude of the short-circuit impedance:
\begin{align*}
|\underline{Z}_{\rm k}|
=
\sqrt{R_{\rm k}^2+X_{\rm k}^2}
\end{align*}

Insert the values:
\begin{align*}
|\underline{Z}_{\rm k}|
&=
\sqrt{(2.0~\Omega)^2+(4.0~\Omega)^2}
\\
&=
4.47~\Omega
\end{align*}

The upper bound for the voltage drop caused by the no-load current is:
\begin{align*}
\Delta U_{1,10}
&\leq
|\underline{Z}_{\rm k}|I_{10}
\\
&=
4.47~\Omega\cdot 0.03~{\rm A}
\\
&=
0.134~{\rm V}
\end{align*}

On the secondary side:
\begin{align*}
\Delta U_{2,10}
&\leq
\frac{0.134~{\rm V}}{10}
\\
&=
0.0134~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2040~@#
\begin{align*}
|\underline{Z}_{\rm k}| &= 4.47~\Omega
\\
\Delta U_{1,10} &\leq 0.134~{\rm V}
\\
\Delta U_{2,10} &\leq 0.0134~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

5. Decide whether the short-circuit equivalent circuit is sufficient for this load estimate.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2041~@#
<WRAP leftalign>
The load-related secondary-side voltage drop is:
\begin{align*}
\Delta U_2\approx 0.20~{\rm V}
\end{align*}

The estimated neglected secondary-side voltage drop caused by the no-load current is at most:
\begin{align*}
\Delta U_{2,10}\leq 0.0134~{\rm V}
\end{align*}

This is small compared with the load-related drop of $0.20~{\rm V}$.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2041~@#
For this engineering estimate, the short-circuit equivalent circuit is sufficient.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#


#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Ideal transformer versus real transformer
#@TaskText_HTML@#

A transformer has the rated primary voltage of $U_1=230~{\rm V}$ and and the turns ratio $n=10$. A resistive load draws $I_2=4.0~{\rm A}$. \\
For the real transformer, the short-circuit equivalent parameters referred to the primary side are $R_{\rm k}=2.4~\Omega$, $X_{\rm k}=3.2~\Omega$. \\
The iron loss is approximately $P_{\rm Fe}=1.5~{\rm W}$. \\
Assume a resistive load with $\cos\varphi=1$.
\\ \\ 
1. Calculate the ideal secondary voltage $U_{2,\rm ideal}$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2042~@#
<WRAP leftalign>
For the ideal transformer:
\begin{align*}
U_{2,\rm ideal}
=
\frac{U_1}{n}
\end{align*}

Insert the values:
\begin{align*}
U_{2,\rm ideal}
&=
\frac{230~{\rm V}}{10}
\\
&=
23.0~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2042~@#
\begin{align*}
U_{2,\rm ideal}=23.0~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

2. Calculate the load-related primary current $I'_2$.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2043~@#
<WRAP leftalign>
The load-related primary current is:
\begin{align*}
I'_2
=
\frac{I_2}{n}
\end{align*}

Insert the values:
\begin{align*}
I'_2
&=
\frac{4.0~{\rm A}}{10}
\\
&=
0.40~{\rm A}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2043~@#
\begin{align*}
I'_2=0.40~{\rm A}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

3. Estimate the real secondary voltage.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2044~@#
<WRAP leftalign>
For a resistive load, the approximate primary-side voltage drop is:
\begin{align*}
\Delta U_1
\approx
I'_2R_{\rm k}
\end{align*}

Insert the values:
\begin{align*}
\Delta U_1
&\approx
0.40~{\rm A}\cdot 2.4~\Omega
\\
&=
0.96~{\rm V}
\end{align*}

The corresponding secondary-side voltage drop is:
\begin{align*}
\Delta U_2
&\approx
\frac{\Delta U_1}{n}
\\
&=
\frac{0.96~{\rm V}}{10}
\\
&=
0.096~{\rm V}
\end{align*}

Thus the real secondary voltage is approximately:
\begin{align*}
U_{2,\rm real}
&\approx
23.0~{\rm V}-0.096~{\rm V}
\\
&=
22.90~{\rm V}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2044~@#
\begin{align*}
\Delta U_1 &\approx 0.96~{\rm V}
\\
\Delta U_2 &\approx 0.096~{\rm V}
\\
U_{2,\rm real} &\approx 22.90~{\rm V}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

4. Estimate the copper losses.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2045~@#
<WRAP leftalign>
The copper losses are:
\begin{align*}
P_{\rm Cu}
\approx
R_{\rm k}(I'_2)^2
\end{align*}

Insert the values:
\begin{align*}
P_{\rm Cu}
&\approx
2.4~\Omega\cdot (0.40~{\rm A})^2
\\
&=
0.384~{\rm W}
\end{align*}

The real transformer also has iron losses:
\begin{align*}
P_{\rm Fe}=1.5~{\rm W}
\end{align*}
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2045~@#
\begin{align*}
P_{\rm Cu}\approx 0.384~{\rm W}
\end{align*}

Additionally:
\begin{align*}
P_{\rm Fe}=1.5~{\rm W}
\end{align*}
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

5. Compare ideal and real transformer behavior.
<WRAP group>
<WRAP half column rightalign>
#@PathBegin_HTML~2046~@#
<WRAP leftalign>
For the ideal transformer:
\begin{align*}
U_{2,\rm ideal}=23.0~{\rm V}
\end{align*}

For the real transformer:
\begin{align*}
U_{2,\rm real}\approx 22.90~{\rm V}
\end{align*}

The real transformer has copper losses and iron losses:
\begin{align*}
P_{\rm Cu}&\approx 0.384~{\rm W}
\\
P_{\rm Fe}&=1.5~{\rm W}
\end{align*}

So the main differences are:

  * the ideal transformer has exactly $U_2=23.0~{\rm V}$, the real transformer has a slightly lower voltage,
  * the ideal transformer has no losses, the real transformer has copper and iron losses,
  * the ideal transformer has no leakage voltage drop, the real transformer has a load-dependent voltage drop.

For this operating point the transformer is close to ideal, but not exactly ideal.
</WRAP>
#@PathEnd_HTML@#
</WRAP>
<WRAP half column>
#@ResultBegin_HTML~2046~@#
The real transformer differs from the ideal transformer by:

  * a slightly lower secondary voltage,
  * copper and iron losses,
  * a load-dependent voltage drop.

For this operating point it is close to ideal, but not exactly ideal.
#@ResultEnd_HTML@#
</WRAP>
</WRAP>

#@TaskEnd_HTML@#

===== Common pitfalls =====

  * **Using a transformer with DC:** A transformer needs changing flux. With DC, after the switching transient, an ideal transformer no longer transfers voltage. A real transformer may overheat because the winding resistance limits the current only weakly.
  * **Forgetting the current ratio sign:** The minus sign in \(\frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}\) comes from reference arrows. Do not interpret it as negative power loss.
  * **Mixing peak values and RMS values:** In AC power and transformer ratings, \(U\) and \(I\) usually mean RMS values. Time functions are written \(u(t)\), \(i(t)\). Instantaneous short-circuit peaks are written here as \(i_{\rm p}\).
  * **Confusing reluctance and resistance:** Magnetic reluctance \(R_{\rm m}\) has the unit \(1/{\rm H}\), not \(\Omega\).
  * **Confusing \(n\) and the technical no-load voltage ratio \(\ddot{\rm u}\):** The ideal ratio is \(n=\frac{N_1}{N_2}\). The measured no-load voltage ratio is close to \(n\), but not exactly equal for a real transformer.
  * **Forgetting the square when referring impedances:** Voltages transform with \(n\), currents with \(\frac{1}{n}\), but impedances transform with \(n^2\).
  * **Ignoring leakage reactance:** Leakage reactance is often the dominant part of short-circuit impedance. It strongly affects fault current and voltage drop.
  * **Treating \(\rm u_k\) as a voltage in volts:** \(\rm u_k\) is normally given in percent. Insert it consistently in formulas.
  * **Using \(2.54\) as a universal law:** The first short-circuit peak depends on the \(R/X\) ratio and on the switching instant. The factor \(2.54\) is an approximation.
  * **Opening a current transformer secondary:** This can create dangerous voltages. Current transformers are operated with a low burden, approximately as a short-circuit.
  * **Assuming ideal isolation at every frequency:** Real transformers have parasitic capacitances between windings. For high-frequency noise and EMC, the “isolated” sides can still be capacitively coupled.

===== Embedded resources =====