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-====== 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
-\[
-\begin{align*}
-\frac{\underline{U}_1}{\underline{U}_2}=\frac{N_1}{N_2}=n,
-\qquad
-\frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}
-\end{align*}
-\]
-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 \(u_{\rm k}\), 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_1:block18|magnetic flux and induction]], [[:electrical_engineering_1:block19|magnetic circuits]], and [[:electrical_engineering_1:block20|inductance and magnetic energy]].
-  * Repeat from EEE2 the use of [[block03|sinusoidal quantities]], [[block04|complex calculation]], and [[block06|complex power]].
-  * For a deeper field-theory view, see also [[:electrical_engineering_2:magnetic_circuits#mutual_induction_and_coupling|Mutual Induction and Coupling]].
-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.
-    * Unit checks for \(j\omega L\), \(j\omega N\Phi\), and \(u_{\rm k}\).
-    * 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_1:block18|induction]] explains why a changing flux induces voltage.
-    * [[:electrical_engineering_1:block19|magnetic circuits]] explains why the iron core guides the flux.
-    * [[:electrical_engineering_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 \(u_{\rm k}\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
-</callout>
-~~PAGEBREAK~~ ~~CLEARFIX~~
-===== Core content =====
-==== From EEE1 induction to an AC transformer ====
-<panel type="info" title="Transition from EEE1 to EEE2">
-In EEE1 we considered magnetic flux \(\Phi\), flux linkage \(\Psi\), and induction.
-For one coil with \(N\) turns the flux linkage is
-\[
-\begin{align*}
-\Psi = N\Phi .
-\end{align*}
-\]
-Faraday's law gives
-\[
-\begin{align*}
-u(t)=\frac{{\rm d}\Psi}{{\rm d}t}=N\frac{{\rm d}\Phi}{{\rm d}t}.
-\end{align*}
-\]
-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>
-<WRAP>
-<panel type="default">
-<imgcaption fig_transformer_principle|Idealized single-phase transformer with primary winding, secondary winding, iron core, and common main flux \(\Phi\).></imgcaption>
-{{drawio>block09_transformer_principle.svg}}
-</panel>
-</WRAP>
-<callout type="info" icon="true">
-**Unit check for induced voltage**
-\[
-\begin{align*}
-\underline{U}=j\omega N\underline{\Phi}
-\end{align*}
-\]
-with
-\[
-\begin{align*}
-[\omega\Phi]={\rm s}^{-1}\cdot {\rm Wb}
-={\rm s}^{-1}\cdot {\rm V\,s}
-={\rm V}.
-\end{align*}
-\]
-The number of turns \(N\) is dimensionless.
-</callout>
-==== Mutual induction: the key idea before the transformer ====
-Before we look at the transformer, we need one important idea:
-<callout icon="fa fa-lightbulb-o" color="blue">
-A changing current in coil \(1\) creates a changing magnetic flux.
-If part of this flux passes through coil \(2\), a voltage is induced in coil \(2\).
-This is called **mutual induction**.
-</callout>
-<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>
-{{:electrical_engineering_2:mutualinductiontwocoils1.svg?650}}
-</panel>
-</WRAP>
-The flux created by coil \(1\) can be split into
-\[
-\begin{align*}
-\Phi_{11}
-=
-{\color{blue}{\Phi_{21}}}
-+
-{\color{orange}{\Phi_{\rm S1}}}.
-\end{align*}
-\]
-  * \(\Phi_{11}\): total flux created by coil \(1\).
-  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\).
-  * \({\color{orange}{\Phi_{\rm S1}}}\): stray or leakage flux that does **not** link coil \(2\).
-The voltage induced in coil \(2\) is
-\[
-\begin{align*}
-u_{{\rm ind},2}(t)
-=
--N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}.
-\end{align*}
-\]
-<panel type="info" title="Analogy: two pendulums connected by a spring">
-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.
-</panel>
-<panel type="info" title="Analogy: a leaky magnetic pipe">
-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>
-==== Linked fluxes and mutual inductance ====
-For a single coil we already know
-\[
-\begin{align*}
-\Psi=L i .
-\end{align*}
-\]
-For two coupled coils, each flux linkage can depend on both currents.
-<panel type="info" title="Color scheme for coupled coils">
-In the following formulas:
-  * green terms: self-linkage of a coil by its own current,
-  * blue terms: mutual linkage between both coils,
-  * orange terms: leakage or stray flux that does not help energy transfer.
-</panel>
-\[
-\begin{align*}
-\Psi_1
-&=
-\underbrace{{\color{green}{L_{11}i_1}}}_{\text{self-linkage of coil 1}}
-+
-\underbrace{{\color{blue}{M_{12}i_2}}}_{\text{mutual linkage from coil 2}},
-\\[4pt]
-\Psi_2
-&=
-\underbrace{{\color{blue}{M_{21}i_1}}}_{\text{mutual linkage from coil 1}}
-+
-\underbrace{{\color{green}{L_{22}i_2}}}_{\text{self-linkage of coil 2}}.
-\end{align*}
-\]
-For most transformer calculations we use the symmetric case
-\[
-\begin{align*}
-M_{12}=M_{21}=M.
-\end{align*}
-\]
-Then
-\[
-\begin{align*}
-\boxed{
-\begin{pmatrix}
-\Psi_1\\
-\Psi_2
-\end{pmatrix}
-=
-\begin{pmatrix}
-{\color{green}{L_{11}}} & {\color{blue}{M}}\\
-{\color{blue}{M}} & {\color{green}{L_{22}}}
-\end{pmatrix}
-\begin{pmatrix}
-i_1\\
-i_2
-\end{pmatrix}
-}
-\end{align*}
-\]
-and
-\[
-\begin{align*}
-M=k\sqrt{L_{11}L_{22}}.
-\end{align*}
-\]
-Here \(k\) is the coupling coefficient.
-<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 |
-| sign of \(k\) | depends on winding direction and reference arrows | dot convention |
-<panel type="info" title="How to read \(M\)">
-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>
-==== Polarity and the dot convention ====
-The sign of the mutual term depends on the winding direction and on the chosen current reference arrows.
-<WRAP>
-<panel type="default">
-<imgcaption fig_direction_of_coupling|Dot convention: the dots indicate corresponding winding ends.></imgcaption>
-{{:electrical_engineering_2:directionofcoupling.svg?650}}
-</panel>
-</WRAP>
-<callout>
-**Rule of thumb**
-  * If both currents enter dotted terminals, the mutual fluxes support each other.
-  * If one current enters a dotted terminal and the other current leaves a dotted terminal, the mutual fluxes oppose each other.
-</callout>
-<WRAP group>
-<WRAP column half>
-<panel type="default">
-<imgcaption fig_positive_coupling|Positive coupling: currents enter corresponding dotted terminals.></imgcaption>
-{{:electrical_engineering_2:poscoupling.svg?500}}
-</panel>
-</WRAP>
-<WRAP column half>
-<panel type="default">
-<imgcaption fig_negative_coupling|Negative coupling: only one current enters a dotted terminal.></imgcaption>
-{{:electrical_engineering_2:negcoupling.svg?500}}
-</panel>
-</WRAP>
-</WRAP>
-For positive coupling:
-\[
-\begin{align*}
-u_1
-&=
-{\color{green}{L_{11}\frac{{\rm d}i_1}{{\rm d}t}}}
-+
-{\color{blue}{M\frac{{\rm d}i_2}{{\rm d}t}}},
-\\[4pt]
-u_2
-&=
-{\color{blue}{M\frac{{\rm d}i_1}{{\rm d}t}}}
-+
-{\color{green}{L_{22}\frac{{\rm d}i_2}{{\rm d}t}}}.
-\end{align*}
-\]
-For negative coupling, the sign of the \(M\)-term changes in the chosen equation system.
-<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>
-~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Magnetic coupling with reluctance ====
-We now use the magnetic circuit model from EEE1.
-The magnetic voltage, also called magnetomotive force, is
-\[
-\begin{align*}
-\Theta=N i .
-\end{align*}
-\]
-For a magnetic path with reluctance \(R_{\rm m}\), Hopkinson's law is
-\[
-\begin{align*}
-\Theta=R_{\rm m}\Phi
-\qquad \Longleftrightarrow \qquad
-\Phi=\frac{\Theta}{R_{\rm m}}.
-\end{align*}
-\]
-<panel type="info" title="Analogy to an electric circuit">
-\[
-\begin{align*}
-\text{electric:}\quad U=R I
-\qquad
-\text{magnetic:}\quad \Theta=R_{\rm m}\Phi
-\end{align*}
-\]
-  * Electric voltage \(U\) pushes electric current \(I\) through resistance \(R\).
-  * Magnetic voltage \(\Theta=N i\) pushes magnetic flux \(\Phi\) through reluctance \(R_{\rm m}\).
-</panel>
-For two windings on the same main magnetic path, the total magnetic voltage is
-\[
-\begin{align*}
-\Theta
-=
-N_1\underline{I}_1
-+
-N_2\underline{I}_2.
-\end{align*}
-\]
-The main flux is
-\[
-\begin{align*}
-\underline{\Phi}
-=
-\frac{\Theta}{R_{\rm mH}}
-=
-\frac{N_1\underline{I}_1+N_2\underline{I}_2}{R_{\rm mH}}.
-\end{align*}
-\]
-The flux linkages are
-\[
-\begin{align*}
-\underline{\Psi}_1
-&=
-N_1\underline{\Phi}
-=
-{\color{green}{\frac{N_1^2}{R_{\rm mH}}\underline{I}_1}}
-+
-{\color{blue}{\frac{N_1N_2}{R_{\rm mH}}\underline{I}_2}},
-\\[4pt]
-\underline{\Psi}_2
-&=
-N_2\underline{\Phi}
-=
-{\color{blue}{\frac{N_1N_2}{R_{\rm mH}}\underline{I}_1}}
-+
-{\color{green}{\frac{N_2^2}{R_{\rm mH}}\underline{I}_2}}.
-\end{align*}
-\]
-Therefore
-\[
-\begin{align*}
-\boxed{
-{\color{green}{L_{1{\rm H}}=\frac{N_1^2}{R_{\rm mH}}}}
-}
-\qquad
-\boxed{
-{\color{green}{L_{2{\rm H}}=\frac{N_2^2}{R_{\rm mH}}}}
-}
-\qquad
-\boxed{
-{\color{blue}{M=\frac{N_1N_2}{R_{\rm mH}}}}
-}
-\end{align*}
-\]
-for an ideal common magnetic path.
-<callout type="info" icon="true">
-**Unit check for mutual inductance**
-\[
-\begin{align*}
-M=\frac{N_1N_2}{R_{\rm mH}}
-\end{align*}
-\]
-with
-\[
-\begin{align*}
-[R_{\rm mH}]
-=
-\frac{{\rm A}}{{\rm Vs}}
-=
-\frac{1}{{\rm H}}.
-\end{align*}
-\]
-Thus
-\[
-\begin{align*}
-[M]=\frac{1}{1/{\rm H}}={\rm H}.
-\end{align*}
-\]
-</callout>
-<panel type="info" title="Example: why iron cores improve coupling">
-Air has a high magnetic reluctance. Iron has a much lower magnetic reluctance.
-If the magnetic path has a lower reluctance \(R_{\rm mH}\), then
-\[
-\begin{align*}
-M=\frac{N_1N_2}{R_{\rm mH}}
-\end{align*}
-\]
-becomes larger.
-A larger \(M\) means that a changing current in one winding induces a stronger voltage in the other winding.
-</panel>
-==== Ideal single-phase transformer ====
-For an ideal transformer we assume:
-  * both windings are linked by the same magnetic flux \(\Phi\),
-  * 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**
-\[
-\begin{align*}
-\boxed{
-\frac{\underline{U}_1}{\underline{U}_2}
-=
-\frac{N_1}{N_2}
-=
-n
-}
-\end{align*}
-\]
-with
-\[
-\begin{align*}
-n=\frac{N_1}{N_2}.
-\end{align*}
-\]
-<WRAP column 100%>
-<panel type="danger" title="Remember: ideal transformer ratios">
-With the indicated reference arrows and a lossless transformer:
-\[
-\begin{align*}
-\underline{U}_1\underline{I}_1+\underline{U}_2\underline{I}_2=0
-\end{align*}
-\]
-and therefore
-\[
-\begin{align*}
-\boxed{
-\frac{\underline{I}_1}{\underline{I}_2}
-=
--\frac{\underline{U}_2}{\underline{U}_1}
-=
--\frac{N_2}{N_1}
-=
--\frac{1}{n}
-}
-\end{align*}
-\]
-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>
-</WRAP>
-<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>
-==== Example: step-down transformer for a robot controller ====
-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:
-\[
-\begin{align*}
-S_1 &= U_1I_1=230~{\rm V}\cdot 0.4~{\rm A}=92~{\rm VA},
-\\
-S_2 &= U_2I_2=23~{\rm V}\cdot 4~{\rm A}=92~{\rm VA}.
-\end{align*}
-\]
-~~PAGEBREAK~~ ~~CLEARFIX~~
-==== 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 **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.
-<WRAP>
-<panel type="default">
-<imgcaption fig_main_and_leakage_flux|Main flux and leakage fluxes in a real transformer.></imgcaption>
-{{drawio>block09_main_and_leakage_flux.svg}}
-</panel>
-</WRAP>
-The real flux linkage equations become
-\[
-\begin{align*}
-\underline{\Psi}_1
-&=
-\underbrace{{\color{blue}{L_{1{\rm H}}\underline{I}_1+M\underline{I}_2}}}_{\text{main magnetic path}}
-+
-\underbrace{{\color{orange}{L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage}},
-\\[4pt]
-\underline{\Psi}_2
-&=
-\underbrace{{\color{blue}{L_{2{\rm H}}\underline{I}_2+M\underline{I}_1}}}_{\text{main magnetic path}}
-+
-\underbrace{{\color{orange}{L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage}}.
-\end{align*}
-\]
-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:
-\[
-\begin{align*}
-P_{\rm Cu,1}=R_1I_1^2,
-\qquad
-P_{\rm Cu,2}=R_2I_2^2.
-\end{align*}
-\]
-<panel type="info" title="Color scheme for the equivalent equations">
-In the following formulas:
-  * blue terms: useful main magnetic coupling,
-  * orange terms: leakage flux,
-  * red terms: winding resistance and copper loss.
-</panel>
-\[
-\begin{align*}
-\underline{U}_1
-&=
-\underbrace{{\color{red}{R_1\underline{I}_1}}}_{\text{primary copper drop}}
-+
-\underbrace{{\color{orange}{j\omega L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage drop}}
-+
-\underbrace{{\color{blue}{j\omega L_{1{\rm H}}\underline{I}_1+j\omega M\underline{I}_2}}}_{\text{main magnetic coupling}},
-\\[6pt]
-\underline{U}_2
-&=
-\underbrace{{\color{red}{R_2\underline{I}_2}}}_{\text{secondary copper drop}}
-+
-\underbrace{{\color{orange}{j\omega L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage drop}}
-+
-\underbrace{{\color{blue}{j\omega L_{2{\rm H}}\underline{I}_2+j\omega M\underline{I}_1}}}_{\text{main magnetic coupling}}.
-\end{align*}
-\]
-<callout>
-The blue terms are responsible for transformer action.
-The orange terms are unwanted but unavoidable.
-The red terms convert electrical energy into heat.
-</callout>
-With the leakage reactances
-\[
-\begin{align*}
-X_{1\sigma}=\omega L_{1\sigma},
-\qquad
-X_{2\sigma}=\omega L_{2\sigma},
-\qquad
-X_{\rm M}=\omega M
-\end{align*}
-\]
-these equations can be represented by an equivalent circuit.
-<callout type="info" icon="true">
-**Unit check for leakage reactance**
-\[
-\begin{align*}
-X_{1\sigma}=\omega L_{1\sigma}
-\end{align*}
-\]
-with
-\[
-\begin{align*}
-[\omega L]={\rm s}^{-1}\cdot{\rm H}
-=
-{\rm s}^{-1}\cdot \frac{{\rm V\,s}}{{\rm A}}
-=
-\Omega .
-\end{align*}
-\]
-So \(jX_{1\sigma}\) is an impedance.
-</callout>
-==== Why leakage flux matters in engineering ====
-<panel type="info" title="Mechatronics example: motor start-up">
-A robot axis may draw a high current during acceleration.
-This high current produces a voltage drop at the leakage reactance and winding resistance of the transformer.
-Possible effects:
-  * the secondary voltage decreases,
-  * a DC link after a rectifier may sag,
-  * controllers may reset due to undervoltage,
-  * cables and protective devices must withstand the fault current.
-So leakage is not only a “field theory detail”. It directly affects the electrical behavior of the machine.
-</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 ====
-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 \(n^2\):
-\[
-\begin{align*}
-\boxed{
-R'_2=n^2R_2
-}
-\qquad
-\boxed{
-X'_{2\sigma}=n^2X_{2\sigma}
-}
-\end{align*}
-\]
-<callout type="info" icon="true">
-**Unit check for referred resistance**
-The turns ratio \(n\) is dimensionless. Therefore
-\[
-\begin{align*}
-[R'_2]=[n^2R_2]=\Omega .
-\end{align*}
-\]
-The value changes, but the unit does not.
-</callout>
-<WRAP>
-<panel type="default">
-<imgcaption fig_reduced_equivalent_circuit|Reduced transformer equivalent circuit with secondary quantities referred to the primary side.></imgcaption>
-{{drawio>block09_reduced_transformer_equivalent_circuit.svg}}
-</panel>
-</WRAP>
-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.
-<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 \(\ddot{u}\):
-\[
-\begin{align*}
-\ddot{u}
-=
-\frac{\text{higher voltage}}{\text{lower voltage}}
-\bigg|_{\rm no~load}.
-\end{align*}
-\]
-For a step-down transformer:
-\[
-\begin{align*}
-\ddot{u}
-=
-\frac{U_{1{\rm N}}}{U_{20}}.
-\end{align*}
-\]
-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,
-\[
-\begin{align*}
-\ddot{u}\neq n,
-\end{align*}
-\]
-but for many practical transformers
-\[
-\begin{align*}
-\ddot{u}\approx n.
-\end{align*}
-\]
-</callout>
-==== Short-circuit operation of the real transformer ====
-In the short-circuit test, the secondary side is shorted:
-\[
-\begin{align*}
-\underline{U}_2=0.
-\end{align*}
-\]
-Because the required primary voltage is small, the magnetizing branch is often neglected:
-\[
-\begin{align*}
-X_{1{\rm H}},\;R_{\rm Fe}
-\gg
-X'_{2\sigma},\;R'_2.
-\end{align*}
-\]
-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
-\[
-\begin{align*}
-\underline{Z}_{\rm k}
-=
-R_{\rm k}+jX_{\rm k}.
-\end{align*}
-\]
-<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>
-<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:
-\[
-\begin{align*}
-\boxed{
-u_{\rm k}
-=
-\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
-}
-\end{align*}
-\]
-Small \(u_{\rm k}\) means: small internal impedance and high possible fault current.
-Large \(u_{\rm k}\) means: stronger current limitation and larger voltage drop under load.
-</panel>
-The continuous short-circuit current for rated primary voltage is
-\[
-\begin{align*}
-\boxed{
-I_{1{\rm k}}
-=
-\frac{U_{1{\rm N}}}{U_{1{\rm k}}}\cdot I_{1{\rm N}}
-=
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
-}
-\end{align*}
-\]
-where \(u_{\rm k}\) is inserted as a percentage value.
-<callout type="info" icon="true">
-**Unit check for \(u_{\rm k}\)**
-\[
-\begin{align*}
-u_{\rm k}
-=
-\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
-\end{align*}
-\]
-is dimensionless. It is usually stated in percent.
-</callout>
-==== Why the first short-circuit peak can be about \(2.54 I_{1{\rm k}}\) ====
-The RMS short-circuit current \(I_{1{\rm k}}\) does not describe the highest instantaneous current.
-When a short circuit starts, the current can contain
-  * a sinusoidal AC component, and
-  * a decaying DC offset.
-The worst case occurs when the fault starts at an unfavorable phase angle. Then the first current peak is larger than the normal sinusoidal peak \(\sqrt{2}I_{1{\rm k}}\).
-A common engineering form is
-\[
-\begin{align*}
-i_{\rm p}
-=
-\kappa\sqrt{2}\,I''_{\rm k}.
-\end{align*}
-\]
-Here
-  * \(i_{\rm p}\) is the instantaneous peak short-circuit current,
-  * \(I''_{\rm k}\) is the initial symmetrical RMS short-circuit current,
-  * \(\kappa\) is a peak factor depending mainly on the \(R/X\) ratio of the short-circuit impedance.
-For a strongly inductive transformer short circuit, a practical approximation is
-\[
-\begin{align*}
-\kappa\approx 1.8.
-\end{align*}
-\]
-Then
-\[
-\begin{align*}
-i_{\rm p}
-\approx
-.8\cdot\sqrt{2}\cdot I_{1{\rm k}}
-=
-.55\,I_{1{\rm k}}
-\approx
-.54\,I_{1{\rm k}}.
-\end{align*}
-\]
-<WRAP column 100%>
-<panel type="danger" title="Do not overinterpret the factor 2.54">
-The factor \(2.54\) is an engineering approximation for the first peak current.
-It is **not** a universal transformer law.
-For real protection design, use the applicable standard, manufacturer data, and the actual \(R/X\) ratio of the installation.
-</panel>
-</WRAP>
-~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Real transformer under load ====
-Under load, the short-circuit equivalent circuit is often sufficient for engineering estimates.
-\[
-\begin{align*}
-\underline{U}_{\rm k}
-=
-\left(R_{\rm k}+jX_{\rm k}\right)\underline{I}_1.
-\end{align*}
-\]
-This voltage drop is subtracted vectorially from the primary-side voltage relation. Therefore the secondary voltage depends on
-  * load current magnitude,
-  * load power factor,
-  * winding resistance,
-  * leakage reactance.
-<WRAP>
-<panel type="default">
-<imgcaption fig_kapp_triangle|Kapp triangle: approximate voltage drop under load using \(R_{\rm k}I\) and \(X_{\rm k}I\).></imgcaption>
-{{drawio>block09_kapp_triangle.svg}}
-</panel>
-</WRAP>
-<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 group>
-<WRAP column half>
-<panel type="default">
-<imgcaption fig_core_transformer|Core-type transformer: usually smaller short-circuit voltage \(u_{\rm k}\).></imgcaption>
-{{drawio>block09_core_type_transformer.svg}}
-</panel>
-</WRAP>
-<WRAP column half>
-<panel type="default">
-<imgcaption fig_shell_transformer|Shell-type transformer: usually larger short-circuit voltage \(u_{\rm k}\).></imgcaption>
-{{drawio>block09_shell_type_transformer.svg}}
-</panel>
-</WRAP>
-</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 \(u_{\rm k}\) ^ 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 |
-<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#~~.1  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 \(I_2=2.0~{\rm A}\).
-  * Calculate the turns ratio \(n\).
-  * Calculate the ideal secondary voltage \(U_2\).
-  * Calculate the magnitude of the ideal primary current \(I_1\).
-  * State whether this is a step-up or step-down transformer.
-#@ResultBegin_HTML~Exercise1~@#
-\[
-\begin{align*}
-n=\frac{N_1}{N_2}
-=
-\frac{1200}{300}
-=
-.
-\end{align*}
-\]
-The secondary voltage is
-\[
-\begin{align*}
-U_2
-=
-\frac{U_1}{n}
-=
-\frac{230~{\rm V}}{4}
-=
-.5~{\rm V}.
-\end{align*}
-\]
-The primary current magnitude is
-\[
-\begin{align*}
-I_1
-=
-\frac{I_2}{n}
-=
-\frac{2.0~{\rm A}}{4}
-=
-.50~{\rm A}.
-\end{align*}
-\]
-Because \(U_2<U_1\), it is a step-down transformer.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  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 \(N_2=100\).
-  * Calculate \(L_{1{\rm H}}\).
-  * Calculate \(L_{2{\rm H}}\).
-  * Calculate \(M\).
-  * Check whether the units are correct.
-#@ResultBegin_HTML~Exercise2~@#
-\[
-\begin{align*}
-L_{1{\rm H}}
-=
-\frac{N_1^2}{R_{\rm mH}}
-=
-\frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
-=
-.125~{\rm H}.
-\end{align*}
-\]
-\[
-\begin{align*}
-L_{2{\rm H}}
-=
-\frac{N_2^2}{R_{\rm mH}}
-=
-\frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
-=
-.0050~{\rm H}
-=
-.0~{\rm mH}.
-\end{align*}
-\]
-\[
-\begin{align*}
-M
-=
-\frac{N_1N_2}{R_{\rm mH}}
-=
-\frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
-=
-.025~{\rm H}
-=
-~{\rm mH}.
-\end{align*}
-\]
-The unit is correct because \(1/(1/{\rm H})={\rm H}\).
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  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.
-#@ResultBegin_HTML~Exercise3~@#
-\[
-\begin{align*}
-R'_2
-=
-n^2R_2
-=
-^2\cdot 0.20~\Omega
-=
-\cdot 0.20~\Omega
-=
-.0~\Omega.
-\end{align*}
-\]
-\[
-\begin{align*}
-X'_{2\sigma}
-=
-n^2X_{2\sigma}
-=
-^2\cdot 0.35~\Omega
-=
-\cdot 0.35~\Omega
-=
-.75~\Omega.
-\end{align*}
-\]
-The unit remains \(\Omega\), because \(n\) is dimensionless.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  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 \(u_{\rm k}=5~\%\).
-  * Calculate the continuous short-circuit current \(I_{1{\rm k}}\) when rated primary voltage is applied.
-  * Estimate the initial peak short-circuit current \(i_{\rm p}\) using \(i_{\rm p}\approx 2.54 I_{1{\rm k}}\).
-#@ResultBegin_HTML~Exercise4~@#
-\[
-\begin{align*}
-I_{1{\rm k}}
-=
-I_{1{\rm N}}\cdot \frac{100~\%}{u_{\rm k}}
-=
-~{\rm A}\cdot \frac{100~\%}{5~\%}
-=
-~{\rm A}.
-\end{align*}
-\]
-\[
-\begin{align*}
-i_{\rm p}
-\approx
-.54\cdot I_{1{\rm k}}
-=
-.54\cdot 200~{\rm A}
-=
-~{\rm A}.
-\end{align*}
-\]
-The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  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.
-  * Calculate \(n\).
-  * Refer \(R_2\) and \(X_{2\sigma}\) to the primary side.
-  * Calculate \(R_{\rm k}\) and \(X_{\rm k}\).
-  * Calculate the primary rated current magnitude \(I_{1{\rm N}}\) using the ideal current ratio.
-  * Estimate the magnitude of the internal voltage drop \(U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}\).
-#@ResultBegin_HTML~Exercise5~@#
-The turns ratio is
-\[
-\begin{align*}
-n
-=
-\frac{U_{1{\rm N}}}{U_{2{\rm N}}}
-=
-\frac{230~{\rm V}}{23~{\rm V}}
-=
-.
-\end{align*}
-\]
-The secondary quantities referred to the primary side are
-\[
-\begin{align*}
-R'_2
-&=
-n^2R_2
-=
-^2\cdot 0.012~\Omega
-=
-.2~\Omega,
-\\
-X'_{2\sigma}
-&=
-n^2X_{2\sigma}
-=
-^2\cdot 0.018~\Omega
-=
-.8~\Omega.
-\end{align*}
-\]
-Therefore
-\[
-\begin{align*}
-R_{\rm k}
-&=
-R_1+R'_2
-=
-.2~\Omega+1.2~\Omega
-=
-.4~\Omega,
-\\
-X_{\rm k}
-&=
-X_{1\sigma}+X'_{2\sigma}
-=
-.8~\Omega+1.8~\Omega
-=
-.6~\Omega.
-\end{align*}
-\]
-The primary current magnitude is
-\[
-\begin{align*}
-I_{1{\rm N}}
-=
-\frac{I_{2{\rm N}}}{n}
-=
-\frac{5.0~{\rm A}}{10}
-=
-.50~{\rm A}.
-\end{align*}
-\]
-The magnitude of the short-circuit impedance is
-\[
-\begin{align*}
-|\underline{Z}_{\rm k}|
-=
-\sqrt{R_{\rm k}^2+X_{\rm k}^2}
-=
-\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2}
-=
-.33~\Omega.
-\end{align*}
-\]
-Thus the internal voltage drop estimate is
-\[
-\begin{align*}
-U_{\rm k}
-\approx
-|\underline{Z}_{\rm k}|I_{1{\rm N}}
-=
-.33~\Omega\cdot 0.50~{\rm A}
-=
-.17~{\rm V}.
-\end{align*}
-\]
-This is a primary-side voltage drop. On the secondary side it corresponds approximately to
-\[
-\begin{align*}
-\frac{2.17~{\rm V}}{10}=0.217~{\rm V}.
-\end{align*}
-\]
-For a \(23~{\rm V}\) actuator supply this is small but not zero.
-#@ResultEnd_HTML@#
-#@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 \(\ddot{u}\):** 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 \(u_{\rm k}\) as a voltage in volts:** \(u_{\rm k}\) 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 =====
-<WRAP group>
-<WRAP column half>
-<panel type="info" title="PhET: Faraday's Law">
-Use this simulation to revisit the EEE1 idea that a changing magnetic flux induces voltage.
-This is the physical basis of the transformer equations in this block.
-{{url>https://phet.colorado.edu/sims/html/faradays-law/latest/faradays-law_en.html 700,500 noborder}}
-</panel>
-</WRAP>
-<WRAP column half>
-<panel type="info" title="Falstad / CircuitJS: transformer circuits">
-Use CircuitJS for qualitative experiments with coupled inductors and transformer circuits.
-Suggested activity: search the example circuits for “transformer”, change the turns ratio, and observe voltage and current.
-{{url>https://www.falstad.com/circuit/circuitjs.html 700,500 noborder}}
-</panel>
-</WRAP>
-</WRAP>
-<WRAP group>
-<WRAP column half>
-<panel type="info" title="Falstad / CircuitJS: coupled inductors">
-Suggested experiment:
-  * build two coupled inductors,
-  * change the coupling coefficient,
-  * observe how the secondary voltage changes.
-This is especially useful for understanding the difference between an iron-core transformer and loosely coupled wireless charging coils.
-</panel>
-</WRAP>
-<WRAP column half>
-<panel type="info" title="Further reading in this wiki">
-Relevant continuity pages:
-  * [[:electrical_engineering_1:block18|EEE1: Induction]]
-  * [[:electrical_engineering_1:block19|EEE1: Magnetic circuits]]
-  * [[:electrical_engineering_1:block20|EEE1: Inductance]]
-  * [[:electrical_engineering_2:magnetic_circuits#mutual_induction_and_coupling|Mutual Induction and Coupling]]
-</panel>
-</WRAP>
-</WRAP>
-~~PAGEBREAK~~ ~~CLEARFIX~~