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electrical_engineering_and_electronics_2:block09 [2026/05/16 22:35] – mexleadminelectrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) – mexleadmin
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 \[ \[
 \begin{align*} \begin{align*}
-\frac{\underline{U}_1}{\underline{U}_2}=\frac{N_1}{N_2}=n,+\frac{\underline{U}_1^\phantom{O}}{\underline{U}_2^\phantom{O}}=\frac{N_1}{N_2}=n,
 \qquad \qquad
 \frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n} \frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}
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   * 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 \(X'_{2\sigma}=n^2X_{2\sigma}\).   * 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.   * 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.+  * 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.   * connect transformer parameters to engineering applications in mechatronics and robotics, such as isolated power supplies, motor current measurement, welding transformers, and safety transformers.
 </callout> </callout>
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   * **Practice (20 min):**   * **Practice (20 min):**
     * Quick ratio calculations for step-up and step-down transformers.     * 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.     * Short-circuit current calculation for a transformer used in an actuator supply.
  
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     * \(R_{\rm Fe}\): iron losses in the core.     * \(R_{\rm Fe}\): iron losses in the core.
     * \(L_{\rm H}\): main magnetizing inductance needed to create the main flux.     * \(L_{\rm H}\): 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.+  * In engineering, transformer data such as \(\rm u_k\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
 </callout> </callout>
  
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 ===== Core content ===== ===== Core content =====
- 
-==== Mutual induction: the key idea before the transformer ==== 
  
 <panel type="info" title="Short Review of the Flux"> <panel type="info" title="Short Review of the Flux">
-In EEE1 we considered magnetic flux \(\Phi\), flux linkage \(\Psi\), and induction.  \\+In EEE1 we considered magnetic flux \(\Phi\), flux linkage / linked flux \(\Psi\), and induction.  \\
 For one coil with \(N\) turns the flux linkage is For one coil with \(N\) turns the flux linkage is
  
Line 103: Line 100:
 \] \]
  
-(Be aware of Lenz law, whenever you want to draw the votage arrows) \\+(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 In sinusoidal steady state this becomes the phasor equation
  
Line 115: Line 112:
 </panel> </panel>
  
-Now, we look onto the situation of two coils nearby each other and expand this formula for the induced voltage. \\ +==== Polarity and the dot convention ====
-For this, we see:+
  
-<callout icon="fa fa-lightbulb-o" color="blue"> +Before we start with the transformer, we have to look at a common convention for the orientation of two coils to each other. \\ 
-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 \(2\).  \\ +<WRAP> 
-This is called **mutual induction**.+<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> </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> <WRAP>
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 {\color{blue}{\Phi_{21}}} {\color{blue}{\Phi_{21}}}
 + +
-{\color{orange}{\Phi_{\rm S1}}}.+{\color{orange}{\Phi_{1 \rm \sigma}}}.
 \end{align*} \end{align*}
 \] \]
  
-  * \(\Phi_{11}\): total flux created by coil \(1\). +  * \(\Phi_{11}\): total flux created by coil \(1\).  
-  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\). +  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\).  
-  * \({\color{orange}{\Phi_{\rm S1}}}\): stray or leakage flux that does **not** link coil \(2\).+  * \({\color{orange}{\Phi_{1 \rm \sigma}}}\): stray or leakage flux that does **not** link coil \(2\). The Greek letter sigma $\sigma$ is used to denote leakage or "stray" quantities.
  
-The voltage induced in coil \(2\) is+For an example, we will have a look at the instantaneous voltage induced in coil \(2\):
  
 \[ \[
 \begin{align*} \begin{align*}
-u_{{\rm ind},2}(t)+u_{2,\rm ind}(t) 
 += 
 +\frac{{\rm d}{\color{blue}{\Psi_{21}}}}{{\rm d}t}
 = =
 N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}. N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}.
 \end{align*} \end{align*}
 \] \]
 +
 +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"> <panel type="info" title="Analogies">
Line 191: Line 531:
 </panel> </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>) 
 +\[ 
 +\begin{align*} 
 +\underline{\Phi}_{1 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H}  
 +\\[4pt] 
 +\underline{\Phi}_{2 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H}  
 +\end{align*} 
 +\]
  
-===== Common pitfalls ===== +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 ===== ===== Exercises =====
-==== Worked examples ==== 
  
-...+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: ideal transformer voltage and current ratio 
 +#@TaskText_HTML@#
  
-===== Embedded resources ===== +A transformer has $N_1=1200$ turns and $N_2=300$ turns.  \\ 
-<WRAP column half> +The primary RMS voltage is $U_1=230~{\rm V}$.  \\ 
-Explanation (video): ...+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> </WRAP>
  
-~~PAGEBREAK~~ ~~CLEARFIX~~+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.
 +
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