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--- electrical_engineering_and_electronics_2:block09 [2026/05/17 08:27] – mexleadmin
+++ electrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) – mexleadmin
@@ Line 19: / Line 19: @@
   * 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.
+  * calculate short-circuit voltage \(\rm u_k\), continuous short-circuit current \(I_{\rm 1k}\), and an estimated initial peak short-circuit current.
   * connect transformer parameters to engineering applications in mechatronics and robotics, such as isolated power supplies, motor current measurement, welding transformers, and safety transformers.
 </callout>
@@ Line 75: / Line 75: @@
     * \(R_{\rm Fe}\): 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.
+  * In engineering, transformer data such as \(\rm u_k\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
 </callout>
@@ Line 277: / Line 277: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Mutual induction: the key idea for the real transformer ====
-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. \\
-For this, we see:
-<callout icon="fa fa-lightbulb-o" color="blue">
-A changing current in coil \(1\) creates a changing magnetic flux.  \\
-(Only) A 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>
-{{drawio>electrical_engineering_and_electronics_2:mutualinductiontwocoils1.svg}}
-</panel>
-</WRAP>
-The flux created by coil \(1\) can be split into
-\[
-\begin{align*}
-\Phi_{1\rm H}
-=
-{\color{blue}{\Phi_{21}}}
-+
-{\color{orange}{\Phi_{1 \rm \sigma}}}.
-\end{align*}
-\]
-  * \(\Phi_{11}\): total flux created by coil \(1\).
-  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\). \\ This part is also \({\color{blue}{\Phi_{21}}} = \Phi_{1\rm H} \). The 'H' denotes the German word "Haupt" (sometimes also given as 'm' for "main").
-  * \({\color{orange}{\Phi_{1 \rm \sigma}}}\): stray or leakage flux that does **not** link coil \(2\). The Greek letter sigma $\sigma$ is used to denote leakage or "stray" quantities.
-For an example, we will have a look at the instantaneous voltage induced in coil \(2\):
-\[
-\begin{align*}
-u_{2}(t)
-=
-\frac{{\rm d}{\color{blue}{\Psi_{21}}}}{{\rm d}t}
-=
-N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}.
-\end{align*}
-\]
-The complex voltage induced in coil \(2\) is
-\[
-\begin{align*}
-\boxed{
-\underline{U}_{2}
-=
-j\omega {\color{blue}{ \underline{\Psi}_{21} }}
-=
-j\omega N_2 {\color{blue}{ \underline{\Phi}_{21} }}
-}.
-\end{align*}
-\]
-We need these complex representations for the next steps into the transformer. \\
-  - __Mutual induction__: \\ First, we will investigate combining both fluxes - the flux in a coil generated by itself plus the flux of the other coil
-  - __Stray flux__: \\ Then, we add the stray flux to our model
-  - __Losses__: \\ Finally, we build a full model including power losses
-\\ \\
-<panel type="info" title="Analogies">
-<fs large>**Analogy 1: two pendulums connected by a spring**</fs> \\
-Imagine two pendulums connected by a weak spring.
-  * If pendulum \(1\) moves, the spring can make pendulum \(2\) move as well.
-  * A strong spring transfers the motion strongly.
-  * A weak spring transfers the motion only weakly.
-  * If the spring is missing, pendulum \(2\) does not react.
-For coupled coils:
-  * the changing motion corresponds to changing current,
-  * the spring corresponds to the magnetic coupling,
-  * the motion transferred to the second pendulum corresponds to the induced voltage,
-  * weak coupling means that only a small part of the magnetic flux links both coils.
-<fs large>**Analogy 2: a leaky magnetic pipe**</fs> \\
-The magnetic core can be imagined as a pipe guiding magnetic flux.
-  * A good iron core is like a wide, low-resistance pipe: most flux reaches the second coil.
-  * A large air gap is like a narrow, difficult path: less flux reaches the second coil.
-  * Leakage flux is like flow escaping through side paths: it belongs to the first coil but does not help the second coil.
-This image is helpful for transformers, wireless charging coils, and current sensors.
-</panel>
-<panel type="info" title="Engineering examples">
-  * **Transformer:** very strong coupling because the iron core guides most of the flux through both windings.
-  * **Wireless charger:** weaker coupling because the flux must cross an air gap and the coils may be misaligned.
-  * **Current transformer:** the measured conductor acts like a one-turn primary winding; the secondary winding detects the changing magnetic field.
-  * **Relay coil near signal wiring:** unwanted coupling can induce noise voltages in nearby loops.
-</panel>
 ==== Linked fluxes and mutual inductance ====
@@ Line 426: / Line 324: @@
 \]
-Often, the self-inductances are shortened:
+Often, the self-inductances are abbreviated:
 \[
 \begin{align*}
@@ Line 473: / Line 371: @@
 | \(0<k<1\)       | 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 (see below)                     |
 </tabcaption>
@@ Line 545: / Line 442: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
+==== Stray and Leakage ====
+The image of the ideal transformer shall be consecutively developed to a more realistic transformer model. \\
+To do so, we look at the situation of two coils near each other and expand this formula for the induced voltage.
+<WRAP>
+<panel type="default">
+<imgcaption fig_mutual_induction_two_coils|Mutual induction of two coils: only part of the flux created by coil \(1\) links coil \(2\).></imgcaption>
+{{drawio>electrical_engineering_and_electronics_2:mutualinductiontwocoils1.svg}}
+</panel>
+</WRAP>
+The flux created by coil \(1\) can be split into
+\[
+\begin{align*}
+\Phi_{11}
+=
+{\color{blue}{\Phi_{21}}}
++
+{\color{orange}{\Phi_{1 \rm \sigma}}}.
+\end{align*}
+\]
+  * \(\Phi_{11}\): total flux created by coil \(1\).
+  * \({\color{blue}{\Phi_{21}}}\): part of this flux that also links coil \(2\).
+  * \({\color{orange}{\Phi_{1 \rm \sigma}}}\): stray or leakage flux that does **not** link coil \(2\). The Greek letter sigma $\sigma$ is used to denote leakage or "stray" quantities.
+For an example, we will have a look at the instantaneous voltage induced in coil \(2\):
+\[
+\begin{align*}
+u_{2,\rm ind}(t)
+=
+\frac{{\rm d}{\color{blue}{\Psi_{21}}}}{{\rm d}t}
+=
+N_2\frac{{\rm d}{\color{blue}{\Phi_{21}}}}{{\rm d}t}.
+\end{align*}
+\]
+The complex voltage induced in coil \(2\) is
+\[
+\begin{align*}
+\underline{U}_{2}
+=
+j\omega {\color{blue}{ \underline{\Psi}_{21} }}
+=
+j\omega N_2 {\color{blue}{ \underline{\Phi}_{21} }}
+.
+\end{align*}
+\]
+\\ \\
+<panel type="info" title="Analogies">
+<fs large>**Analogy 1: two pendulums connected by a spring**</fs> \\
+Imagine two pendulums connected by a weak spring.
+  * If pendulum \(1\) moves, the spring can make pendulum \(2\) move as well.
+  * A strong spring transfers the motion strongly.
+  * A weak spring transfers the motion only weakly.
+  * If the spring is missing, pendulum \(2\) does not react.
+For coupled coils:
+  * the changing motion corresponds to changing current,
+  * the spring corresponds to the magnetic coupling,
+  * the motion transferred to the second pendulum corresponds to the induced voltage,
+  * weak coupling means that only a small part of the magnetic flux links both coils.
+<fs large>**Analogy 2: a leaky magnetic pipe**</fs> \\
+The magnetic core can be imagined as a pipe guiding magnetic flux.
+  * A good iron core is like a wide, low-resistance pipe: most flux reaches the second coil.
+  * A large air gap is like a narrow, difficult path: less flux reaches the second coil.
+  * Leakage flux is like flow escaping through side paths: it belongs to the first coil but does not help the second coil.
+This image is helpful for transformers, wireless charging coils, and current sensors.
+</panel>
+<panel type="info" title="Engineering examples">
+  * **Transformer:** very strong coupling because the iron core guides most of the flux through both windings.
+  * **Wireless charger:** weaker coupling because the flux must cross an air gap and the coils may be misaligned.
+  * **Current transformer:** the measured conductor acts like a one-turn primary winding; the secondary winding detects the changing magnetic field.
+  * **Relay coil near signal wiring:** unwanted coupling can induce noise voltages in nearby loops.
+</panel>
 ==== Real transformer: leakage and losses ====
@@ Line 550: / Line 535: @@
 In a real transformer, not all flux links both windings.
-  * The **main flux** \(\Phi_{\rm H}\) links primary and secondary winding.
+  * 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.
@@ Line 562: / Line 547: @@
 </panel>
 </WRAP>
+The total main flux linking is given by (see <imgref fig_main_and_leakage_flux>)
+\[
+\begin{align*}
+\underline{\Phi}_{1 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H}
+\\[4pt]
+\underline{\Phi}_{2 \rm H} = \underline{\Phi}_{21} + \underline{\Phi}_{12} = \underline{\Phi}_{\rm H}
+\end{align*}
+\]
+For the flux linkage also the leakage flux has to be considered:
+\[
+\begin{align*}
+\underline{\Psi}_{1} =N_1 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{1\sigma})
+\\[4pt]
+\underline{\Psi}_{2} =N_2 ( \underline{\Phi}_{21} + \underline{\Phi}_{12} + \underline{\Phi}_{2\sigma})
+\end{align*}
+\]
 The real flux linkage equations become
@@ Line 567: / Line 570: @@
 \[
 \begin{align*}
 \underline{\Psi}_1
-&=
+&= \underbrace{{\color{blue  }{L_{1{\rm H}}\underline{I}_1 + M\underline{I}_2}}  }_{\text{main magnetic path}}
-\underbrace{{\color{blue}{L_{1{\rm H}}\underline{I}_1+M\underline{I}_2}}}_{\text{main magnetic path}}
++  \underbrace{{\color{orange}{L_{1\sigma }\underline{I}_1}}                     }_{\text{primary leakage}   },
-+
-\underbrace{{\color{orange}{L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage}},
 \\[4pt]
 \underline{\Psi}_2
-&=
+&= \underbrace{{\color{blue  }{M\underline{I}_1 + L_{2{\rm H}}\underline{I}_2 }} }_{\text{main magnetic path}}
-\underbrace{{\color{blue}{L_{2{\rm H}}\underline{I}_2+M\underline{I}_1}}}_{\text{main magnetic path}}
++  \underbrace{{\color{orange}{L_{2\sigma }\underline{I}_2}}                     }_{\text{secondary leakage} }.
-+
-\underbrace{{\color{orange}{L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage}}.
 \end{align*}
 \]
@@ Line 626: / Line 625: @@
 +
 \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 $N_1$, $N_2$:
+\[
+\begin{align*}
+\boxed{
+L_{1 \rm H} = \frac{N_1^2}{R_{\rm mFe} }  \\
+L_{2 \rm H} = \frac{N_2^2}{R_{\rm mFe} }  \\
+M = \frac{N_1 N_2}{R_{\rm mFe} }  \\
+}
 \end{align*}
 \]
@@ Line 697: / Line 708: @@
 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 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.
@@ Line 742: / Line 753: @@
 </WRAP>
-The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{u}\):
+The technical voltage ratio is often defined from the no-load voltages. Here it is denoted by \(\ddot{\rm u}\):
 \[
@@ Line 772: / Line 783: @@
 \[
 \begin{align*}
-\ddot{\ rm u}\neq n,
+\ddot{\rm u}\neq n,
 \end{align*}
 \]
@@ Line 795: / Line 806: @@
 \[
 \begin{align*}
-| jX_{1{\rm H}} \; || \;R_{\rm Fe} |
+\Big|\; jX_{1{\rm H}} \; || \;R_{\rm Fe} \;\Big|
 \;\; \gg \;\;
-| jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 | .
+\Big|\; jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 \;\Big| .
 \end{align*}
 \]
@@ Line 867: / Line 878: @@
 \]
-where \(u_{\rm k}\) is inserted as a percentage value.
+where \(\rm u_k\) is inserted as a percentage value.
 <WRAP column 100%>
@@ Line 911: / Line 922: @@
 \begin{align*}
 \boxed{
-\underline{U_1}
+\underline{U}_1
 -
 \underline{U}_{\rm k}
 =
-\underline{U'_2}
+\underline{U}'_2
 =
-n \underline{U_2}
+n \underline{U}_2
 }
 \end{align*}
@@ Line 1012: / Line 1023: @@
 ===== Exercises =====
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: ideal transformer voltage and current ratio
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: ideal transformer voltage and current ratio
 #@TaskText_HTML@#
-A transformer has \(N_1=1200\) turns and \(N_2=300\) turns.
+A transformer has $N_1=1200$ turns and $N_2=300$ turns.  \\
-The primary RMS voltage is \(U_1=230~{\rm V}\).
+The primary RMS voltage is $U_1=230~{\rm V}$.  \\
-The secondary side supplies a load current \(I_2=2.0~{\rm A}\).
+The secondary side supplies a load current $I_2=2.0~{\rm A}$.
-  * Calculate the turns ratio \(n\).
+. Calculate the turns ratio $n$.
-  * Calculate the ideal secondary voltage \(U_2\).
+<WRAP group>
-  * Calculate the magnitude of the ideal primary current \(I_1\).
+<WRAP half column rightalign>
-  * State whether this is a step-up or step-down transformer.
+#@PathBegin_HTML~2001~@#
+<WRAP leftalign>
-#@ResultBegin_HTML~Exercise1~@#
+The turns ratio of an ideal transformer is defined as:
-\[
 \begin{align*}
 n=\frac{N_1}{N_2}
-=
+\end{align*}
+Insert the given values:
+\begin{align*}
+n
+&=
 \frac{1200}{300}
-=
+\\
-.
+&=
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2001~@#
+\begin{align*}
+n=4
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-The secondary voltage is
+. 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*}
-\[
+Therefore:
 \begin{align*}
 U_2
-=
+&=
 \frac{U_1}{n}
-=
+\\
+&=
 \frac{230~{\rm V}}{4}
-=
+\\
-.5~{\rm V}.
+&=
+.5~{\rm V}
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2002~@#
+\begin{align*}
+U_2=57.5~{\rm V}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-The primary current magnitude is
+. 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*}
-\[
+Insert the values:
 \begin{align*}
 I_1
-=
+&=
-\frac{I_2}{n}
-=
 \frac{2.0~{\rm A}}{4}
-=
+\\
-.50~{\rm A}.
+&=
+.50~{\rm A}
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2003~@#
+\begin{align*}
+I_1=0.50~{\rm A}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-Because \(U_2<U_1\), it is a step-down transformer.
+. 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#~~.1  Quick check: mutual inductance from reluctance
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: mutual inductance from reluctance
 #@TaskText_HTML@#
@@ Line 1075: / Line 1163: @@
 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 \(N_2=100\).
+The number of turns is $N_1=500$ and $N_2=100$.
-  * Calculate \(L_{1{\rm H}}\).
+. Calculate $L_{1{\rm H}}$.
-  * Calculate \(L_{2{\rm H}}\).
+<WRAP group>
-  * Calculate \(M\).
+<WRAP half column rightalign>
-  * Check whether the units are correct.
+#@PathBegin_HTML~2005~@#
+<WRAP leftalign>
-#@ResultBegin_HTML~Exercise2~@#
+The main-flux inductance of coil 1 is:
-\[
 \begin{align*}
 L_{1{\rm H}}
 =
 \frac{N_1^2}{R_{\rm mH}}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+L_{1{\rm H}}
+&=
 \frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
-=
+\\
-.125~{\rm H}.
+&=
+.125~{\rm H}
 \end{align*}
-\]
+</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>
-\[
+. Calculate $L_{2{\rm H}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2006~@#
+<WRAP leftalign>
+The main-flux inductance of coil 2 is:
 \begin{align*}
 L_{2{\rm H}}
 =
 \frac{N_2^2}{R_{\rm mH}}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+L_{2{\rm H}}
+&=
 \frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
+\\
+&=
+.0050~{\rm H}
+\\
+&=
+.0~{\rm mH}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2006~@#
+\begin{align*}
+L_{2{\rm H}}
 =
 .0050~{\rm H}
 =
-.0~{\rm mH}.
+.0~{\rm mH}
 \end{align*}
-\]
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-\[
+. Calculate $M$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2007~@#
+<WRAP leftalign>
+The mutual inductance is:
 \begin{align*}
 M
 =
 \frac{N_1N_2}{R_{\rm mH}}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+M
+&=
 \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
+\\
+&=
+.025~{\rm H}
+\\
+&=
+~{\rm mH}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2007~@#
+\begin{align*}
+M
 =
 .025~{\rm H}
 =
-~{\rm mH}.
+~{\rm mH}
 \end{align*}
-\]
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-The unit is correct because \(1/(1/{\rm H})={\rm H}\).
+. 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#~~.1  Quick check: referring secondary quantities to the primary side
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Mutual inductance and leakage from a magnetic path
 #@TaskText_HTML@#
-A transformer has \(n=5\).
+Two coils are wound on the same magnetic core. The shared main magnetic path has the reluctance
-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.
+\begin{align*}
+R_{\rm mH}=1.6\cdot 10^6~\frac{1}{\rm H}.
+\end{align*}
-#@ResultBegin_HTML~Exercise3~@#
+The numbers of turns are
-\[
 \begin{align*}
-R'_2
+N_1=400,
+\qquad
+N_2=100.
+\end{align*}
+The leakage inductances are
+\begin{align*}
+L_{1\sigma}=4.0~{\rm mH},
+\qquad
+L_{2\sigma}=0.30~{\rm mH}.
+\end{align*}
+. Calculate $L_{1{\rm H}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2009~@#
+<WRAP leftalign>
+The main-flux self-inductance of coil 1 is:
+\begin{align*}
+L_{1{\rm H}}
 =
-n^2R_2
+\frac{N_1^2}{R_{\rm mH}}
+\end{align*}
+Insert the values:
+\begin{align*}
+L_{1{\rm H}}
+&=
+\frac{400^2}{1.6\cdot 10^6~1/{\rm H}}
+\\
+&=
+.100~{\rm H}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2009~@#
+\begin{align*}
+L_{1{\rm H}}=0.100~{\rm H}=100~{\rm mH}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate $L_{2{\rm H}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2010~@#
+<WRAP leftalign>
+The main-flux self-inductance of coil 2 is:
+\begin{align*}
+L_{2{\rm H}}
 =
-^2\cdot 0.20~\Omega
+\frac{N_2^2}{R_{\rm mH}}
+\end{align*}
+Insert the values:
+\begin{align*}
+L_{2{\rm H}}
+&=
+\frac{100^2}{1.6\cdot 10^6~1/{\rm H}}
+\\
+&=
+.00625~{\rm H}
+\\
+&=
+.25~{\rm mH}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2010~@#
+\begin{align*}
+L_{2{\rm H}}
 =
-\cdot 0.20~\Omega
+.00625~{\rm H}
 =
-.0~\Omega.
+.25~{\rm mH}
 \end{align*}
-\]
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-\[
+. Calculate $M$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2011~@#
+<WRAP leftalign>
+The mutual inductance is:
 \begin{align*}
-X'_{2\sigma}
+M
 =
-n^2X_{2\sigma}
+\frac{N_1N_2}{R_{\rm mH}}
+\end{align*}
+Insert the values:
+\begin{align*}
+M
+&=
+\frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}}
+\\
+&=
+.025~{\rm H}
+\\
+&=
+~{\rm mH}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2011~@#
+\begin{align*}
+M
 =
-^2\cdot 0.35~\Omega
+.025~{\rm H}
 =
-\cdot 0.35~\Omega
+~{\rm mH}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the total self-inductances $L_1$ and $L_2$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2012~@#
+<WRAP leftalign>
+The total self-inductance is the sum of main-flux inductance and leakage inductance.
+For coil 1:
+\begin{align*}
+L_1
+&=
+L_{1{\rm H}}+L_{1\sigma}
+\\
+&=
+~{\rm mH}+4.0~{\rm mH}
+\\
+&=
+~{\rm mH}
+\end{align*}
+For coil 2:
+\begin{align*}
+L_2
+&=
+L_{2{\rm H}}+L_{2\sigma}
+\\
+&=
+.25~{\rm mH}+0.30~{\rm mH}
+\\
+&=
+.55~{\rm mH}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2012~@#
+\begin{align*}
+L_1 &= 104~{\rm mH}
+\\
+L_2 &= 6.55~{\rm mH}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Calculate the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2013~@#
+<WRAP leftalign>
+The coupling coefficient is:
+\begin{align*}
+k
 =
-.75~\Omega.
+\frac{M}{\sqrt{L_1L_2}}
 \end{align*}
-\]
-The unit remains \(\Omega\), because \(n\) is dimensionless.
+Insert the values in henry:
+\begin{align*}
+k
+&=
+\frac{0.025~{\rm H}}{\sqrt{0.104~{\rm H}\cdot 0.00655~{\rm H}}}
+\\
+&\approx
+.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#~~.1  Quick check: short-circuit voltage and fault current
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Quick check: referring secondary quantities to the primary side
 #@TaskText_HTML@#
-A transformer has a rated primary current \(I_{1{\rm N}}=10~{\rm A}\) and a short-circuit voltage \(u_{\rm k}=5~\%\).
+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 continuous short-circuit current \(I_{1{\rm k}}\) when rated primary voltage is applied.
+Calculate the values $R'_2$ and $X'_{2\sigma}$ referred to the primary side.
-  * Estimate the initial peak short-circuit current \(i_{\rm p}\) using \(i_{\rm p}\approx 2.54 I_{1{\rm k}}\).
-#@ResultBegin_HTML~Exercise4~@#
+. 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
+&=
+^2\cdot 0.20~\Omega
+\\
+&=
+\cdot 0.20~\Omega
+\\
+&=
+.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>
+. 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}
+&=
+^2\cdot 0.35~\Omega
+\\
+&=
+\cdot 0.35~\Omega
+\\
+&=
+.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>
+. 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~\%$.
+. 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~\%}{u_{\rm k}}
+I_{1{\rm N}}\cdot \frac{100~\%}{\rm u_k}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+I_{1{\rm k}}
+&=
 ~{\rm A}\cdot \frac{100~\%}{5~\%}
-=
+\\
-~{\rm A}.
+&=
+~{\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>
-\[
+. Estimate the initial peak short-circuit current $i_{\rm p}$ using $i_{\rm p}\approx 2.54 I_{1{\rm k}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2018~@#
+<WRAP leftalign>
+The initial peak current is estimated by:
 \begin{align*}
 i_{\rm p}
 \approx
 .54\cdot I_{1{\rm k}}
-=
+\end{align*}
+Insert the continuous short-circuit current:
+\begin{align*}
+i_{\rm p}
+&\approx
 .54\cdot 200~{\rm A}
-=
+\\
-~{\rm A}.
+&=
+~{\rm A}
 \end{align*}
-\]
 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#~~.1  Longer exercise: transformer equivalent circuit for an actuator supply
+#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~  Longer exercise: transformer equivalent circuit for an actuator supply
 #@TaskText_HTML@#
@@ Line 1223: / Line 1744: @@
 Rated data and equivalent circuit data are:
-\[
 \begin{align*}
 U_{1{\rm N}}&=230~{\rm V},
@@ Line 1239: / Line 1759: @@
 X_{2\sigma}&=0.018~\Omega.
 \end{align*}
-\]
-Assume \(n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}\). The magnetizing branch is neglected for the loaded operating point.
+Assume $n=\frac{U_{1{\rm N}}}{U_{2{\rm N}}}$. The magnetizing branch is neglected for the loaded operating point.
-  * Calculate \(n\).
+. Calculate $n$.
-  * Refer \(R_2\) and \(X_{2\sigma}\) to the primary side.
+<WRAP group>
-  * Calculate \(R_{\rm k}\) and \(X_{\rm k}\).
+<WRAP half column rightalign>
-  * Calculate the primary rated current magnitude \(I_{1{\rm N}}\) using the ideal current ratio.
+#@PathBegin_HTML~2019~@#
-  * Estimate the magnitude of the internal voltage drop \(U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}\).
+<WRAP leftalign>
+The turns ratio is estimated from the rated voltages:
-#@ResultBegin_HTML~Exercise5~@#
-The turns ratio is
-\[
 \begin{align*}
 n
 =
 \frac{U_{1{\rm N}}}{U_{2{\rm N}}}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+n
+&=
 \frac{230~{\rm V}}{23~{\rm V}}
-=
+\\
-.
+&=
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2019~@#
+\begin{align*}
+n=10
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
-The secondary quantities referred to the primary side are
+. 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*}
-\[
+With $n=10$:
 \begin{align*}
 R'_2
 &=
-n^2R_2
-=
 ^2\cdot 0.012~\Omega
-=
-.2~\Omega,
 \\
+&=
+.2~\Omega
+\\[4pt]
 X'_{2\sigma}
 &=
-n^2X_{2\sigma}
-=
 ^2\cdot 0.018~\Omega
-=
+\\
-.8~\Omega.
+&=
+.8~\Omega
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
-Therefore
+</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>
-\[
+. Calculate $R_{\rm k}$ and $X_{\rm k}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2021~@#
+<WRAP leftalign>
+The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities:
 \begin{align*}
 R_{\rm k}
 &=
 R_1+R'_2
-=
-.2~\Omega+1.2~\Omega
-=
-.4~\Omega,
 \\
 X_{\rm k}
 &=
 X_{1\sigma}+X'_{2\sigma}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+R_{\rm k}
+&=
+.2~\Omega+1.2~\Omega
+\\
+&=
+.4~\Omega
+\\[4pt]
+X_{\rm k}
+&=
 .8~\Omega+1.8~\Omega
-=
+\\
-.6~\Omega.
+&=
+.6~\Omega
 \end{align*}
-\]
+</WRAP>
+#@PathEnd_HTML@#
-The primary current magnitude is
+</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>
-\[
+. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2022~@#
+<WRAP leftalign>
+For an ideal transformer, the primary current magnitude is:
 \begin{align*}
 I_{1{\rm N}}
 =
 \frac{I_{2{\rm N}}}{n}
-=
+\end{align*}
+Insert the values:
+\begin{align*}
+I_{1{\rm N}}
+&=
 \frac{5.0~{\rm A}}{10}
+\\
+&=
+.50~{\rm A}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2022~@#
+\begin{align*}
+I_{1{\rm N}}=0.50~{\rm A}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2023~@#
+<WRAP leftalign>
+First calculate the magnitude of the short-circuit impedance:
+\begin{align*}
+|\underline{Z}_{\rm k}|
 =
-.50~{\rm A}.
+\sqrt{R_{\rm k}^2+X_{\rm k}^2}
 \end{align*}
-\]
-The magnitude of the short-circuit impedance is
+Insert the values:
+\begin{align*}
+|\underline{Z}_{\rm k}|
+&=
+\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2}
+\\
+&=
+.33~\Omega
+\end{align*}
-\[
+Now calculate the internal voltage drop:
+\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:
+\begin{align*}
+\frac{2.17~{\rm V}}{10}
+=
+.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*}
+. 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}
+\\
+&=
+.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>
+. 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}}
 =
-\sqrt{(2.4~\Omega)^2+(3.6~\Omega)^2}
+|\underline{Z}_{\rm k}|I_{1{\rm N}}
+\end{align*}
+Insert the values:
+\begin{align*}
+U_{1{\rm k}}
+&=
+.27~\Omega\cdot 2.0~{\rm A}
+\\
+&=
+.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>
+. 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}
 =
-.33~\Omega.
+\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
 \end{align*}
-\]
-Thus the internal voltage drop estimate is
+Insert the values:
+\begin{align*}
+{\rm u_k}
+&=
+\frac{8.54~{\rm V}}{230~{\rm V}}\cdot 100~\%
+\\
+&=
+.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>
-\[
+. 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*}
-U_{\rm k}
+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}}
+&=
+.0~{\rm A}\cdot \frac{100}{3.71}
+\\
+&=
+.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>
+. 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
-|\underline{Z}_{\rm k}|I_{1{\rm N}}
+.54 I_{1{\rm k}}
+\end{align*}
+Insert the short-circuit current:
+\begin{align*}
+i_{\rm peak}
+&\approx
+.54\cdot 53.9~{\rm A}
+\\
+&=
+~{\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*}
+. 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
 =
-.33~\Omega\cdot 0.50~{\rm A}
+\frac{I_2}{n}
+\end{align*}
+Insert the values:
+\begin{align*}
+I_1
+&=
+\frac{4.0~{\rm A}}{10}
+\\
+&=
+.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>
+. 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
 =
-.17~{\rm V}.
+\sqrt{1-\cos^2\varphi}
 \end{align*}
-\]
-This is a primary-side voltage drop. On the secondary side it corresponds approximately to
+Insert the value:
+\begin{align*}
+\sin\varphi
+&=
+\sqrt{1-0.8^2}
+\\
+&=
+.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>
-\[
+. Estimate the primary-side voltage drop $\Delta U_1$.
+<WRAP group>
+<WRAP half column rightalign>
+#@PathBegin_HTML~2031~@#
+<WRAP leftalign>
+Use the Kapp approximation:
 \begin{align*}
-\frac{2.17~{\rm V}}{10}=0.217~{\rm V}.
+\Delta U_1
+&\approx
+I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
 \end{align*}
-\]
-For a \(23~{\rm V}\) actuator supply this is small but not zero.
+Insert the values:
+\begin{align*}
+\Delta U_1
+&\approx
+.40~{\rm A}
+\left(
+.4~\Omega\cdot 0.8
++
+.6~\Omega\cdot 0.6
+\right)
+\\
+&=
+.40~{\rm A}
+\left(
+.92~\Omega+2.16~\Omega
+\right)
+\\
+&=
+.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>
+. 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}
+\\
+&=
+.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.
+\\ \\ \\
+. 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}}
+&=
+.06\cdot 230~{\rm V}
+\\
+&=
+.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>
+. 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}
+&=
+.12~{\rm A}\cdot \frac{13.8~{\rm V}}{230~{\rm V}}
+\\
+&=
+.0072~{\rm A}
+\\
+&=
+.2~{\rm mA}
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2034~@#
+\begin{align*}
+I_{\rm m,k}
+=
+.0072~{\rm A}
+=
+.2~{\rm mA}
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. 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}}}
+&=
+.0024
+\\
+&=
+.24~\%
+\end{align*}
+</WRAP>
+#@PathEnd_HTML@#
+</WRAP>
+<WRAP half column>
+#@ResultBegin_HTML~2035~@#
+\begin{align*}
+\frac{I_{\rm m,k}}{I_{1{\rm N}}}
+=
+.24~\%
+\end{align*}
+#@ResultEnd_HTML@#
+</WRAP>
+</WRAP>
+. 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 $I_{10}=0.03~{\rm A}$.
+\\ \\ \\
+. 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}
+\\
+&=
+.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>
+. 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}
+\\
+&=
+.6
+\end{align*}
+Insert the values:
+\begin{align*}
+\Delta U_1
+&\approx
+.50~{\rm A}
+\left(
+.0~\Omega\cdot 0.8
++
+.0~\Omega\cdot 0.6
+\right)
+\\
+&=
+.50~{\rm A}
+\left(
+.6~\Omega+2.4~\Omega
+\right)
+\\
+&=
+.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>
+. 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}
+\\
+&=
+.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>
+. 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}
+\\
+&=
+.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}
+\\
+&=
+.47~\Omega\cdot 0.03~{\rm A}
+\\
+&=
+.134~{\rm V}
+\end{align*}
+On the secondary side:
+\begin{align*}
+\Delta U_{2,10}
+&\leq
+\frac{0.134~{\rm V}}{10}
+\\
+&=
+.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>
+. 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 $\cos\varphi=1$.
+\\ \\
+. 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}
+\\
+&=
+.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>
+. 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}
+\\
+&=
+.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>
+. 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
+.40~{\rm A}\cdot 2.4~\Omega
+\\
+&=
+.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}
+\\
+&=
+.096~{\rm V}
+\end{align*}
+Thus the real secondary voltage is approximately:
+\begin{align*}
+U_{2,\rm real}
+&\approx
+.0~{\rm V}-0.096~{\rm V}
+\\
+&=
+.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>
+. 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
+.4~\Omega\cdot (0.40~{\rm A})^2
+\\
+&=
+.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>
+. 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@#
@@ Line 1370: / Line 2954: @@
   * **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.
+  * **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 \(u_{\rm k}\) as a voltage in volts:** \(u_{\rm k}\) is normally given in percent. Insert it consistently in formulas.
+  * **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.