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electrical_engineering_and_electronics_2:block09 [2026/05/17 08:29] – mexleadminelectrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) – mexleadmin
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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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     * \(R_{\rm Fe}\): 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>
  
Line 277: Line 277:
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~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 ==== ==== Linked fluxes and mutual inductance ====
  
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 \] \]
  
-Often, the self-inductances are shortened:+Often, the self-inductances are abbreviated:
 \[ \[
 \begin{align*} \begin{align*}
Line 544: Line 442:
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~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 ==== ==== Real transformer: leakage and losses ====
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 In a real transformer, not all flux links both windings. 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 **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 **secondary leakage flux** \(\Phi_{2\sigma}\) mainly links only the secondary winding.
Line 561: Line 547:
 </panel> </panel>
 </WRAP> </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 The real flux linkage equations become
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 \[ \[
 \begin{align*} \begin{align*}
-\underline{\Psi}_1 +\underline{\Psi}_1  
-&= +&= \underbrace{{\color{blue  }{L_{1{\rm H}}\underline{I}_1 + M\underline{I}_2}}  }_{\text{main magnetic path}} 
-\underbrace{{\color{blue}{L_{1{\rm H}}\underline{I}_1+M\underline{I}_2}}}_{\text{main magnetic path}} ++  \underbrace{{\color{orange}{L_{1\sigma }\underline{I}_1}}                     }_{\text{primary leakage}   },
-+ +
-\underbrace{{\color{orange}{L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage}},+
 \\[4pt] \\[4pt]
-\underline{\Psi}_2 +\underline{\Psi}_2  
-&= +&= \underbrace{{\color{blue  }{M\underline{I}_1 + L_{2{\rm H}}\underline{I}_2 }} }_{\text{main magnetic path}} 
-\underbrace{{\color{blue}{L_{2{\rm H}}\underline{I}_2+M\underline{I}_1}}}_{\text{main magnetic path}} ++  \underbrace{{\color{orange}{L_{2\sigma }\underline{I}_2}}                     }_{\text{secondary leakage} }.
-+ +
-\underbrace{{\color{orange}{L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage}}.+
 \end{align*} \end{align*}
 \] \]
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 + +
 \underbrace{{\color{blue}{j\omega L_{2{\rm H}}\underline{I}_2+j\omega M\underline{I}_1}}}_{\text{main magnetic coupling}}. \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 $N_1$, $N_2$:
 +
 +\[
 +\begin{align*}
 +\boxed{
 +L_{1 \rm H} = \frac{N_1^2}{R_{\rm mFe} }  \\
 +L_{2 \rm H} = \frac{N_2^2}{R_{\rm mFe} }  \\
 +M = \frac{N_1 N_2}{R_{\rm mFe} }  \\
 +}
 \end{align*} \end{align*}
 \] \]
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 The phasor diagram is shown in <imgref fig_phasor_eq_circ01>. \\  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.    * The lower one shows all voltages and currents. 
  
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 </WRAP> </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 \(\ddot{\rm u}\):
  
 \[ \[
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 \[ \[
 \begin{align*} \begin{align*}
-\ddot{\ rm u}\neq n,+\ddot{\rm u}\neq n,
 \end{align*} \end{align*}
 \] \]
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 \[ \[
 \begin{align*} \begin{align*}
-| jX_{1{\rm H}} \; || \;R_{\rm Fe} |+\Big|\; jX_{1{\rm H}} \; || \;R_{\rm Fe} \;\Big|
 \;\; \gg \;\; \;\; \gg \;\;
-| jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 | .+\Big|\; jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 \;\Big| .
 \end{align*} \end{align*}
 \] \]
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 \] \]
  
-where \(u_{\rm k}\) is inserted as a percentage value.+where \(\rm u_k\) is inserted as a percentage value.
  
 <WRAP column 100%> <WRAP column 100%>
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 \begin{align*} \begin{align*}
 \boxed{ \boxed{
-\underline{U_1}+\underline{U}_1
 - -
 \underline{U}_{\rm k} \underline{U}_{\rm k}
 = =
-\underline{U'_2}+\underline{U}'_2
 = =
-n \underline{U_2}+n \underline{U}_2
 } }
 \end{align*} \end{align*}
Line 1011: Line 1023:
 ===== Exercises ===== ===== 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@# #@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 \(n\). +1. 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:
- +
-\[+
 \begin{align*} \begin{align*}
 n=\frac{N_1}{N_2} n=\frac{N_1}{N_2}
-=+\end{align*} 
 + 
 +Insert the given values: 
 +\begin{align*} 
 +n 
 +&=
 \frac{1200}{300} \frac{1200}{300}
-= +\\ 
-4.+&= 
 +4
 \end{align*} \end{align*}
-\]+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2001~@# 
 +\begin{align*} 
 +n=4 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The secondary voltage is+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*}
  
-\[+Therefore:
 \begin{align*} \begin{align*}
 U_2 U_2
-=+&=
 \frac{U_1}{n} \frac{U_1}{n}
-=+\\ 
 +&=
 \frac{230~{\rm V}}{4} \frac{230~{\rm V}}{4}
-= +\\ 
-57.5~{\rm V}.+&= 
 +57.5~{\rm V}
 \end{align*} \end{align*}
-\]+</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+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*}
  
-\[+Insert the values:
 \begin{align*} \begin{align*}
 I_1 I_1
-= +&=
-\frac{I_2}{n} +
-=+
 \frac{2.0~{\rm A}}{4} \frac{2.0~{\rm A}}{4}
-= +\\ 
-0.50~{\rm A}.+&= 
 +0.50~{\rm A}
 \end{align*} \end{align*}
-\]+</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.+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@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@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@# #@TaskText_HTML@#
  
Line 1074: Line 1163:
 The main magnetic reluctance is The main magnetic reluctance is
  
-\[ 
 \begin{align*} \begin{align*}
 R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}. R_{\rm mH}=2.0\cdot 10^6~\frac{1}{\rm H}.
 \end{align*} \end{align*}
-\] 
  
-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 \(L_{1{\rm H}}\). +1. 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:
- +
-\[+
 \begin{align*} \begin{align*}
 L_{1{\rm H}} L_{1{\rm H}}
 = =
 \frac{N_1^2}{R_{\rm mH}} \frac{N_1^2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{1{\rm H}} 
 +&=
 \frac{500^2}{2.0\cdot 10^6~1/{\rm H}} \frac{500^2}{2.0\cdot 10^6~1/{\rm H}}
-= +\\ 
-0.125~{\rm H}.+&= 
 +0.125~{\rm H}
 \end{align*} \end{align*}
-\]+</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>
  
-\[+2. Calculate $L_{2{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2006~@# 
 +<WRAP leftalign> 
 +The main-flux inductance of coil 2 is:
 \begin{align*} \begin{align*}
 L_{2{\rm H}} L_{2{\rm H}}
 = =
 \frac{N_2^2}{R_{\rm mH}} \frac{N_2^2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{2{\rm H}} 
 +&=
 \frac{100^2}{2.0\cdot 10^6~1/{\rm H}} \frac{100^2}{2.0\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.0050~{\rm H}
 +\\
 +&=
 +5.0~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2006~@#
 +\begin{align*}
 +L_{2{\rm H}}
 = =
 0.0050~{\rm H} 0.0050~{\rm H}
 = =
-5.0~{\rm mH}.+5.0~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+3. Calculate $M$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2007~@# 
 +<WRAP leftalign> 
 +The mutual inductance is:
 \begin{align*} \begin{align*}
 M M
 = =
 \frac{N_1N_2}{R_{\rm mH}} \frac{N_1N_2}{R_{\rm mH}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +M 
 +&=
 \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}} \frac{500\cdot 100}{2.0\cdot 10^6~1/{\rm H}}
 +\\
 +&=
 +0.025~{\rm H}
 +\\
 +&=
 +25~{\rm mH}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2007~@#
 +\begin{align*}
 +M
 = =
 0.025~{\rm H} 0.025~{\rm H}
 = =
-25~{\rm mH}.+25~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-The unit is correct because \(1/(1/{\rm H})={\rm H}\).+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@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@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@# #@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
  
-\[ 
 \begin{align*} \begin{align*}
-R'_2+N_1=400, 
 +\qquad 
 +N_2=100. 
 +\end{align*} 
 + 
 +The leakage inductances are 
 + 
 +\begin{align*} 
 +L_{1\sigma}=4.0~{\rm mH}, 
 +\qquad 
 +L_{2\sigma}=0.30~{\rm mH}. 
 +\end{align*} 
 + 
 +1. Calculate $L_{1{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2009~@# 
 +<WRAP leftalign> 
 +The main-flux self-inductance of coil 1 is: 
 +\begin{align*} 
 +L_{1{\rm H}}
 = =
-n^2R_2+\frac{N_1^2}{R_{\rm mH}} 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{1{\rm H}} 
 +&= 
 +\frac{400^2}{1.6\cdot 10^6~1/{\rm H}} 
 +\\ 
 +&= 
 +0.100~{\rm H} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2009~@# 
 +\begin{align*} 
 +L_{1{\rm H}}=0.100~{\rm H}=100~{\rm mH} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Calculate $L_{2{\rm H}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2010~@# 
 +<WRAP leftalign> 
 +The main-flux self-inductance of coil 2 is: 
 +\begin{align*} 
 +L_{2{\rm H}}
 = =
-5^2\cdot 0.20~\Omega+\frac{N_2^2}{R_{\rm mH}} 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +L_{2{\rm H}} 
 +&= 
 +\frac{100^2}{1.6\cdot 10^6~1/{\rm H}} 
 +\\ 
 +&= 
 +0.00625~{\rm H} 
 +\\ 
 +&= 
 +6.25~{\rm mH} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2010~@# 
 +\begin{align*} 
 +L_{2{\rm H}}
 = =
-25\cdot 0.20~\Omega+0.00625~{\rm H}
 = =
-5.0~\Omega.+6.25~{\rm mH}
 \end{align*} \end{align*}
-\]+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+3. Calculate $M$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2011~@# 
 +<WRAP leftalign> 
 +The mutual inductance is:
 \begin{align*} \begin{align*}
-X'_{2\sigma}+M
 = =
-n^2X_{2\sigma}+\frac{N_1N_2}{R_{\rm mH}} 
 +\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +M 
 +&= 
 +\frac{400\cdot 100}{1.6\cdot 10^6~1/{\rm H}} 
 +\\ 
 +&= 
 +0.025~{\rm H} 
 +\\ 
 +&= 
 +25~{\rm mH} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2011~@# 
 +\begin{align*} 
 +M
 = =
-5^2\cdot 0.35~\Omega+0.025~{\rm H}
 = =
-25\cdot 0.35~\Omega+25~{\rm mH} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +4. Calculate the total self-inductances $L_1$ and $L_2$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2012~@# 
 +<WRAP leftalign> 
 +The total self-inductance is the sum of main-flux inductance and leakage inductance. 
 + 
 +For coil 1: 
 +\begin{align*} 
 +L_1 
 +&= 
 +L_{1{\rm H}}+L_{1\sigma} 
 +\\ 
 +&= 
 +100~{\rm mH}+4.0~{\rm mH} 
 +\\ 
 +&= 
 +104~{\rm mH} 
 +\end{align*} 
 + 
 +For coil 2: 
 +\begin{align*} 
 +L_2 
 +&= 
 +L_{2{\rm H}}+L_{2\sigma} 
 +\\ 
 +&= 
 +6.25~{\rm mH}+0.30~{\rm mH} 
 +\\ 
 +&= 
 +6.55~{\rm mH} 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2012~@# 
 +\begin{align*} 
 +L_1 &= 104~{\rm mH} 
 +\\ 
 +L_2 &= 6.55~{\rm mH} 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +5. Calculate the coupling coefficient $k=\frac{M}{\sqrt{L_1L_2}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2013~@# 
 +<WRAP leftalign> 
 +The coupling coefficient is: 
 +\begin{align*} 
 +k
 = =
-8.75~\Omega.+\frac{M}{\sqrt{L_1L_2}}
 \end{align*} \end{align*}
-\] 
  
-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 
 +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@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@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@# #@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~@#+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*} \begin{align*}
 I_{1{\rm k}} 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}} 
 +&=
 10~{\rm A}\cdot \frac{100~\%}{5~\%} 10~{\rm A}\cdot \frac{100~\%}{5~\%}
-= +\\ 
-200~{\rm A}.+&= 
 +200~{\rm A}
 \end{align*} \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>
  
-\[+2. Estimate the initial peak short-circuit current $i_{\rm p}$ using $i_{\rm p}\approx 2.54 I_{1{\rm k}}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2018~@# 
 +<WRAP leftalign> 
 +The initial peak current is estimated by:
 \begin{align*} \begin{align*}
 i_{\rm p} i_{\rm p}
 \approx \approx
 2.54\cdot I_{1{\rm k}} 2.54\cdot I_{1{\rm k}}
-=+\end{align*} 
 + 
 +Insert the continuous short-circuit current: 
 +\begin{align*} 
 +i_{\rm p} 
 +&\approx
 2.54\cdot 200~{\rm A} 2.54\cdot 200~{\rm A}
-= +\\ 
-508~{\rm A}.+&= 
 +508~{\rm A}
 \end{align*} \end{align*}
-\] 
  
 The short-circuit current is much larger than the rated current. Protection devices must be selected accordingly. 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@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@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@# #@TaskText_HTML@#
  
Line 1222: Line 1744:
 Rated data and equivalent circuit data are: Rated data and equivalent circuit data are:
  
-\[ 
 \begin{align*} \begin{align*}
 U_{1{\rm N}}&=230~{\rm V}, U_{1{\rm N}}&=230~{\rm V},
Line 1238: Line 1759:
 X_{2\sigma}&=0.018~\Omega. X_{2\sigma}&=0.018~\Omega.
 \end{align*} \end{align*}
-\] 
  
-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 \(n\). +1. 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 +
- +
-\[+
 \begin{align*} \begin{align*}
 n n
 = =
 \frac{U_{1{\rm N}}}{U_{2{\rm N}}} \frac{U_{1{\rm N}}}{U_{2{\rm N}}}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +n 
 +&=
 \frac{230~{\rm V}}{23~{\rm V}} \frac{230~{\rm V}}{23~{\rm V}}
-= +\\ 
-10.+&= 
 +10
 \end{align*} \end{align*}
-\]+</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+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*}
  
-\[+With $n=10$:
 \begin{align*} \begin{align*}
 R'_2 R'_2
 &= &=
-n^2R_2 
-= 
 10^2\cdot 0.012~\Omega 10^2\cdot 0.012~\Omega
-= 
-1.2~\Omega, 
 \\ \\
 +&=
 +1.2~\Omega
 +\\[4pt]
 X'_{2\sigma} X'_{2\sigma}
 &= &=
-n^2X_{2\sigma} 
-= 
 10^2\cdot 0.018~\Omega 10^2\cdot 0.018~\Omega
-= +\\ 
-1.8~\Omega.+&= 
 +1.8~\Omega
 \end{align*} \end{align*}
-\] +</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>
  
-\[+3. Calculate $R_{\rm k}$ and $X_{\rm k}$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2021~@# 
 +<WRAP leftalign> 
 +The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities:
 \begin{align*} \begin{align*}
 R_{\rm k} R_{\rm k}
 &= &=
 R_1+R'_2 R_1+R'_2
-= 
-1.2~\Omega+1.2~\Omega 
-= 
-2.4~\Omega, 
 \\ \\
 X_{\rm k} X_{\rm k}
 &= &=
 X_{1\sigma}+X'_{2\sigma} X_{1\sigma}+X'_{2\sigma}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +R_{\rm k} 
 +&= 
 +1.2~\Omega+1.2~\Omega 
 +\\ 
 +&= 
 +2.4~\Omega 
 +\\[4pt] 
 +X_{\rm k} 
 +&=
 1.8~\Omega+1.8~\Omega 1.8~\Omega+1.8~\Omega
-= +\\ 
-3.6~\Omega.+&= 
 +3.6~\Omega
 \end{align*} \end{align*}
-\] +</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>
  
-\[+4. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2022~@# 
 +<WRAP leftalign> 
 +For an ideal transformer, the primary current magnitude is:
 \begin{align*} \begin{align*}
 I_{1{\rm N}} I_{1{\rm N}}
 = =
 \frac{I_{2{\rm N}}}{n} \frac{I_{2{\rm N}}}{n}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{1{\rm N}} 
 +&=
 \frac{5.0~{\rm A}}{10} \frac{5.0~{\rm A}}{10}
 +\\
 +&=
 +0.50~{\rm A}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2022~@#
 +\begin{align*}
 +I_{1{\rm N}}=0.50~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2023~@#
 +<WRAP leftalign>
 +First calculate the magnitude of the short-circuit impedance:
 +\begin{align*}
 +|\underline{Z}_{\rm k}|
 = =
-0.50~{\rm A}.+\sqrt{R_{\rm k}^2+X_{\rm k}^2}
 \end{align*} \end{align*}
-\] 
  
-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} 
 +\\ 
 +&= 
 +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*} \begin{align*}
 |\underline{Z}_{\rm k}| |\underline{Z}_{\rm k}|
 = =
 \sqrt{R_{\rm k}^2+X_{\rm k}^2} \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}}
 = =
-\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}} 
 +&= 
 +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}
 = =
-4.33~\Omega.+\frac{U_{1{\rm k}}}{U_{1{\rm N}}}\cdot 100~\%
 \end{align*} \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~\% 
 +\\ 
 +&= 
 +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*} \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}} 
 +&= 
 +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 \approx
-|\underline{Z}_{\rm k}|I_{1{\rm N}}+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
 = =
-4.33~\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} 
 +\\ 
 +&= 
 +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
 = =
-2.17~{\rm V}.+\sqrt{1-\cos^2\varphi}
 \end{align*} \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} 
 +\\ 
 +&= 
 +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>
  
-\[+3. Estimate the primary-side voltage drop $\Delta U_1$. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2031~@# 
 +<WRAP leftalign> 
 +Use the Kapp approximation:
 \begin{align*} \begin{align*}
-\frac{2.17~{\rm V}}{10}=0.217~{\rm V}.+\Delta U_1 
 +&\approx 
 +I_1\left(R_{\rm k}\cos\varphi+X_{\rm k}\sin\varphi\right)
 \end{align*} \end{align*}
-\] 
  
-For a \(23~{\rm V}\) actuator supply this is small but not zero.+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@# #@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@# #@TaskEnd_HTML@#
  
Line 1369: 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 \(i_{\rm p}\).   * **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 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 \(n^2\).   * **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.   * **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.   * **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.   * **Opening a current transformer secondary:** This can create dangerous voltages. Current transformers are operated with a low burden, approximately as a short-circuit.