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electrical_engineering_and_electronics_2:block09 [2026/05/17 07:41] – mexleadminelectrical_engineering_and_electronics_2:block09 [2026/05/19 02:55] (current) – mexleadmin
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 \[ \[
 \begin{align*} \begin{align*}
-\frac{\underline{U}_1}{\underline{U}_2}=\frac{N_1}{N_2}=n,+\frac{\underline{U}_1^\phantom{O}}{\underline{U}_2^\phantom{O}}=\frac{N_1}{N_2}=n,
 \qquad \qquad
 \frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n} \frac{\underline{I}_1}{\underline{I}_2}=-\frac{1}{n}
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 \(X'_{2\sigma}=n^2X_{2\sigma}\).   * 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>
Line 50: Line 50:
   * **Practice (20 min):**   * **Practice (20 min):**
     * Quick ratio calculations for step-up and step-down transformers.     * Quick ratio calculations for step-up and step-down transformers.
-    * Unit checks for \(j\omega L\), \(j\omega N\Phi\), and \(u_{\rm k}\). 
     * Short-circuit current calculation for a transformer used in an actuator supply.     * Short-circuit current calculation for a transformer used in an actuator supply.
  
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     * \(R_{\rm Fe}\): iron losses in the core.     * \(R_{\rm Fe}\): iron losses in the core.
     * \(L_{\rm H}\): main magnetizing inductance needed to create the main flux.     * \(L_{\rm H}\): main magnetizing inductance needed to create the main flux.
-  * In engineering, transformer data such as \(u_{\rm k}\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.+  * In engineering, transformer data such as \(\rm u_k\) are not abstract: they determine voltage drop, fault current, thermal stress, and protection design.
 </callout> </callout>
  
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 \] \]
  
-(Be aware of Lenz law, whenever you want to draw the votage arrows) \\+(Be aware of Lenz law: Here u(t) is the terminal voltage according to the chosen voltage reference arrow. The induced voltage $u_{\rm ind}$ according to Faraday–Lenz would have the opposite sign) \\
 In sinusoidal steady state this becomes the phasor equation In sinusoidal steady state this becomes the phasor equation
  
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 <panel type="danger" title="Remember: ideal transformer ratios"> <panel type="danger" title="Remember: ideal transformer ratios">
-With the indicated reference arrows and a lossless transformer, the resulting power as a sum of input and output power must be zero:+With the indicated reference arrows and a lossless transformer, the resulting complex power as a sum of input and output power $\underline{S}$ must be zero:
  
 \[ \[
 \begin{align*} \begin{align*}
-\underline{U}_1\underline{I}^*_1+\underline{U}^*_2\underline{I}_2=0+\underline{S}_1+\underline{S}_2= \underline{U}_1\underline{I}^*_1+\underline{U}_2\underline{I}^*_2=0
 \end{align*} \end{align*}
 \] \]
Line 278: 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 onto the situation of two coils nearby 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 
- 
-\[ 
-\begin{align*} 
-\Phi_{1\rm H} 
-= 
-{\color{blue}{\Phi_{21}}} 
-+ 
-{\color{orange}{\Phi_{1 \rm \sigma}}}. 
-\end{align*} 
-\] 
- 
-  * \(\Phi_{1\rm H}\): total flux created by coil \(1\). The 'H' denotes the German word "Haupt" (sometimes also given as 'm' for "main"). 
-  * \({\color{blue}{\Phi_{21}}}\): 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 sigma $\sigma$ is used to depict the term "stray" 
- 
-For an example, we will have a look onto 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 
- 
-\[ 
-\begin{align*} 
-\boxed{ 
-\underline{U}_{2}(t) 
-= 
-j\omega {\color{blue}{ \underline{\Psi}_{21} }} 
-= 
-j\omega N_2 {\color{blue}{ \underline{\Phi}_{21} }} 
-}. 
-\end{align*} 
-\] 
- 
-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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 \underline{\Psi}_1 \underline{\Psi}_1
 &= &=
-\underbrace{{\color{green}{L_{1\rm H}\underline{I}_1}}}_{\text{self-linkage of coil 1}}+\underbrace{{\color{green}{L_{11}\underline{I}_1}}}_{\text{self-linkage of coil 1}}
 + +
 \underbrace{{\color{blue}{M_{12}\underline{I}_2}}}_{\text{mutual linkage from coil 2}}, \underbrace{{\color{blue}{M_{12}\underline{I}_2}}}_{\text{mutual linkage from coil 2}},
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 \underbrace{{\color{blue}{M_{21}\underline{I}_1}}}_{\text{mutual linkage from coil 1}} \underbrace{{\color{blue}{M_{21}\underline{I}_1}}}_{\text{mutual linkage from coil 1}}
 + +
-\underbrace{{\color{green}{L_{2\rm H}\underline{I}_2}}}_{\text{self-linkage of coil 2}}.+\underbrace{{\color{green}{L_{22}\underline{I}_2}}}_{\text{self-linkage of coil 2}}.
 \end{align*} \end{align*}
 \] \]
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 \] \]
  
 +Often, the self-inductances are abbreviated:
 +\[
 +\begin{align*}
 +\color{green}{L_{11}} \rightarrow \color{green}{L_{1}} \\
 +\color{green}{L_{22}} \rightarrow \color{green}{L_{2}} \\
 +\end{align*}
 +\]
  
 Then Then
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 = =
 \begin{pmatrix} \begin{pmatrix}
-{\color{green}{L_{1\rm H}}} & {\color{blue}{M}}\\ +{\color{green}{L_{1}}} & {\color{blue}{M}}\\ 
-{\color{blue}{M}} & {\color{green}{L_{2\rm H}}}+{\color{blue}{M}} & {\color{green}{L_{2}}}
 \end{pmatrix} \end{pmatrix}
 \begin{pmatrix} \begin{pmatrix}
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 \[ \[
 \begin{align*} \begin{align*}
-M=k\sqrt{L_{1\rm H}L_{2\rm H}}.+M=k\sqrt{L_{1}L_{2}}.
 \end{align*} \end{align*}
 \] \]
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 | \(0<k<1\)       | partial coupling                                   | wireless charger with air gap or misalignment  | | \(0<k<1\)       | partial coupling                                   | wireless charger with air gap or misalignment  |
 | \(k\approx 1\)  | almost all useful flux links both coils            | transformer with iron core                     | | \(k\approx 1\)  | 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> </tabcaption>
  
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 Imagine two people lying on a bed holding the same bed sheet at different positions. Imagine two people lying on a bed holding the same bed sheet at different positions.
  
-  * \(L_{1{\rm H}}\): \\ how strongly winding 1 couples its current into the shared magnetic path, \\ like person 1 moving the sheet at their hand. +  * \(L_{1}\): \\ how strongly winding 1 couples its current into the shared magnetic path, \\ like person 1 moving the sheet at their hand. 
-  * \(L_{2{\rm H}}\): \\ how strongly winding 2 couples its current into the shared magnetic path, \\ like person 2 moving the sheet at their hand.+  * \(L_{2}\): \\ how strongly winding 2 couples its current into the shared magnetic path, \\ like person 2 moving the sheet at their hand.
   * \(M\): \\ how strongly the motion from one hand is felt at the other hand through the same sheet.   * \(M\): \\ how strongly the motion from one hand is felt at the other hand through the same sheet.
-  * \(L_{1\sigma}\), \(L_{2\sigma}\): \\ local wrinkles near one hand. They move locally but do not effectively reach the other hand. 
  
-In transformer language, \(L_{1{\rm H}}\) and \(L_{2{\rm H}}\) describe each winding's connection to the shared main flux path. \\+In transformer language, \(L_{1}\) and \(L_{2}\) describe each winding's connection to the shared main flux path. \\
 The mutual inductance \(M\) describes the transfer between the two windings through this shared path. The mutual inductance \(M\) describes the transfer between the two windings through this shared path.
 </panel> </panel>
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 \[ \[
 \begin{align*} \begin{align*}
-\underline{U}_1 &= R_1 \underline{I}_1 + {\color{green} {j\omega L_{1\rm H} \underline{I}_1 }} + {\color{blue}{j\omega M       \underline{I}_2 }},+\underline{U}_1 &= R_1 \underline{I}_1 + {\color{green} {j\omega L_{1} \underline{I}_1 }} + {\color{blue}{j\omega M       \underline{I}_2 }},
 \\[4pt] \\[4pt]
-\underline{U}_2 &= R_2 \underline{I}_2 + {\color{blue}  {j\omega M      \underline{I}_1 }} + {\color{green}{j\omega L_{2\rm H} \underline{I}_2 }}.+\underline{U}_2 &= R_2 \underline{I}_2 + {\color{blue}  {j\omega M      \underline{I}_1 }} + {\color{green}{j\omega L_{2} \underline{I}_2 }}.
 \end{align*} \end{align*}
 \] \]
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 <panel type="info" title="Tunnel Analogy for AC circuits"> <panel type="info" title="Tunnel Analogy for AC circuits">
 The dots are like matching openings for magnetic action. \\ The dots are like matching openings for magnetic action. \\
-A positive sinusoidal current (e.g. $i_1$) entering the dotted terminal of one winding produces a positive sinusoidal induced voltage (e.g. aligned with $u_2$) at the dotted terminal of the other winding. \\+A positive reference current (e.g. $i_1$) entering the dotted terminal of one winding produces a positive induced voltage (e.g. aligned with $u_2$) at the dotted terminal of the other winding. \\
 With only a load $R_2$ connected to the secondary side, this voltage tends to drive current out of the dotted terminal into the load. \\  With only a load $R_2$ connected to the secondary side, this voltage tends to drive current out of the dotted terminal into the load. \\ 
 </panel> </panel>
  
 ~~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.
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 </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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 \[ \[
 \begin{align*} \begin{align*}
-\underline{\Psi}_1 +\underline{\Psi}_1  
-&= +&= \underbrace{{\color{blue  }{L_{1{\rm H}}\underline{I}_1 + M\underline{I}_2}}  }_{\text{main magnetic path}} 
-\underbrace{{\color{blue}{L_{1{\rm H}}\underline{I}_1+M\underline{I}_2}}}_{\text{main magnetic path}} ++  \underbrace{{\color{orange}{L_{1\sigma }\underline{I}_1}}                     }_{\text{primary leakage}   },
-+ +
-\underbrace{{\color{orange}{L_{1\sigma}\underline{I}_1}}}_{\text{primary leakage}},+
 \\[4pt] \\[4pt]
-\underline{\Psi}_2 +\underline{\Psi}_2  
-&= +&= \underbrace{{\color{blue  }{M\underline{I}_1 + L_{2{\rm H}}\underline{I}_2 }} }_{\text{main magnetic path}} 
-\underbrace{{\color{blue}{L_{2{\rm H}}\underline{I}_2+M\underline{I}_1}}}_{\text{main magnetic path}} ++  \underbrace{{\color{orange}{L_{2\sigma }\underline{I}_2}}                     }_{\text{secondary leakage} }.
-+ +
-\underbrace{{\color{orange}{L_{2\sigma}\underline{I}_2}}}_{\text{secondary leakage}}.+
 \end{align*} \end{align*}
 \] \]
Line 621: Line 625:
 + +
 \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 $N_1$, $N_2$:
 +
 +\[
 +\begin{align*}
 +\boxed{
 +L_{1 \rm H} = \frac{N_1^2}{R_{\rm mFe} }  \\
 +L_{2 \rm H} = \frac{N_2^2}{R_{\rm mFe} }  \\
 +M = \frac{N_1 N_2}{R_{\rm mFe} }  \\
 +}
 \end{align*} \end{align*}
 \] \]
Line 692: Line 708:
  
 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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 \[ \[
 \begin{align*} \begin{align*}
-\ddot{\ rm u}\neq n,+\ddot{\rm u}\neq n,
 \end{align*} \end{align*}
 \] \]
Line 790: Line 806:
 \[ \[
 \begin{align*} \begin{align*}
-jX_{1{\rm H}} \; || \;R_{\rm Fe}+\Big|\; jX_{1{\rm H}} \; || \;R_{\rm Fe} \;\Big|
 \;\; \gg \;\; \;\; \gg \;\;
-jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2.+\Big|\; jX_{1\sigma} + \;R_1 + jX'_{2\sigma} + \;R'_2 \;\Big| .
 \end{align*} \end{align*}
 \] \]
Line 862: 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%> <WRAP column 100%>
Line 906: Line 922:
 \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 932: Line 948:
  
  
-The Kapp triangle (see yellow triangle in <imgref fig_kapp_triangle>) is reflecting the formula for $\underline{U}_{\rm k}$. +The Kapp triangle (see yellow triangle in <imgref fig_kapp_triangle>) represents the formula for $\underline{U}_{\rm k}$. 
  
 <panel> <panel>
Line 986: Line 1002:
  
 ^ Name / use ^ Typical \({\rm u_k}\) ^ Secondary voltage \(U_2\) ^ Important note ^ ^ Name / use ^ Typical \({\rm u_k}\) ^ Secondary voltage \(U_2\) ^ Important note ^
-| Power transformer | \(4\ldots 12~\%\) | application-dependent | low voltage drop, high fault currents possible | +| Power transformer      | \(4\ldots 12~\%\)   | application-dependent                    | low voltage drop, high fault currents possible            | 
-| Isolating transformer | \(\approx 10~\%\) | max. \(250~{\rm V}\) | galvanic isolation for safety and measurement | +| Isolating transformer  | \(\approx 10~\%\)   | max. \(250~{\rm V}\)                     | galvanic isolation for safety and measurement             | 
-| Toy transformer | \(\approx 20~\%\) | max. \(24~{\rm V}\) | current limitation is desired | +| Toy transformer        | \(\approx 20~\%\)   | max. \(24~{\rm V}\)                      | current limitation is desired                             | 
-| Doorbell transformer | \(\approx 40~\%\) | max. \(12~{\rm V}\), often several taps | simple robust low-voltage supply | +| Doorbell transformer   | \(\approx 40~\%\)   | max. \(12~{\rm V}\), often several taps  | simple robust low-voltage supply                          | 
-| Ignition transformer | \(\approx 100~\%\) | \(\leq 14~{\rm kV}\) | high voltage, limited current | +| Ignition transformer   | \(\approx 100~\%\)  | \(\leq 14~{\rm kV}\)                     | high voltage, limited current                             | 
-| Welding transformer | \(\approx 100~\%\) | max. \(70~{\rm V}\) | large current, strong current limitation | +| Welding transformer    | \(\approx 100~\%\)  | max. \(70~{\rm V}\)                      | large current, strong current limitation                  | 
-| Voltage transformer | \(<1~\%\) | \(100~{\rm V}\) | operate with high load resistance, approximately no-load | +| Voltage transformer    | \(<1~\%\)           | \(100~{\rm V}\)                          | operate with high load resistance, approximately no-load  | 
-| Current transformer | \(100~\%\) | \(0~{\rm V}\) ideal secondary voltage | operate with low burden, approximately short-circuit |+| Current transformer    | \(100~\%\)          | \(0~{\rm V}\) ideal secondary voltage    | operate with low burden, approximately short-circuit      |
 </tabcaption> </tabcaption>
  
Line 1007: 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 1070: 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 1218: Line 1744:
 Rated data and equivalent circuit data are: Rated data and equivalent circuit data are:
  
-\[ 
 \begin{align*} \begin{align*}
 U_{1{\rm N}}&=230~{\rm V}, U_{1{\rm N}}&=230~{\rm V},
Line 1234: Line 1759:
 X_{2\sigma}&=0.018~\Omega. X_{2\sigma}&=0.018~\Omega.
 \end{align*} \end{align*}
-\] 
  
-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@# 
 +</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>
  
-Therefore +3. Calculate $R_{\rm k}$ and $X_{\rm k}$. 
- +<WRAP group> 
-\[+<WRAP half column rightalign> 
 +#@PathBegin_HTML~2021~@# 
 +<WRAP leftalign> 
 +The short-circuit equivalent values are the sums of the primary quantities and the referred secondary quantities:
 \begin{align*} \begin{align*}
 R_{\rm k} R_{\rm k}
 &= &=
 R_1+R'_2 R_1+R'_2
-= 
-1.2~\Omega+1.2~\Omega 
-= 
-2.4~\Omega, 
 \\ \\
 X_{\rm k} X_{\rm k}
 &= &=
 X_{1\sigma}+X'_{2\sigma} X_{1\sigma}+X'_{2\sigma}
-= 
-1.8~\Omega+1.8~\Omega 
-= 
-3.6~\Omega. 
 \end{align*} \end{align*}
-\] 
  
-The primary current magnitude is+Insert the values: 
 +\begin{align*} 
 +R_{\rm k} 
 +&= 
 +1.2~\Omega+1.2~\Omega 
 +\\ 
 +&= 
 +2.4~\Omega 
 +\\[4pt] 
 +X_{\rm k} 
 +&= 
 +1.8~\Omega+1.8~\Omega 
 +\\ 
 +&= 
 +3.6~\Omega 
 +\end{align*} 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 +<WRAP half column> 
 +#@ResultBegin_HTML~2021~@# 
 +\begin{align*} 
 +R_{\rm k} &= 2.4~\Omega 
 +\\ 
 +X_{\rm k} &= 3.6~\Omega 
 +\end{align*} 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-\[+4. Calculate the primary rated current magnitude $I_{1{\rm N}}$ using the ideal current ratio. 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~2022~@# 
 +<WRAP leftalign> 
 +For an ideal transformer, the primary current magnitude is:
 \begin{align*} \begin{align*}
 I_{1{\rm N}} I_{1{\rm N}}
 = =
 \frac{I_{2{\rm N}}}{n} \frac{I_{2{\rm N}}}{n}
-=+\end{align*} 
 + 
 +Insert the values: 
 +\begin{align*} 
 +I_{1{\rm N}} 
 +&=
 \frac{5.0~{\rm A}}{10} \frac{5.0~{\rm A}}{10}
 +\\
 +&=
 +0.50~{\rm A}
 +\end{align*}
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +<WRAP half column>
 +#@ResultBegin_HTML~2022~@#
 +\begin{align*}
 +I_{1{\rm N}}=0.50~{\rm A}
 +\end{align*}
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +5. Estimate the magnitude of the internal voltage drop $U_{\rm k}\approx |\underline{Z}_{\rm k}|I_{1{\rm N}}$.
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~2023~@#
 +<WRAP leftalign>
 +First calculate the magnitude of the short-circuit impedance:
 +\begin{align*}
 +|\underline{Z}_{\rm k}|
 = =
-0.50~{\rm A}.+\sqrt{R_{\rm k}^2+X_{\rm k}^2}
 \end{align*} \end{align*}
-\] 
  
-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 1365: 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.