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electrical_engineering_and_electronics_1:block19 [2025/12/02 18:38] – mexleadminelectrical_engineering_and_electronics_1:block19 [2026/01/10 10:22] (aktuell) – mexleadmin
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 ====== Block 19 — Magnetic Circuits and Inductance ====== ====== Block 19 — Magnetic Circuits and Inductance ======
  
-===== Learning objectives =====+===== 19.0 Intro  ===== 
 +==== 19.0.1 Learning objectives ====
 <callout> <callout>
 After this 90-minute block, you can After this 90-minute block, you can
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 </callout> </callout>
  
-===== Preparation at Home =====+==== 19.0.2 Preparation at Home ====
  
 Well, again  Well, again 
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 For checking your understanding please do the following exercises: For checking your understanding please do the following exercises:
-  * ...+  * Exercise E2 Magnetic Circuit 
 +  * Exercise 5.1.4 Magnetic Voltage 
 +  * Exercise 5.1.7 Comparison with simplified Calculation
  
-===== 90-minute plan =====+==== 19.0.3 90-minute plan ====
   - Warm-up (x min):    - Warm-up (x min): 
     - ....      - .... 
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   - Wrap-up (x min): Summary box; common pitfalls checklist.   - Wrap-up (x min): Summary box; common pitfalls checklist.
  
-===== Conceptual overview =====+==== 19.0.4 Conceptual overview ====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
   - ...   - ...
 </callout> </callout>
  
-===== Core content =====+===== 19.1 Core content =====
  
 <callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout> <callout> For this and the following chapter the online Book 'DC Electrical Circuit Analysis - A Practical Approach' is strongly recommended as a reference. In detail this is chapter [[https://eng.libretexts.org/Bookshelves/Electrical_Engineering/Electronics/DC_Electrical_Circuit_Analysis_-_A_Practical_Approach_(Fiore)/10%3A_Magnetic_Circuits_and_Transformers/10.3%3A_Magnetic_Circuits|10.3 Magnetic Circuits]] </callout>
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 In this chapter, we will investigate how far we have come with such an analogy and where it can be practically applied. In this chapter, we will investigate how far we have come with such an analogy and where it can be practically applied.
  
-==== Basics for Linear Magnetic Circuits ====+==== 19.1.2 Basics for Linear Magnetic Circuits ====
  
 For the upcoming calculations, the following assumptions are made For the upcoming calculations, the following assumptions are made
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-==== Reluctance - the magnetic Resistance ====+==== 19.1.2 Reluctance - the magnetic Resistance ====
  
 Let's have a look at the simplest situation:  Let's have a look at the simplest situation: 
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-==== Applications of Flux and Reluctance ====+==== 19.1.3 Applications of Flux and Reluctance ====
  
  
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 </callout> </callout>
  
-===== Common pitfalls =====+===== 19.2 Common pitfalls =====
   * ...   * ...
  
-===== Exercises =====+===== 19.3 Exercises =====
  
 {{page>electrical_engineering_and_electronics:task_yp4rbdlj8kktyrhp_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_yp4rbdlj8kktyrhp_with_calculation&nofooter}}
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 {{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_0j7accfimmemytq9_with_calculation&nofooter}}
  
 +<panel type="info" title="Exercise 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. 
 +The $N=400$ windings are evenly distributed along the circumference. 
 +The diameter on the cross-section of the plastic ring is $d = 10 ~\rm mm$. In the windings, a current of $I=500 ~\rm mA$ is flowing.
 +
 +Calculate
 +
 +  - the magnetic field strength $H$ in the middle of the ring cross-section.
 +  - the magnetic flux density $B$ in the middle of the ring cross-section.
 +  - the magnetic resistance $R_{\rm m}$ of the plastic ring.
 +  - the magnetic flux $\Phi$.
 +
 +<button size="xs" type="link" collapse="Solution_5_1_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_1_1_Result" collapsed="true">
 +
 +  - $H = 667 ~\rm {{A}\over{m}}$
 +  - $B = 0.84 ~\rm mT$
 +  - $R_m = 3 \cdot 10^9 ~\rm {{1}\over{H}}$
 +  - $\Phi = 66 ~\rm nVs$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions:
 +
 +  - $l=35.8~\rm cm$, $d=1.90~\rm cm$
 +  - $l=11.1~\rm cm$, $d=1.50~\rm cm$
 +
 +#@HiddenBegin_HTML~5_1_2s,Solution~@#
 +
 +The magnetic resistance is given by:
 +\begin{align*}
 +\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} 
 +\end{align*}
 +
 +With
 +  * the area $ A = \left({{d}\over{2}}\right)^2 \cdot \pi $
 +  * the vacuum magnetic permeability $\mu_{0}=4\pi\cdot 10^{-7} ~\rm H/m$, and 
 +  * the relative permeability $\mu_{\rm r}=1$.
 +
 +#@HiddenEnd_HTML~5_1_2s,Solution ~@#
 +
 +#@HiddenBegin_HTML~5_1_2r,Result~@#
 +  - $1.00\cdot 10^9 ~\rm {{1}\over{H}}$
 +  - $0.50\cdot 10^9 ~\rm {{1}\over{H}}$
 +#@HiddenEnd_HTML~5_1_2r,Result~@#
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +Calculate the magnetic resistances of an airgap with the following dimensions:
 +
 +  - $\delta=0.5~\rm mm$, $A=10.2~\rm cm^2$
 +  - $\delta=3.0~\rm mm$, $A=11.9~\rm cm^2$
 +
 +<button size="xs" type="link" collapse="Solution_5_1_3_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_3_1_Result" collapsed="true">
 +
 +  - $3.9\cdot 10^5 ~\rm {{1}\over{H}}$
 +  - $2.0\cdot 10^6 ~\rm {{1}\over{H}}$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions:
 +
 +  - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$
 +  - $\delta=5.0~\rm mm$, $A=7.1~\rm cm^2$
 +
 +<button size="xs" type="link" collapse="Solution_5_1_4_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_4_1_Result" collapsed="true">
 +
 +  - $\theta = 1.5\cdot 10^3 ~\rm A$, or $1000$ windings with $1.5~\rm A$
 +  - $\theta = 2.8\cdot 10^3 ~\rm A$, or $1000$ windings with $2.8~\rm A$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +Calculate the magnetic flux created on a magnetic resistance of $R_m = 2.5 \cdot 10^6 ~\rm {{1}\over{H}}$ with the following magnetic voltages:
 +
 +  - $\theta = 35   ~\rm A$
 +  - $\theta = 950  ~\rm A$
 +  - $\theta = 2750 ~\rm A$
 +
 +<button size="xs" type="link" collapse="Solution_5_1_5_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_5_1_Result" collapsed="true">
 +
 +  - $\Phi =14  ~\rm µVs$
 +  - $\Phi =0.38~\rm mVs$
 +  - $\Phi =1.1 ~\rm mVs$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A core shall consist of two parts, as seen in <imgref ImgExNr08>. 
 +In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$.
 +
 +The cross sections are $A_1=530 ~\rm mm^2$ and $A_2=460 ~\rm mm^2$. 
 +The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$.
 +
 +The air gaps on the coupling joint between both parts have the length $\delta = 0.23 ~\rm mm$ each. 
 +The permeability of the ferrite is $\mu_r = 3000$. 
 +The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$
 +
 +<WRAP> <imgcaption ImgExNr08 | Two-parted ferrite Core> </imgcaption> {{drawio>TwoPartedCoil.svg}} </WRAP>
 +
 +  - Draw the lumped circuit of the magnetic system
 +  - Calculate all magnetic resistances $R_{\rm m,i}$
 +  - Calculate the flux in the circuit
 +
 +<button size="xs" type="link" collapse="Solution_5_1_6_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_6_1_Result" collapsed="true">
 +
 +  - -
 +  - magnetic resistances: $R_{\rm m,1} = 100 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{\rm m,1} = 75 \cdot 10^3 ~\rm {{1}\over{H}}$, $R_{{\rm m},\delta} = 400 \cdot 10^3 ~\rm {{1}\over{H}}$
 +  - magnetic flux: $\Phi = 0.80 ~\rm mVs$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +The magnetic circuit in <imgref ImgExNr09> passes a magnetic flux density of $0.4 ~\rm T$ given by an excitation current of $0.50 ~\rm A$ in $400$ windings. 
 +At position $\rm A-B$, an air gap will be inserted. After this, the same flux density will be reached with $3.70 ~\rm A$
 +
 +<WRAP> <imgcaption ImgExNr09 | Example of a magnetic circuit> </imgcaption> {{drawio>ExMagncirc01.svg}} </WRAP>
 +
 +  - Calculate the length of the airgap $\delta$ with the simplification $\mu_{\rm r} \gg 1$
 +  - Calculate the length of the airgap $\delta$ exactly with $\mu_{\rm r} = 1000$
 +
 +<button size="xs" type="link" collapse="Solution_5_1_7_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_7_1_Result" collapsed="true">
 +
 +  - $\delta = 4.02(12) ~\rm mm$
 +  - $\delta = 4.02(52) ~\rm mm$
 +
 +</collapse>
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +The choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. 
 +The number of windings shall be $N$ and the current through a single winding $I$.
 +
 +<WRAP> <imgcaption ImgExNr10 | Example for a Choke Coil> </imgcaption> {{drawio>ChokeCoilEx1.svg}} </WRAP>
 +
 +  - Draw the lumped circuit of the magnetic system
 +  - Calculate all magnetic resistances $R_{{\rm m},i}$
 +  - Calculate the partial fluxes in all the legs of the circuit
 +
 +</WRAP></WRAP></panel>
 +
 +<panel type="info" title="Exercise 5.3.3 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 +
 +A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. 
 +At the core, there are two coils $N_1=500$ and $N_2=250$ wound. The currents on the coils are $I_1 = 250 ~\rm mA$ and $I_2=300 ~\rm mA$.
 +
 +  - The coils shall pass the currents with positive polarity (see the image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ in the coil?
 +  - The coils shall pass the currents with negative polarity (see the image **B** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm B}$ in the coil?
 +
 +<WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP>
 +
 +#@HiddenBegin_HTML~5_3_2s,Solution~@#
 +
 +The resulting flux can be derived from a superposition of the individual fluxes $\Phi_1(I_1)$ and $\Phi_2(I_2)$, or alternatively by summing the magnetic voltages in the loop ($\sum_x \theta_x = 0$).
 +
 +**Step 1 - Draw an equivalent magnetic circuit**
 +
 +Since there are no branches, all of the core can be lumped into a single magnetic resistance (see <imgref ImgEx14circ>).
 +<WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>
 +
 +**Step 2 - Get the absolute values of the individual fluxes**
 +
 +Hopkinson's Law can be used here as a starting point. \\
 +It connects the magnetic flux $\Phi$ and the magnetic voltage $\theta$ on the single magnetic resistor $R_\rm m$. \\
 +It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\
 +
 +\begin{align*} 
 +\theta_x             &= R_{\rm m}                                  \cdot \Phi_x \\
 +N_x \cdot I_x        &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}} \cdot \Phi_x \\
 +\rightarrow \Phi_x   &= \mu_0 \mu_{\rm r} {{A}\over{l}} \cdot N_x \cdot I_x 
 +                      = {{1}\over{R_{\rm m} }}          \cdot N_x \cdot I_x \\
 +\end{align*}
 +
 +With the given values we get: $R_{\rm m} = 495 {\rm {kA}\over{Vs}}$
 +
 +**Step 3 - Get the signs/directions of the fluxes**
 +
 +The <imgref5_3_2_Solution> shows how to get the correct direction for every single flux by use of the right-hand rule. \\
 +The fluxes have to be added regarding these directions and the given direction of the flux in question.
 +<WRAP> <imgcaption 5_3_2_Solution| toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg_solution.svg}} </WRAP>
 +
 +Therefore, the formulas are
 +\begin{align*} 
 +\Phi_{\rm A}   &= \Phi_{1} - \Phi_{2} \\
 +               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  -  N_2 \cdot I_2 \right) \\
 +               & = 0.25 ~{\rm mVs} - 0.15 ~{\rm mVs} \\
 +\Phi_{\rm B}   &= \Phi_{1} + \Phi_{2} \\
 +               &={{1}\over{R_{\rm m} }} \cdot \left( N_1 \cdot I_1  +  N_2 \cdot I_2 \right) \\
 +               & = 0.25 ~{\rm mVs} + 0.15 ~{\rm mVs} 
 +\end{align*}
 +
 +#@HiddenEnd_HTML~5_3_2s,Solution~@#
 +
 +
 +#@HiddenBegin_HTML~5_3_2r,Result~@#
 +  - $0.10 ~\rm mVs$
 +  - $0.40 ~\rm mVs$
 +#@HiddenEnd_HTML~5_3_2r,Result~@#
 +
 +</WRAP></WRAP></panel>
  
 ===== Embedded resources ===== ===== Embedded resources =====
 +<WRAP>
 +Practical Application of calculation in magnetics \\
 +{{youtube>rE0CdZUWDFU?start=498}}
 +</WRAP>
 +
 +\\ \\ \\
 +
 <WRAP column half> <WRAP column half>
-Explanation (video): ...+Practical Application of calculation in magnetics (examples) \\ 
 +  - {{electrical_engineering_and_electronics_1:investigation_of_permanent_magnet_synchronous_mach.pdf}} 
 +  - [[https://ieeexplore.ieee.org/abstract/document/7732749|Estimation of losses in the stator and rotor of interior permanent magnets synchronous (IPMs) machines using reluctance network Reluctance]] 
 +  - [[https://lammotor.com/comparison-of-motor-of-tesla-byd-and-huawei/|Comparison of Motor Stator and Rotor Technologies of Tesla, BYD, and Huawei]]
 </WRAP> </WRAP>
 +
  
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