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electrical_engineering_2:magnetic_circuits [2023/06/12 10:47] – ottelectrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (aktuell) – mexleadmin
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-====== 5. Magnetic Circuits ======+====== 5 Magnetic Circuits ======
  
 <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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 <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP> <WRAP> <imgcaption ImgNr01 | Similarities magnetic Circuit vs electric Circuit> </imgcaption> {{drawio>CompMagElCircuit.svg}} </WRAP>
  
-In this chapter, we will investigate, how far we 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.
  
 ===== 5.1 Linear Magnetic Circuits ===== ===== 5.1 Linear Magnetic Circuits =====
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   - The fields inside of airgaps are homogeneous. This is true for small air gaps.   - The fields inside of airgaps are homogeneous. This is true for small air gaps.
  
-One can calculate a lot of simple magnetic circuits when these assumptions and focusing on the average field line are applied.+One can calculate a lot of simple magnetic circuits when these assumptions are applied and focusing on the average field line are applied.
  
 <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP> <WRAP> <imgcaption ImgNr03 | Simplifications and Linearization> </imgcaption> {{drawio>SimplificationLin.svg}} </WRAP>
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 |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) | |Simplifications |The simplifications often work for good results \\ (small wire diameter, relatively constant resistivity) |The simplification is often too simple \\ (widespread beyond the mean magnetic path length, non-linearity of the permeability) |
  
-<panel type="info" title="Task 5.1.1 Coil on a plastic Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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 $l_R = 300 ~\rm mm$.  A coil is set up onto a toroidal plastic ring ($\mu_{\rm r}=1$) with an average circumference of $l_R = 300 ~\rm mm$. 
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.2 magnetic Resistance of a cylindrical coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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: Calculate the magnetic resistances of cylindrical coreless (=ironless) coils with the following dimensions:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.3 magnetic Resistance of an airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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: Calculate the magnetic resistances of an airgap with the following dimensions:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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: Calculate the magnetic voltage necessary to create a flux of $\Phi=0.5 ~\rm mVs$ in an airgap with the following dimensions:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.5 Magnetic Flux"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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: 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:
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.6 Two-parted ferrite Core"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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>. +A core shall consist of two parts, as seen in <imgref ImgExNr08>. 
 In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$. In the coil, with $600$ windings shall pass the current $I=1.30 ~\rm A$.
  
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 The mean magnetic path lengths are $l_1 = 200 ~\rm mm$ and $l_2 = 130 ~\rm mm$. 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 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 permeability of the ferrite is $\mu_r = 3000$. 
 The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$ The cross-section area $A_{\delta}$ of the airgap can be considered the same as $A_2$
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.7 Comparison with simplified Calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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.  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. 
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 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.8 Coil on a ferrite Core with airgap"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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 choke coil shown in <imgref ImgExNr10> shall be given, with a constant cross-section in all legs $l_0$, $l_1$, $l_2$. 
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 ===== 5.3 Mutual Induction and Coupling ===== ===== 5.3 Mutual Induction and Coupling =====
  
-Situation: Two coils $1$ and $2$ near each other. \\ Questions:+Imagine charging your phone wirelessly by simply placing it on a charging pad.  
 +This seamless experience is made possible by the fascinating phenomenon of mutual induction and coupling between two coils. 
 + 
 +This situation is depicted in <imgref ImgNr09>:  
 +When an alternating current flows through one coil (Coil $1$), it creates a time-varying magnetic field that induces a voltage in the nearby coil (Coil $2$), even though they are not physically connected.  
 +This mutual influence is governed by the principle of electromagnetic induction. 
 + 
 +<WRAP center 35%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP> 
 + 
 +The key factor determining the strength of mutual induction is the mutual inductance ($M$) between the coils.  
 +It quantifies the magnetic flux linkage and depends on factors like the number of turns, current, and relative orientation of the coils. 
 + 
 +While geometric properties play a role, the fundamental principle can be described using electric properties alone, making mutual induction a versatile concept with numerous applications, including: 
 + 
 +  * Wireless power transfer 
 +  * Transformers 
 +  * Inductive coupling in communication systems 
 +  * Inductive sensors 
 + 
 +As we explore this chapter, we'll delve into the mathematical models, equations, and practical considerations of mutual induction and coupling, unlocking a world of innovative technologies that shape our modern lives. 
 +We explicitly try to answer the following questions:
  
   * Which effect do the coils have on each other?   * Which effect do the coils have on each other?
   * Can we describe the effects with mainly electric properties (i.e. no geometric properties)   * Can we describe the effects with mainly electric properties (i.e. no geometric properties)
  
-<WRAP center 35%> <imgcaption ImgNr09 | Mutual Induction of two Coils> </imgcaption> {{drawio>MutualInductionTwoCoils1.svg}} </WRAP> 
  
 ==== Effect of Coils on each other ==== ==== Effect of Coils on each other ====
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 For the single coil, we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$. \\  For the single coil, we got the relationship between the linked flux $\Psi$ and the current $i$ as: $\Psi = L \cdot i$. \\ 
-Now the coils also are interacting with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$: +Now the coils also interact with each other. This must also be reflected in the relationship $\Psi_1 = f(i_1, i_2)$, $\Psi_2 = f(i_1, i_2)$: 
 \begin{align*}  \begin{align*} 
 \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\  \Psi_1 &= &\Psi_{11}        &+ \Psi_{12} \\ 
Zeile 344: Zeile 363:
  
 The formula can also be described as:  The formula can also be described as: 
-\begin{align*}  +{{drawio>VectorialformulaOfInduction.svg}}
-\left( \begin{array}{c} \Psi_1 \\          \Psi_2 \end{array}           \right) =  +
-\left( \begin{array}{c} L_{11} & M_{12} \\  M_{21} & L_{22} \end{array} \right)  +
-\cdot  +
-\left( \begin{array}{c} i_1 \\              i_2 \end{array}             \right)  +
-\end{align*}+
  
-The view of the magnetic flux is sometimes good when effects like an acting Lorentz force in of interest.  +The view of the magnetic flux is advantageous when effects like an acting Lorentz force is of interest.  
-More often the coils are coupling two electric circuits linked in a transformer or a wireless charger. +However, more often the coils couple electric circuits, like in a transformer or a wireless charger. 
 Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit.  Here, the effect on the circuits is of interest. This can be calculated with the induced electric voltages $u_{\rm ind,1}$ and $u_{\rm ind,2}$ in each circuit. 
 They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$: They are given by the formula $u_{{\rm ind},x} = -{\rm d}\Psi_x /{\rm d}t$:
Zeile 372: Zeile 386:
 <WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP> <WRAP> <imgcaption ImgNr08 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils.svg}} </WRAP>
  
-The <imgref ImgNr08> shows the fluxes on each part. The black dots nearby the windings mark the direction of the windings, and therefore the sign of the generated flux. \\ +The <imgref ImgNr08> shows the fluxes on each part. The black dots near the windings mark the direction of the windings, and therefore the sign of the generated flux. \\ 
 All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\ All the fluxes caused by currents flowing into the __marked pins__ are summed up __positively__ in the core. \\
 When there is a current flowing into a __non-marked pin__, its flux has to be __subtracted__ from the others. When there is a current flowing into a __non-marked pin__, its flux has to be __subtracted__ from the others.
Zeile 405: Zeile 419:
 $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$): $k_{21}$ describes how much of the flux from coil $1$ is acting on coil $2$ (similar for $k_{12}$):
  
-\begin{align*} k_{21} = {{\Phi_{21} \cdot \Phi_{22} }\over{|\Phi_{21} \cdot \Phi_{22}|}} {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}+\begin{align*} k_{21} = \pm {{\Phi_{21}}\over{\Phi_{11}}} \\ \end{align*}
  
-When $k_{21}=100~\%$, there is no flux in the middle leg but only in the second coil. \\ +The sign of $k_{21}$ depends on the direction of $\Phi_{21}$ relative to $\Phi_{22}$! If the directions are the same, the positive sign applies, if the directions are opposite, the minus sign applies. 
 + 
 +When $k_{21}=+100~\%$, there is no flux in the middle leg but only in the second coil and in the same direction as the flux that originates from the second coil. \\  
 +When $k_{21}=-100~\%$, there is no flux in the middle leg but only in the second coil and in the opposite direction as the flux that originates from the second coil. \\ 
 For  $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling. For  $k_{21}=0~\%$ all the flux is in the middle leg circumventing the second coil, i.e. there is no coupling.
  
Zeile 421: Zeile 438:
        &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\         &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\ 
 \end{align*} \end{align*}
 +
 +Note, that also $M_{21}$ and $M_{12}$ can be either positive or negative, depending on the sign of the coupling coefficients.
  
 The formula is finally:  The formula is finally: 
Zeile 431: Zeile 450:
 \end{align*} \end{align*}
  
-<panel type="info" title="Task 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+For most of the applications the induction matrix has to be symmetric((This can be derived from energy considerations, when only electric circuits are coupled without additional flow of mechanical energy. This is, for example  not the case for motors with a mechanical load.)). \\ Therefore, the following applies:
  
-The magnetical configuration in <imgref ExImgNr01> shall be given. \\  +  * In General: the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ 
-The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.  +  * For symmetric induction matrix: The mutual inductances are equal: $M_{12} = M_{21} = M$ 
-The coupling factors are $k_{12}=0.6$ and $k_{21}=0.8$.+  * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}
  
-Calculate $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$.+ 
 + 
 +<panel type="info" title="Exercise 5.3.1 Example for magnetic Circuit with two Sources"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +The magnetical configuration in <imgref ExImgNr01> shall be given. \\  
 +The area of the cross-section is $A=9 ~\rm cm^2$ in all parts, the permeability is $\mu_r=800$, the length $l=12 ~\rm cm$ and the number of windings $N_1 = 400$, $N_2=300$.
  
 <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP> <WRAP> <imgcaption ExImgNr01 | Example for Iron Core with two Coils> </imgcaption> {{drawio>CoreWithTwoCoils2.svg}} </WRAP>
  
-=== Step 1: Draw the problem as a network ===+1.  Simplify the configuration into three magnetic resistors and 2 voltage sources. Draw the problem as an equivalent circuit 
 + 
 +#@HiddenBegin_HTML~5311,Result~@# 
 +<WRAP> <imgcaption ExImgNr11 | Equivalent Network> </imgcaption> {{drawio>CoreWithTwoCoils2network.svg}} </WRAP> 
 +#@HiddenEnd_HTML~5311,Result~@# 
 + 
 +2. Calculate all magnetic resistances. Additionally, calculate the magnetic resistances $R_{\rm m1}$  and $R_{\rm m2}$ seen from the magnetic voltage source $1$ and $2$. 
 + 
 +#@HiddenBegin_HTML~5312,Path~@#
  
-=== Step 2: Calculate the magnetic resistances ===+<WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>
  
-The magnetic resistance is summed up by looking at the circuit from the source $1$: +The magnetic resistance is summed up by looking at the circuit from the source $1$ (see <imgref ExImgNr13>): 
 \begin{align*}  \begin{align*} 
 R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\  R_{\rm m1} &= R_{\rm m,11} + R_{\rm m,ss} || R_{\rm m,22} \\ 
Zeile 452: Zeile 484:
 where the parts are given as  where the parts are given as 
 \begin{align*}  \begin{align*} 
-R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} \\  +R_{\rm m,11} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{3\cdot l}\over{A}} &&= 398 \cdot 10^{3} ~\rm {{1}\over{H}} \\  
-R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} \\  +R_{\rm m,ss} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{1\cdot l}\over{A}} &&= 133 \cdot 10^{3} ~\rm {{1}\over{H}} \\  
-R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} \\ +R_{\rm m,22} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{2\cdot l}\over{A}} &&= 265 \cdot 10^{3} ~\rm {{1}\over{H}} \\ 
 \end{align*} \end{align*}
  
-With the given geometry this leads to +With the given geometry, this leads to 
 \begin{align*}  \begin{align*} 
-R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot (3 + {{1\cdot 2}\over{1 + 2}}) \\  +R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\  
-           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} \\  +           &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{3}} &&= 486 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
-           &= 133 \cdot 10^{3}                          \cdot {{11}\over{3}} ~\rm {{1}\over{H}}\\ +
 \end{align*} \end{align*}
  
 Similarly, the magnetic resistance $R_{m2}$ is  Similarly, the magnetic resistance $R_{m2}$ is 
 \begin{align*}  \begin{align*} 
-R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} \\  +R_{\rm m2} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot {{11}\over{4}} &&= 365 \cdot 10^{3} ~\rm {{1}\over{H}}\\ 
-           &= 133 \cdot 10^{3}                          \cdot {{11}\over{4}} ~\rm {{1}\over{H}}\\ +
 \end{align*} \end{align*}
  
-== Step 3: Calculate the magnetic inductances ==+#@HiddenEnd_HTML~5312,Path~@#
  
 +3. Calculate the self-inductions $L_{11}$ and $L_{22}$
 +
 +#@HiddenBegin_HTML~5313,Path~@#
 +For the self-induction the effect on the electrical circuit is relevant. That is why the number of windings has to be considered.
 \begin{align*}  \begin{align*} 
 L_{11} &= {{N_1^2}\over{R_{\rm m1}}}                    &= 329 ~\rm mH\\ \\  L_{11} &= {{N_1^2}\over{R_{\rm m1}}}                    &= 329 ~\rm mH\\ \\ 
 L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\  L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\ 
-M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &= 197 ~\rm mH\\ \\  
-M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &= 197 ~\rm mH\\  
 \end{align*} \end{align*}
 +#@HiddenEnd_HTML~5313,Path~@#
 +
 +4. Calculate the coupling factors $k_{12}$ and $k_{21}$.
 +
 +#@HiddenBegin_HTML~5314,Path~@#
 +<WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
 +
 +The coupling factor $k_{21}$ is defined as "how much of the flux created by one coil ($\Phi_{11}$) crosses the other coil ($\Phi_{21}$) ":
 +\begin{align*} 
 +k_{21} &= {{\Phi_{21}}\over{\Phi_{11}}}
 +\end{align*}
 +
 +For this, we look at the circuit considering only one coil ("magnetic voltage source"), see <imgref ExImgNr12>. Based on the image, we see that the flux $\Phi_{11}$ divides into $\Phi_{S1}$ and $\Phi_{21}$ based on the magnetic resistance $\Phi_{\rm m,SS}$ and $\Phi_{\rm m,22}$. 
 +In step 2, we have calculated that $R_{\rm m,22}$ is twice $R_{\rm m,SS}$.  So from the incoming flux, only $1/3$ reaches $R_{\rm m,22}$ and - by this - "crosses the other coil". 
 +
 +Therefore, the coupling factor $k_{21}$ is: $k_{21}= 1/3$.
 +
 +A similar approach leads to $k_{12}$ with $k_{12}= 1/4$.
 +#@HiddenEnd_HTML~5314,Path~@#
 +
 +5. Calculate the mutual inductions $M_{12}$, and $M_{21}$,
 +
 +#@HiddenBegin_HTML~5315,Path~@#
 +\begin{align*} 
 +M_{21} &= k_{21}\cdot{{N_1 \cdot N_2}\over{R_{\rm m1}}} &&= {{1}\over{3}}\cdot{{400 \cdot 300}\over{ 486 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ 
 +M_{12} &= k_{12}\cdot{{N_1 \cdot N_2}\over{R_{\rm m2}}} &&= {{1}\over{4}}\cdot{{400 \cdot 300}\over{ 365 \cdot 10^{3} ~\rm {{1}\over{H}} }} &&= 82.2 ~\rm mH\\ \\ 
 +\end{align*}
 +#@HiddenEnd_HTML~5315,Path~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-For symmetrical magnetic structures and $\mu_{\rm r} = \rm const.$ the following applies:+#@TaskTitle_HTML@# 5.3.2 Wireles Charging #@TaskText_HTML@# 
 + 
 +For Electric vehicles, sometimes wireless charging systems are employed. These use the principle of mutual inductance to transfer power from a charging pad on the ground to the vehicle's battery pack. \\ 
 +This system consists of two coils: a transmitter coil embedded in the charging pad and a receiver coil mounted on the underside of the vehicle. 
 + 
 +  * The transmitter coil has a self-inductance of $L_{\rm T} = 200 ~\rm \mu H$.  
 +  * The receiver coil has a self-inductance of $L_{\rm R} = 150 ~\rm \mu H$.  
 +  * The mutual inductance between the coils at this distance is measured to be  $M = 20 ~\rm \mu H$ - when the vehicle is properly aligned over the charging pad. 
 + 
 +1. Calculate the coupling coefficient $k$ between the transmitter and receiver coils when the vehicle is properly aligned over the charging pad. 
 + 
 +#@HiddenBegin_HTML~53211,Path ~@# 
 + 
 +The given self-inductances are $L_{\rm T} = L_{11}$, $L_{\rm R} = L_{22}$. \\ 
 +By this, the following formula can be applied: 
 + 
 +\begin{align*} 
 +M = k \cdot \sqrt{L_{\rm T} \cdot L_{\rm R}} 
 +\end{align*} 
 + 
 +Therefore, $k$ is given as: 
 +\begin{align*} 
 +k = {{M}\over{ \sqrt{ L_{\rm T} \cdot L_{\rm R} } }} 
 +\end{align*} 
 + 
 +#@HiddenEnd_HTML~53211,Path ~@# 
 + 
 +2. If the vehicle is misaligned by 10 cm from the center of the charging pad, the mutual inductance drops to $M = 12 ~\rm \mu H$. Calculate the new coupling coefficient in this misaligned position. 
 + 
 +#@TaskEnd_HTML@# 
  
-  * the mutual inductances are equal: $M_{12} = M_{21} = M$ 
-  * the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ 
-  * The resulting *total coupling* $k$ is given as \begin{align*} k = \sqrt{k_{12}\cdot k_{21}} \end{align*} 
  
 ==== Effects in the electric Circuits ==== ==== Effects in the electric Circuits ====
Zeile 501: Zeile 588:
  
   * the direction of the windings, and   * the direction of the windings, and
-  * the orientation/counting of the current in the circuit.+  * The orientation/counting of the current in the circuit.
  
 <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP> <WRAP center 50%> <imgcaption ImgNr11 | Polarity of Coupling> </imgcaption> {{drawio>DirectionOfCoupling.svg}} </WRAP>
Zeile 511: Zeile 598:
 <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}} <imgcaption ImgNr12 | Example Circuits with positive Polarity> </imgcaption> {{drawio>posCoupling.svg}}
  
-In this case, the **mutual induction added positively**.+In this case, the **mutual induction is positive $(M>0)$**.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
Zeile 519: Zeile 606:
 \end{align*} \end{align*}
  
-=== negative Polarity ===+=== Negative Polarity ===
  
-The polarity is negative when only one current either flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).+The polarity is negative when only one current flows into the dotted pin and the other one out of the dotted pin (see <imgref ImgNr13>).
  
 <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP> <WRAP> <imgcaption ImgNr13 | Example Circuits with negative Polarity> </imgcaption> {{drawio>negCoupling.svg}} </WRAP>
  
-In this case, the **mutual induction added negatively**.+In this case, the **mutual induction is negative $(M<0)$***.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
 \begin{align*}  \begin{align*} 
-u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\  +u_1 &= R_1 \cdot i_1 &+ L_{11} \cdot {{{\rm d}i_1}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_2}\over{{\rm d}t}} & \\  
-u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} &- M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ +u_2 &= R_2 \cdot i_2 &+ L_{22} \cdot {{{\rm d}i_2}\over{{\rm d}t}} & + M \cdot {{{\rm d}i_1}\over{{\rm d}t}} & \\ 
 \end{align*} \end{align*}
  
-<panel type="info" title="Task 5.3.2 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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 $l=280 ~\rm mm$.  A toroidal core (ferrite, $\mu_{\rm r} = 900$) has a cross-sectional area of $A = 500 ~\rm mm^2$ and an average circumference of $l=280 ~\rm mm$. 
-On 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 $I_2=300 ~\rm mA$.+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 positive polarity (see the image **A** in <imgref ImgEx14>). What is the resulting magnetic flux $\Phi_{\rm A}$ in the coil?
Zeile 549: Zeile 636:
 **Step 1 - Draw an equivalent magnetic circuit** **Step 1 - Draw an equivalent magnetic circuit**
  
-Since there are no branches all of the core can be lumped to a single magnetic resistance (see <imgref ImgEx14circ>).+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> <WRAP> <imgcaption ImgEx14circ | equivalent magnetic circuit> </imgcaption> {{drawio>torCoilPosNegCirc.svg}} </WRAP>
  
Zeile 555: Zeile 642:
  
 Hopkinson's Law can be used here as a starting point. \\ 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 $R_m$. \\+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 $\theta_x$. \\ It also connects the single magnetic fluxes $\Phi_x$ (with $x = {1,2}$) and the single magnetic voltages $\theta_x$. \\
  
Zeile 587: Zeile 674:
  
 #@HiddenBegin_HTML~5_3_2r,Result~@# #@HiddenBegin_HTML~5_3_2r,Result~@#
-  - $0.10 ~\rm  mVs$+  - $0.10 ~\rm mVs$
   - $0.40 ~\rm mVs$   - $0.40 ~\rm mVs$
 #@HiddenEnd_HTML~5_3_2r,Result~@# #@HiddenEnd_HTML~5_3_2r,Result~@#
Zeile 596: Zeile 683:
  
 The magnetic field of a coil stores magnetic energy.  The magnetic field of a coil stores magnetic energy. 
-The energy transfer from the electric circuit to the magnetic field is also the cause of the "current dampening" effect of the inductor. +The energy transfer from the electric circuit to the magnetic field is also the cause of the "current-damping" effect of the inductor. 
 The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by: The energetic turnover for charging an conductor from $i(t_0=0)=0$ to $i(t_1)=I$ is given by:
  
Zeile 622: Zeile 709:
 ==== magnetic Energy of a toroid Coil ==== ==== magnetic Energy of a toroid Coil ====
  
-The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. +The formula can also be used for calculating the stored energy of a toroid coil with $N$ windings, the cross-section $A$, and an average length $l$ of a field line. \\
 By this, the following formulas can be used:  By this, the following formulas can be used: 
-\begin{align*}  +  - For the magnetic voltage: $\theta = H \cdot l = N \cdot I $ \\  
-\theta = H \cdot l = N \cdot I \\  +  - For the magnetic flux: $\Phi   = B \cdot A $
-\Phi   = B \cdot A  +
-\end{align*}+
  
 With the above-mentioned formulas of the magnetic circuit, we get:  With the above-mentioned formulas of the magnetic circuit, we get: 
Zeile 648: Zeile 733:
 ==== generalized magnetic Energy ==== ==== generalized magnetic Energy ====
  
-The general term to find the magnetic energy (e.g. for inhomogeneous magnetic fields) is given by +The general term to find the magnetic energy (e.g., for inhomogeneous magnetic fields) is given by 
 \begin{align*}  \begin{align*} 
 W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\  W_{\rm m} &= \iiint_V{w_{\rm m}            {\rm d}V} \\ 
Zeile 668: Zeile 753:
 \end{align*} \end{align*}
  
-Multiplying with $i$ and with $dt$ we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step.+Multiplying with $i$ and with $dt$, we get the principle of conservation of energy $dw = u \cdot i \cdot {\rm d}t$ for each small time step.
  
 \begin{align*}  \begin{align*} 
Zeile 678: Zeile 763:
 \begin{align*}  \begin{align*} 
 dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}}          \cdot i \cdot {\rm d}t \\  dW_{\rm m} &= N {{{\rm d}\Phi}\over{{\rm d}t}}          \cdot i \cdot {\rm d}t \\ 
- W_{\rm d} &=   \int                                                  {\rm d}W_{\rm m} \\ + W_{\rm m} &=   \int                                                  {\rm d}W_{\rm m} \\ 
            &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\             &= N \int_0^t {{{\rm d}\Phi}\over{{\rm d}t}} \cdot i \cdot {\rm d}t \\ 
            &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\             &= N \int_0^                                {\Phi} i \cdot {\rm d}\Phi \\ 
Zeile 700: Zeile 785:
  
 In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown.  In <imgref ImgNr15> the situation for a magnetic material with a linear relationship between $B$ and $H$ is shown. 
-Given by the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. +Given the maximum current $I_{\rm max}$ the maximum field strength $H_{\rm max}$ can be derived. 
 In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current.  In the circuit in <imgref ImgNr14>, the inductor will experience increasing and decreasing current. 
 Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$. Therefore, the $B$-$H$-curve gets passed through positive and negative values of $H$ and $H$ along the line of $B=\mu H$.
Zeile 707: Zeile 792:
  
 The situation for integrating the area in the graph is also shown:  The situation for integrating the area in the graph is also shown: 
-For each step ${\rm d}B$ the corresponding value of the field strength $H$ has to be integrated. +For each step ${\rm d}B$, the corresponding value of the field strength $H$ has to be integrated. 
 For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is For $B_0=0$ to $B=B_{\rm max}$ the magnetic energy is
  
Zeile 730: Zeile 815:
 As an example, the situation of the field strength $H(t_1)=H_1$ is shown.  As an example, the situation of the field strength $H(t_1)=H_1$ is shown. 
 This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow).  This shall be the field strength after magnetizing the ferrite material to $H_{\rm max}$ (yellow arrows) and then partly demagnetizing the material again (blue arrow). 
-The magnetization corresponds to an energy intake to the magnetic field and the demagnetization to an energy outtake.+The magnetization corresponds to an energy intake into the magnetic field, and the demagnetization to an energy outtake.
  
-Moving along the $H$-$B$-curve, one can see, that the energy intake and outtake are the same when coming back to a start point. +Moving along the $H$-$B$-curve, one can see that the energy intake and outtake are the same when coming back to a start point. 
 This means that the magnetization and demagnetization take place lossless in this example.  This means that the magnetization and demagnetization take place lossless in this example. 
-This is a good approximation for magnetically soft materials, however, does not work for magnetically hard materials like a permanent magnet. +This is a good approximation for magnetically soft materials; however, it does not work for magnetically hard materials like a permanent magnet. 
 Here, hysteresis also has to be considered. Here, hysteresis also has to be considered.
  
Zeile 741: Zeile 826:
 <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP> <WRAP> <imgcaption ImgNr17 | H-B-Curve material with Hysteresis> </imgcaption> {{drawio>HBcurvehyst.svg}} </WRAP>
  
-===== Tasks =====+===== Exercise =====
  
-<panel type="info" title="Task 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.9 Application: Shaded Pole Motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The <imgref ImgTask01> and <imgref ImgTask01> show a shaded pole motor of a commercial oven.+The <imgref ImgTask01> and <imgref ImgTask02> show a shaded pole motor of a commercial oven.
  
   * Find out how this motor works - explicitly: why is there a preferred direction of the motor?   * Find out how this motor works - explicitly: why is there a preferred direction of the motor?
Zeile 758: Zeile 843:
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
  
-<panel type="info" title="Task 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<panel type="info" title="Exercise 5.1.10 Further exercise"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercise for beginning in the topic of magnetic circuits.+The book [[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|DC Electrical Circuit Analysis - A Practical Approach (Fiore)]] has some nice exercises for beginning in the topic of magnetic circuits.
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 767: Zeile 852:
  
 An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}.  An alternative interpretation of the magnetic circuits is the {{https://en.wikipedia.org/wiki/Gyrator–capacitor model|Gyrator–capacitor model}}. 
-The big difference there is, that there the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$.  +The big difference there is that the magnetic flux $\Phi$ is not interpreted as an analogy to the electric current $I$ but to the electric charge $Q$.  
-This model can solve more questions, however, is a bit less intuitive based on this course and less commonly used compared to the {{https://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter.+This model can solve more questions; however, it is a bit less intuitive based on this course and less commonly used compared to the {{https://en.wikipedia.org/wiki/Magnetic_circuit#Resistance–reluctance_model|Magnetic_circuit}}, which was also presented in this chapter.
  
 ==== Moving a Plate into an Air Gap ==== ==== Moving a Plate into an Air Gap ====