Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Beide Seiten der vorigen Revision Vorhergehende Überarbeitung
Nächste Überarbeitung		 Vorhergehende Überarbeitung
electrical_engineering_2:magnetic_circuits [2024/05/10 20:02] – [Magnetic Circuit with two Sources] mexleadminelectrical_engineering_2:magnetic_circuits [2025/05/27 07:56] (aktuell) – mexleadmin
Zeile 7: Zeile 7:
 <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 =====
Zeile 17: Zeile 17:
   - 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>
Zeile 140: Zeile 140:
 |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 $l_R = 300 ~\rm mm$. 
Zeile 164: Zeile 164:
 </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:
Zeile 192: Zeile 192:
 </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:
Zeile 208: Zeile 208:
 </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:
Zeile 224: Zeile 224:
 </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:
Zeile 242: Zeile 242:
 </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 $I=1.30 ~\rm A$.
  
Zeile 250: Zeile 250:
 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 $A_2$
Zeile 270: Zeile 270:
 </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. 
Zeile 289: Zeile 289:
 </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 $l_0$, $l_1$, $l_2$. 
Zeile 348: Zeile 348:
  
 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 363: 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 455: Zeile 450:
 \end{align*} \end{align*}
  
-Based on energy considerations, the induction Matrix has to be symmetric. Therefore, the following applies:+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 mutual inductances are equal: $M_{12} = M_{21} = M$ +  * In General: the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$ 
-  * the mutual inductance $M$ is:     $M = \sqrt{M_{12}\cdot M_{21}} = k \cdot \sqrt {L_{11}\cdot L_{22}}$+  * For symmetric induction matrix: The mutual inductances are equal: $M_{12} = M_{21} = M$
   * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}   * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \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%>+<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 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 $N_1 = 400$, $N_2=300$. 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 $N_1 = 400$, $N_2=300$.
- 
-Calculate  
-  * the self inductions $L_{11}$, $L_{22}$,  
-  * the mutual inductions $M_{12}$, and $M_{21}$, 
-  * the coupling factors $k_{12}$ and $k_{21}$. 
  
 <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> <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 $2$.
  
-=== Step 2: Calculate the magnetic resistances === +#@HiddenBegin_HTML~5312,Path~@#
-\\+
  
 <WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP> <WRAP right> <imgcaption ExImgNr13 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkSingleVolt.svg}} </WRAP>
Zeile 497: Zeile 489:
 \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 \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\  R_{\rm m1} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}\cdot \left(3 + {{1\cdot 2}\over{1 + 2}}\right) \\ 
Zeile 508: Zeile 500:
 \end{align*} \end{align*}
  
-== Step 3: Calculate the self-induction == +#@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. 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*} 
Zeile 515: Zeile 510:
 L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\  L_{22} &= {{N_2^2}\over{R_{\rm m2}}}                    &= 247 ~\rm mH\\ \\ 
 \end{align*} \end{align*}
 +#@HiddenEnd_HTML~5313,Path~@#
 +
 +4. Calculate the coupling factors $k_{12}$ and $k_{21}$.
  
-== Step 4: Calculate the coupling factors == +#@HiddenBegin_HTML~5314,Path~@#
-\\+
 <WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP> <WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
  
Zeile 531: Zeile 528:
  
 A similar approach leads to $k_{12}$ with $k_{12}= 1/4$. 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*}  \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_{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\\ \\  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*} \end{align*}
 +#@HiddenEnd_HTML~5315,Path~@#
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
 +
 +#@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@#
 +
 +
 +
 ==== Effects in the electric Circuits ==== ==== Effects in the electric Circuits ====
  
Zeile 553: 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 563: 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 is positiv $(M>0)$**.+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 571: 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 is negativ $(M<0)$***.+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: 
Zeile 585: Zeile 620:
 \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 601: 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 607: 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 639: 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 648: 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 674: 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 700: 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 720: 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 730: 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 752: 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 759: 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 782: 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 793: 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 810: 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 819: 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 ====