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electrical_engineering_2:magnetic_circuits [2023/05/09 13:46]
mexleadmin [Application and Limitations of the Circuit Interpretation] 		electrical_engineering_2:magnetic_circuits [2024/05/03 16:02] (aktuell)
mexleadmin [Magnetic Circuit with two Sources]
Zeile 1: Zeile 1:
-====== 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>
Zeile 9: Zeile 9:
 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 come with such an analogy and where it can be practically applied.
  
-===== 5.1 Linear magnetic Circuits =====+===== 5.1 Linear Magnetic Circuits =====
  
 For the upcoming calculations, the following assumptions are made For the upcoming calculations, the following assumptions are made
Zeile 25: Zeile 25:
   * a current-carrying coil   * a current-carrying coil
   * a ferrite core   * a ferrite core
-  * an airgap (in the picture (2) ).+  * an airgap (in picture (2) ).
  
 <WRAP> <imgcaption ImgNr02 | A simple magnetic Circuit> </imgcaption> {{drawio>SimpleMagnCircuit.svg}} </WRAP> <WRAP> <imgcaption ImgNr02 | A simple magnetic Circuit> </imgcaption> {{drawio>SimpleMagnCircuit.svg}} </WRAP>
Zeile 55: Zeile 55:
 With the above-mentioned assumptions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{\rm core}=\rm const.$. \\  With the above-mentioned assumptions the magnetic flux $\Phi$ must remain constant along the ferrite core, so $\Phi_{\rm core}=\rm const.$. \\ 
 Since the magnetic field lines neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$ This can also be seen in <imgref ImgNr04> (1) ). \\  Since the magnetic field lines neither show sources nor sinks, also the flux passing over to the airgap must be $\Phi_{\rm airgap}=\Phi_{\rm core}=\rm const.$ This can also be seen in <imgref ImgNr04> (1) ). \\ 
-A different view onto this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[:electrical_engineering_2:the_time-dependent_magnetic_field#recap_of_magnetic_field|Recap of magnetic Field]] we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.+A different view of this is the closed surface $\vec{A}$ (<imgref ImgNr04> (2)): Based on the examination in [[:electrical_engineering_2:the_time-dependent_magnetic_field#recap_of_magnetic_field|Recap of magnetic Field]] we know that the flux into the volume must be equal the flux out of the volume, or $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot d \vec{A} = 0$.
  
 The relationship $B=\mu \cdot H$, and $\mu_{\rm core}\gg\mu_{\rm airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (<imgref ImgNr04> (3)). The relationship $B=\mu \cdot H$, and $\mu_{\rm core}\gg\mu_{\rm airgap}$ lead to the fact that the ${H}$-Field must be much stronger within the airgap (<imgref ImgNr04> (3)).
Zeile 93: Zeile 93:
 \end{align*} \end{align*}
  
-Comparing the formula $5.2.3$ with the ohmic resistance and resistivity of two resistors in series, shows something interesting: +Comparing the formula $5.2.3$ with the ohmic resistance and resistivity of two resistors in series shows something interesting: 
 \begin{align*}  \begin{align*} 
 u &= &R_1 \cdot                    &i &+ &R_2 \cdot i \\  u &= &R_1 \cdot                    &i &+ &R_2 \cdot i \\ 
Zeile 168: Zeile 168:
 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:
  
-  - $l=35.8~\rm cm$, $d=1.9~\rm cm$ +  - $l=35.8~\rm cm$, $d=1.90~\rm cm$ 
-  - $l=22.5~\rm cm$, $d=1.5~\rm cm$+  - $l=11.1~\rm cm$, $d=1.50~\rm cm$
  
-<button size="xs" type="link" collapse="Solution_5_1_2_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_1_2_1_Result" collapsed="true">+#@HiddenBegin_HTML~5_1_2s,Solution~@#
  
-  - $1.5\cdot 10^5 ~\rm {{1}\over{H}}$ +The magnetic resistance is given by: 
-  - $3.0\cdot 10^5 ~\rm {{1}\over{H}}$+\begin{align*} 
 +\ R_{\rm m} &= {{1}\over{\mu_0 \mu_{\rm r}}}{{l}\over{A}}  
 +\end{align*}
  
-</collapse>+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> </WRAP></WRAP></panel>
Zeile 198: Zeile 210:
 <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="Task 5.1.4 Magnetic Voltage"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Calculate the magnetic voltage necessary in order 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:
  
   - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$   - $\delta=1.7~\rm mm$, $A=4.5~\rm cm^2$
Zeile 339: Zeile 351:
 \end{align*} \end{align*}
  
-The view onto the magnetic flux is sometimes good when effects like an acting Lorentz force in of interest. +The view of the magnetic flux is sometimes good when effects like an acting Lorentz force in of interest. 
 More often the coils are coupling two electric circuits linked in a transformer or a wireless charger.  More often the coils are coupling two electric circuits linked 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. 
Zeile 353: Zeile 365:
 The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$? The main question is now: How do we get $L_{11}$, $M_{12}$, $L_{22}$, $M_{21}$?
  
-==== Magnetic Circuit with 2 Sources ====+==== Magnetic Circuit with two Sources ====
  
-In order to get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08>). \\ +To get the self-induction and mutual induction of two interacting coils, we are going to investigate two coils on an iron core with a middle leg (see <imgref ImgNr08>). \\ 
 There the stray flux of the previous situation is only located in the middle leg. This also means, that there is no stray flux outside of the iron core. There the stray flux of the previous situation is only located in the middle leg. This also means, that there is no stray flux outside of the iron core.
  
 <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 winding: \\  +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. \\  
-When there is one current for each winding ingoing into the marked pin, the fluxes will get added up positively.+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.
  
-In order to get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ +To get $L_{11}$ and $L_{22}$, we look back to the inductance $L$ of a long coil with the length $l$. \\ 
 This was given in the chapter [[:electrical_engineering_2:the_time-dependent_magnetic_field#self-induction1|Self-Induction]] as This was given in the chapter [[:electrical_engineering_2:the_time-dependent_magnetic_field#self-induction1|Self-Induction]] as
  
Zeile 389: Zeile 402:
 \end{align*} \end{align*}
  
-In order to get the effect of the mutual induction, a coupling coefficient $k$ is introduced. +To get the effect of the mutual induction, a coupling coefficient $k$ is introduced. 
 $k_{21}$ 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}}\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 oposite, 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 second coil. \\  
 +When $k_{21}=-100~\%$, there is no flux in the middle leg but only in the second coil and in 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 408: Zeile 424:
        &= 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 positiv or negative, depending on the sign of the coupling coefficients.
  
 The formula is finally:  The formula is finally: 
Zeile 472: Zeile 490:
   * the mutual inductances are equal: $M_{12} = M_{21} = M$   * 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 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*}+  * The resulting **total coupling** $k$ is given as \begin{align*} k = \rm{sgn}(k_{12}) \sqrt{k_{12}\cdot k_{21}} \end{align*}
  
 ==== Effects in the electric Circuits ==== ==== Effects in the electric Circuits ====
  
-  * Whenever two coils are magnetically coupled, not only the self-induction $L$, but also the mutual induction $M$ applies.+  * Whenever two coils are magnetically coupled, not only the self-induction $L$ but also the mutual induction $M$ applies.
   * Based on the currents $i_1$, $i_2$ in the two circuits, the induced voltages are given by:   * Based on the currents $i_1$, $i_2$ in the two circuits, the induced voltages are given by:
  
Zeile 498: Zeile 516:
 <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 positiv $(M>0)$**.
  
 The formula of the shown circuitry is then:  The formula of the shown circuitry is then: 
Zeile 512: Zeile 530:
 <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 negativ $(M<0)$***.
  
 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 5.3.1 toroidal Core with two Coils"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>+<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%>
  
 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 $I_2=300 ~\rm mA$. 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 $I_2=300 ~\rm mA$.
  
-  - The coils shall pass the currents with positive polarity (see the top image in <imgref ImgEx14>).    What is the magnetic flux 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? 
-  - The coils shall pass the currents with negative polarity (see the bottom image in <imgref ImgEx14>). What is the magnetic flux 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> <WRAP> <imgcaption ImgEx14 | toroidal core with two coils in positive and negative polarity> </imgcaption> {{drawio>torCoilPosNeg.svg}} </WRAP>
  
-<button size="xs" type="link" collapse="Solution_5_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_5_3_1_1_Result" collapsed="true">+#@HiddenBegin_HTML~5_3_2s,Solution~@#
  
-  - $0.40 ~\rm  mVs$ +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 ($\sum_x \theta_x = 0$).
-  - $0.10 ~\rm mVs$+
  
-</collapse>+**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>). 
 +<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 $R_m$. \\ 
 +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> </WRAP></WRAP></panel>
Zeile 557: Zeile 616:
 \boxed{W_m = {{1}\over{2}}L\cdot I^2 }  \boxed{W_m = {{1}\over{2}}L\cdot I^2 } 
 \end{align*} \end{align*}
 +
 +
 +
 +
  
 ==== magnetic Energy of a magnetic Circuit ==== ==== magnetic Energy of a magnetic Circuit ====
  
-With this formula also the stored energy in a magnetic circuit can be calculated.  +With this formula also the stored energy in a magnetic circuit can be calculated. For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit: \begin{align*} \boxed{W_{\rm m} = {{1}\over{2}} \Psi \cdot I = {{1}\over{2}} {{\Psi^2}\over{L}}= {{1}\over{2}}{{\Phi^2 }\over{N^2 \cdot L}} = {{1}\over{2}} \Phi^2 \cdot R_{\rm m} = {{1}\over{2}}{{\theta^2 }\over{R_{\rm m}}}} \end{align*}
-For this, the formula be rewritten by the properties linked flux $\Psi = N \cdot \Phi = L \cdot I$ and magnetic voltage $\theta=N \cdot I = \Phi \cdot R_{\rm m}$ of the magnetic circuit:  +
-\begin{align*}  +
-\boxed{W_{\rm m} = {{1}\over{2}}  \Psi\cdot I  +
-                 = {{1}\over{2}}{{\Psi^2 }\over{L}}}  +
-\end{align*}+
  
 ==== magnetic Energy of a toroid Coil ==== ==== magnetic Energy of a toroid Coil ====
Zeile 675: Zeile 733:
 In this case, the permeability $\mu_{\rm r}$ is not a constant but can be represented as a function: $\mu_{\rm r}= f(B)$.  In this case, the permeability $\mu_{\rm r}$ is not a constant but can be represented as a function: $\mu_{\rm r}= f(B)$. 
 Here, the formula $W_{\rm m} = V\int_0^{B} H(B) \cdot {\rm d}B$ also applies - so the magnetic energy is again the area between the curve and the $B$-axis.  Here, the formula $W_{\rm m} = V\int_0^{B} H(B) \cdot {\rm d}B$ also applies - so the magnetic energy is again the area between the curve and the $B$-axis. 
-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 to the magnetic field and the demagnetization to an energy outtake.
Zeile 716: Zeile 774:
 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 there 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, 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 ====
 +
 +<WRAP> <imgcaption ImgNr81 | a magnetic circuit with a moving plate> {{electrical_engineering_2:plate_in_airgap_50_.gif}} {{electrical_engineering_2:stiftt_in_luftspalt_2.3.gif}} </imgcaption> </WRAP>
  
 ==== Switch Reluctance Motor ==== ==== Switch Reluctance Motor ====