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electrical_engineering_2:magnetic_circuits [2024/05/10 20:20] – mexleadminelectrical_engineering_2:magnetic_circuits [2024/07/11 18:54] (aktuell) – [Effects in the electric Circuits] mexleadmin
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 \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*}
  
Zeile 462: Zeile 462:
 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>
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 \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*} 
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 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~@#
  
-== Step 4: Calculate the coupling factors == +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> <WRAP right> <imgcaption ExImgNr12 | Equivalent Network for coupling> </imgcaption> {{drawio>CoreWithTwoCoils2networkCoupling.svg}} </WRAP>
  
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 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 ====
  
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 \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="Task 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 $l=280 ~\rm mm$. 
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 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 634: 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 792: Zeile 832:
 <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="Task 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?