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--- electrical_engineering_2:magnetic_circuits [2023/05/09 14:30]
mexleadmin
+++ 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>
@@ 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.
-===== 5.1 Linear magnetic Circuits =====
+===== 5.1 Linear Magnetic Circuits =====
 For the upcoming calculations, the following assumptions are made
@@ Zeile 168: / Zeile 168: @@
 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>
@@ Zeile 393: / Zeile 405: @@
 $k_{21}$ 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.
@@ Zeile 409: / Zeile 424: @@
        &= k_{21}                  \cdot {{N_1 \cdot N_2 }\over {R_{\rm m1}}} \\
 \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:
@@ Zeile 473: / Zeile 490: @@
   * 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*}
+  * 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 ====
@@ Zeile 499: / Zeile 516: @@
 <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:
@@ Zeile 513: / Zeile 530: @@
 <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:
 \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*}
-<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 $l=280 ~\rm mm$.
 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>
-<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>
@@ Zeile 558: / Zeile 616: @@
 \boxed{W_m = {{1}\over{2}}L\cdot I^2 }
 \end{align*}
 ==== 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 ====
@@ Zeile 717: / 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 $Q$.
 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 ====