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--- electrical_engineering_and_electronics_1:block20 [2025/12/02 18:45] – mexleadmin
+++ electrical_engineering_and_electronics_1:block20 [2025/12/02 18:50] (aktuell) – mexleadmin
@@ Zeile 167: / Zeile 167: @@
+==== 6 Inductances in Circuits ====
+Focus here: uncoupled inductors!
+=== Series Circuits ===
-===== Common pitfalls =====
+Based on $L = {{ \Psi(t)}\over{i}}$ and Kirchhoff's mesh law ($i=\rm const$) the series circuit of inductions can be interpreted as a single current $i$ which generates multiple linked fluxes $\Psi$. Since the current must stay constant in the series circuit, the following applies for the equivalent inductor of a series connection of single ones:
-  * ...
-===== Exercises =====
-<panel type="info" title="Exercise 4.1.4 Effects of induction I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A change of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> show different flux-time-diagrams as examples.
-  * Create for each $\Phi(t)$-diagram the corresponding $u_{\rm ind}(t)$-diagram!
-  * Write down each maximum value of $u_{\rm ind}(t)$
-<WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia1.svg}} \\ </WRAP></WRAP>
+\begin{align*} L_{\rm eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}
-<button size="xs" type="link" collapse="Solution_4_1_4_1_Solution">{{icon>eye}} Solution for (a)</button><collapse id="Solution_4_1_4_1_Solution" collapsed="true">
+A similar result can be derived from the induced voltage $u_{ind}= L {{{\rm d}i}\over{{\rm d}t}}$, when taking the situation of a series circuit (i.e. $i_1 = i_2 = i_1 = ... = i_{\rm eq}$ and $u_{\rm eq}= u_1 + u_2 + ...$):
-For partwise linear $u_{\rm ind}$ one can derive:
 \begin{align*}
-u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\
+& u_{\rm eq}                                       & = &u_1                                    & + &u_2                        &+ ... \\
-            &= -{{\Delta \Phi}\over{\Delta t}}
+& L_{\rm eq} {{{\rm d}i_{\rm eq} }\over{{\rm d}t}} & = &L_{1} {{{\rm d}i_{1} }\over{{\rm d}t}} & + &L_{2} {{di_{2} }\over{dt}} &+ ... \\
+& L_{\rm eq} {{{\rm d}i }\over{{\rm d}t}}          & = &L_{1} {{{\rm d}i     }\over{{\rm d}t}} & + &L_{2} {{di     }\over{dt}} &+ ... \\
+& L_{\rm eq}                                       & = &L_{1}                                  & + &L_{2}                      &+ ... \\
 \end{align*}
-For diagram (a):
+===Parallel Circuits ===
-  * $t= 0.0 ... 0.6 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$
+For parallel circuits, one can also start with the principles based on Kirchhoff's mesh law:
-  * $t= 0.6 ... 1.5 ~\rm s$: $u_{\rm ind} = -{{-3.75\cdot 10^{-3} ~\rm Vs}\over{0.9 ~\rm s}}= +4.17 ~\rm mV$
-  * $t= 1.5 ... 2.1 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$
-</collapse>
+\begin{align*} u_{\rm eq}= u_1 = u_2 = ... \\ \end{align*}
-<button size="xs" type="link" collapse="Solution_4_1_4_1_Finalresult">
+and Kirchhoff's nodal law:
-{{icon>eye}} Final result for (a)</button><collapse id="Solution_4_1_4_1_Finalresult" collapsed="true">
-<WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams Solution> </imgcaption> <WRAP> {{drawio>FluxTimeDia1Solution.svg}} \\
-</WRAP></WRAP> \\
-</collapse>
-</WRAP></WRAP></panel>
+\begin{align*} i_{\rm eq}= i_1 + i_2 + ... \\ \end{align*}
-<panel type="info" title="Exercise 4.1.5 Effects of induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+Here, the formula for the induced voltage has to be rearranged:
-A changing of magnetic flux is passing a coil with a single winding and induces the voltage $u_{\rm ind}(t)$.
-The following pictures <imgref ImgNrEx02> show different voltage-time diagrams as examples.
-  * Create for each $u_{\rm ind}(t)$-diagram the corresponding $\Phi(t)$-diagram!
-  * Write down each maximum value of $\Phi(t)$
-Note the given start value $\Phi_0$ for each flux.
-<WRAP> <imgcaption ImgNrEx02| Voltage-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia2.svg}} \\ </WRAP></WRAP>
-#@HiddenBegin_HTML~415_1S,Solution for (a)~@#
-For partwise linear $u_{\rm ind}$ one can derive:
 \begin{align*}
-u_{\rm ind}        &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\
+     u_{\rm ind}          &= L {{{\rm d}i}\over{{\rm d}t}} \quad \quad \quad \quad \bigg| \int(){\rm d}t \\
-\rightarrow  \Phi  &= -\int_0^t{ u_{\rm ind} \;{\rm d}t} \\
+\int u_{\rm ind} {\rm d}t &= L \cdot i \\
-\Phi               &= \Phi_0 -\sum_k {u_{{\rm ind},~k} \; \Delta t} \\
+                        i &= {{1}\over{L}} \cdot \int u_{\rm ind} {\rm d}t \\
 \end{align*}
-For diagram (a):
+By this, we get:
-  * $t= 0.00 ... 0.04 ~\rm s\quad$: $\quad \Phi = \Phi_0 - {0 \cdot \; \Delta t} \quad\quad\quad\quad\quad\quad\quad= 0 ~\rm Wb$
-  * $t= 0.04 ... 0.10 ~\rm s\quad$: $\quad \Phi =   0 {~\rm Wb} - {{30 ~\rm mV} \cdot \; (t - 0.04 ~\rm s)} = \quad {1.2 ~\rm mWb} - t \cdot 30 ~\rm mV$
-  * $t= 0.10 ... 0.14 ~\rm s\quad$: $\quad \Phi =   {1.2 ~\rm mWb} - {0.10 ~\rm s} \cdot 30 ~\rm mV \quad = - {1.8 ~\rm mWb}$
-#@HiddenEnd_HTML~415_1S,Solution ~@#
-#@HiddenBegin_HTML~415_1R,Result for (a)~@#
-{{drawio>FluxTimeDia2Res.svg}}
-#@HiddenEnd_HTML~415_1R,Result~@#
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 4.1.6 Coil in magnetic Field I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A single winding is located in a homogenous magnetic field ($B = 0.5 ~\rm T$) between the pole pieces.
-The winding has a length of $150 ~\rm mm$ and a distance between the conductors of $50 ~\rm mm$ (see <imgref ImgNrEx03>).
-  * Determine the function $u_{\rm ind}(t)$, when the coil is rotating with $3000 ~\rm min^{-1}$.
-  * Given a current of $20 ~\rm A$ through the winding: What is the torque $M(\varphi)$ depending on the angle between the surface vector of the winding and the magnetic field?
-<WRAP> <imgcaption ImgNrEx03| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces.svg}} \\ </WRAP></WRAP>
-<button size="xs" type="link" collapse="Solution_4_1_6_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_6_1_Solution" collapsed="true">
 \begin{align*}
-u_{\rm ind} &= -   {{{\rm d}\Phi}\over{{\rm d}t}} \\
+                                  i_{\rm eq}          &=& i_1                                       &+& i_2                                       &+& ... \\
-            &= -         {{\rm d}\over{{\rm d}t}} B \cdot A \\
+{{1}\over{L_{\rm eq}}} \cdot \int u_{\rm eq} {\rm d}t &=& {{1}\over{L_1}} \cdot \int u_{1} {\rm d}t &+& {{1}\over{L_2}} \cdot \int u_{2} {\rm d}t &+& ... \\
-            &= - B \cdot {{\rm d}\over{{\rm d}t}} A\\
+{{1}\over{L_{\rm eq}}} \cdot \int u          {\rm d}t &=& {{1}\over{L_1}} \cdot \int u     {\rm d}t &+& {{1}\over{L_2}} \cdot \int u     {\rm d}t &+& ... \\
-            &= - B \cdot {{\rm d}\over{{\rm d}t}} \left(l \cdot d \cdot \cos(\omega t) \right)\\
+{{1}\over{L_{\rm eq}}}                                &=& {{1}\over{L_1}}                           &+& {{1}\over{L_2}}                           &+& ... \\
-            &= + B \cdot l \cdot d \cdot \omega \cdot \sin(\omega t)\\
 \end{align*}
-</collapse> </WRAP></WRAP></panel>
+<callout icon="fa fa-exclamation" color="red" title="Notice:"> The inductor behaves in the parallel and series circuit similar to the resistor. </callout>
-<panel type="info" title="Exercise 4.1.7 Coil in magnetic Field II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-A rectangular coil is given by the sizes $a=10 ~\rm cm$, $b=4 ~\rm cm$, and the number of windings $N=200$.
-This coil moves with a constant speed of $v=2 ~\rm m/s$ perpendicular to a homogeneous magnetic field ($B=1.3 ~\rm T$ on a length of $l=5 ~\rm cm$).
-The area of the coil is tilted with regard to the field in $\alpha=60°$ and enters the field from the left side (see <imgref ImgNrEx04>).
-  * Determine the function $u_{\rm ind}(t)$ on the coil along the given path. Sketch of the $u_{\rm ind}(t)$ diagram.
-  * What is the maximum induced voltage $u_{\rm ind,Max}$?
-<WRAP> <imgcaption ImgNrEx04| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces2.svg}} \\ </WRAP></WRAP>
+===== Common pitfalls =====
+  * ...
-<button size="xs" type="link" collapse="Solution_4_1_7_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_7_1_Solution" collapsed="true">
-Let assume, that $l$ is in the $x$-axis, $\vec{B}$ in the $y$-axis and $a$.
-\\ \\
-**Step 1**: Calculate the effective area, perpendicular to the $\vec{B}$-field (independent from whether the area is in the $\vec{B}$-field or not).
-For this $b$ has to be projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$
-\begin{align*}
-A_{\rm eff} &= a \cdot b \cdot \cos \alpha
-\end{align*}
-**Step 2**: Focus on entering and exiting the $\vec{B}$-field. \\
-Induction only occurs for ${{\rm d}\over{{\rm d}t}}(A\cdot B)\neq 0$, so here: when the area $A_{\rm eff}$ enters and leave the constant $\vec{B}$-field.
-When entering the $\vec{B}$-field the area $A$ with $0<A<A_{eff}$ is in the field.
-The area moves with $v$. Therefore, after $\Delta t = b_{\rm eff} \cdot v$ the full $\vec{B}$-field is provided onto the area $A_{\rm eff}$:
-\begin{align*}
-u_{\rm ind} &= -        {{{\rm d}\Psi}\over{{\rm d}t}} \\
-            &= -N \cdot       {{\rm d}\over{{\rm d}t}} B \cdot A \\
-            &= -N \cdot           {{1}\over{\Delta t}} B \cdot A_{\rm eff} \\
-            &= -N \cdot {{1}\over{b \cdot \cos \alpha \cdot v}} B \cdot a \cdot b \cdot \cos \alpha \\
-            &= -N \cdot B \cdot {{a}\over{v}}\\
-\end{align*}
-The following diagram shows ...
-  * ... how one can derive the effective width $b_{\rm eff}$, which is projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$
-  * ... what happens on the effective area $A_{\rm eff}$: there is a change of the field lines in the area only for entering and leaving the $\vec{B}$-field.
-  * ... how the $u_{\rm ind}(t)$ looks as a graph: the part of $A_{\rm eff}$, where the $\vec{B}$-field passes through increase linearly due to the constant speed $v$
-Be aware, that the task did not provide a clue for the direction of windings and therefore it provides no clue for the polarization of the induced voltage. \\
-So, the course of the voltage when entering or exiting is not uniquely given.
-<WRAP> <imgcaption ImgNrEx04s| Solution> </imgcaption> <WRAP>{{drawio>WindingPolePieces2solution.svg}}  \\ </WRAP></WRAP>
+===== Exercises =====
-</collapse> </WRAP></WRAP></panel>
 {{page>electrical_engineering_and_electronics:task_ljxf80q7vxywehqf_with_calculation&nofooter}}