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electrical_engineering_and_electronics_1:block20 [2025/12/02 18:45] – mexleadminelectrical_engineering_and_electronics_1:block20 [2026/01/20 15:39] (aktuell) – [20.3 Exercises] mexleadmin
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-====== Block 20 — Electromagnetic Induction and Energy ======+====== Block 20 — Inductance and Energy ======
  
-===== Learning objectives =====+===== 20.0 Intro ===== 
 + 
 +==== 20.0.1 Learning objectives ====
 <callout> <callout>
 After this 90-minute block, you can After this 90-minute block, you can
Zeile 7: Zeile 9:
 </callout> </callout>
  
-===== Preparation at Home =====+==== 20.0.2 Preparation at Home ====
  
 Well, again  Well, again 
Zeile 16: Zeile 18:
   * ...   * ...
  
-===== 90-minute plan =====+==== 20.0.3 90-minute plan ====
   - Warm-up (x min):    - Warm-up (x min): 
     - ....      - .... 
Zeile 24: Zeile 26:
   - Wrap-up (x min): Summary box; common pitfalls checklist.   - Wrap-up (x min): Summary box; common pitfalls checklist.
  
-===== Conceptual overview =====+==== 20.0.4 Conceptual overview ====
 <callout icon="fa fa-lightbulb-o" color="blue"> <callout icon="fa fa-lightbulb-o" color="blue">
   - ...   - ...
 </callout> </callout>
  
-===== Core content =====+===== 20.1 Core content =====
  
-==== Self-Induction ====+==== 20.1.1 Self-Induction ====
  
 Up to now, we investigated the induction of electric voltages and currents based on the change of an external flux ${\rm d}\Psi / {\rm d}t$.  Up to now, we investigated the induction of electric voltages and currents based on the change of an external flux ${\rm d}\Psi / {\rm d}t$. 
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 <WRAP> <imgcaption ImgNr15 | Self-Induction of a Coil> </imgcaption> {{drawio>SelfInductionCoil.svg}} </WRAP> <WRAP> <imgcaption ImgNr15 | Self-Induction of a Coil> </imgcaption> {{drawio>SelfInductionCoil.svg}} </WRAP>
  
-The created field density of the coil can be derived from Ampere's Circuital Law +Given by the [[block16#Recap of the fieldline images]] in Block16, we know that the $H$-field is given by magnetic voltage $\theta(t) = N \cdot i$ as:
 \begin{align*}  \begin{align*} 
-\theta(t) &= \int & \vec{H}(t) \cdot {\rm d}\vec{s} \\  +   {H}(t) &=                         {{N \cdot i }\over {l}} \\ 
-          &= \int & \vec{H}_{\rm inner}(t) \cdot {\rm d}\vec{s} & + & \int \vec{H}_{\rm outer}(t) \cdot {\rm d} \vec{s} \\  +
-          &= \int & \vec{H}(t) \cdot {\rm d}\vec{s}             & + &   0 \\  +
-          &=      & {H}(t) \cdot l \\ +
 \end{align*} \end{align*}
  
-With magnetic voltage $\theta(t) = N \cdot i$ this lead to the magnetic flux density $B(t)$+This lead to the magnetic flux density $B(t)$
  
 \begin{align*}  \begin{align*} 
-N \cdot i &= {H}(t) \cdot l \\  
-   {H}(t) &=                         {{N \cdot i }\over {l}} \\  
    {B}(t) &= \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \\     {B}(t) &= \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \\ 
 \end{align*} \end{align*}
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 \end{align*} \end{align*}
  
-The changing flux $\Phi$ is now creating an induced electric voltage and current, which counteracts the initial change of the current. +The changing flux $\Phi$ is now creating an induced electric voltage and current, which counteracts the initial change of the current. \\
 This effect is called **Self Induction**. The induced electric voltage $u_{\rm ind}$ is given by: This effect is called **Self Induction**. The induced electric voltage $u_{\rm ind}$ is given by:
  
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 \end{align*} \end{align*}
  
-The result means that the induced electric voltage $u_{\rm ind}$ is proportional to the change of the current ${{\rm d}\over{{\rm d}t}}i$. +The result means that the induced electric voltage $u_{\rm ind}$ is proportional to the change of the current ${{\rm d}\over{{\rm d}t}}i$. \\
 The proportionality factor is also called **Self-inductance**  $L$ (or often simply called inductance). The proportionality factor is also called **Self-inductance**  $L$ (or often simply called inductance).
  
-===== 4.5 Inductance =====+==== 20.1.2 Inductance ====
  
 The inductance is another passive basic component of the electric circuit.  The inductance is another passive basic component of the electric circuit. 
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 \end{align*} \end{align*}
  
-==== Inductance of different Components ====+==== 20.1.3 Inductance of different Components ====
  
 === Long Coil === === Long Coil ===
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 +==== 20.1.4  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 =====+\begin{align*} L_{\rm eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}
  
 +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 $u_{\rm eq}= u_1 + u_2 + ...$):
  
-<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> 
- 
-<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"> 
- 
-For partwise linear $u_{\rm ind}$ one can derive:  
 \begin{align*}  \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*} \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*}  \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*} \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*}  \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*} \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%>+==== 20.1.5 Energy of the magnetic Field ====
  
-A rectangular coil is given by the sizes $a=10 ~\rm cm$, $b=4 ~\rm cm$, and the number of windings $N=200$.  +not covered  
-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 $l=5 ~\rm cm$).  +===== 20.2 Common pitfalls ===== 
-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> +
- +
-<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>+
  
 +===== 20.3 Exercises =====
  
-</collapse> </WRAP></WRAP></panel> 
  
 {{page>electrical_engineering_and_electronics:task_ljxf80q7vxywehqf_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_ljxf80q7vxywehqf_with_calculation&nofooter}}
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 \begin{align*}  \begin{align*} 
 L_1 &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \\ L_1 &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \\
-    &= 4\pi \cdot 10^{-7} {\rm {{H}\over{m}}} \cdot 1 \cdot (390)^2 \cdot {{\pi \cdot (0.03~\rm m)^2 }\over {0.18 ~\rm m}}+    &= 4\pi \cdot 10^{-7} {\rm {{H}\over{m}}} \cdot 1 \cdot (390)^2 \cdot {{\pi \cdot ({{0.03~\rm m}\over{2}})^2 }\over {0.18 ~\rm m}}
 \end{align*} \end{align*}
  
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 #@HiddenBegin_HTML~4511R,Result~@# #@HiddenBegin_HTML~4511R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 3.0 ~\rm mH+L_1 &= 0.75 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4511R,Result~@# #@HiddenEnd_HTML~4511R,Result~@#
Zeile 350: Zeile 256:
 #@HiddenBegin_HTML~4512R,Result~@# #@HiddenBegin_HTML~4512R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 12 ~\rm mH+L_1 &= 4 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4512R,Result~@# #@HiddenEnd_HTML~4512R,Result~@#
Zeile 363: Zeile 269:
 #@HiddenBegin_HTML~4513R,Result~@# #@HiddenBegin_HTML~4513R,Result~@#
 \begin{align*}  \begin{align*} 
-L_1 &= 6.0 ~\rm mH+L_1 &= 1.5 ~\rm mH
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4513R,Result~@# #@HiddenEnd_HTML~4513R,Result~@#
Zeile 379: Zeile 285:
 #@HiddenBegin_HTML~4514R,Result~@# #@HiddenBegin_HTML~4514R,Result~@#
 \begin{align*}  \begin{align*} 
-L_4 &= 3.0 ~\rm H+L_4 &= 0.75 ~\rm H
 \end{align*} \end{align*}
 #@HiddenEnd_HTML~4514R,Result~@# #@HiddenEnd_HTML~4514R,Result~@#
Zeile 387: Zeile 293:
 <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A cylindrical air coil (length $l=40 ~\rm cm$, diameter $d=5.0 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2.0 ~\rm ms$ down to $0.0 ~\rm A$.+A cylindrical air coil (length $l=40 ~\rm cm$, radius $r=5.0 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2.0 ~\rm ms$ down to $0.0 ~\rm A$.
  
 What is the amount of the induced voltage $u_{\rm ind}$?  What is the amount of the induced voltage $u_{\rm ind}$?