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--- electrical_engineering_1:network_analysis [2021/10/16 14:16]
slinn
+++ electrical_engineering_1:network_analysis [2023/11/28 00:45] (aktuell)
mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 4. Analysis of direct current networks ======
+====== 4 Analysis of Networks ======
-====== 4. Analysis of direct current networks ======
+<callout> <WRAP> <imgcaption imageNo1 | examples for networks> </imgcaption> {{drawio>Beispiele Netzwerke.svg}} </WRAP>
-<callout> <WRAP right> <imgcaption imageNo1 | examples for networks> </imgcaption> {{drawio>Beispiele Netzwerke}} </WRAP>
+Network analysis plays a central role in electrical engineering.
+It is so important because it can be used to simplify what at first sight appear to be complicated circuits and systems to such an extent that they can be understood and results derived from them.
-Network analysis plays a central role in electrical engineering. It is so important because it can be used to simplify what at first sight appear to be complicated circuits and systems to such an extent that they can be understood and results derived from them.
+In addition, networks also occur in other areas, for example, the momentum flux through a truss or the heat flux through individual hardware elements (<imgref imageNo1>, or [[https://www.onsemi.com/pub/Collateral/AND9596-D.PDF#page=5|an example for heat flow through electronics]]). The concepts shown below can also be applied to these networks.
-In addition, networks also occur in other areas, for example the force flow through a truss or the heat flow through individual hardware elements (<imgref imageNo1 >). The concepts shown below can also be applied to these networks.
+On the {{https://en.wikipedia.org/wiki/Network_analysis_(electrical_circuits)|wiki page for network analysis}}  the different methods are described very well in a compact way </callout>
-On the {{https://de.wikipedia.org/wiki/network_analysis_(electrical_engineering)|wiki page on network analysis}}  the different methods are described very well in a compact way </callout>
 <callout>
-=== Goals ===
+=== Learning Objectives ===
-=== Goals ===
+By the end of this section, you will be able to:
+  - <del>determine the number of nodes, number of (tree and connecting) branches, and number of meshes.</del>
-After this lesson, you should:
+  - <del>construct a complete tree from an electrical network.</del>
+  - <del>understand the mesh current method and node potential method.</del>
-  - <del>be able to determine the number of nodes, number of (tree and connecting) branches, and number of meshes.</del>
+  - understand and be able to apply the superposition procedure.
-  - <del>be able to construct a complete tree from an electrical network.</del>
-  - <del>be able to understand the branch current method, mesh current method, and node potential method.</del>
-  - Understand and be able to apply the superposition procedure.
 </callout>
-<callout type="danger" icon="true">
+===== 4.1 Preliminary Work for Network Analysis =====
-Due to the shortened semester, only the subchapter [[:electrical_engineering_1:analysis_of_dc_grids#overlay_procedure_superposition_principle|4.5]] is relevant for WiSe2020.
+==== Preparation of the Circuit ====
-</callout>
+<WRAP> <imgcaption imageNo10 | Preparing the circuit> </imgcaption> {{drawio>VorbereitungDerSchaltung.svg}} </WRAP>
-===== 4.1 Preliminary work on network analysis =====
-===== 4.1 Preliminary work on network analysis =====
-==== Preparation of the circuit ====
-==== Preparation of the circuit ====
-<WRAP right> <imgcaption imageNo10 | Preparing the circuit> </imgcaption> {{drawio>VorbereitungDerSchaltung}} </WRAP>
 Before the network analysis can be tackled, the circuit must be suitably prepared (cf. <imgref imageNo10 >):
@@ Zeile 46: / Zeile 32: @@
   - Clarify what is given and what is sought
   - Draw a circuit
-  - Add counting arrows. If not already given, then:
+  - Add voltage and current arrows. If not already given, then:
-      - First draw current and voltage arrows at all sources according to the generator arrow system.
+      - First, draw current and voltage arrows at all sources according to the generator arrow system.
-      - Afterwards define the current arrows at the remaining branches as you like.
+      - Afterwards, define the current arrows at the remaining branches as you like.
       - Finally, draw the voltage arrows at the loads according to the load arrow system.
   - Select suitable current and voltage designations. If not already given, then:
@@ Zeile 54: / Zeile 40: @@
       - Do not insert any signs in front of the designators in the circuit.
-In real applications it is useful to specify the number of variables ("what is wanted?"), parameters ("what can be adjusted?", e.g. potentiometer) and known quantities ("what is given?"). \\ This makes it clear how many equations are needed. This seems to become difficult for larger networks - but a trick for this is presented below.
+In real applications, it is useful to specify the number of variables ("what is wanted?"), parameters ("what can be adjusted?", e.g. potentiometer) and known quantities ("what is given?"). \\
+This makes it clear how many equations are needed. This seems to become difficult for larger networks - but a trick for this is presented below.
 It often helps to draw the drawing several times (at least in your head) to have enough space for the identifiers (cf. <imgref imageNo10> below).
@@ Zeile 60: / Zeile 47: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Graph and Trees ====
+==== Graphs and Trees ====
-==== Graph and Trees ====
-<WRAP right> <imgcaption imageNo11 | Graph of a network> </imgcaption> {{drawio>GraphEinesNetzwerks}} </WRAP>
+<WRAP> <imgcaption imageNo11 | Graph of a network> </imgcaption> {{drawio>GraphEinesNetzwerks.svg}} </WRAP>
-In the chapter [[:electrical_engineering_1:simple_circuits#nodes_branches_and_loops|2. simple dc_circuits]] the terms nodes, branches and loops have already been explained. These will now be expanded here to better explain the various network analysis methods in the following. In <imgref imageNo11 > the **graph**  of the example network is drawn. We had already seen this one too, but without knowing that this is called a graph! \\ But the important thing is: In this graph only the (real) nodes are drawn. Nodes are by definition the connection of __more than two__  branches. Accordingly, the connection between $R_{10}$ and $R_7$ is __not a node__  ((sometimes such connections are called "fake nodes")) ! For this reason also the blue circle as sign for nodes is omitted here.
+In the chapter [[:electrical_engineering_1:simple_circuits#nodes_branches_and_loops|2. simple dc_circuits]] the terms nodes, branches, and loops have already been explained. These will now be expanded here to better explain the various network analysis methods in the following. In <imgref imageNo11 > the **graph**  of the example network is drawn. We had already seen this one too, but without knowing that this is called a graph! \\
+But the important thing is: In this graph, only the (real) nodes are drawn. Nodes are by definition the connection of __more than two__  branches. Accordingly, the connection between $R_{10}$ and $R_7$ is __not a node__  ((sometimes such connections are called "fake nodes")) ! For this reason, also the blue circle as a sign for nodes is omitted here.
-A concept that has not yet appeared is that of the complete tree. For this, some (mathematical) graph theory is needed. There, too, the terms nodes and loops are used as before. A **tree**  here is a special kind of graph. The graph in <imgref imageNo11> shows several loops. \\ Now a tree is characterized precisely by the fact that it contains __no__  loops. Three different trees are drawn in the picture. From a given network, many different trees can be created (depending on the number of nodes). \\ Among the different trees, there are now some in which each node connects two or fewer loops.((Here we now depart from the previous electrotechnical notion of node (= connecting more than 2 branches). The mathematical notion of node does not have this restriction))  These are called **complete trees**  (occasionally also {{https://de.wikipedia.org/wiki/Hamilton circle problem|Hamilton way}}  ). Complete trees can also be understood as this shows a path through the network where all nodes are visited only exactly once.
+A concept that has not yet appeared is that of the complete tree. For this, some (mathematical) graph theory is needed. There, too, the terms nodes and loops are used as before. A **tree**  here is a special kind of graph. The graph in <imgref imageNo11> shows several loops. \\
+Now a tree is characterized precisely by the fact that it contains __no__  loops. Three different trees are drawn in the picture. From a given network, many different trees can be created (depending on the number of nodes). \\ Among the different trees, there are now some in which each node connects two or fewer loops. ((Here we now depart from the previous electrotechnical notion of node (= connecting more than 2 branches). The mathematical notion of a node does not have this restriction))  These are called **complete trees**  (occasionally also [[https://en.wikipedia.org/wiki/Hamiltonian_path|Hamiltonian path]] ). Complete trees can also be understood as this shows a path through the network where all nodes are visited only exactly once.
 Tree 3 in <imgref imageNo11> is now just one of the possible complete trees of this network.
@@ Zeile 74: / Zeile 61: @@
 The branches in complete trees are now distinguished according to their membership:
-  * **tree branches**  belong to the complete tree (solid lines in <imgref imageNo11>).
+  * **tree branches** belong to the complete tree (solid lines in <imgref imageNo11>).
   * **Connecting branches**  do not belong to the complete tree (dotted lines in <imgref imageNo11>).
-Why does the swing to graph theory make sense now? The trick is that by defining the complete tree, all loops have just been removed. Conversely, a new (independent) loop can be created by each connecting branch. So if the number of independent loop equations $m$ is sought, this is just equal to the number of connecting branches.
+Why does the excursion to graph theory make sense now? The trick is that by defining the complete tree, all loops have just been removed. Conversely, a new (independent) loop can be created by each connecting branch. So if the number of independent loop equations $m$ is sought, this is just equal to the number of connecting branches.
 To do this, proceed as follows:
@@ Zeile 87: / Zeile 74: @@
 Thus, the number of independent loop equations $m$ is findable by counting the nodes $k$ and branches $z$ over $m = v = z - k + 1$.
-This explanation can also be heard again in [[https://www.youtube.com/watch?v=c7z1pRCzEuw|this video]] and is explained again clearly via [[https://studyflix.de/informatik/euler-und-hamiltonkreis-1287|StudyFlix]].
+This explanation can also be heard again in [[https://www.youtube.com/watch?v=BpoDAfHnOkk|this video]] and is explained again clearly via [[https://www.youtube.com/watch?v=AwsMTEl79wI|this video]].
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-===== 4.2 Branch Current Procedure =====
+===== 4.2 Branch Current Method=====
-===== 4.2 Branch Current Procedure =====
+<WRAP> <imgcaption imageNo12 | example circuit> </imgcaption> {{drawio>Beispielschaltung.svg}} </WRAP>
-===== 4.2 Branch Current Procedure =====
+The branch current method (also called branch current method) now "simply times" (almost) all equations of the circuit.
+Specifically, for each node and each __independent__ loop, the node and loop equations are written down:
-===== 4.2 Branch Current Procedure =====
-===== 4.2 Branch Current Procedure =====
-<WRAP right> <imgcaption imageNo12 | example circuit> </imgcaption> {{drawio>Beispielschaltung}} </WRAP>
-In the branch current method now "simply times" (almost) all equations of the circuit. Specifically, for each node and each __independent__  loop, the node and loop equations are written down:
   * for all nodes $k$ respectively the equation: $\sum_{k=0}^{N_k}{I_k}=0$
@@ Zeile 110: / Zeile 90: @@
 This forms a linear system of equations. This can then be considered as a matrix equation and solved with the rules of (mathematical) art.
-<WRAP onlyprint> For the example (<imgref imageNo12>), these would be the equations:
+<WRAP onlyprint>
+For the example (<imgref imageNo12>), these would be the equations:
 ~~PAGEBREAK~~ ~~CLEARFIX~~
 The matrices still need to be corrected for the voltage and current sources!!!
-<WRAP group><WRAP column half>
 === Example of nodal equations ===
-=== Example of nodal equations ===
+\begin{align*} \sum\limits_{k=0}^{N_k}{I_k}=0 \end{align*}
-=== Example of nodal equations ===
+Setting up the individual equations:
+\begin{align*}
+\scriptsize\text{node 'a'} & \scriptsize : -I_0 - I_9 - I_7 = 0 \\
+\scriptsize\text{node 'b'} & \scriptsize : +I_0 - I_1 - I_3 = 0 \\
+\scriptsize\text{node 'c'} & \scriptsize : + I_1 - I_2 - I_4 = 0  \\
+\scriptsize\text{node 'd'} & \scriptsize : - I_5 + I_4 - I_{11} = 0  \\
+\scriptsize\text{node 'e'} & \scriptsize : + I_5 + I_6 - I_7 = 0  \\
+\scriptsize\text{node 'f'} & \scriptsize : - I_2 + I_3 - I_6 + I_9 + I_{11} = 0  \\
+\end{align*}
-=== Example of nodal equations ===
+Sorting currents into columns:
+\begin{align*} \begin{smallmatrix}
+\text{node 'a'}: & -I_0 & & & & & & & - I_7 & - I_9 & & = 0  \\
+\text{node 'b'}: & +I_0 & - I_1 & & - I_3 & & & & & & = 0  \\
+\text{node 'c'}: & & + I_1 &- I_2 & & - I_4 & & & & & = 0  \\
+\text{node 'd'}: & & & & + I_4 & - I_5 & & & - I_{11} & = 0 \\
+\text{node 'e'}: & & & & & + I_5 & + I_6 & - I_7 & & = 0  \\
+\text{node 'f'}: & & & - I_2 & + I_3 & & - I_6 & & + I_9 & + I_{11} & = 0 \\
+\end{smallmatrix} \end{align*}
-=== Example of nodal equations ===
+Setting up the matrix:
+\begin{align*} \left( \begin{smallmatrix}
+-1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 \\
++1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\
+& +1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\
+& 0 & 0 & 0 & +1 & -1 & 0 & 0 & 0 & -1 \\
+& 0 & 0 & 0 & 0 & +1 & +1& -1 & 0 & 0 \\
+& 0 & -1 & +1 & 0 & 0 & -1& 0 & +1 & +1 \\
+\end{smallmatrix} \right)
+\cdot
+\left( \begin{smallmatrix}
+I_0 \\ I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5 \\ I_6 \\ I_7 \\ I_9 \\ I_{11}
+\end{smallmatrix} \right) = \vec{0} \end{align*}
-=== Example of nodal equations ===
-=== Example of nodal equations ===
-=== Example of nodal equations ===
-\begin{align*} \sum\limits_{k=0}^{N_k}{I_k}=0 \. \end{align*}
-Setting up the individual equations: \begin{align*} \scriptsize\text{node 'a'} & \scriptsize : -I_0 - I_9 - I_7 = 0 \. \scriptsize\text{node 'b'} & \scriptsize : +I_0 - I_1 - I_3 = 0 \. \scriptsize\text{node 'c'} & \scriptsize : + I_1 - I_2 - I_4 = 0 \. \scriptsize\text{node 'd'} & \scriptsize : - I_5 + I_4 - I_{11} = 0 \. \scriptsize\text{node 'e'} & \scriptsize : + I_5 + I_6 - I_7 = 0 \. \scriptsize\text{node 'f'} & \scriptsize : - I_2 + I_3 - I_6 + I_9 + I_{11} = 0 \end{align*}
-Sorting streams into columns: \begin{align*} \begin{smallmatrix} \text{node 'a'}: & -I_0 & & & & & & & - I_7 & - I_9 & & = 0 \. \text{node 'b'}: & +I_0 & - I_1 & & - I_3 & & & & & & = 0 \. \text{node 'c'}: & & + I_1 &- I_2 & & - I_4 & & & & & = 0 \. \text{node 'd'}: & & & & + I_4 & - I_5 & & & - I_{11} & = 0 \\ \text{node 'e'}: & & & & & + I_5 & + I_6 & - I_7 & & = 0 \. \text{node 'f'}: & & & - I_2 & + I_3 & & - I_6 & & + I_9 & + I_{11} & = 0 \\ \end{smallmatrix} \end{align*}
-Setting up the matrix: \begin{align*} \left( \begin{smallmatrix} -1 & 0 & 0 & 0 & 0 & 0 & 0 & -1 & -1 & 0 \\ +1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & +1 & -1 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & +1 & -1 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 & +1 & +1& -1 & 0 & 0 \\ 0 & 0 & -1 & +1 & 0 & 0 & -1& 0 & +1 & +1 \\ \end{smallmatrix} \right) \cdot \left( \begin{smallmatrix} I_0 \\ I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5 \\ I_6 \\ I_7 \\ I_9 \\ I_{11} \end{smallmatrix} \right) = \vec{0} \end{align*}
-</WRAP><WRAP column half>
 === Example of loop equations ===
-=== Example of loop equations ===
+\begin{align*} \sum\limits_{m=0}^{N_m}{U_m}=0  \end{align*}
-=== Example of loop equations ===
+Setting up the individual equations:
+\begin{align*}
+\scriptsize\text{Masche 'abf'} & \scriptsize : -U_0 + U_3  - U_9    = 0 \\
+\scriptsize\text{Masche 'bcf'} & \scriptsize : +U_1  - U_2   - U_3          = 0 \\
+\scriptsize\text{Masche 'cdf'} & \scriptsize :   + U_2 + U_4   - U_{11}         = 0 \\
+\scriptsize\text{Masche 'def'} & \scriptsize :    + U_5  - U_6        + U_{11}  = 0 \\
+\scriptsize\text{Masche 'eaf'} & \scriptsize :     + U_6    - U_7 - U_{10} + U_9      = 0 \\
+\quad \\
+\end{align*}
-=== Example of loop equations ===
-=== Example of loop equations ===
+Sorting voltages into columns:
+\begin{align*}
+\begin{smallmatrix}
+\text{Masche 'abf'}: &-U_0 & & & + U_3 & & & & & & - U_9 & & &  = 0 \\
+\text{Masche 'bcf'}: & & + U_1 & - U_2 & - U_3 & & & & & & & & & = 0 \\
+\text{Masche 'cdf'}: & & & + U_2 & & + U_4 & & & & & & & - U_{11}& = 0 \\
+\text{Masche 'def'}: & & & & & & + U_5 & - U_6 & & & & & + U_{11} & = 0 \\
+\text{Masche 'eaf'}: & & & & & & & + U_6 & - U_7 - U_{10} & & - U_9 & & & = 0 \\
+\quad \\ \quad \\
+\end{smallmatrix}
+\end{align*}
-=== Example of loop equations ===
+Set up the matrix, but note $U_m = R_x \cdot I_m$:
+\begin{align*}
+\left( \begin{smallmatrix}
+-R_0 & 0    &  0   & +R_3 & 0    &  0   & 0   & -R_9 & 0    & 0 \\
+    & +R_1 & -R_2 & -R_3 & 0    &  0   & 0   & 0    & 0    & 0  \\
+    & 0    & +R_2 & 0    & +R_4 &  0   & 0   & 0    & 0    & -R_{11}  \\
+    & 0    &  0   & 0    & 0    & +R_5 &-R_6 & 0    & 0    & +R_{11}   \\
+    & 0    &  0   & 0    & 0    & 0    &+R_6 &-R_7-U_{10}& -R_9 & 0  \\
+\end{smallmatrix} \right) \cdot
+\left( \begin{smallmatrix}
+I_0 \\ I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5 \\ I_6 \\ I_7 \\ I_9 \\ I_{11}
+\end{smallmatrix} \right) = \vec{0}
+\quad \\
+\end{align*}
-=== Example of loop equations ===
+These matrices can be solved using, for example, the [[https://en.wikipedia.org/wiki/Gaussian_elimination|Gaussian elimination]].
-=== Example of loop equations ===
+~~PAGEBREAK~~ ~~CLEARFIX~~
-\begin{align*} \sum\limits_{m=0}^{N_m}{U_m}=0 \. \end{align*}
+</WRAP>
-Setting up the individual equations: \begin{align} \scriptsize\text{loop 'abf'} & \scriptsize : -U_0 + U_3 - U_9 = 0 \. \scriptsize\text{loop 'bcf'} & \scriptsize : +U_1 - U_2 - U_3 = 0 \. \scriptsize\text{loop 'cdf'} & \scriptsize : + U_2 + U_4 - U_{11} = 0 \. \scriptsize\text{loop 'def'} & \scriptsize : + U_5 - U_6 + U_{11} = 0 \. \scriptsize\text{loop 'eaf'} & \scriptsize : + U_6 - U_7 - U_{10} + U_9 = 0 \. \quad \ \end{align*}
-Sorting stresses into columns: \begin{align*} \begin{mallmatrix} \text{loop 'abf'}: &-U_0 & & + U_3 & & & & & - U_9 & & = 0 \. \text{loop 'bcf'}: & & + U_1 & - U_2 & - U_3 & & & & & & & & = 0 \. \text{loop 'cdf'}: & & + U_2 & & + U_4 & & & & & & & - U_{11}& = 0 \. \text{loop 'def'}: & & & & & + U_5 & - U_6 & & & & + U_{11} & = 0 \\ \text{loop 'eaf'}: & & & & & & + U_6 & - U_7 - U_{10} & & - U_9 & & & = 0 \\ \quad \\ \quad \ \end{smallmatrix} \end{align*}
+=== Another example in videos ===
-Set up the matrix, but note $U_m = R_x \cdot I_m$: \begin{align*} \left( \begin{smallmatrix} -R_0 & 0 & 0 & +R_3 & 0 & 0 & -R_9 & 0 & 0 \\. 0 & +R_1 & -R_2 & -R_3 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & +R_2 & 0 & +R_4 & 0 & 0 & 0 & -R_{11} \\ 0 & 0 & 0 & 0 & +R_5 &-R_6 & 0 & 0 & +R_{11} \\ 0 & 0 & 0 & 0 & 0 &+R_6 &-R_7-U_{10}& -R_9 & 0 \ \end{smallmatrix} \right) \cdot \left( \begin{smallmatrix} I_0 \\ I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5 \\ I_6 \\ I_7 \\ I_9 \\ I_{11} \end{smallmatrix} \right) = \vec{0} \quad \\ \end{align*} </WRAP></WRAP> </WRAP>
-These matrices can be solved using, for example, the {{https://de.wikipedia.org/wiki/Gaußsches_Eliminationsverfahren#Example|Gaußes Eliminationsverfahren}}. ~~PAGEBREAK~~ ~~CLEARFIX~~
-=== another examples in videos ===
 <WRAP group> <WRAP half column>
@@ Zeile 178: / Zeile 194: @@
  \\ 1. writing down the given circuit and sizes
-. drawing in and designating the knots
+. drawing in and designating the nodes
 . draw in and label the loops
-. draw and label the branch currents
+. draw and label the mesh currents
-. drawing in and designating the branch voltages
+. drawing in and designating the mesh voltages
 for a simple circuit
-</WRAP> <WRAP half column> Branch current analysis Example 1/3
+</WRAP> <WRAP half column> Mesh current analysis Example 1/3
 {{youtube>6sVeFqlSV4A}}
@@ Zeile 198: / Zeile 214: @@
  \\ 1. writing down the given circuit and sizes
-. drawing in and designating the knots
+. drawing in and designating the nodes
 . draw in and label the loops
@@ Zeile 208: / Zeile 224: @@
 for a more complex circuit
-</WRAP> <WRAP half column> Branch current analysis 2/4
+</WRAP> <WRAP half column> Branch current analysis 2/3
 {{youtube>8gGEmrbURsA}}
@@ Zeile 216: / Zeile 232: @@
 Video 3 describes the following steps:
-. writing down the given circuit and sizes
+. write down the given circuit and sizes
-. drawing in and designating the knots
+. draw in and designate the nodes
 . draw in and label the loops
@@ Zeile 224: / Zeile 240: @@
 . draw and label the branch currents
-. drawing in and designating the branch voltages
+. draw in and designate the branch voltages
 . set up node equations and loop equations
@@ Zeile 232: / Zeile 248: @@
 . solve the matrix
-</WRAP> <WRAP half column> Branch current analysis 3/4
+</WRAP> <WRAP half column> Mesh current analysis 3/3
 {{youtube>_xomX-d8XU4}}
@@ Zeile 238: / Zeile 254: @@
 </WRAP> </WRAP> <WRAP group> <WRAP half column>
-In the [[https://www.youtube.com/watch?v=75NnTVDKHNA|Video 4]] (not embedded here), will teache you how to solve a matrix by using a calculator: \\
+In the [[https://www.youtube.com/watch?v=75NnTVDKHNA|Video 4]] (not embedded here), will teach you how to solve a matrix by using a calculator: \\
  \\
-. inserting the numerical values to a calculator \\
+. inserting the numerical values into a calculator
-. calculating the matrix with a calculator \
+. calculating the matrix with a calculator
 </WRAP> <WRAP half column> </WRAP> </WRAP>
 <panel type="info" title="Exercise 4.2.1"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>Z2QDXjG2ynU}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 4.2.2"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>14Rcbm5m3dQ}}
@@ Zeile 257: / Zeile 268: @@
 </WRAP></WRAP></panel>
-===== 4.3 loop flow method =====
+===== 4.3 Mesh Current Method =====
-===== 4.3 loop flow method =====
+In the [[https://en.wikipedia.org/wiki/Mesh_analysis|mesh current method]], only for all loops $m$ each equation: $\sum\limits_{j=0}^{N_j}{U_j}=0$ are considered. However, these are represented in the form $R\cdot I = U $.
-In the loop flow method, only for all loops $m$ each equation: $\sum\limits_{j=0}^{N_j}{U_j}=0$ are considered. However, these are represented in the form $R\cdot I = U $.
 The advantage here is that the number of equations to be solved is reduced to the number of independent loop currents.
-These can also be considered as matrix equations and can be solved with the rules of (mathematical) art.
+These can also be considered matrix equations and can be solved with the rules of (mathematical) art.
 <WRAP group> <WRAP half column>
-In video 1, the loop flow method is applied by means of an example.
+In video 1, the mesh current method is applied using an example.
-<WRAP important>Important: Although the video explains the application super, but contains a small error at minute 6:50. The sign of the voltages on the right side must be inverted in each case. This was also explained correctly a few seconds before.</WRAP>
+</WRAP> <WRAP half column> mesh current method
-</WRAP> <WRAP half column> loop stream analysis
 {{youtube>j8LHrm3_brk}}
@@ Zeile 279: / Zeile 286: @@
 </WRAP> </WRAP> <WRAP group> <WRAP half column>
-Also in video 2, the loop flow method is applied by means of an example.
+Also in video 2, the mesh current method is applied using an example.
-</WRAP> <WRAP half column> loop stream analysis
+</WRAP> <WRAP half column> mesh current method
 {{youtube>7I3-HAW8rmM}}
@@ Zeile 291: / Zeile 298: @@
 </WRAP> <WRAP half column> </WRAP> </WRAP>
-===== 4.4 Nodal potential method =====
+===== 4.4 Nodal Potential Method =====
-===== 4.4 Nodal potential method =====
-In the nodal potential method, only the equation: $\sum\limits_{i=0}^{N_i}{I_i}=0$ are considered for all nodes k respectively. However, these are expressed in the form ${1\over R} \cdot U = I $ and $G \cdot U = I $ respectively.
+In the [[https://en.wikipedia.org/wiki/Nodal_analysis|nodal potential method]], only the equation: $\sum\limits_{i=0}^{N_i}{I_i}=0$ are considered for all nodes k respectively. However, these are expressed in the form ${1\over R} \cdot U = I $ and $G \cdot U = I $ respectively.
 The advantage here is that the number of equations to be solved is reduced to the number of existing nodes (minus 1).
-These can also be considered as matrix equations and can be solved with the rules of (mathematical) art.
+These can also be considered matrix equations and can be solved with the rules of (mathematical) art.
 <WRAP group> <WRAP half column>
@@ Zeile 323: / Zeile 328: @@
 </WRAP> <WRAP half column> </WRAP> </WRAP>
-===== 4.5 Superposition method / Superposition principle =====
+===== 4.5 Superposition Method / Superposition Principle =====
-===== 4.5 Superposition method / Superposition principle =====
+The superposition principle shall first be illustrated by some examples:
-The superposition principle shall first be illustrated by some examples
+<callout title="Example 1 - from an interview of a consulting company">
-<callout title="Example 1 - from consulting industry interviews">
+**Task**: Three students are to fill a pool. If Alice has to fill it alone, she would need 2 days. Bob would need 3 days and Carol would need 4 days. How long would it take all three to fill a pool if they helped together?
-**Task**: Three students are to fill a pool. If Alice were to fill it alone, she would need 2 days. Bob would need 3 days and Carol would need 4 days. How long would it take all three to fill a pool if they helped together?
+The question sounds far off-topic at first but is directly related. The point is that to solve it, filling the pool is assumed to be linear. So Alice will fill $1 \over 2$, Bob $1 \over 3$, and Carol $1 \over 4$ of the pool per day. So on the first day, ${1 \over 2}+{1 \over 3}+{1 \over 4} = {{6 + 4 + 3} \over 12} = {13 \over 12}$ of the pool filled. \\ So the three of them need ${12 \over 13}$ of a day. \\  \\ However, this solution path is only possible because in linear systems the partial results can be added. </callout>
-The question sounds far off topic at first, but is directly related. The point is that to solve it, filling the pool is assumed to be linear. So Alice will fill $1 \over 2$, Bob $1 \over 3$, and Carol $1 \over 4$ of the pool per day. So on the first day, ${1 \over 2}+{1 \over 3}+{1 \over 4} = {{6 + 4 + 3} \over 12} = {13 \over 12}$ of the pool filled. \\ So the three of them need ${12 \over 13}$ of a day. \\  \\ However, this solution path is only possible because in linear systems the partial results can be added. </callout>
+<callout title="Example 2 - Spring Force and Displacement">
-<callout title="Example 2 - Spring Force and Travel">
+<WRAP> <imgcaption imageNo02 | mechanical spring> </imgcaption> {{drawio>mechanischeFeder.svg}} </WRAP>
-<WRAP right> <imgcaption imageNo02 | mechanical spring> </imgcaption> {{drawio>mechanischeFeder}} </WRAP>
+**Task**:A mechanical, linear spring is displaced due to masses $m_1$ and $m_2$ in the Earth's gravitational field (see <imgref imageNo02 >). What is the magnitude of the deflection if both masses are attached simultaneously?
-**Task**:A mechanical, linear spring is deflected with masses $m_1$ and $m_2$ in the Earth's gravitational field (see <imgref imageNo02 >). What is the magnitude of the deflection if both masses are attached simultaneously?
 Again, a linear law is used here: \begin{align*} \vec{s}= f(\vec{F}) = - D \cdot \vec{F} \end{align*}
-The (seemingly trivial) approach applies here: \begin{align*} \vec{s}_{1+2} = f(\vec{F_1} + \vec{F_2}) &= - D \cdot (\vec{F_1} + \vec{F_2}) \\ &= - D \cdot \vec{F_1} - D \cdot \vec{F_2} \\ &= f(\vec{F_1}) + f(\vec{F_2}) \ &= \vec{s_1} + \vec{s_2} \end{align*} </callout>
+The (seemingly trivial) approach applies here:
+\begin{align*}
+\vec{s}_{1+2} = f(\vec{F_1} + \vec{F_2}) &= - D \cdot (\vec{F_1} + \vec{F_2}) \\
+             &= - D \cdot \vec{F_1} - D \cdot \vec{F_2} \\
+             &= f(\vec{F_1}) + f(\vec{F_2}) \\
+             &= \vec{s_1} + \vec{s_2}
+\end{align*} </callout>
-<callout icon="fa fa-exclamation" color="red" title="Notice:"> In a physical system in which effect and cause are linearly related, the effect of each cause can first be determined separately. The total effect is then the sum of the individual effects. </callout>
+<callout icon="fa fa-exclamation" color="red" title="Notice:"> In a physical system in which effect and cause are linearly related, the effect of each cause can first be determined separately.
+The total effect is then the sum of the individual effects. </callout>
-For electrical engineering this principle was described by {{https://de.wikipedia.org/wiki/Hermann_von_Helmholtz|Hermann_von_Helmholtz}}:
+For electrical engineering this principle was described by [[https://en.wikipedia.org/wiki/Hermann_von_Helmholtz|Hermann_von_Helmholtz]]:
-> The currents in the branches in a linear network are equal to the sum of the partial currents in the branches concerned caused by the individual sources.
+> The currents in the branches of a linear network are equal to the sum of the partial currents in the branches concerned caused by the individual sources.
 <WRAP group> <WRAP half column> Thus, in the superposition method, the current (or voltage) sought in a circuit with multiple sources can be viewed as a superposition of the resulting currents (or voltages) of the individual sources.
@@ Zeile 355: / Zeile 365: @@
 The "recipe" for the overlay is as follows:
-  - Choose next source ''x''
+  - Choose the next source ''x''
   - Replace all ideal sources with their respective equivalent resistors:
       - ideal voltage sources by short circuits
       - ideal current sources by an open line
   - Calculate the partial currents sought in the branches considered.
-  - Go to the next source ''x=x+1'' ((x=x+1 is not meant mathematically, but procedurally as in the programming language C))  and to point 2, as long as the partial currents of all sources have not been calculated.
+  - Go to the next source ''x=x+1'' ((x=x+1 is not meant mathematically, but procedurally as in the programming language C)), and to point 2, as long as the partial currents of all sources have not been calculated.
   - Add up the partial currents in the branches under consideration, observing the correct sign.
@@ Zeile 369: / Zeile 379: @@
 {{youtube>w4N9CBc_nkA}}
-more complex example of the superposition method
+A more complex example of the superposition method
 {{youtube>-48-4qWnhjA}}
@@ Zeile 379: / Zeile 389: @@
 === Example ===
-=== Example ===
+<WRAP> <imgcaption imageNo03 | example circuit with superposition> </imgcaption> {{drawio>BeispielschaltungSuperposition.svg}} <WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
+=== Introduction to Nodal, Mesh and Superposition Method ===
+{{youtube>8f-2yXiYmRI}}
-<WRAP right> <imgcaption imageNo03 | example circuit with superposition> </imgcaption> {{drawio>BeispielschaltungSuperposition}} <WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<panel type="info" title="Exercise 4.5.1 Converting a bipolar signal to a unipolar signal"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+===== Exercises =====
-<WRAP right>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&cct=$+1+3e-7+63.8+50+5+43%0Ar+-48+224+64+224+0+10000%0Ar+64+224+64+304+0+50000%0Ar+64+224+64+160+0+2000%0Ag+64+304+64+320+0%0AR+64+160+64+128+0+0+40+5+0+0+0.5%0Av+-192+304+-192+240+0+1+40+20+0+0+0.5%0Ag+-192+304+-192+320+0%0Aw+-192+240+-192+224+0%0Ar+-192+224+-96+224+0+1000%0A368+64+224+224+224+0+0%0Ab+-256+144+-112+341+0%0Ax+-252+367+-84+397+4+24+bipolare%5CsQuelle%5Cs%5C%5Cn(z.B.%5CsSensor)%0Ax+92+193+109+196+4+24+R%0Ax+110+207+123+210+4+24+1%0Ax+112+273+125+276+4+24+2%0Ax+94+259+111+262+4+24+R%0Ax+-18+207+-5+210+4+24+3%0Ax+-36+193+-19+196+4+24+R%0Ax+182+361+355+391+4+24+unipolare%5CsSenke%5C%5Cn(z.B.%5CsuC))%0A370+-96+224+-48+224+1+0%0Ax+-188+184+-171+187+4+24+R%0Ax+-169+193+-156+196+4+24+q%0Ao+9+1024+0+4098+20+6.4+0+2+5+0%0A 730,400 noborder}} </WRAP>
+#@TaskTitle_HTML@#4.5.1 Converting a bipolar signal to a unipolar signal <fs medium>(not from written test)</fs>#@TaskText_HTML@#
-Imagine you want to develop a circuit that conditions a sensor signal so that it can be processed by a microcontroller. The sensor signal is in the range $U_{sens} \in [-15...15V]$, the microcontroller input can read values in the range $U_{uC} \in [0...3.3V]$. The sensor can supply a maximum current of $I_{sens,max}=1mA$. For the internal resistance of the microcontroller input applies: $R_{uC} \rightarrow \infty$
+Imagine you want to develop a circuit that conditions a sensor signal so that it can be processed by a microcontroller. The sensor signal is in the range $U_{\rm sens} \in [-15...15~\rm V]$, and the microcontroller input can read values in the range $U_{\rm uC} \in [0...3.3~\rm V]$. The sensor can supply a maximum current of $I_{\rm sens, max}=1~\rm mA$. For the internal resistance of the microcontroller, input applies: $R_{\rm uC} \rightarrow \infty$
-For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ against $U_{uC,max}$ (similar circuit see in simulation on the right).
+For conditioning, the input signal is to be fed via the series resistor $R_3$ to the center potential of a voltage divider $R_1 - R_2$ with $R_1$ to a supply voltage $U_{\rm s}$.
-  - Find the relationship between $R_1$, $R_2$ and $R_3$ using superposition.
+The following simulation shows roughly the situation (the resistor values are not correct).
-  - Find the relationship between $R_1$, $R_2$ and $R_3$ using star-delta transformation.
-  - What is the input resistance $R_{in}(R_1, R_2,R_3)$ of the circuit (viewed from the sensor)?
-  - What is the maximum allowed input resistance $R_{in}(R_1, R_2,R_3)$ for the sensor to still deliver current?
-  - Determine suitable values for $R_1$, $R_2$ and $R_3$.
-  - What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this?
-</WRAP></WRAP></panel>
+<WRAP>{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCDMCmC0DsIBskB08CcGkA4AsYGADEQKxFIYjnUh6TVxhgBQATiLHjiAEy95kg-oKLgSJdkL4DpkIqOoSiUpMNlrwSMWN4SWAczkK5ekDpYAlaWG03ePHXTGlzboqlIsAbp0K8oE1h-PjwnCDC+Jx1PQz8MAPlBYISoMwsAd3iAgTEUnNkVDnyZZOxStzB9SFxpEQr6iwAjTl5SJHA8ZOZEgnMWAA9W0kSkRFgMCEgMRGFBJoBLAAcAewAbAEM2AB0AZwBlFYBXNgBjaD3t7YA7AApoVANUPY293ehr3ZW2AEpBkFShAYVSohA6cxAln+zF0REQYF4DF4VTooXA0IRfHgwLaWPBaN4-wwwlIoOYfCQAQhUKGwR4enGrmRYghkH+sBq4AwDBSXPxshp4Bwoymtigk1RsiO12W6y2bwW1wA1lc7g8nnsjgBhH5-SDwPLlepcemyCAWWlgHA8K3deAQK2zNGC4KULk8sCkYHlCH7FgrAHiQrODD0sRIVCKAKuFRAA noborder}} </WRAP>
-{{page>task_4.5.2_with_calculation&nofooter}}{{page>aufgabe_4.5.3&nofooter}}{{page>aufgabe_4.5.4&nofooter}}
+Questions:
+. Find the relationship between $R_1$, $R_2$, and $R_3$ using superposition. \\
+  * Determine suitable values for $R_1$, $R_2$, and $R_3$.
+  * What values for $R^0_1$, $R^0_2$, and $R^0_3$ from the [[https://de.wikipedia.org/wiki/E-Reihe|E24 series]] can be used to do this?
+#@HiddenBegin_HTML~1,Solution~@#
+Using superposition, we create two separate circuits where one source is considered.
+For these two circuits, we calculate $U_\rm A^{(1)}$ and $U_\rm A^{(2)}$. \\
+To make the calculation simpler, the resistors $R_3$ and $R_{\rm s}$ will be joined to $R_4 =R_3 +R_{\rm s}$.
+<callout>
+=== Circuit 1 : only consider $U_{\rm S}$, ignore $U_{\rm I}$ ===
+{{drawio>electrical_engineering_1:exc541circ1.svg}}
+\begin{align*}
+U_{\rm O}^{(1)}  &= U_{\rm S} \cdot {{R_2||R_4}\over{R_1 + R_2||R_4}}
+                  = U_{\rm S} \cdot {{ {{R_2  R_4}\over{R_2 + R_4}} }\over{R_1 + {{R_2  R_4}\over{R_2 + R_4}} }} \\
+                 &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1  (R_2 + R_4)+ R_2 R_4 }} \\
+                 &= U_{\rm S} \cdot {{ R_2 R_4 }\over{R_1  R_2 + R_1 R_4 + R_2 R_4 }} \\
+\end{align*}
+</callout>
+<callout>
+=== Circuit 2 : only consider $U_{\rm I}$, ignore $U_{\rm S}$ ===
+{{drawio>electrical_engineering_1:exc541circ2.svg}}
+\begin{align*}
+U_{\rm O}^{(2)}  &= U_{\rm I} \cdot {{R_1||R_2}\over{R_4 + R_1||R_2}}
+                  = U_{\rm I} \cdot {{ {{R_1 R_2}\over{R_1 + R_2}} }\over{R_4 + {{R_1 R_2}\over{R_1 + R_2}} }} \\
+                 &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 (R_1 + R_2)+ R_1 R_2 }} \\
+                 &= U_{\rm I} \cdot {{ R_1 R_2 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \\
+\end{align*}
+</callout>
+<callout>
+=== Superposition: Let's sum it up! ===
+\\
+These two intermediate voltages for the single sources have to be summed up as $U_{\rm O}= U_{\rm O}^{(1)} + U_{\rm O}^{(2)}$. \\
+When deeper investigated, one can see that the denominator for both $U_{\rm O}^{(1)}$ and $U_{\rm O}^{(2)}$ is the same. \\
+We can also simplify further when looking at often-used sub-terms (here: $R_2$)
+\begin{align*}
+U_{\rm O}                                            &=  {{ 1 }\over{R_4 R_1 + R_4 R_2+ R_1 R_2 }} \cdot (U_{\rm S} \cdot R_2 R_4  + U_{\rm I} \cdot R_1 R_2 ) \\
+U_{\rm O} \cdot (R_4 R_1 + R_4 R_2+ R_1 R_2 )        &=   U_{\rm S} \cdot R_2 R_4  + U_{\rm I} \cdot R_1 R_2  \\ \\
+U_{\rm O} \cdot ({{R_1 R_4}\over{R_2}} + R_4 + R_1 ) &=   U_{\rm S} \cdot     R_4  + U_{\rm I} \cdot R_1      \tag 1 \\
+\end{align*}
+The formula $(1)$ is the general formula to calculate the output voltage $U_{\rm O}$ for a changing input voltage $U_{\rm I}$, where the supply voltage $U_{\rm S}" is constant. \\
+</callout>
+Now, we can use the requested boundaries:
+  - For the minimum input voltage $U_{\rm I}= -15 ~\rm V$, the output voltage shall be $U_{\rm O} =   0 ~\rm V$
+  - For the maximum input voltage $U_{\rm I}= +15 ~\rm V$, the output voltage shall be $U_{\rm O} = 3.3 ~\rm V$
+This leads to two situations:
+<callout>
+=== Situation I : $U_{\rm I,min}= -15 ~\rm V$ shall create $U_{\rm O,min} = 0 ~\rm V$ ===
+\\
+We put $U_{\rm A} = 0 ~\rm V$ in the formula $(1)$ :
+\begin{align*}
+                          &=   U_{\rm S} \cdot     R_4  + U_{\rm I,min} \cdot R_1     \\
+- U_{\rm I,min} \cdot R_1  &=   U_{\rm S} \cdot     R_4       \\
+ {{R_1}\over{R_4}}         &=-{{U_{\rm S}}\over {U_{\rm I,min}}} = k_{14} \tag 2 \\
+\end{align*}
+So, with formula $(2)$, we already have a relation between $R_1$ and $R_4$. Yeah 😀 \\
+The next step is situation 2
+</callout>
+<callout>
+=== Situation II : $U_{\rm I,max}= +15 ~\rm V$ shall create $U_{\rm O,max} = 3.3 ~\rm V$ ===
+\\
+We use formula $(2)$ to substitute $R_1 = k_{14} \cdot R_4 $ in formula $(1)$, and:
+\begin{align*}
+U_{\rm O,max} \cdot (k_{14}{{ R_4^2}\over{R_2}} + R_4 + k_{14} R_4 ) &=   U_{\rm S} \cdot     R_4  + U_{\rm I,max} \cdot k_{14} R_4 \\
+U_{\rm O,max} \cdot (k_{14}{{ R_4  }\over{R_2}} + 1   + k_{14}     ) &=   U_{\rm S}                + U_{\rm I,max} \cdot k_{14}  \\
+                     k_{14}{{ R_4  }\over{R_2}}  + 1   + k_{14}      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{        U_{\rm O,max} }} \\
+                     k_{14}{{ R_4  }\over{R_2}}                      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{        U_{\rm O,max} }}        - (1   + k_{14})\\
+                           {{ R_4  }\over{R_2}}                      &= {{U_{\rm S}                + U_{\rm I,max} \cdot k_{14} }\over{ k_{14} U_{\rm O,max} }} - {{1   + k_{14} }\over{k_{14}}} \tag 3 \\
+\end{align*}
+So, another relation for $R_4$ and $R_2$.  😀 \\
+</callout>
+So, to get values for the relations, we have to put in the values for the input and output voltage conditions. For $k_{14}$ we get by formula $(2)$:
+\begin{align*}
+k_{14} = {{R_1}\over{R_4}}  =-{{5 ~\rm V}\over {-15 ~\rm V }} = {{1}\over{3}} \\
+\end{align*}
+This value $k_{14}$ we can use for formula $(3)$:
+\begin{align*}
+{{ R_4  }\over{R_2}} &= {{5 ~\rm V + 15 ~\rm V \cdot {{1}\over{3}} }\over{ 3.3 ~\rm V \cdot {{1}\over{3}} }} - {{1   + {{1}\over{3}} }\over{ {{1}\over{3}} }} \\
+                     &= {{10}\over{1.1}} - 4 \\
+k_{42}               &\approx 5.09
+\end{align*}
+We could now - theoretically - arbitrarily choose one of the resistors, e.g., $R_2$, and then calculate the other two. \\
+But we must consider another boundary, a boundary for $R_{\rm S}$. The maximum voltage and the maximum current are given for the sensor. By this, we can calculate $R_{\rm S}$:
+\begin{align*}
+R_{\rm S}   &= {{ U_{\rm OC} }\over{ I_{\rm SC} }} = {{ U_{\rm S,max} }\over{ I_{\rm S,max} }} = {{ 15 ~\rm V }\over{ 1 ~\rm mA }} \\
+            &= 15 ~\rm k\Omega
+\end{align*}
+Therefore, $R_4 = R_{\rm S} + R_3$ must be larger than this. \\
+#@HiddenEnd_HTML~1,Solution ~@#
+#@HiddenBegin_HTML~Result1,Result~@#
+The sensor resistance is
+\begin{align*}
+R_S &= 15 {~\rm k\Omega}\\
+\end{align*}
+We can choose $R_3$ arbitrarily. Here I choose a nice value to get integer values for $R_3$ and $R_1$:
+\begin{align*}
+R_3 &= 45 {~\rm k\Omega}\\
+R_1 &= {{1}\over{3}}   (R_3 + 15 {~\rm k\Omega}) = 20   {~\rm k\Omega} \\
+R_2 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 11.8 {~\rm k\Omega}
+\end{align*}
+Based on the E24 series, the following values are next to the calculated ones:
+\begin{align*}
+R_3^0 &= 43 {~\rm k\Omega}\\
+R_1^0 &= {{1}\over{3}}   (R_3 + 15 {~\rm k\Omega}) = 20 {~\rm k\Omega} \\
+R_2^0 &= {{1}\over{5.09}}(R_3 + 15 {~\rm k\Omega}) = 12 {~\rm k\Omega}
+\end{align*}
+#@HiddenEnd_HTML~Result1,Result~@#
+. Find the relationship between $R_1$, $R_2$, and $R_3$ by investigating Kirchhoff's nodal rule for the node where $R_1$, $R_2$, and $R_3$ are interconnected.
+#@HiddenBegin_HTML~Solution2,Solution~@#
+The potential of the node is $U_\rm O$. Therefore the currents are:
+  - the current $I_2$ over $R_2$ is flowing to ground: $I_2 = - {{U_\rm O}\over{R_2}} $
+  - the current $I_1$ over $R_1$ is coming from the supply voltage $U_{\rm S}$ to the nodal voltage $U_{\rm O}$:  $I_1 = {{U_{\rm S} - U_{\rm O}}\over{R_1}}$
+  - the current $I_4$ over $R_4$ is coming from the input voltage  $U_{\rm I}$ to the nodal voltage $U_{\rm O}$:  $I_4 = {{U_{\rm I} - U_{\rm O}}\over{R_4}}$
+This led to the formula based on the Kirchhoff's nodal rule:
+\begin{align*}
+\Sigma I = 0 &= I_1 + I_2 + I_3 \\
+&= {{U_{\rm S} - U_{\rm O}}\over{R_1}} + {{U_{\rm I} - U_{\rm O}}\over{R_4}} - {{U_\rm O}\over{R_2}}
+\end{align*}
+The formula can be rearranged, with all terms containing $ U_{\rm O}$ on the left side:
+\begin{align*}
+    {{U_{\rm O}}\over{R_1}} + {{U_{\rm O}}\over{R_2}} + {{U_{\rm O}}\over{R_4}}         &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\
+      U_{\rm O}\cdot \left( {{1}\over{R_1}} + {{1}\over{R_2}} + {{1}\over{R_4}} \right) &=  {{U_{\rm S}}\over{R_1}}  + {{U_{\rm I}}\over{R_4}}  \\
+\end{align*}
+Both sides can be multiplied by $\cdot R_1$, $\cdot R_2$, $\cdot R_4$  - in order to get rid of the fractions :
+\begin{align*}
+      U_{\rm O}\cdot \left( {{R_1 R_2 R_4 }\over{R_1}} + {{R_1 R_2 R_4 }\over{R_2}} + {{R_1 R_2 R_4 }\over{R_4}} \right) &=  R_1 R_2 R_4 \cdot {{U_{\rm S}}\over{R_1}}  + R_1 R_2 R_4 \cdot {{U_{\rm I}}\over{R_4}}  \\
+      U_{\rm O}\cdot \left( R_2 R_4 + R_1 R_4 + R_1 R_2 \right) &=  R_2 R_4 \cdot U_{\rm S}  + R_1 R_2 \cdot U_{\rm I} \\
+      U_{\rm O} &= {{R_2}\over{R_2 R_4 + R_1 R_4 + R_1 R_2 }}  \left( R_4 \cdot U_{\rm S}  + R_1 \cdot U_{\rm I} \right)\\
+\end{align*}
+The last formula was just the result we also got by the superposition but by more thinking. \\
+So, sometimes there is an easier way...
+  * Unluckily, there is no simple way to know before, what way is the easiest.
+  * Luckily, all ways lead to the correct result.
+#@HiddenEnd_HTML~Solution2,Solution ~@#
+. What is the input resistance $R_{\rm in}(R_1, R_2, R_3)$ of the circuit (viewed from the sensor)?
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+#@HiddenBegin_HTML~Solution3,Solution~@#
-''
+\begin{align*}
+R_{\rm in}(R_1, R_2, R_3) &= R_3 + R_1 || R_2 \\
+                          &= R_3 + {{R_1  R_2}\over{R_1 + R_2}}
+\end{align*}
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+#@HiddenEnd_HTML~Solution3,Solution~@#
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+. What is the minimum allowed input resistance ($R_{\rm in, min}(R_1, R_2, R_3)$) for the sensor to still deliver current?
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+#@HiddenBegin_HTML~Solution4,Solution~@#
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+\begin{align*}
+R_{\rm in, min} &= {{U_{\rm sense}}\over{I_{\rm sense, max}}} \\
+                &= \rm {{15 V}\over{1 mA}} \\
+                &= 15 k\Omega \\
+\end{align*}
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+#@HiddenEnd_HTML~Solution4,Solution~@#
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+#@TaskEnd_HTML@#
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