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--- electrical_engineering_1:simple_circuits [2021/10/06 08:58] – tfischer
+++ electrical_engineering_1:simple_circuits [2024/10/24 08:13] (aktuell) – mexleadmin
@@ Zeile 1: / Zeile 1: @@
-====== 2. Simple DC circuits ======
+====== 2 Simple DC circuits ======
-<WRAP right>
+So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\
+In the following, more complicated circuit arrangements will be analyzed. These initially contain only one source, but several lines and many ohmic loads (cf. <imgref BildNr91>).
+<WRAP>
 <imgcaption BildNr91 | Example of a circuit>
 </imgcaption>
-{{drawio>BeispieleStromkreis}}
+{{drawio>BeispieleStromkreis.svg}}
 </WRAP>
-So far, only simple circuits consisting of a source and a load connected by wires have been considered. \\ In the following, more complicated circuit arrangements will be analysed. These initially contain only one source, but several lines and many ohmic loads (cf. <imgref BildNr91>).
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-===== 2.1 ideal components =====
+===== 2.1 Idealized Components =====
 <callout>
-=== goals ===
+=== Learning Objectives ===
-After this lesson you should:
+By the end of this section, you will be able to:
   - Know the representation of ideal current and voltage sources in the U-I diagram.
   - Know the internal resistance of ideal current and voltage sources.
@@ Zeile 26: / Zeile 27: @@
 Every electrical circuit consists of three elements:
-  - **Consumers**: consumers convert electrical energy into energy that is not purely electrical. \\ e.g.
+  - **Consumers** also called **Loads** (in German: Verbraucher or Last): consumers convert electrical energy into energy that is not purely electrical. \\ e.g.
     - into electrostatic energy (capacitor)
     - into magnetostatic energy (magnet)
@@ Zeile 32: / Zeile 33: @@
     - into mechanical energy (loudspeaker, motor)
     - into chemical energy (charging an accumulator)
-  - **sources (generators)**: sources convert energy from another form of energy into electrical energy. (e.g. generator, battery, photovoltaic).
+  - **Sources** also called **Generator** (in German: Quellen): sources convert energy from another form of energy into electrical energy. (e.g. generator, battery, photovoltaic).
-  - **wires (interconnections)**: the wires of interconnection lines link consumers to sources.
+  - **Wires** also called **Interconnections** (in German Leitungen or Verbindungen): The wires of interconnection lines link consumers to sources.
 These elements will be considered in more detail below.
@@ Zeile 39: / Zeile 40: @@
 ==== Consumer ====
-  * The colloquial term 'consumer' in electrical engineering stands for an electrical consumer - i.e. a component which converts electrical energy into another form of energy.
+  * The colloquial term 'consumer' in electrical engineering stands for an electrical consumer - i.e. a component that converts electrical energy into another form of energy.
   * A resistor is often also referred to as a consumer. In addition to pure ohmic consumers, however, there are also ohmic-inductive consumers (e.g. coils in a motor) or ohmic-capacitive consumers (e.g. various power supplies using capacitors at the output). Correspondingly the equation "consumer is a resistor" is wrong.
-  * Current-voltage characteristics (vgl. <imgref BildNr4>)
+  * Current-voltage characteristics (see <imgref BildNr4>)
-    * Current-voltage characteristics of a load always run through the origin, because without current there is no voltage and vice versa.
+    * Current-voltage characteristics of a load always run through the origin, because without current there is no voltage, and vice versa.
     * Ohmic loads have a linear current-voltage characteristic which can be described by a single numerical value. \\ The slope in the $U$-$I$-characteristic is the conductance: $I = G \cdot U = {{U}\over{R}}$
@@ Zeile 48: / Zeile 49: @@
 <imgcaption BildNr4 | Examples of current-voltage characteristics>
 </imgcaption>
-{{drawio>BeispieleStromSpannungsKennlinie}}
+{{drawio>BeispieleStromSpannungsKennlinie.svg}}
 </WRAP>
@@ Zeile 55: / Zeile 56: @@
 ==== Sources ====
-<WRAP>
+  * Sources act as generators of electrical energy
+  * A distinction is made between ideal and real sources. \\ The real sources are described in the following chapter "[[non-ideal_sources_and_two_terminal_networks]]".
-Ideal Sources
+The **ideal voltage source** generates a defined constant output voltage $U_\rm s$ (in German often $U_\rm q$ for Quellenspannung).
-{{youtube>8_AWiueI4Qg}}
+In order to maintain this voltage, it can supply any current.
+The current-voltage characteristic also represents this (see <imgref BildNr6>). \\
+The circuit symbol shows a circle with two terminals. In the circuit, the two terminals are short-circuited. \\
+Another circuit symbol shows the negative terminal of the voltage source as a "thick minus symbol", the positive terminal is drawn wider.
+The **ideal current source** produces a defined constant output current $I_\rm s$ (in German often $I_\rm q$ for Quellenstrom).
+For this current to flow, any voltage can be applied to its terminals.
+The current-voltage characteristic also represents this (see <imgref BildNr7>). \\
+The circuit symbol shows again a circle with two connections. This time the two connections are left open in the circle and a line is drawn perpendicular to them.
+<WRAP>
-\\
 <WRAP group><WRAP column 45%>
 <imgcaption BildNr6 | ideal voltage source>
 </imgcaption>
-{{drawio>IdealeSpannungsquelle}}
+{{drawio>IdealeSpannungsquelle.svg}}
 </WRAP>
 <WRAP column 45%>
@@ Zeile 70: / Zeile 81: @@
 <imgcaption BildNr7 | ideal current source>
 </imgcaption>
-{{drawio>IdealeStromquelle}}
+{{drawio>IdealeStromquelle.svg}}
 </WRAP></WRAP></WRAP>
-  * Sources act as generators of electrical energy
+Another Explanation of Ideal Sources
-  * A distinction is made between ideal and real sources. \\ The real sources are described in the following chapter "[[non-ideal sources and two pole networks]]".
+{{youtube>8_AWiueI4Qg}}
-The **ideal voltage source** generates a defined constant output voltage $U_s$ (in German often $U_q$ for Quellenspannung).
+\\
-In order to maintain this voltage, it can supply any current.
-The current-voltage characteristic also represents this (see <imgref BildNr6>). \\
-The circuit symbol shows a circle with two terminals. In the circuit, the two terminals are short-circuited. \\
-Another circuit symbol shows the negative terminal of the voltage source as a "thick minus symbol", the positive terminal is drawn wider.
-The **ideal current source** produces a defined constant output current $I_s$ (in German often $I_q$ for Quellenstrom).
-For this current to flow, any voltage can be applied to its terminals.
-The current-voltage characteristic also represents this (see <imgref BildNr7>). \\
-The circuit symbol shows again a circle with two connections. This time the two connections are left open in the circle and a line is drawn perpendicular to them.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== wire connection ====
+==== Wire Connection ====
   * The ideal connection line is resistance-free and transmits current and voltage instantaneously.
   * Real existing influences (e.g. voltage drop) of connections are considered via separately drawn components (e.g. ohmic resistance).
-===== 2.2 Reference-arrow Systems, sign Convention and first consideration of a DC circuit =====
+===== 2.2 Reference-Arrow Systems, Sign Convention, and first Consideration of a DC Circuit =====
 <callout>
-=== Goals ===
+=== Learning Objectives ===
-After this lesson you should:
+By the end of this section, you will be able to:
-  - Be able to apply and distinguish between the producer and consumer reference arrow systems (German: Erzeuger-Pfeilsystem und Verbraucher-Pfeilsystem).
+  - apply and distinguish between the producer and consumer reference arrow systems (German: Erzeuger-Pfeilsystem und Verbraucher-Pfeilsystem).
-  - similarly be able to use passive and active sign convention.
+  - similarly use passive and active sign conventions.
 </callout>
-In the chapter [[preparation_properties_proportions|1. Preparation and Proportions]] the  direction of conventional current and voltages has already been discussed. Unfortunately, with meshed networks it is often not clear ahead of the calculation in which direction the conventional sense of direction of all currents and voltages runs.
+In the chapter [[preparation_properties_proportions|1. Preparation and Proportions]] the direction of conventional current and voltages has already been discussed. Unfortunately, with meshed networks it is often not clear ahead of the calculation in which direction the conventional sense of direction of all currents and voltages runs.
-In <imgref BildNr5> such a meshed net is shown. In this circuit a switch $S_1$ and a current $I_2$ are marked. Once the state of the switch is swapped, the direction of the current changes.
+In <imgref BildNr5> such a meshed net is shown. In this circuit, a switch $S_1$ and a current $I_2$ are marked. Once the state of the switch is swapped, the direction of the current changes.
 <WRAP>
 <imgcaption BildNr5 | Example of a circuit>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxEJRCQBZsAoAJxDqsJvA0JGO8hBRpITTt14hsacMSr9Mw5pP5gZEwnnA0OyhHBGsKHGuJQI+LDHubHup7kumyQBKxKkqqNMPfVRwQhRYTMxZvNQ1+dUCUGmVVbzlYvzAAkQcPASTsbHMzVxj+bO4vHwiLEQKJHP8NTCcweHoANxqBDCowLh5zfiT+JH4YMARiYeM6SBl6AGUBPFr8TLlF5RAAMwBDABsAZwBTPxR6AHc5hY101WikwQ4EtqdsYmFT+8qU2vaoE-BCOS+Pg9vgAPAQjaQQTBIFRGTIgab0UHYGjkFIQbAIDQpWFSI5Ip7gFISFGE2gSfgASXo2Es4U0HGweDuKCxfheEiZrQM7BBbTuxFqeCQaAgHHc9CAA 600,500 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxEJRCQBZsAoAJxDqsJvA0JGO8hBRpITTt14hsacMSr9Mw5pP5gZEwnnA0OyhHBGsKHGuJQI+LDHubHup7kumyQBKxKkqqNMPfVRwQhRYTMxZvNQ1+dUCUGmVVbzlYvzAAkQcPASTsbHMzVxj+bO4vHwiLEQKJHP8NTCcweHoANxqBDCowLh5zfiT+JH4YMARiYeM6SBl6AGUBPFr8TLlF5RAAMwBDABsAZwBTPxR6AHc5hY101WikwQ4EtqdsYmFT+8qU2vaoE-BCOS+Pg9vgAPAQjaQQTBIFRGTIgab0UHYGjkFIQbAIDQpWFSI5Ip7gFISFGE2gSfgASXo2Es4U0HGweDuKCxfheEiZrQM7BBbTuxFqeCQaAgHHc9CAA noborder}}
 </WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Sign and Arrow Systems ====
+==== Sign and Arrow-Systems ====
 For the direction of the arrows different conventions are available. Here (and quite often in Germany) the [[https://en.wikipedia.org/wiki/Passive_sign_convention#Alternative_convention_in_power_engineering|convention of power engineering]] is used.
 This convention is
-=== Generator Reference Arrow System / Active sign convention ===
+=== Generator Reference Arrow System / Active Sign Convention ===
 <WRAP group><WRAP column 45%>
@@ Zeile 129: / Zeile 131: @@
 <imgcaption BildNr2 | Generator Arrow System>
 </imgcaption>
-{{drawio>Erzeugerpfeilsystem}}
+{{drawio>Erzeugerpfeilsystem.svg}}
 </WRAP>
 With **sources** (or generators), energy is taken __from__ the environment and made available to the circuit. \\
 For generators, the arrow__foot__ of the current is attached to the arrow__head__ of the voltage. Voltage and current arrows are antiparallel ($\uparrow \downarrow$). \\
-Similarly, the active sign convention for one component states: The current enters the component on the more negative terminal. Or vise versa: The current exits the component on the positive terminal.
+Similarly, the active sign convention for one component states: The current enters the component on the more negative terminal. Or vice versa: The current exits the component on the positive terminal.
-Both expressions "generator arrow system" and "active sign convention" come to te same result, when drawing the arrows.
+Both expressions "generator arrow system" and "active sign convention" come to the same result, when drawing the arrows.
 For generators holds:
@@ Zeile 147: / Zeile 149: @@
 <callout color="grey">
-<WRAP left>
+<WRAP>
-<imgcaption BildNr3 | Verbraucherpfeilsystem>
+<imgcaption BildNr3 | Load Arrow System>
 </imgcaption>
-{{drawio>Verbraucherpfeilsystem}}
+{{drawio>Verbraucherpfeilsystem.svg}}
 </WRAP>
 === Load Reference Arrow System ===
@@ Zeile 158: / Zeile 160: @@
 Here we have to use the passive sign convention: The current enters the component on the more positive terminal. Or vise versa: The current exits the component on the negative terminal.
-Both expressions again come to te same result, when drawing the arrows.
+Both expressions again come to the same result, when drawing the arrows.
 For consumers, the following holds:
@@ Zeile 167: / Zeile 169: @@
 </callout>
 </WRAP></WRAP>
 <callout icon="fa fa-exclamation" color="red" title="Note:">
 <WRAP>
-<imgcaption BildNr1 | Reference arrows>
-</imgcaption>
-{{drawio>Bezugspfeile1}}
-</WRAP>
   * **Before the calculation,** the __reference arrows__ for currents and voltages are set arbitrarily, with the following conditions:
-    * the active sign convention / generator arrow system is used for all sources (e.g. all voltage and current sources): the current is antiparallel to the voltage arrow.
+    * the active sign convention/generator arrow system is used for all sources (e.g. all voltage and current sources): the current is antiparallel to the voltage arrow.
-    * the passive sign convention / motor arrow system is used for all consumers (e.g. all passives like resistors, capacitors, inductors, diodes etc.): the current is parallel to the voltage arrow.
+    * the passive sign convention/motor arrow system is used for all consumers (e.g. all passives like resistors, capacitors, inductors, diodes, etc.): the current is parallel to the voltage arrow.
-    * for loads, where the direction of the power is not known, the motor arrow system is recommented (e.g. passives, in case what these are part of a machine, like inductors of a motor)
+    * for loads, where the direction of the power is not known, the motor arrow system is recommended (e.g. passives, in case these are part of a machine, like inductors of a motor)
   * **After the calculation** means
     * $I>0$: The reference arrow reflects the conventional directional sense of the current
     * $I<0$: The reference arrow points in the opposite direction to the conventional directional sense of the current
   * Reference arrows of the current are drawn **in** the wire if possible.
+<imgcaption BildNr1 | Reference arrows>
+</imgcaption>
+{{drawio>Bezugspfeile1.svg}}
+</WRAP>
 </callout>
 <WRAP>
-The reference arrow system (in the clip '+' and '-' is shown in the component terminal. We will instead use voltage arrows from plus to minus)
+The reference arrow system (in the clip '+' and '-' is shown in the component terminal. \\
+We will instead use voltage arrows from plus to minus
 {{youtube>nXyCffP8PXQ}}
 </WRAP>
@@ Zeile 194: / Zeile 200: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
+===== 2.3 Nodes, Branches, and Loops =====
-===== 2.3 Nodes, Branches and Loops =====
 <WRAP>
 Explanation of the different network structures \\
 (Graphs and trees are only needed in later chapters)
-nodes
 {{youtube>-82UNytyrCQ}}
 </WRAP>
 <callout>
-=== Goals ===
+=== Learning Objectives ===
-After this lesson you should:
+By the end of this section, you will be able to:
-  - Be able to identify the nodes, branches and loops in a circuit.
+  - identify the nodes, branches, and loops in a circuit.
-  - Be able to use them to make a circuit clearer.
+  - use them to reshape a circuit.
 </callout>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
+Electrical circuits typically have the structure of networks. Networks consist of two elementary structural elements:
+  - <fc #cd5c5c>**Branches**</fc> (German: Zweige): Connections between two nodes.
+  - <fc #6495ed>**Node**</fc> (German: Knoten): Connection "point" of several branches.
 <WRAP>
 <imgcaption BildNr0 | circuitry and mesh>
 </imgcaption> \\
-{{drawio>Stromkreise_Stromnetze}}
+{{drawio>Stromkreise_Stromnetze.svg}}
-<imgcaption BildNr8 | nodes, branches and loops>
-</imgcaption> \\
-{{drawio>KnotenZweigeMaschen}}
 </WRAP>
-Electrical circuits typically have the structure of networks. Networks consist of two elementary structural elements:
-  - <fc #cd5c5c>**Branches**</fc> (German: Zweige): Connections between two nodes
-  - <fc #6495ed>**Node**</fc> (German: Knoten): Connection points of several branches
 Please note in the case of electrical circuits, we will use the following definition:
   - <fc #cd5c5c>**Branches**</fc> contain at least one component.
-  - <fc #6495ed>**Nodes**</fc> connect __more than two branches__ and can also be spatially extended .
+  - <fc #6495ed>**Nodes**</fc> connect __more than two branches__. Since the wire in a circuit diagram is an ideal conductor, all connected wires to a node are at the same voltage level. Therefore the node in the circuit diagram can also be spatially extended by the wires.
-Sometimes there is a differentiation between "simple nodes" (only connecting 2 branches) and "principal nodes" (connecting more than 2 branches). We will in the followin only mark the connection of more than two branches with a node.
+<WRAP>
+<imgcaption BildNr8 | nodes, branches and loops>
+</imgcaption> \\
+{{drawio>KnotenZweigeMaschen.svg}}
+</WRAP>
+Sometimes there is a differentiation between "simple nodes" (only connecting 2 branches) and "principal nodes" (connecting more than 2 branches). We will in the following often only mark the connection of more than two branches with a node.
-Branches in electrical networks are also called two-pole.
+Branches in electrical networks are also called two-terminal networks.
-Their behaviour is described by current-voltage characteristics and explained in more detail in the chapter [[non-ideal_sources_and_two_pole_networks]] .
+Their behavior is described by current-voltage characteristics and explained in more detail in the chapter [[non-ideal_sources_and_two_terminal_networks]].
 In addition, another term is to be explained: \\
@@ Zeile 245: / Zeile 254: @@
 Since a voltmeter can also be present as a component between two nodes, it is also possible to close a loop by a drawn voltage arrow (cf. $U_1$ in <imgref BildNr8>).
-A loop whicht does not contain other (smaller) loops is called a mesh.
+A loop that does not contain other (smaller) loops is called a mesh.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-Please keep in mind, that usually the entire behaviour of networked circuits almost always changes when a change occurs in one branch or at one node. This is in contrast to other cause-effect relationships, but comparable to changes in other larger networks, e.g. a traffic jam in the road network, due to which other roads experience a higher load. For electrical engineering, this means that in the case of changing circuits, the focus is often on determining the interrelationships (formulas, current-voltage characteristics) and not on a single numerical value.
+Please keep in mind, that usually the entire behavior of networked circuits almost always changes when a change occurs in one branch or at one node. This is in contrast to other cause-effect relationships, but comparable to changes in other larger networks, e.g. a traffic jam in the road network, due to which other roads experience a higher load. For electrical engineering, this means that in the case of changing circuits, the focus is often on determining the interrelationships (formulas, current-voltage characteristics) and not on a single numerical value.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Simplifications ====
+==== Reshaping Circuits ====
+With the knowledge of nodes, branches, and meshes, circuits can be simplified.
+Circuits can be reshaped arbitrarily as long as all branches remain at the same nodes after reshaping
+The <imgref BildNr9> shows how such a transformation is possible.
 <WRAP>
@@ Zeile 259: / Zeile 272: @@
 </imgcaption>
 </WRAP>
-With the knowledge of nodes, branches and meshes, circuits can be simplified.
-Circuits can be reshaped arbitrarily as long as all branches remain at the same nodes after reshaping
-The <imgref BildNr9> shows how such a transformation is possible.
 For practical tasks, repeated trial and error can be useful.
-It is important to check afterwards that the same components are connected to each node as before the transformation.
+It is important to check afterward that the same components are connected to each node as before the transformation.
 Further examples can be found in the following video
@@ Zeile 273: / Zeile 282: @@
 <panel type="info" title="Exercise 2.3.1 Branches and Nodes"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP right>
+<WRAP>
 <imgcaption BildNr70 | Branches and Nodes>
 </imgcaption>
-{{drawio>ZweigeundKnoten}}
+{{drawio>ZweigeundKnoten.svg}}
 </WRAP>
@@ Zeile 283: / Zeile 292: @@
 </WRAP></WRAP></panel>
-<panel type="info" title="Exercise 2.3.3 Branches and Nodes (with explanation)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.3.2 Branches and Nodes (with explanation)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>GNumiT_Y4B8}}
@@ Zeile 289: / Zeile 298: @@
-<panel type="info" title="Exercise 2.3.3 Simplifications of circuits"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.3.3 Reshaping circuits"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP right>
+<WRAP>
 <imgcaption BildNr71 | more Branches and Nodes>
 </imgcaption>
-{{drawio>SchaltungenVereinfachen}}
+{{drawio>SchaltungenVereinfachen.svg}}
 </WRAP>
-Simplify the circuits in <imgref BildNr71>.
+Reshape the circuits in <imgref BildNr71>.
 </WRAP></WRAP></panel>
-===== 2.4 Kirchhoff's circuit laws =====
+===== 2.4 Kirchhoff's Circuit Laws =====
+<callout>
+=== Learning Objectives ===
+By the end of this section, you will be able to:
+ Know and apply Kirchhoff's circuit laws (Kirchhoff's current law and Kirchhoff's voltage law).
+</callout>
 <WRAP>
-{wp>Kirchhoff's circuit laws
+{{wp>Kirchhoff's circuit laws}}
 {{youtube>d0O-KUKP4nM}}
 </WRAP>
-<callout>
-=== Goals ==
-After this lesson you should:
- Know and be able to apply Kirchhof's circuit laws (Kirchhoff's current law and Kirchhoff's voltage law).
-</callout>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Kirchhoff's current law  ====
+==== Kirchhoff's Current Law  ====
-The Kirchhoff's current law (Kirchhoff's first law, Kirchhoff's nodal rule, in German: Knotensatz) formulates in the language of mathematics the experience that no charge "accumulations" occur in electrical wires.
+Kirchhoff's current law (Kirchhoff's first law, Kirchhoff's nodal rule, in German: Knotensatz) formulates in the language of mathematics the experience that no charge "accumulations" occur in electrical wires.
 This is of particular relevance at a network node (<imgref BildNr10>).
 To formulate the equation at this node, the reference arrows of the currents are all set in the same way.
@@ Zeile 324: / Zeile 333: @@
 <callout icon="fa fa-exclamation" color="red" title="Note:">
-<WRAP right>
+<WRAP>
 <imgcaption BildNr10 | Kirchhoff's current law>
 </imgcaption>
-{{drawio>Knotensatz}}
+{{drawio>Knotensatz.svg}}
 </WRAP>
@@ Zeile 335: / Zeile 344: @@
 From now on, the following definition applies:
-  * Currents whose current arrows point towards the node are added in the calculation.
+  * Currents whose current arrows point towards the node are added to the calculation.
   * Currents whose current arrows point away from the node are subtracted in the calculation.
 </callout>
+=== Parallel circuit of resistors ===
+From Kirchhoff's current law, the total resistance for resistors connected in parallel can be derived (<imgref BildNr11>):
 <WRAP>
 <imgcaption BildNr11 | Parallel circuit>
 </imgcaption>
-{{drawio>Parallelschaltung}}
+{{drawio>Parallelschaltung.svg}}
 </WRAP>
-=== Parallel circuit of resistors ===
+Since the same voltage $U_{ab}$ is dropped across all resistors, using Kirchhoff's current law:
-From the Kirchhoff's current law, the total resistance for resistors connected in parallel can be derived (<imgref BildNr11>):
-Since the same voltage $U_{ab}$ is dropped across all resistors, using the Kirchhoff's current law:
-$\large{{U_{ab}}\over{R_1}}+ {{U_{ab}}\over{R_2}}+ ... + {{U_{ab}}\over{R_n}}= {{U_{ab}}\over{R_{substitute}}}$
+$\large{{U_{ab}}\over{R_1}}+ {{U_{ab}}\over{R_2}}+ ... + {{U_{\rm ab}}\over{R_n}}= {{U_{\rm ab}}\over{R_{\rm eq}}}$
-$\rightarrow \large{{{1}\over{R_1}}+ {{1}\over{R_2}}+ ... + {{1}\over{R_n}}= {{1}\over{R_{substitute}}} = \sum_{x=1}^{n} {{1}\over{R_x}}}$
+$\rightarrow \large{{{1}\over{R_1}}+ {{1}\over{R_2}}+ ... + {{1}\over{R_n}}= {{1}\over{R_{\rm eq}}} = \sum_{x=1}^{n} {{1}\over{R_x}}}$
-Thus, for resistors connected in parallel, the equivalent conductance $G_{eq}$ (German: Ersatzleitwert) is the sum of the individual conductances: $G_{eq} = \sum_{x=1}^{n} {G_x}$
+Thus, for resistors connected in parallel, the equivalent conductance $G_{\rm eq}$ (German: //Ersatzleitwert//) is the sum of the individual conductances: $G_{\rm eq} = \sum_{x=1}^{n} {G_x}$
 __In general__: the equivalent resistance of a parallel circuit is always smaller than the smallest resistance.
-Especially for two parallel resistors $R_1$ and $R_2$ applies: $R_{eq}= \large{{R_1 \cdot R_2}\over{R_1 + R_2}}$
+Especially for two parallel resistors $R_1$ and $R_2$ applies: $R_{\rm eq}= \large{{R_1 \cdot R_2}\over{R_1 + R_2}}$
 === Current divider ===
@@ Zeile 367: / Zeile 376: @@
 {{youtube>VojwBoSHc8U}}
 </WRAP>
+\\ \\
+The current divider rule shows in which way an incoming current on a node will be divided into two outgoing branches.
+The rule states that the currents $I_1, ... I_n$ on parallel resistors $R_1, ... R_n$ behave just like their conductances $G_1, ... G_n$ through which the current flows. \\
-The current divider rule can also be derived from the Kirchhoff's current law. \\
+$\large{{I_1}\over{I_{\rm res}} = {{G_1}\over{G_{\rm res}}}$
-This states that, for resistors $R_1, ... R_n$ their currents $I_1, ... I_n$ behave just like the conductances $G_1, ... G_n$ through which they flow.
-$\large{{I_1}\over{I_g}} = {{G_1}\over{G_g}}$
 $\large{{I_1}\over{I_2}} = {{G_1}\over{G_2}}$
+The rule also be derived from Kirchhoff's current law: \\
+  - The voltage drop $U$ on parallel resistors $R_1, ... R_n$ is the same.
+  - When $U_1 = U_2 = ... = U$, then the following equation is also true: $R_1 \cdot I_1 = R_2 \cdot I_2 = ... = R_{\rm eq} \cdot I_{\rm res}$. \\
+  - Therefore, we get with the conductance: ${{I_1} \over {G_1}} = {{I_2} \over {G_2}}= ... = {{I_{\rm eq}} \over {G_{\rm res}}}$
 ~~PAGEBREAK~~ ~~CLEARFIX~~
+<wrap anchor #exercise_2_4_1 />
 <panel type="info" title="Exercise 2.4.1 Current divider"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
@@ Zeile 383: / Zeile 398: @@
 <imgcaption BildNr85| Current divider>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsBMA2AzAgnAdjBgBxhgK7q64gKTXWQBQATiCgmuLh2xxpACxRwcBmHKt2nbpLApCQqkiS0V0JADUA9gBsALgEMA5gFMGhibwzTeCFFAYB3CyD6CeLq+GYvbUn3dchMHhHfxcBMIxPMDNIzywA30YnQLEbOzT7J3dM1K4s50ywNFpMxhZi2hQUQUrnUpCnOtSSjw4Y82bo1oSCupyeiMYAZ3BW6trWsvAQADN9bWHTIA 600,400 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsBMA2AzAgnAdjBgBxhgK7q64gKTXWQBQATiCgmuLh2xxpACxRwcBmHKt2nbpLApCQqkiS0V0JADUA9gBsALgEMA5gFMGhibwzTeCFFAYB3CyD6CeLq+GYvbUn3dchMHhHfxcBMIxPMDNIzywA30YnQLEbOzT7J3dM1K4s50ywNFpMxhZi2hQUQUrnUpCnOtSSjw4Y82bo1oSCupyeiMYAZ3BW6trWsvAQADN9bWHTIA noborder}}
 </WRAP>
 In the simulation in <imgref BildNr85> a current divider can be seen. The resistances are just inversely proportional to the currents flowing through it.
-  - What currents would you expect in each branch if the input voltage were lowered from $5V$ to $3.3V$? __After__ thinking about your result, you can adjust the ''Voltage'' (bottom right of the simulation) accordingly by moving the slider.
+  - What currents would you expect in each branch if the input voltage were lowered from $5~\rm V$ to $3.3V~\rm $? __After__ thinking about your result, you can adjust the ''Voltage'' (bottom right of the simulation) accordingly by moving the slider.
   - Think about what would happen if you flipped the switch __before__ you flipped the switch. \\ After you flip the switch, how can you explain the current in the branch?
@@ Zeile 395: / Zeile 410: @@
 <panel type="info" title="Exercise 2.4.2 two resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-Two resistors of $18\Omega$ and $2 \Omega$ are connected in parallel. The total current of the resistors is $3A$. \\
+Two resistors of $18~\Omega$ and $2~\Omega$ are connected in parallel. The total current of the resistors is $3~\rm A$. \\
-Calculate the total resistance and how the currents is split to the branches.
+Calculate the total resistance and how the currents are split to the branches.
+<button size="xs" type="link" collapse="Loesung_2_4_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_4_2_1_Lösungsweg" collapsed="true">
+The substitute resistor can be calculated to
+\begin{equation*}
+R_{eq} = \frac{R_1R_2}{R_1+R_2} = \frac{18~\Omega \cdot 2~\Omega}{18~\Omega+2~\Omega}
+\end{equation*}
+The current through resistor $R_1$ is
+\begin{equation*}
+I_1 = \frac{R_{eq}}{R_1} I =\frac{1.8~\Omega}{18~\Omega} \cdot 3~\rm A
+\end{equation*}
+The current through resistor $R_2$ is
+\begin{equation*}
+I_2 = \frac{R_{eq}}{R_2}I = \frac{1.8~\Omega}{2~\Omega} \cdot 3~\rm A
+\end{equation*}
+</collapse>
+<button size="xs" type="link" collapse="Loesung_2_4_1_4_2_Lösungsweg">{{icon>eye}} Final result</button><collapse id="Loesung_2_4_1_4_2_Lösungsweg" collapsed="true">
+The values of the substitute resistor and the currents in the branches are
+\begin{equation*}
+R_{eq} = 1.8~\Omega \qquad I_1 = 0.3~{\rm A} \qquad I_2 = 2.7~\rm A
+\end{equation*}
+</collapse>
 </WRAP></WRAP></panel>
 \\
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== Kirchhoff's voltage law ====
+==== Kirchhoff's Voltage Law ====
-Also the Kirchhoff's voltage law (also called: Kirchhoff's second law, or loop law) describes in mathematical language another practical experience:
+Also, Kirchhoff's voltage law (also called: Kirchhoff's second law, or loop law) describes in mathematical language another practical experience:
-Between two points $1$ and $2$ of a network there is only one potential difference.
+Between two points $1$ and $2$ of a network, there is only one potential difference.
 Thus the potential difference is in particular independent of the way a network is traversed between the two points $1$ and $2$.
 This can be described by considering the meshes.
 <callout icon="fa fa-exclamation" color="red" title="Note:">
-<WRAP right>
+<WRAP>
 <imgcaption BildNr12 | loop law>
 </imgcaption>
-{{drawio>Maschensatz}}
+{{drawio>Maschensatz.svg}}
 </WRAP>
@@ Zeile 419: / Zeile 455: @@
 $\boxed{U_{1} + U_{2} + ... + U_{n} = \sum_{x=1}^{n} U_x = 0}$
-To calculate this, a convention for the loop direction must be specified. Theoretically this can be chosen arbitrarily. In practice, often a specific direction (e.g. [[https://en.wikipedia.org/wiki/Mesh_analysis#Mesh_currents_and_essential_meshes|clockwise]]) is used. \\
+To calculate this, a convention for the loop direction must be specified. Theoretically, this can be chosen arbitrarily. In practice, often a specific direction (e.g. [[https://en.wikipedia.org/wiki/Mesh_analysis#Mesh_currents_and_essential_meshes|clockwise]]) is used. \\
-Independently, the following specification always applies: For voltage drop the inverse sign of a voltage risde has to be taken into account.
+Independently, the following specification always applies: For voltage drop, the inverse sign of a voltage rise has to be taken into account.
 For example:
   * Voltages, whose voltage arrows point __in__ the direction of circulation are __added__ in the calculation.
@@ Zeile 428: / Zeile 464: @@
 === Proof of Kirchhoff's voltage law ===
-If one expresses the voltage in <imgref BildNr12> dby the potentials in the nodes, we get:
+If one expresses the voltage in <imgref BildNr12> by the potentials in the nodes, we get:
 $U_{12}= \varphi_1 - \varphi_2 $ \\
 $U_{23}= \varphi_2 - \varphi_3 $ \\
@@ Zeile 443: / Zeile 479: @@
 <imgcaption BildNr13 | series circuit>
 </imgcaption>
-{{drawio>Reihenschaltung}}
+{{drawio>Reihenschaltung.svg}}
 </WRAP>
-Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: Reihenschaltung, see <imref FigNo13>) can be easily determined:
+Using Kirchhoff's voltage law, the total resistance of a series circuit (in German: //Reihenschaltung//, see <imgref BildNr13>) can be easily determined:
-$U_1 + U_2 + ... + U_n = U_g$
+$U_1 + U_2 + ... + U_n = U_{\rm res}$
-$R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{ersatz} \cdot I $
+$R_1 \cdot I_1 + R_2 \cdot I_2 + ... + R_n \cdot I_n = R_{\rm eq} \cdot I $
-Since in series ciruit the current through all resistors must be the same - i.e. $I_1 = I_2 = ... = I$ - it follows that:
+Since in a series circuit, the current through all resistors must be the same - i.e. $I_1 = I_2 = ... = I$ - it follows that:
-$R_1 + R_2 + ... + R_n = R_{eq} =  \sum_{x=1}^{n} R_x $
+$R_1 + R_2 + ... + R_n = R_{\rm eq} =  \sum_{x=1}^{n} R_x $
-__In general__: The equivalent resistance of a series circuit is always greater than the greatest resistance..
+__In general__: The equivalent resistance of a series circuit is always greater than the greatest resistance.
+==== Application ====
+=== Kelvin-Sensing ===
+Often resistors are used to measure a current $I$ via the voltage drop on the resistor $U = R \cdot I$. Applications include the measurement of motor currents in the range of $0.1 ... 500 ~\rm A$. \\
+Those resistors are called //shunt resistors// and are commonly in the range of some $\rm m\Omega$.
+This measurement can be interfered by the resistor of the supply lines.
+To get an accurate measurement often Kelvin sensing, also known as {{wp>Four-terminal sensing}} or 4 wire sensing, is used.
+This is a method of measuring electrical resistance avoiding errors caused by wire resistances. \\
+The simulation in <imgref BildNr005> shows such a setup.
+Four-terminal sensing involves using:
+  * a pair of //current leads// or //force leads// (with the resistances $R_{\rm cl1}$ and $R_{\rm cl2}$) to supply current to the circuit and
+  * a pair of //voltage leads// or //sense leads// (with the resistances $R_{\rm vl1}$ and $R_{\rm vl2}$) to measure the voltage drop across the impedance to be measured.
+The sense connections via the voltage leads are made immediately adjacent to the target impedance $R_{\rm s}$ at the device under test $\rm DUT$.
+By this, they do not include the voltage drop in the force leads or contacts. \\
+Since almost no current flows to the measuring instrument, the voltage drop in the sense leads is negligible.
+This method can be a practical tool for finding poor connections or unexpected resistance in an electrical circuit.
+<WRAP>
+<imgcaption BildNr005 | Example of a circuit>
+</imgcaption> \\
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsYBsAOdAmSYBOdAdkwUkxRB2JCRQGY6BTAWjDACgAnEIkFJnBgcAoZGRdeOBOmGiZcsCioSp86hT5y84yZwDuGwdVljqh00pVXN4y4rsbdFo444KUE-pAdnlVDhe4OgSvgBuIOzEOlrRciZhAmqiSTAIlqyYEi78Lr4ADlHZ5lk5WkxhmWAxTmVOONW1JvUmjUXecmqhIJVQPNqDHoNh0FxG-D4jA-wYaiIhaWOW3Z2L-UatQvyJDsFzGge+bsGzCz6cAB5RHIEoogy0OJi0VGBCAGoA9gA2AC4AW2Yf2Y3AAOgBnACWADtIWCwTCcKwDFDuMxIRDmDDoTCAOaQgCGMIAJgiYSgUWiMRCsTjYXirnwIEEFBxqOgEuAhKj0RDIai-gALeGI3k0vlQiF-YkAY2YTJwDCYrI5BGoKHo3JAABEAKoAFU4jxyZja-gWECqeJutRc8WoQSg-WuCCQDCQJF6QjeQgASkz0OqGOrKAJRG9RLKflxrkHNBIw3gmJGQNHGq6kIQ5F7COrfSAA3HQ7QwwQJKnwjGmW6QJgmF7sAJtUW6FmTLmEBAC6342AEKIw-2tQsqxmOUJ+yqSv2ue9C4rMKJ+09MJOeqmIZwgA noborder}}
+</WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 462: / Zeile 524: @@
 <panel type="info" title="Exercise 2.4.3 Three Resistors"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-Three equal resistors of $20k\Omega$ each are given. \\
+Three equal resistors of $20~k\Omega$ each are given. \\
-Which values are realizable by arbitrary interconnection of one to three resistors?
+Which values are realizable by the arbitrary interconnection of one to three resistors?\\
+<button size="xs" type="link" collapse="Loesung_2_4_3_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_4_3_1_Lösungsweg" collapsed="true">
+The resistors can be connected in series:
+\begin{equation*}
+R_{\rm series} = 3\cdot R = 3\cdot20~k\Omega
+\end{equation*}
+The resistors can also be connected in parallel:
+\begin{equation*}
+R_{\rm parallel} = \frac{R}{3} = \frac{20~k\Omega}{3}
+\end{equation*}
+On the other hand, they can also be connected in a way that two of them are in parallel and those are in series to the third one:
+\begin{equation*}
+R_{\rm res} = R + \frac{R\cdot R}{R+R} = \frac{3}{2}R = \frac{3}{2} \cdot 20~k\Omega
+\end{equation*}
+</collapse>
+<button size="xs" type="link" collapse="Loesung_2_4_3_2_Lösungsweg">{{icon>eye}} Final values</button><collapse id="Loesung_2_4_3_2_Lösungsweg" collapsed="true">
+\begin{equation*}
+R_{series} = 60~k\Omega\qquad R_{\rm parallel} = 6.7~k\Omega\qquad R_{\rm res} = 30~k\Omega
+\end{equation*}
+</collapse>
 </WRAP></WRAP></panel>
-===== 2.5  Unloaded and loaded voltage divider =====
+===== 2.5 Voltage Divider =====
 <WRAP>
@@ Zeile 473: / Zeile 553: @@
 {{youtube>xSRe_4TQbuo?end=655}}
 </WRAP>
-<callout>
-==== The unloaded voltage divider ====
+==== The unloaded Voltage Divider ====
-=== Goals ===
+<callout>
+=== Learning Objectives ===
-After this lesson you should be able to:
+By the end of this section, you will be able to:
   - to distinguish between the loaded and unloaded voltage divider.
   - to describe the differences between loaded and unloaded voltage dividers.
@@ Zeile 485: / Zeile 565: @@
 </callout>
+Especially the series circuit of two resistors $R_1$ and $R_2$ shall be considered now.
+This situation occurs in many practical applications e.g. in a {{wp>potentiometer}}.
 <WRAP>
-<imgcaption BildNr14 | unloaded voltage divider>
+<imgcaption BildNr114 | potentiometer>
 </imgcaption>
-{{drawio>unbelasteterSpannungsteiler}}
+{{drawio>Poti.svg}}
 </WRAP>
-Especially the series ciruit of two resistors $R_1$ and $R_2$ shall be considered now.
-This situation occurs in many practical applications (e.g. {{wp>potentiometer}}).
 In <imgref BildNr14> this circuit is shown.
-Via the Kirchhoff's voltage law we get
+<WRAP>
+<imgcaption BildNr14 | unloaded voltage divider>
+</imgcaption>
+{{drawio>unbelasteterSpannungsteiler.svg}}
+</WRAP>
+Via Kirchhoff's voltage law, we get
-$\boxed{ {{U_1}\over{U}} = {{R_1}\over{R_1 + R_2}} }$
+$\boxed{ {{U_1}\over{U}} = {{R_1}\over{R_1 + R_2}} \rightarrow U_1 = k \cdot U}$
 The ratio $k={{R_1}\over{R_1 + R_2}}$ also corresponds to the position on a potentiometer.
@@ Zeile 508: / Zeile 595: @@
 <imgcaption BildNr81| unloaded voltage divider>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClgFABu4O1RGL1YaOUfK1JNWHQE9bihBZIk7rywY2CXhFJxkSAEoBTAM4BLXQBcAhgDsAxtvoB3EIoAc9zFJnOOkW68nspDny6edgpsGJJY-vZhUPQADn4+0REBHBDCcfaQTnKZTr5pMXZgrOA8uaW8QZzu5b6eAObeNdI+Dk6eimgJUZICob0gADq6w6O6AKoA+gD2AK5GXo4VPctVLcv5ZZ66KzmbleAgAGYmADa61kA 500,400 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsAsB2ATAZgGw4QhmimJGOiApJSClpQKYC0YYAUAG7g7VEbfUwaHFFEpqSalOgI2QlCCyQFQkVgy8EIiKTjIkAJQYBnAJbGALgEMAdgGMGbAO4gNADleZFyz-0jPvBT5FNyCvfxd1XgwFLFDXGKg2AAcQoMS4sP4IKRTXSA9VfI9gnKSXMB5wYWLqkQiBX1rg-wBzQKalILcPXI00NISFEhEMRIUAHWMpmeMAVQB9AHkA9zqh9Yau9dKa-2MNot368BAAMysAG2NHIA noborder}}
 </WRAP>
-In the simulation in <imgref BildNr81> an unloaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5V$. The potentiometer has a total resistance of $1K\Omega$. In the configuration shown, this is divided out to $500 \Omega$ and $500 \Omega$.
+In the simulation in <imgref BildNr81> an unloaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5~\rm V$. The potentiometer has a total resistance of $1~\rm k\Omega$. In the configuration shown, this is divided into $500 ~\Omega$ and $500 ~\Omega$.
-  - What voltage $U_out$ would you expect if the switch were closed? After thinking about your result, you can check it by closing the switch.
+  - What voltage $U_{\rm O}$ would you expect if the switch were closed? After thinking about your result, you can check it by closing the switch.
-  - First think about what would happen if you would change the distribution of the resistors by moving the wiper ("intermediate terminal")? \\ You can check your assumption by using the slider at the bottom right of the simulation.
+  - First, think about what would happen if you would change the distribution of the resistors by moving the wiper ("intermediate terminal"). \\ You can check your assumption by using the slider at the bottom right of the simulation.
-  - At which position do you get a $U_out = 3.5V$?
+  - At which position do you get a $U_{\rm O} = 3.5~\rm V$?
 </WRAP></WRAP></panel>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-==== The loaded voltage divider ====
+==== The loaded Voltage Divider ====
+If - in contrast to the abovementioned, unloaded voltage divider - a load $R_{\rm L}$ is connected to the output terminals (<imgref BildNr15>), this load influences the output voltage.
 <WRAP>
 <imgcaption BildNr15 | loaded voltage divider>
 </imgcaption>
-{{drawio>belasteterSpannungsteiler}}
+{{drawio>belasteterSpannungsteiler.svg}}
 </WRAP>
-If - in contrast to the above-mendtioned, unloaded voltage divider - a load $R_L$ is connected to the output terminals (<imgref BildNr15>), this load influences the output voltage.
 A circuit analysis yields:
@@ Zeile 532: / Zeile 619: @@
 $ U_1 = \LARGE{{U} \over {1 + {{R_2}\over{R_L}} + {{R_2}\over{R_1}} }}$
-or on a potentiometer with $k$ and the sum of resistors $R_s = R_1 + R_2$:
+or on a potentiometer with $k$ and the sum of resistors $R_{\rm s} = R_1 + R_2$:
-$ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_s}\over{R_L}}  }}$
+$ U_1 = \LARGE{{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_{\rm s}}\over{R_{\rm L}}} }}$
+<imgref BildNr65> shows the ratio of the output voltage $U_1$ to the input voltage $U$ (y-axis), in relation to the ratio $k={{R_1}\over{R_1 + R_2}}$.
+In principle, this is similar to <imgref BildNr14>, but here it has another dimension: multiple graphs are plotted. These differ by the ratio ${{R_{\rm s}}\over{R_{\rm L}}}$.
 <WRAP>
@@ Zeile 542: / Zeile 632: @@
 </WRAP>
-<imgref BildNr65> shows the ratio of the output voltage $U_1$ to the input voltage $U$ (y-axis), in relation to the ratio $k={{R_1}\over{R_1 + R_2}}$.
+What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4.0 ~\rm k\Omega$ and $R_1 = 6.0 ~\rm k\Omega$, and an input voltage of $10~\rm V$. Thus $k = 0.60$, $R_s = 10~\rm k\Omega$ and $U_1 = 6.0~\rm V$.
-In principle, this is similar to <imgref BildNr14>, but here it has another dimension: multiple graphs are plotted. These differ by the ratio ${{R_s}\over{R_L}}$.
+Now this voltage divider is loaded with a load resistor. If this is at $R_{\rm L} = R_1 = 10 ~\rm k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8~\rm V$ - so the output voltage drops. For $R_{\rm L} = 4.0~\rm k\Omega$, $k$ becomes even smaller to $k=0.375$ and $U_1 = 3.75~\rm V$. If the load $R_{\rm L}$ is only one-tenth of the resistor $R_{\rm s}=R_1 + R_2$, the result is $k = 0.18$ and $U_1 = 1.8~\rm V$. The output voltage of the unloaded voltage divider ($6.0~\rm V$) thus became less than one-third.
-What does this diagram tell us? This shall be investigated by an example. First, assume an unloaded voltage divider with $R_2 = 4 k\Omega$ and $R_1 = 6 k\Omega$, and an input voltage of $10V$. Thus $k = 0.6$, $R_s = 10k\Omega$ and $U_1 = 6V$.
-Now this voltage divider is loaded with a load resistor. If this is at $R_L = R_1 = 10 k\Omega$, $k$ reduces to about $0.48$ and $U_1$ reduces to $4.8V$ - so the output voltage drops. For $R_L = 4k\Omega$, $k$ becomes even smaller to $k=0.375$ and $U_1 = 3.75V$. If the load $R_L$ is only one tenth of the resistor $R_s=R_1 + R_2$, the result is $k=0.18$ and $U_1=1.8V$. The output voltage of the unloaded voltage divider ($6V$) thus became less than one third.
-What is the practical use of the (loaded) voltage divider? \\ Here some examples:
+What is the practical use of the (loaded) voltage divider? \\ Here are some examples:
-  * Voltage dividers are in use for controlling the output of power supply IC's (see [[https://www.analog.com/en/technical-articles/a101121-voltage-dividers-in-power-supplies.html|Voltage Dividers in Power Supplies]]). In order not to create a loaded voltage divider, a range for the resistance is given here.
+  * Voltage dividers are in use for controlling the output of power supply ICs (see [[https://www.analog.com/en/technical-articles/a101121-voltage-dividers-in-power-supplies.html|Voltage Dividers in Power Supplies]]). In order not to create a loaded voltage divider, a range for the resistance is given here.
-  * Another "invisible" voltage divider is for example in the electical system of a car. As we will learn in the next chapters, voltage supplies have in internal resistance (and therefore batteries, too). The other consumer in the car also represent a resistance. By this, the electical system states an unloaded voltage divider. Given another, additional low-resistance load (e.g. the spark of the starter system) one can understand that there will be a voltage drop when starting the car.
+  * Another "invisible" voltage divider is for example in the electrical system of a car. As we will learn in the next chapters, voltage supplies have internal resistance (and therefore batteries, too). The other consumer in the car also represents a resistance. By this, the electrical system states an unloaded voltage divider. Given another, additional low-resistance load (e.g. the spark or the starter motor of the starter system) one can understand that there will be a voltage drop when starting the car.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 556: / Zeile 643: @@
 <panel type="info" title="Exercise 2.5.2 loaded voltage divider I "> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-Determine from the circuit in <imgref BildNr15> the above equation $ U_1 = {{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_s}\over{R_L}}}}$ where $k={{R_1}\over{R_1 + R_2}}$ and $R_s = R_1 + R_2$.
+Determine from the circuit in <imgref BildNr15> the equation $ U_1 = {{k \cdot U} \over { 1 + k \cdot (1-k) \cdot{{R_{\rm s}}\over{R_{\rm L}}}}}$ where $k={{R_1}\over{R_1 + R_2}}$ and $R_{\rm s} = R_1 + R_2$.\\
+<button size="xs" type="link" collapse="Loesung_2_5_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_2_5_2_1_Lösungsweg" collapsed="true">
+According to the voltage division rule, the loaded voltage is
+\begin{align*}
+U_1 &=\frac{\frac{R_1 R_L}{R_1+R_L}}{R_2+\frac{R_1 R_L}{R_1+R_L}}U \\
+    &=\frac{R_1 R_{\rm L}}{R_2 (R_1 + R_{\rm L}) + R_1 R_{\rm L}} U \\
+    &=\frac{R_1 R_{\rm L}}{R_1 R_2 + R_2 R_{\rm L} + R_1 R_{\rm L}} U \\
+    &=\frac{R_1 R_{\rm L}}{R_1 R_2 + (R_1+R_2) R_{\rm L}} U
+\end{align*}
+The divided resistor $R_1$ and $R_2$ are put together to form $R_{\rm s}=R_1 + R_2$.
+\begin{equation*}
+U_1=\frac{R_1 R_{\rm L}}{R_1 R_2 + R_{\rm s} R_{\rm L}} U
+\end{equation*}
+With the equations given there is also $R_1=k(R_1+R_2)=k R_{\rm s}$ and $R_2 = R_{\rm s} - R_1 = R_{\rm s} - k R_{\rm s} = (1-k) R_{\rm s}$.
+\begin{equation*}
+U_1=\frac{k R_{\rm s} R_{\rm L}}{k R_{\rm s} (1-k) R_{\rm s} + R_{\rm s} R_{\rm L}}U
+\end{equation*}
+Dividing the numerator and denominator by $R_{\rm s} R_{\rm L}$ yields to
+\begin{equation*}
+U_1=\frac{k}{k(1-k)\frac{R_{\rm s}}{R_{\rm L}}+1}U
+\end{equation*}
+</collapse>
 </WRAP></WRAP></panel>
@@ Zeile 562: / Zeile 670: @@
 <panel type="info" title="Exercise 2.5.3 loaded voltage divider II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP right>
+In the simulation in <imgref BildNr82> a loaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5.00~\rm V$. The potentiometer has a total resistance of $1.00~k\Omega$. In the configuration shown, this is divided into $500 ~\Omega$ and $500 ~\Omega$. The load resistance has $R_{\rm L} = 1.00 ~\rm k\Omega$.
+  - What voltage ''U_O'' would you expect if the switch were closed? This is where you need to do some math! __After__ you calculated your result, you can check it by closing the switch.
+  - At which position of the wiper do you get $3.50~\rm V$ as an output? Determine the result first by means of a calculation. \\ Then check it by moving the slider at the bottom right of the simulation.
+<WRAP>
 <imgcaption BildNr82| loaded voltage divider>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsAoAN3BppExRbbA0Kn5pUkVEdAT0ebXN14hsadgj4Qw8KGJAAlAKYBnAJa6ALgEMAdgGNt9AO7t87DJ2mPOkW6-CyXPPu7suHHIInEH+cgooKFIh7NFQ9AAOcnhsUVKpnhAi9ABOwaHx2LFhyHAexaFOKWnV4ZVxGWnx4WCsnmh4nq3tvvZdfe7JnV58I0HZCXYjfemjCbqN8xOyVBAAZiYANrrWaOTYmXOCVMcgADq6l9e6AKoA+gD2AK5G9ADmcpC1zt9fbO4gA 500,400 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3cGmkTFbrzAZCUMTSpIq06AjbDeuISJDY0fBKIhh4UWSABKDAM4BLYwBcAhgDsAxgzYB3Pvj4YBS9wMjPv4FS9hUV8XL35VBAEI0NV1FBRFKL5EqDYAB1U8XgTFbP8IaTYAJ0jo1OxkmOQ4P0rojyycxtj6lLyc1NiwHn80PH9u3uDXAZHfTP6A0SmIwrSXKZHc6bTjdtW5lSoIADMrABtjRzRybHyVmjAZ1N4AHWMHp+MAVQB9AHsAVws2AHNVJBmp4gYDeEUgA noborder}}
 </WRAP>
-In the simulation in <imgref BildNr82> a loaded voltage divider in the form of a potentiometer can be seen. The ideal voltage source provides $5V$. The potentiometer has a total resistance of $1K\Omega$. In the configuration shown, this is divided out to $500 \Omega$ and $500 \Omega$. The load resistance has $R_L = 1 k\Omega$.
-  - What voltage ''U_OUT'' would you expect if the switch were closed? This is where you need to do some math! __After__ you calculated your result, you can check it by closing the switch.
-  - At which division you get $3.5V$. Determine the result first for a calculation. \\ Then check it by moving the slider at the bottom right of the simulation.
 </WRAP></WRAP></panel>
 <panel type="info" title="Exercise 2.5.4 Application of the loaded voltage divider - motor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP right>
+<WRAP>
-<imgcaption BildNr16| Skizze des Aufbaus>
+<imgcaption BildNr16| Sketch of the setup>
 </imgcaption> \\
-{{drawio>MotorAmSpannungsteiler}}
+{{drawio>MotorAmSpannungsteiler.svg}}
 </WRAP>
-You wanted to test a micromotor for a small robot. Using the maximum current and the internal resistance ($R_M = 5\Omega$) you calculate that this can be operated with a maximum of $U_{M,max}=4V$. A colleague said that you can get $4V$ using the setup in <imgref BildNr16> from a $9V$ block battery.
+You wanted to test a micromotor for a small robot. Using the maximum current and the internal resistance ($R_{\rm M} = 5~\Omega$) you calculate that this can be operated with a maximum of $U_{\rm M, max}=4~\rm V$. A colleague said that you can get $4~\rm V$ using the setup in <imgref BildNr16> from a $9~\rm V$ block battery.
-  - First, calculate the maximum current $I_{M,max}$ of the motor.
+  - First, calculate the maximum current $I_{\rm M,max}$ of the motor.
   - Draw the corresponding electrical circuit with the motor connected as an ohmic resistor.
-  - At the maximum current, the motor should be able to deliver a torque of $M= 100mNm$. What torque would the motor deliver if you implement the setup like this? (Assumption: The torque of the motor increases proportionally to the motor current).
+  - At the maximum current, the motor should be able to deliver a torque of $M_{\rm max}=M(I_{\rm M, max})= 100~\rm mNm$. What torque would the motor deliver if you implement the setup like this? (Assumption: The torque of the motor increases proportionally to the motor current).
-  - What might a setup with a potentiometer look like that would actually allow you to set a voltage between $0.5V$ to $4V$ on the motor? What resistance value should the potentiometer have?
+  - What might a setup with a potentiometer look like that would actually allow you to set a voltage between $0.5~\rm V$ to $4~\rm V$ on the motor? What resistance value should the potentiometer have?
   - Build and test your circuit in the simulation below. For an introduction to online simulation, see: [[circuit_design:0_tools#online_circuit_simulator]]. \\ You will essentially need the following tips for this setup:
-    - Routing connections can be activated via the menu: ''Draw'' >> ''add wire''. Afterwards you have to click on the start point and then drag to the end mode.
+    - Routing connections can be activated via the menu: ''Draw'' >> ''add wire''. Afterward, you have to click on the start point and then drag it to the end mode.
-    - Note that connections can only ever be connected at nodes. A red marked node (e.g. at the $5 \Omega$ resistor) indicates that it is not connected. This could be moved one grid step to the left, as there is a node point there.
+    - Note that connections can only ever be connected at nodes. A red-marked node (e.g. at the $5 ~\Omega$ resistor) indicates that it is not connected. This could be moved one grid step to the left, as there is a node point there.
     - Pressing the ''<ESC>'' key will disable the insertion of components.
     - With a right click on a component it can be copied or values like the resistor can be changed via ''Edit....''
@@ Zeile 595: / Zeile 703: @@
 <imgcaption BildNr83| Simulation for motor setup>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIKQhKE4BQAbiIeiHoQ0wEzuvnnEgCcUITATUATuH7sQnVu0wyuQ0QHM2zSK0bMcANijUA7upYgsG7kfWyz5G+2pgUrJ-tOvmCaRDDxkSACUAUwBnAEsQgBcAQwA7AGMgq3ZIDhRpFI4lSCswKUUXFH0bHKA 800,400 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIKQhKE4BQAbiIeiHoQ0wEzuvnnEgCcUITATUATuH7sQnVu0wyuQ0QHM2zSK0bMcANijUA7upYgsG7kfWyz5G+2pgUrJ-tOvmCaRDDxkSACUAUwBnAEsQgBcAQwA7AGMgq3ZIDhRpFI4lSCswKUUXFH0bHKA noborder}}
 </WRAP>
@@ Zeile 602: / Zeile 710: @@
 <panel type="info" title="Exercise 2.5.5 Examples of the calculation of loaded voltage dividers"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP group><WRAP column half>
-Voltage divider, series circuit (series resistance) and shunt resistor
-{{youtube>XRBiYdMhiF0}}
-</WRAP><WRAP column half>
 Exercise on the voltage divider
 {{youtube>6NG4uOMDU7Y}}
-</WRAP></WRAP>
 </WRAP></WRAP></panel>
-<panel type="info" title="Exercise 2.5.6 Example of a loaded voltage divider: Explanation without calculation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.5.6 Example of a loaded voltage divider: Explanation without calculation (German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>hRVOZIb8-6g}}
@@ Zeile 621: / Zeile 723: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-===== 2.6 Star-delta circuit =====
+===== 2.6 Circuits with three Connections =====
-<WRAP>
-<imgcaption BildNr98 | Example of a circuit>
-</imgcaption>
-{{drawio>BeispielStromkreis}}
-<imgcaption BildNr17|Conversion of parallel circuit to series circuit>
-</imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIIRKE4BQATiALQ4BM4T6Dz4KAbFON92gDhVAOYcWTJoQkhpkKFQBKssGBaMWATl4KFxEEj3IEVAO6r1qtuHMMwWlmo0OnhGWGX3H4K-VcgOnz6CkbB0KYW-j5EMv7cCrGKUVIe7t5uMpBiGXKQ7NGSBtlRAUz5DKl57CX2CdW55TW0dYnp8QpNfGDwLYXg7WUVib109FVJ49INI3B2jAhOfjiLAx5eC0saK9q6fAZhehE5mw2n8opjO2scq10KOI-Qj9RXTjyyVYkCQsLcVEA 600,300 noborder}}
-</WRAP>
 <callout>
-=== Goals ===
+=== Learning Objectives ===
-After this lesson you should:
+By the end of this section, you will be able to:
-  - be able to convert triangular loops into a star shape (and vice versa)
+  - convert triangular loops into a star shape (and vice versa)
 </callout>
-At the beginning of the chapter an example of a network was shown (<imgref BildNr91>). Here, however, one does not come directly to the solution with the set of nodes and loops. It is visible, that there are many triangle-shaped loops resp. star-shaped nodes (<imgref BildNr98>). These will be discussed in more detail now.
+At the beginning of the chapter, an example of a network was shown (<imgref BildNr91>). Here, however, one does not come directly to the solution with the set of nodes and loops.
+It is visible, that there are many $\Delta$-shaped (or triangle-shaped) loops resp. $\rm Y$-shaped (or star-shaped) nodes, see <imgref BildNr98>.
+A method to calculate these will be discussed in more detail now.
-First of all a summary of the previous findings. Using the node and loop rule it became clear that an equivalent resistance can be determined from a series as well as from a parallel circuit. If one considers the equivalent resistance as a black box - i.e. the internals are unknown - it could be interpreted by both types of circuit (<imgref BildNr17>).
+<WRAP>
+<imgcaption BildNr98 | Example of a circuit>
+</imgcaption>
+{{drawio>BeispielStromkreis,svg}}
+</WRAP>
+First of all a summary of the previous findings: Using the node and loop rule it became clear that an equivalent resistance can be determined from a series as well as from a parallel circuit. If one considers the equivalent resistance as a black box - i.e. the internals are unknown - it could be interpreted by both types of a circuit (<imgref BildNr17>).
+<WRAP>
+<imgcaption BildNr17|Conversion of a parallel circuit to series circuit>
+</imgcaption> \\
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWKsDsKAc64YMyRWACyRIIRKE4BQATiALQ4BM4T6Dz4KAbFON92gDhVAOYcWTJoQkhpkKFQBKssGBaMWATl4KFxEEj3IEVAO6r1qtuHMMwWlmo0OnhGWGX3H4K-VcgOnz6CkbB0KYW-j5EMv7cCrGKUVIe7t5uMpBiGXKQ7NGSBtlRAUz5DKl57CX2CdW55TW0dYnp8QpNfGDwLYXg7WUVib109FVJ49INI3B2jAhOfjiLAx5eC0saK9q6fAZhehE5mw2n8opjO2scq10KOI-Qj9RXTjyyVYkCQsLcVEA noborder}}
+</WRAP>
-Now how does this help us in the case of a triangular loop?
+Now how does this help us in the case of a $\Delta$-load (= triangular loop)?
-Also in this case one can provide a black box. However, this should always behave in the same way as the triangular loop, i.e. any voltages applied should produce the same currents as the known structure. \\
+Also in this case one can provide a black box. However, this should always behave in the same way as the $\Delta$-load, i.e. any voltages applied should produce the same currents as the known structure. \\
-In other words: The resistances measured between two terminals must be identical for the blackbox and the known circuit.
+In other words: The resistances measured between two terminals must be identical for the black box and for the known circuit.
-For this purpose, the different resistances between the individual nodes $a$, $b$ and $c$ are now to be considered, see <imgref BildNr18>. The aim is to find out how a delta circuit can be developed from a star circuit (and vice versa).
+For this purpose, the different resistances between the individual nodes $\rm a$, $\rm b$, and $\rm c$ are now to be considered, see <imgref BildNr18>. The aim is to find out how a $\Delta$-load (triangular circuit) can be developed from a $\rm Y$-load (star circuit) and vice versa.
 ~~PAGEBREAK~~ ~~CLEARFIX~~
@@ Zeile 659: / Zeile 763: @@
 <imgcaption BildNr18| Star-delta transformation>
 </imgcaption> \\
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3KgUwFowwAoAJRCaKPAkqbAAmXpWgwQJCeLFwkLAO6diPSPzDkxq8CwBOnQRGFaGgwTzMzwiXUr4j9h+5ut6md46fOTnCBZ3wqdiZY9qwcDAH2EbiUWpaSKJayfiZmIAbCqeYU2uGe6Y5Z6SHxYsJJsHLhyiC4GDwCwnU8pVIVcopcPM1KGrX12q75ROr6aUTkLbW4uLAz8zM2RSP8Gf1T83MLi65aPcujYpvbi52R+4IhPWH+0N0D0ZQ9rYnSlSnD2GM8Eyrs3xIvssAJyicSScpvDq9I4PPYDaAfK4PNbPJEYKjAzJrJBYmApdRiXGZLTE-F5NIQO4AwxJCRlaSoaFFKkNfLkEI3CLUsmPTGQsQJRnJM48vHBDFk1i7Oy89lpI4zLbbJbytmOMmK2YnXCqtLkNJFciOLXKhYE7j8mFWrk1XnMYSa8GC4XvAAeDloSCEvSg3uo4CQYkAAaQsD0mYG0UzA-Qc9K4YReEBsACdYYcKgNsczAbAGIAhuioms4gS+sYtKXw5daHmYyZy3ifmJADOk6aYGBjdd6sSb9JAberMXS3v0YCjSB+-cHSgwKHUtDc8-IlBUGIARu3BGAykhsdAwCPcxiAMYfNKCHJFCz46uOQTDkxIKO4GNrkAnwvhwnxtQjeNvuAG5nt+GCHtGSg5A+ibpDwqZbjQ4BEBiJh5oGh5JvB1amIGKAmNARhIBi775puQ5+hg2I+kRAZpFhY6UER9ZgKCVrvuuiLVsCPAGqs5AxgagF0Wmd60OQMz6M+IDiauQEgPmnH6AQ0mTLOwgrlAmEibOlDGoeTDGtJEBTmhJ6KcEUAaGyqkaJp05bpgVDoLGkpfCR7Z5l6EDMRAXqUVIaFkbOlmkL0PHQMR9nVsh0kYKsr6xYBplbsO-FsuO0nAnZkhtp07KFOyJTtmBVD8eATkwWhADOAAu+Y6AAOlVJ4AJY6CeACuLU1e2jnqIeGXqCZGIACZ0AANnVTWte1XU9eGdi4BAeEoEtkWBeN+YngA1uuAD2bosEAA 600,800 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3KgUwFowwAoAJRCaKPAkqbAAmXpWgwQJCeLFwkLAO6diPSPzDkxq8CwBOnQRGFaGgwTzMzwiXUr4j9h+5ut6md46fOTnCBZ3wqdiZY9qwcDAH2EbiUWpaSKJayfiZmIAbCqeYU2uGe6Y5Z6SHxYsJJsHIc5DwkYgCclHXSEmI8FXKKNa1tGD3aruoSuJlDRNjiuFOwU7NTNkxNI5zdzWKzM3PzesvNi8PlIBtb84oMjf0ME81hnBd7Qn1r4gktyWdjF-vjmuxKTV8hkghC1JIcZJU-ON+rsEDAUtc4QxVnDoH5lKENE4-Bl7IIcnE-oJoH1BCFiX1IqV0m9IYoKcVSSSCkY-o97Bc4i8xIlpHTwGTQtxQjZ8bE7AyGetpidcKLmQz1Jo7NLcJstvKmVr0jlVeq5jiQlpcVpbmS+loRsrRNypB0WAAPFZkQVEMBUXDulRIMSAANJHfp6rRTPV9ORSbt0jw2ABOgMmCAqciZQTddTUcB9ACGKXNURNKpSSqipsLTuCtDAGFDJixYHqwlqYkAM6Txquh9v-WINqSSVvlwQxHXukxgYNIWqtED9pQYFDqWhuefkJqZkAAI3jgjAZSQKeg7vxGarIAAxrm0mL9JfvFvHIP+Pjg7hQyo+qec+Xiw+lG70i+pBPddzy-DBD0EGswByQdGyjEBYy3GhwGhUcLWBKQ0gQgdTHAJAUBMaAjCQJ41yzTcnXrZdKFMTRyFfOCsPSIZIGfcVVyAtEKNwUkd3AchgzHDDozjCikFDbcoEgcTiUA7NOPSfEo1JH1hyE+CRKjai8PScgUHxEiT1PeTiHnI1w3AeovSnGcoKTd11CTScT0-KMO34nU3PYvpyKjUzxPM+tYL7AMq3MCBNCDApaDfM8QouAwePEsDexbHEFWghVHGMuijgA-F3VwIhYJPABnAAXLMdAAHRK08AEsdFPABXOqyq3bopjUCBaCmeiTwAEzoAAbCqavqxqWra8sVMKqBtyOFSYvXIas1PABrdcAHsHRYIA noborder}}
 \\ \\
-Calculation of the transformation formulae: Star connection in delta connection
+Calculation of the transformation formulae: Star connection in delta connection (Alternatively in [[https://www.youtube.com/watch?v=AFSWn5xR8tE|German]])
-{{youtube>AFSWn5xR8tE}}
+{{youtube>eBUWWU0Ekq4}}
 </WRAP>
-==== Delta circuit ====
+==== Delta Circuit ====
-In the delta connection, the 3 resistors $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$ are connected in a loop this terminals on each node.
+In the delta circuit, the 3 resistors $R_{\rm ab}^1$, $R_{\rm bc}^1$, and $R_{\rm ca}^1$ are connected in a loop. At the connection of the resistors, an additional terminal is implemented. \\
+The labeling with a superscript $\square^1$ refers to the three resistors in the next paragraphs.
-For the resistors between the two terminals (e.g. $a$ and $b$), the third one ($c$) is considered as not connected to anything outside. This results in a parallel circuit of the direct star resistor $R_{ab}^1$ with the series connection of the other two star resistors $R_{ca}^1 + R_{bc}^1$:
+For the measurable resistance between two terminals (e.g. $R_{\rm ab}$ between $\rm a$ and $\rm b$), the third terminal (here: $\rm c$) is considered as not connected to anything outside. This results in a parallel circuit of the direct delta resistor $R_{\rm ab}^1$ with the series connection of the other two delta resistors $R_{\rm ca}^1 + R_{\rm bc}^1$:
-$R_{ab} = R_{ab}^1 || (R_{ca}^1 + R_{bc}^1) $ \\
+<WRAP><imgcaption BildNr80 | measurable resistance between two terminals>
-$R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + (R_{ca}^1 + R_{bc}^1)}} =  {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} $ \\
+</imgcaption>
+{{drawio>Resistancebetweentwoterminals.svg}}
+</WRAP>
+$R_{\rm ab} = R_{\rm ab}^1 || (R_{\rm ca}^1 + R_{\rm bc}^1) $ \\
+$R_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + (R_{\rm ca}^1 + R_{\rm bc}^1)}} =  {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} $ \\
 The same applies to the other connections. This results in:
 \begin{align*}
-R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}}  \\
+R_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}}  \\
-R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}}  \\
+R_{\rm bc} = {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}}  \\
-R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \tag{2.6.1}  \end{align*}
+R_{\rm ca} = {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \tag{2.6.1}  \end{align*}
-==== Star circuit ====
+==== Star Circuit ====
-Given the idea, that the star ciruict shall behave equal to the delta circuit, the resistance measured between the terminals must be similar. Also in the star connection 3 resistors are connected, but in a star shape. The star resistors are all connected with another node $0$ in the middle: $R_{a0}^1$, $R_{b0}^1$ and $R_{c0}^1$.
+Given the idea, that the star circuit shall behave equally to the delta circuit, the resistance measured between the terminals must be similar.
+Also in the star circuit, 3 resistors are connected, but now in a star (or $\rm Y$) shape. The star resistors are all connected with another node $0$ in the middle:
+$R_{\rm a0}^1$, $R_{\rm b0}^1$ and $R_{\rm c0}^1$.
-Again, the procedure is the same as for the delta connection: the resistance between two terminals (e.g. $a$ and $b$) is determined, and the further terminal ($c$) is considered open. The resistance of the further terminal ($R_{c0}^1$) is only connected at one node. Therefore, no current flows through it - it is thus not to be considered. It results in:
+Again, the procedure is the same as for the delta connection: the resistance between two terminals (e.g. $\rm a$ and $\rm b$) is determined, and the further terminal ($\rm c$) is considered to be open.
+The resistance of the further terminal ($R_{\rm c0}^1$) is only connected at one node. Therefore, no current flows through it - it is thus not to be considered. It results in:
 \begin{align*}
-R_{ab} = R_{a0}^1 + R_{b0}^1  \\
+R_{\rm ab} = R_{\rm a0}^1 + R_{\rm b0}^1  \\
-R_{bc} = R_{b0}^1 + R_{c0}^1  \\
+R_{\rm bc} = R_{\rm b0}^1 + R_{\rm c0}^1  \\
-R_{ca} = R_{c0}^1 + R_{a0}^1  \tag{2.6.2}
+R_{\rm ca} = R_{\rm c0}^1 + R_{\rm a0}^1  \tag{2.6.2}
 \end{align*}
 From equations $(2.6.1)$ and $(2.6.2)$ we get:
-\begin{align} R_{ab} = {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{a0}^1 + R_{b0}^1 \tag{2.6.3} \end{align}
+\begin{align} R_{\rm ab} = {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm a0}^1 + R_{\rm b0}^1 \tag{2.6.3} \end{align}
-\begin{align} R_{bc} = {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{b0}^1 + R_{c0}^1 \tag{2.6.4} \end{align}
+\begin{align} R_{\rm bc} = {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm b0}^1 + R_{\rm c0}^1 \tag{2.6.4} \end{align}
-\begin{align} R_{ca} = {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} = R_{c0}^1 + R_{a0}^1 \tag{2.6.5} \end{align}
+\begin{align} R_{\rm ca} = {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} = R_{\rm c0}^1 + R_{\rm a0}^1 \tag{2.6.5} \end{align}
 Equations $(2.6.3)$ to $(2.6.5)$ can now be cleverly combined so that there is only one resistor on one side. \\
-A variation is to write the formulas as ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ or ${{1}\over{2}} \cdot \left(R_{ab} + R_{bc} - R_{ca}\right)$ to combine. This gives $R_{b0}^1$ \\
+A variation is to write the formulas as ${{1}\over{2}} \cdot \left( (2.6.3) + (2.6.4) - (2.6.5) \right)$ or ${{1}\over{2}} \cdot \left(R_{\rm ab} + R_{\rm bc} - R_{\rm ca}\right)$ to combine. This gives $R_{\rm b0}^1$ \\
 \begin{align*}
-{{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} + {{R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} - {{R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= {{1}\over{2}} \cdot \left( R_{a0}^1 + R_{b0}^1 + R_{b0}^1 + R_{c0}^1 - R_{c0}^1 - R_{a0}^1 \right) \\
+{{1}\over{2}} \cdot \left( {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}}
+                         + {{R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}}
+                         - {{R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) &=
+{{1}\over{2}} \cdot \left(   R_{\rm a0}^1 + R_{\rm b0}^1 + R_{\rm b0}^1 + R_{\rm c0}^1 - R_{\rm c0}^1 - R_{\rm a0}^1 \right) \\
-{{1}\over{2}} \cdot \left( {{R_{ab}^1 \cdot (R_{ca}^1 + R_{bc}^1)} + {R_{bc}^1 \cdot (R_{ab}^1 + R_{ca}^1)} - {R_{ca}^1 \cdot (R_{bc}^1 + R_{ab}^1)}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &= {{1}\over{2}} \cdot \left( 2 \cdot  R_{b0}^1  \right) \\
+{{1}\over{2}} \cdot \left( {{R_{\rm ab}^1 \cdot (R_{\rm ca}^1 + R_{\rm bc}^1)} + {R_{\rm bc}^1 \cdot (R_{\rm ab}^1 + R_{\rm ca}^1)}
+                          - {R_{\rm ca}^1 \cdot (R_{\rm bc}^1 + R_{\rm ab}^1)}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) &=
+                            {{1}\over{2}} \cdot \left( 2 \cdot  R_{\rm b0}^1  \right) \\
-{{1}\over{2}} \cdot \left( {{R_{ab}^1 R_{ca}^1 + R_{ab}^1 R_{bc}^1 + R_{bc}^1 R_{ab}^1 + R_{bc}^1 R_{ca}^1 - R_{ca}^1 R_{bc}^1 - R_{ca}^1 R_{ab}^1}\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &=  R_{b0}^1  \\
+{{1}\over{2}} \cdot \left( {{R_{\rm ab}^1 R_{\rm ca}^1 + R_{\rm ab}^1 R_{\rm bc}^1 + R_{\rm bc}^1 R_{\rm ab}^1 + R_{\rm bc}^1 R_{\rm ca}^1 - R_{\rm ca}^1 R_{\rm bc}^1 - R_{\rm ca}^1 R_{\rm ab}^1}\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) &=  R_{\rm b0}^1  \\
-{{1}\over{2}} \cdot \left( {{ 2 \cdot R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \right) &=  R_{b0}^1  \\
+{{1}\over{2}} \cdot \left( {{ 2 \cdot R_{\rm ab}^1 R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \right) &=  R_{\rm b0}^1  \\
-{{ R_{ab}^1 R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} &=  R_{b0}^1  \\
+{{ R_{\rm ab}^1 R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} &=  R_{\rm b0}^1  \\
 \end{align*}
-Similarly, one can resolve to $R_{a0}^1$ and $R_{c0}^1$, and with a slightly modified approach to $R_{ab}^1$, $R_{bc}^1$ and $R_{ca}^1$.
+Similarly, one can resolve to $R_{\rm a0}^1$ and $R_{\rm c0}^1$, and with a slightly modified approach to $R_{\rm ab}^1$, $R_{\rm bc}^1$ and $R_{\rm ca}^1$.
-==== star-triangle-transformation ====
+==== Y-Δ-Transformation  ====
-<callout icon="fa fa-exclamation" color="red" title="Merke:">
+<callout icon="fa fa-exclamation" color="red" title="Notice:">
-<WRAP group><WRAP column half>
+<WRAP>
-If a **delta connection is to be converted into a star connection**, the star resistors can be determined via:
+If a **delta circuit is to be converted into a star circuit**, the star resistors can be determined via:
 \begin{align*}
@@ Zeile 732: / Zeile 850: @@
 \text{therefore:}\quad\quad\quad\quad\quad\quad
-R_{a0}^1 &= {{ R_{ca}^1 \cdot R_{ab}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\
+R_{\rm a0}^1 &= {{ R_{\rm ca}^1 \cdot R_{\rm ab}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \\
-R_{b0}^1 &= {{ R_{ab}^1 \cdot R_{bc}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}} \\
+R_{\rm b0}^1 &= {{ R_{\rm ab}^1 \cdot R_{\rm bc}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}} \\
-R_{c0}^1 &= {{ R_{bc}^1 \cdot R_{ca}^1 }\over{R_{ab}^1 + R_{ca}^1 + R_{bc}^1}}
+R_{\rm c0}^1 &= {{ R_{\rm bc}^1 \cdot R_{\rm ca}^1 }\over{R_{\rm ab}^1 + R_{\rm ca}^1 + R_{\rm bc}^1}}
 \end{align*}
-</WRAP><WRAP column half>
+</WRAP><WRAP>
-Soll von einer **Sternschaltung in eine Dreieckschaltung** umgewandelt werden, so sind die Dreieckwiderstände ermittelbar über:
+If a **star circuit is to be converted into a delta circuit**, the star resistors can be determined via:
 \begin{align*}
@@ Zeile 747: / Zeile 865: @@
 \text{therefore:}\quad\quad\quad\quad\quad\quad
-R_{ab}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{c0}^1}} \\
+R_{\rm ab}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm c0}^1}} \\
-R_{bc}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{a0}^1}} \\
+R_{\rm bc}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm a0}^1}} \\
-R_{ca}^1 &= {{ R_{a0}^1 \cdot R_{b0}^1 +R_{b0}^1 \cdot R_{c0}^1 +R_{c0}^1 \cdot R_{a0}^1 }\over{ R_{b0}^1}}
+R_{\rm ca}^1 &= {{ R_{\rm a0}^1 \cdot R_{\rm b0}^1 +R_{\rm b0}^1 \cdot R_{\rm c0}^1 +R_{\rm c0}^1 \cdot R_{\rm a0}^1 }\over{ R_{\rm b0}^1}}
 \end{align*}
-</WRAP></WRAP>
+</WRAP>
 </callout>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<panel type="info" title="Task 2.6.1 Application of triangle-star conversion"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.6.1 Application of delta-Y conversion"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>YDpNFEWkN9U}}{{youtube>s7NqWI_ZSt4}}
+{{youtube>hM7oSuvdmyw}}
+{{youtube>tOCp2qAV8BI}}
 </WRAP></WRAP></panel>
-<panel type="info" title="Task 2.6.2 more difficult task with star-triangle transformation"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.6.2 more difficult exercise with Y-delta transformation (German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>poP4a0y0oLU}}
@@ Zeile 769: / Zeile 887: @@
 </WRAP></WRAP></panel>
-===== 2.7 Group circuit of resistors =====
+===== 2.7 Circuits with multiple Resistors =====
 <callout>
-=== Goals ===
+=== Learning Objectives ===
-After this lesson, you should:
-  - be able to simplify circuits consisting only of resistors.
+By the end of this section, you will be able to:
-  - Be able to calculate the voltages and currents in circuits with a voltage source and several resistors.
+  - simplify circuits consisting only of resistors.
-  - Be able to simplify symmetrical circuits.
+  - calculate the voltages and currents in circuits with a voltage source and several resistors.
+  - simplify symmetrical circuits.
 </callout>
-In this subchapter a methodology is discussed, which should help to reshape circuits. In subchapter [[#2.6 Star-Delta-Circuit]] towards the end a network was already transformed in a way, that it does not contain triangular meshes any more. Now this procedure shall be systematized.
+In this subchapter, a methodology is discussed, which should help to reshape circuits. In subchapter [[#2.6 Star-Delta-Circuit]] towards the end a network was already transformed in a way, that it does not contain triangular meshes anymore. Now, this procedure shall be systematized.
-Starting point are tasks, where for a resistor network the total resistance, total current or total voltage has to be calculated.
+Starting points are tasks, where for a resistor network the total resistance, total current, or total voltage has to be calculated.
-==== simple example ====
+==== Simple Example ====
-<WRAP right>
+An example of such a circuit is given in <imgref imageNo89>. Here $I_0$ is wanted.
+This current can be found by the (given) voltage $U_0$ and the total resistance between the terminals $\rm a$ and $\rm b$. So we are looking for $R_{\rm ab}$.
+<WRAP>
 <imgcaption imageNo89 | example of a circuit>
 </imgcaption>
-{{drawio>BeispielStromkreis2}}
+{{drawio>BeispielStromkreis2.svg}}
 </WRAP>
+As already described in the previous subchapters, partial circuits can also be converted into equivalent resistors step by step.
+It is important to note that these partial circuits for conversion into equivalent resistors may only ever have two connections (= two nodes to the "outside world").
-An example of such a circuit is given in <imgref imageNo89>. Here $I_0$ is wanted. This current can be found by the (given) voltage $U_0$ and the total resistance between the terminals $a$ and $b$. So we are looking for $R_{ab}$.
-As already described in the previous subchapters, partial circuits can also be converted into equivalent resistors step by step. It is important to note that these partial circuits for conversion into equivalent resistors may only ever have two connections (= two nodes to the "outside world").
+<WRAP>
+<imgcaption imageNo88 | Step-by-step solution of the example >
-<WRAP right>
-<imgcaption imageNo88 | Step by step solution of the example >
 </imgcaption>
-{{drawio>BeispielStromkreis2Loesung}}
+{{drawio>BeispielStromkreis2Loesung.svg}}
 </WRAP>
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<imgref imageNo88 > shows the step-by-step conversion of the equivalent resistors in this example. \\ As a result of the equivalent resistance one gets:
+<imgref imageNo88 > shows the step-by-step conversion of the equivalent resistors in this example. \\
+As a result of the equivalent resistance one gets:
 \begin{align*}
-R_g = R_{12345} &= R_{12}||R_{345} = R_{12}||(R_3+R_{45}) =  (R_1||R_2)||(R_3+R_4||R_5) \\
+R_{\rm  eq} = R_{12345} &= R_{12}||R_{345} = R_{12}||(R_3+R_{45}) =  (R_1||R_2)||(R_3+R_4||R_5) \\
 &= {{ {{R_1 \cdot R_2}\over{R_1 + R_2}} \cdot (R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}}) }\over{ {{R_1 \cdot R_2}\over{R_1 + R_2}} +R_3 + {{R_4 \cdot R_5}\over{R_4 + R_5}} }} \quad \quad \quad \quad \quad \quad \bigg\rvert \cdot{{(R_1 + R_2) \cdot (R_4 + R_5)}\over{(R_1 + R_2) \cdot (R_4 + R_5)}} \\
@@ Zeile 817: / Zeile 936: @@
   \end{align*}
-==== Example with triangle-star transformation ====
+==== Example of Δ-Y-Transformation ====
-With the triangle-star-transformation now also the initial example can be transformed. For more complicated circuits, the repeated triangle-star-transformation with subsequent combining of the resistors is useful, until the resulting circuit is easily calculable with node and mesh theorem (<imgref imageNo92>). Here a calculation is omitted - it is recommended to calculate here with intermediate results for the transformed resistors.
+With the Δ-Y-transformation now also the initial example can be transformed. For more complicated circuits, the repeated Δ-Y-transformation with a subsequent combining of the resistors is useful, until the resulting circuit is easily calculable with node and mesh theorem (<imgref imageNo92>). Here a calculation is omitted - it is recommended to calculate here with intermediate results for the transformed resistors.
-<WRAP right>
+<WRAP>
 <imgcaption imageNo92 | example circuit conversion>
 </imgcaption>
-{{drawio>BeispielStromkreisUmgewandelt}}
+{{drawio>BeispielStromkreisUmgewandelt.svg}}
 <WRAP>
-==== Example with symmetries in the circuit ====
+==== Example with Symmetries in the Circuit ====
-A certain special case concerns possible symmetries in circuits. If these are present, a further simplification can be made.
+A certain special case concerns possible symmetries in circuits. If these are present, further simplification can be made.
-<WRAP right>
+<WRAP>
 <imgcaption imageNo40| Example with symmetries in the circuit>
-</imgcaption> <WRAP>.
+</imgcaption> <WRAP>
-{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3AKACcQBaSS15gJgg-AkunCIGzduxZ8QXHgLBCA7s3xFelJhywrwtAEqLoysblxsJAgSSowrcJLQVqNYqZujC13Te57OZQxuMpnL0kPXwQ3dmMMTg8oq1lw-2cg5ziw1392IMjyaPTaAHNmbBA4phKiMH5aAGdOCw4LNWhojhb4kAAzAEMAGxqAUztOdrbogJdhtQam5NC3EqCStMFE4oEynLzVjPWQngmfHYWZCQn2fP8SsUXQ4-tGjeMYnjSpsdL9EQkfWvqn1QfJQdHr9Ib2YjKR7fQLzCFgbztcqnap1FhEKFNMAIyTtGRdPqDKaQ0rPCZveGvL7I-YwP7o5RGVTYqnKfGgokAD2Y5BQHFwyiYSFwklwUAM0W0tG5TF5pQ4KCFItw6hAEpAUpl+A2HAO6miKuo4El0r0An5YE4Gn5ynVms4HAAnKVHXrIC6jWATTKwAKXSKWLqXZQ7aaWNjJOQA47nRxyCHjRqw76UFVFQiRVVoqGZfyZORBRwveByOLE-bxJa2oKwEhYwQ1eXk-HPumkIz2jmRlCoyJXZGE8WpZsJGJRDizLQytczirwKrJ9PzXcjq4ZeQvnGA3XvC2u0KW-hFUh259s02ZUhCLjBSf62Wh2Go-nBeRi1UH97mMLU3KhfyS3PR9LwjLdmCIGMB0bYDvz9XBHW3IN4MHL9ZSdF0DjfKB4M9VCIAQUVLVla0-X3fCdUwjRDWg1CiDlaimCIYV5SA2i5RtcDmP5T8k38IgZ1Ufjl2kY4+IE79R1MUTwNuA4r1hESEl2RjllyCTKBWJS3CE2kVI2NT8iKPTSjU4zKmqfx5N0qzNL8HkPCCUtXgM6ShQkMonJM7YtMsn4PFlBy7h8+yeDETyji0+xcmiMKPAuHRmHwgwzggZ4LisCwUDMVBbHsJLPGizQwDcKyxCsiK7LckxVE89KtLiJlSRDKpJAoawpzUuYeHM+dtlcOIPiBL4WXa7k00kR0oEgbCCK7fBlXdXBa3lcwLyg3UoUwSQuBopMxowVMkBkSbwGYrsuGdSBWkdS0vUtOaWq4ZVVS4FC9tKOi1WgZ1cCQbMcXOjFSggPk2rFVj3t9ULIJJNpurWtoRXUc1lvUEU5sg37xQjYxVpg4xYzrUoMCrch4cfRhfu8RpSiQbwNEuFbwH7BU2H7Rmqcka1ue8uzcAwRkBaZ-m2WkkXhaF8X8gUMB+1l0KN22tlCmJwW1bVOddn1LnWioun4js+XWrZ142gNtZtc5jbacUuzWeN5nvDaxmqnzN2lfN1wZaB12S3d1xdARGLFcdNhFeyzLrBy4Yg798AfcV4rGHto37ftxnrd9rPw-uD2s7l9mY4LxFWjxYZrczlnnfLmmK7ao2vZ1pvrc5xu07arq6QUduTaZxuVWOp3AmrhQrdrnmu5tqfxYHyfOfns2peGK2qMXs3+7X809ekcvRlGWuabbwL6d1vJl83prL-755Gvn-XXCAA 1000,400 noborder}}
+{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjA7CAMB00IQVhvCEAcGGYMzQjABZoUkA2EFI3AKACcQBaSS15gJgg-AkunCIGzduxZ8QXHgLBCA7s3xFelJhywrwtAEqLoysblxsJAgSSowrcJLQVqNYqZujC13Te57OZQxuMpnL0kPXwQ3dmMMTg8oq1lw-2cg5ziw1392IMjyaPTaAHNmbBA4phKiMH5aAGdOCw4LNWhojhb4kAAzAEMAGxqAUztOdrbogJdhtQam5NC3EqCStMFE4oEynLzVjPWQngmfHYWZCQn2fP8SsUXQ4-tGjeMYnjSpsdL9EQkfWvqn1QfJQdHr9Ib2YjKR7fQLzCFgbztcqnap1FhEKFNMAIyTtGRdPqDKaQ0rPCZveGvL7I-YwP7o5RGVTYqnKfGgokAD2Y5BQHFwyiYSFwklwUAM0W0tG5TF5pQ4KCFItw6hAEpAUpl+A2HAO6miKuo4El0r0An5YE4Gn5ynVms4HAAnKVHXrIC6jWATTKwAKXSKWLqXZQ7aaWNjJOQA47nRxyCHjRqw76UFVFQiRVVoqGZfyZORBRwveByOLE-bxJa2oKwEhYwQ1eXk-HPumkIz2jmRlCoyJXZGE8WpZsJGJRDizLQytczirwKrJ9PzXcjq4ZeQvnGA3XvC2u0KW-hFUh259s02ZUhCLjBSf62Wh2Go-nBeRi1UH97mMLU3KhfyS3PR9LwjLdmCIGMB0bYDvz9XBHW3IN4MHL9ZSdF0DjfKB4M9VCIAQUVLVla0-X3fCdUwjRDWg1CiDlaimCIYV5SA2i5RtcDmP5T8k38IgZ1Ufjl2kY4+IE79R1MUTwNuA4r1hESEl2RjllyCTKBWJS3CE2kVI2NT8iKPTSjU4zKmqfx5N0qzNL8HkPCCUtXgM6ShQkMonJM7YtMsn4PFlBy7h8+yeDETyji0+xcmiMKPAuHRmHwgwzggZ4LisCwUDMVBbHsJLPGizQwDcKyxCsiK7LckxVE89KtLiJlSRDKpJAoawpzUuYeHM+dtlcOIPiBL4WXa7k00kR0oEgbCCK7fBlXdXBa3lcwLyg3UoUwSQuBopMxowVMkBkSbwGYrsuGdSBWkdS0vUtOaWq4ZVVS4FC9tKOi1WgZ1cCQbMcXOjFSggPk2rFVj3t9ULIJJNpurWtoRXUc1lvUEU5sg37xQjYxVpg4xYzrUoMCrch4cfRhfu8RpSiQbwNEuFbwH7BU2H7Rmqcka1ue8uzcAwRkBaZ-m2WkkXhaF8X8gUMB+1l0KN22tlCmJwW1bVOddn1LnWioun4js+XWrZ142gNtZtc5jbacUuzWeN5nvDaxmqnzN2lfN1wZaB12S3d1xdARGLFcdNhFeyzLrBy4Yg798AfcV4rGHto37ftxnrd9rPw-uD2s7l9mY4LxFWjxYZrczlnnfLmmK7ao2vZ1pvrc5xu07arq6QUduTaZxuVWOp3AmrhQrdrnmu5tqfxYHyfOfns2peGK2qMXs3+7X809ekcvRlGWuabbwL6d1vJl83prL-755Gvn-XXCAA noborder}}
 </WRAP>
-~~PAGEBREAK~~ ~~CLEARFIX~~~
+~~PAGEBREAK~~ ~~CLEARFIX~~
-<imgref imageNo40> shows on the left a symmetrical construction of a network of equal resistors $R$. For better understanding, in the middle of the same circuit additional switches and test points (TP) are installed, which indicate the voltage to ground.
+<imgref imageNo40> shows on the left a symmetrical construction of a network of equal resistors $R$. For better understanding, in the middle of the same circuit, additional switches and test points (TP) are installed, which indicate the voltage to the ground.
-The switches can be used to check whether a current flows if the respective nodes are connected. In the simulation it can be seen that this is not the case. In the symmetrical setup, these nodes are each at the same potential.
+The switches can be used to check whether a current flows if the respective nodes are connected. In the simulation, it can be seen that this is not the case. In the symmetrical setup, these nodes are each at the same potential.
-This also allows the circuit to take the form shown in <imgref FigNo40> on the right. This circuit is again easy to calculate:
+This also allows the circuit to take the form shown in <imgref imageNo40> on the right. This circuit is again easy to calculate:
 \begin{align*}
-R_g = R || R + R || R || R || R + R || R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1.5\cdot R
+R_{\rm eq} = R || R + R || R || R || R + R || R || R || R + R || R = {{1}\over{2}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{4}}\cdot R + {{1}\over{2}}\cdot R = 1.5\cdot R
   \end{align*}
-~~PAGEBREAK~~ ~~CLEARFIX~~~
+~~PAGEBREAK~~ ~~CLEARFIX~~
-<panel type="info" title="Task 2.7.1 Circuit Simplification Task I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.1 Circuit Simplification Exercise I (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>8nhzwwRYaUI}}
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-<panel type="info" title="Task 2.7.2 Circuit Simplification Task II + III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.2 Circuit Simplification Exercise II + III (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>QqUQF3ky7gk}}
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-<panel type="info" title="Task 2.7.3 Circuit Simplification Task IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.3 Circuit Simplification Exercise IV (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>SzXWWrPRsDU}}
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-<panel type="info" title="Task 2.7.4 Circuit Simplification Task IV"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.4 Circuit Simplification Exercise V (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>MhaO6kiB4dk}}
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-<panel type="info" title="Task 2.7.5 Circuit Simplification Task V"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.5 Circuit Simplification Exercise VI (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
 {{youtube>9eIRRUNba4A}}
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-<panel type="info" title="Task 2.7.6 Circuit Simplification VI Task"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.6 Circuit Simplification VII Exercise (in German)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
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-<panel type="info" title="other tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+<panel type="info" title="Exercise 2.7.11 - Simplifying symmetric Circuits"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
+well explained example of a simplification due to symmetry:
+{{youtube>Eoh-JKVQZwg}}
-More exercises can be found online on the pages of [[https://www.eit.hs-karlsruhe.de/hertz/teil-b-gleichstromtechnik/zusammenschaltung-von-widerstaenden-und-idealen-quellen/uebungsaufgaben-zusammenschaltung-von-widerstaenden/berechnung-von-ersatzwiderstaenden.html|HErTZ]] (selection on the left in the menu).
 </WRAP></WRAP></panel>