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--- electrical_engineering_and_electronics_1:block06 [2025/09/29 00:23] – angelegt mexleadmin
+++ electrical_engineering_and_electronics_1:block06 [2025/09/29 00:57] (aktuell) – mexleadmin
@@ Zeile 171: / Zeile 171: @@
 <WRAP> <imgcaption imageNob7 | Resistance of linear sources> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCsICMZICYaoOyYMxgByoBsAnCZkiAgCw5UCmAtIogFABuIJRIqcP3JJh4QI1CAlFQ4rAE5ceiYUMXKIyGKwDuC3vxV6RrVIh6pU1Q7wsH1265b6qz+sPaXOHt1gEsQOdCsA9TUoeHtggxx8Sw9wOX8YqKSncCRYCJT9SNS3HWjY5WoiQqN8rJ5ix1djUxAqqwa4uwBzL1TzR3x8NLc2htSB7t7WACMQIkQkOFiiOixUePHUcmme6hZeTEW3AA9eElicGjgexGj6pB6AGx8AO3oAQ1kAHQBnAGMAV1lZejuAC7vN4Aex+H3orH2JSQODMeFhJEuiBu9yerze7BB1wBjxa9GBYNkENYQA noborder}} </WRAP>
-In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[elektrotechnik_labor:1_widerstaende|resistor]]s in the 3rd semester.)) Let us have a look at the properties of the ohmmeter in the simulation by double-clicking on the ohmmeter. Here, a very large measuring current of $I_\Omega=1~\rm A$ is used. This could lead to high voltages or the destruction of components in real setups. \\
+In the simulation, a measuring current $I_\Omega$ is used to determine the resistance value. ((This concept will also be used in an electrical engineering lab experiment on [[elektrotechnik_labor:1_widerstaende|resistor]]s in the 2nd semester.)) Let us have a look at the properties of the ohmmeter in the simulation by double-clicking on the ohmmeter. Here, a very large measuring current of $I_\Omega=1~\rm A$ is used. This could lead to high voltages or the destruction of components in real setups. \\
  \\
 In order to understand why is this nevertheless chosen so high in the simulation, do the following: Set the measuring current for both linear sources to (more realistic) $1~\rm mA$. What do you notice?
@@ Zeile 184: / Zeile 184: @@
 <WRAP> <imgcaption imageNo8 | circuit with two current sources> </imgcaption> {{drawio>SchaltungZeiStromquellen.svg}} </WRAP>
-<callout icon="fa fa-exclamation" color="red" title="Note:"> If resistors are to be measured in a circuit, at least one terminal of the resistor must be disconnected from the circuit. Otherwise, other sources or resistors may falsify the measurement result. </callout>
+Any interconnection of __linear__  voltage sources, current sources, and __ohmic__  resistors can be seen as
+  * as a single, linear voltage source \\ ({{https://en.wikipedia.org/wiki/Thévenin_theorem|Thévenin theorem}}  ) or
+  * as a single, linear current source \\ ({{https://en.wikipedia.org/wiki/Norton_theorem|Norton theorem}}  )
+In <imgref imageNo63 > it can be seen that the three circuits give the same result (voltage/current) with the same load. This is also true when an (AC) source is used instead of the load.
+<WRAP> <imgcaption imageNo63 | Equivalent voltage and current source> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-sd7jMEdzhfLyKr1MkZGGcTBuh4MCwYhodcYjQcBcgoNAUAoMAGAPAAlAB9AAZOlWDfXQP1EL9wHMSU2U4ABlPh4mkFDiEOToQAAM27YdmHRQCSLoVhSF0Gw+GIOgGxABDFw7AATej3hQJioFYSg2K8TjuL4l13EqHc2lMGTP3QtCSlCag5KI51-FQmMSNyL8XVQsJ9IUtTfBM3xDNtRI9JCayQzMvTDIItNGN8N5KOo2iLGQ+zRBMzCoJWHCBXoogQN0YDIOg554KQ6J2ESYMY0wn8-wA1xcQsZ4wm9LFVNxAJ+myDlUu5YqZAIhkYyKvKsKorsaLonQivkMr5HIkL+VOQDVCYUZ3noTj4q4LVcq5TomnS-9shqsI5Ly7JFo0xV8rKsJhhjdaKqDbkloAI0Iyo6n4PAoCCRSjv4YUQDqMAmEumRrowUi+CQLx+ECDxnWuyDciQKDgOYzRy2FITiGCNBcmYySeP4xcGGCI40GA39VFQaAiSUbpDybWALyvWAlxXNdEbArCOOgVHwFgDG4CxrDd2Aphm1gcdJ2Jp8X0Ri6UbRum1ngCCyhAKAmAolsuy7EdRzDZ8237TggA noborder}} </WRAP>
+~~PAGEBREAK~~ ~~CLEARFIX~~
 <panel type="info" title="Example / micro-exercise">
@@ Zeile 203: / Zeile 213: @@
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-Any interconnection of __linear__  voltage sources, current sources, and __ohmic__  resistors can be.
-  * as a single, linear voltage source \\ ({{https://en.wikipedia.org/wiki/Thévenin_theorem|Thévenin theorem}}  ) or
-  * as a single, linear current source \\ ({{https://en.wikipedia.org/wiki/Norton_theorem|Norton theorem}}  )
-In <imgref imageNo63 > it can be seen that the three circuits give the same result (voltage/current) with the same load. This is also true when an (AC) source is used instead of the load.
-<WRAP> <imgcaption imageNo63 | Equivalent voltage and current source> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-sd7jMEdzhfLyKr1MkZGGcTBuh4MCwYhodcYjQcBcgoNAUAoMAGAPAAlAB9AAZOlWDfXQP1EL9wHMSU2U4ABlPh4mkFDiEOToQAAM27YdmHRQCSLoVhSF0Gw+GIOgGxABDFw7AATej3hQJioFYSg2K8TjuL4l13EqHc2lMGTP3QtCSlCag5KI51-FQmMSNyL8XVQsJ9IUtTfBM3xDNtRI9JCayQzMvTDIItNGN8N5KOo2iLGQ+zRBMzCoJWHCBXoogQN0YDIOg554KQ6J2ESYMY0wn8-wA1xcQsZ4wm9LFVNxAJ+myDlUu5YqZAIhkYyKvKsKorsaLonQivkMr5HIkL+VOQDVCYUZ3noTj4q4LVcq5TomnS-9shqsI5Ly7JFo0xV8rKsJhhjdaKqDbkloAI0Iyo6n4PAoCCRSjv4YUQDqMAmEumRrowUi+CQLx+ECDxnWuyDciQKDgOYzRy2FITiGCNBcmYySeP4xcGGCI40GA39VFQaAiSUbpDybWALyvWAlxXNdEbArCOOgVHwFgDG4CxrDd2Aphm1gcdJ2Jp8X0Ri6UbRum1ngCCyhAKAmAolsuy7EdRzDZ8237TggA noborder}} </WRAP>
-~~PAGEBREAK~~ ~~CLEARFIX~~
 ==== Simplified Determination of the internal Resistance ====
@@ Zeile 240: / Zeile 237: @@
 ===== Exercises =====
 ~~PAGEBREAK~~ ~~CLEARFIX~~
-<panel type="info" title="Exercise 3.2.1 Solving a circuit simplification I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>xtOPwmUgPjc}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 3.2.2 Solving a circuit simplification II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>UU_RJJ6ne4I}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 3.2.3 Solution sketch for a more difficult circuit simplification"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>In3NF8f-mzg}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 3.2.4 Interesting circuit tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>zTDgziJC-q8}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 3.1.1 Convert current source to voltage source"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>ZqohGL-40a4}}
-</WRAP></WRAP></panel>
-<panel type="info" title="Exercise 3.1.2 Convert voltage source to current source"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-{{youtube>vVDNsztDmAk}}
-</WRAP></WRAP></panel>
-{{page>task_3.1.3_with_calculation&nofooter}}
-~~PAGEBREAK~~ ~~CLEARFIX~~
-===== Common pitfalls =====
-  * **Wrong deactivation:** do **not** set an ideal voltage source to open or an ideal current source to short; the rules are: $U$-source→**short**, $I$-source→**open**.
-  * **Confusing goals:** **max power** ($R_{\rm L}=R_{\rm i}$, $\eta=50\%$) vs. **high efficiency** ($R_{\rm L}\gg R_{\rm i}$). Don’t equate them.
-  * **Ignoring ratings:** not every real source is short-circuit-proof—$I_{\rm SC}$ is a **model parameter**, not a recommended experiment.
-  * **Mixed conventions:** keep the **passive sign convention** for loads; use conventional current ($+$ to $-$).
-===== Exercises =====
 ==== Quick checks ====
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~.1  From $U_{\rm OC}$, $I_{\rm SC}$ to $R_{\rm i}$ and $U_{\rm L}$
-#@TaskText_HTML@#
-A source has $U_{\rm OC}=12.0~\rm V$, $I_{\rm SC}=3.0~\rm A$. Find $R_{\rm i}$ and, for $R_{\rm L}=9.0~\Omega$, compute $U_{\rm L}$ and $\eta$.
-#@ResultBegin_HTML~Exercise1~@#
-$R_{\rm i}=U_{\rm OC}/I_{\rm SC}=4.00~\Omega$.
-$U_{\rm L}=U_{\rm OC}\dfrac{R_{\rm L}}{R_{\rm L}+R_{\rm i}}=12.0~{\rm V}\cdot \dfrac{9.0~\Omega}{13.0~\Omega}=8.31~{\rm V}$.
-$\eta=\dfrac{R_{\rm L}}{R_{\rm L}+R_{\rm i}}=\dfrac{9.0}{13.0}=0.692\;(-)$.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~.2  Thevenin ↔ Norton conversion
@@ Zeile 309: / Zeile 247: @@
 #@ResultEnd_HTML@#
 #@TaskEnd_HTML@#
 ==== Longer exercises ====
-#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee1_taskctr#~~.1  Loaded divider as Thevenin
-#@TaskText_HTML@#
-A divider $R_1=3.3~\rm k\Omega$, $R_2=6.8~\rm k\Omega$ is fed from $U=10.0~\rm V$ and loaded by $R_{\rm L}=10.0~\rm k\Omega$. Replace the divider by its Thevenin equivalent, then compute $U_{\rm L}$ and the **loading error** relative to the ideal (no-load) divider output.
-#@ResultBegin_HTML~LongerExercise1~@#
-$R_{\rm ie}=R_1\parallel R_2=\dfrac{(3.3)(6.8)}{3.3+6.8}~\rm k\Omega=2.22~\rm k\Omega$.
-$U_{0\rm e}=\dfrac{R_2}{R_1+R_2}U=6.8/(3.3+6.8)\cdot 10.0~\rm V=6.80~\rm V$.
-$U_{\rm L}=U_{0\rm e}\dfrac{R_{\rm L}}{R_{\rm L}+R_{\rm ie}}=6.80~{\rm V}\cdot \dfrac{10.0}{12.22}=5.56~{\rm V}$.
-Ideal (no-load) output would be $6.80~\rm V$ ⇒ loading error $=1.24~\rm V$.
-#@ResultEnd_HTML@#
-#@TaskEnd_HTML@#
 <panel type="info" title="Exercise 3.1.1 Convert current source to voltage source"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
@@ Zeile 345: / Zeile 272: @@
 </WRAP></WRAP></panel>
-~~PAGEBREAK~~ ~~CLEARFIX~~
-===== Embedded resources =====
+<panel type="info" title="Exercise 3.3.1 Simplification by Norton / Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
-<WRAP> DC Voltage & Current Source Theory
+Simplify the following circuits by the Norton theorem to a linear current source (circuits marked with NT) or by Thevenin theorem to a linear voltage source (marked with TT).
-{{youtube>AQK7RyecVW0}}
+<imgcaption BildNr3_0 | Simplification by Norton / Thevenin theorem> </imgcaption> {{drawio>BildNr3_0.svg}} </WRAP>
-</WRAP>
+<button size="xs" type="link" collapse="Loesung_3_3_1_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_3_3_1_1_Lösungsweg" collapsed="true">
+To substitute the circuit in $a)$ first we determine the inner resistance.
+Shutting down all sources leads to
+\begin{equation*}
+R_{\rm i}= 8~\Omega \end{equation*}
+Next, we figure out the current in the short circuit.
+In case of a short circuit, we have $2~V$ in a branch which in turn means there must be $−2~V$ on the resistor.
+The current through that branch is
+\begin{equation*} I_R=\frac{2~V}{8~\Omega} \end{equation*}
+The current in question is the sum of both the other branches
+\begin{equation*} I_S= I_R + 1~A \end{equation*}
+To substitute the circuit in $b)$ first we determine the inner resistance.
+Shutting down all sources leads to
+\begin{equation*} R_{\rm i}= 4 ~\Omega \end{equation*}
+Next, we figure out the voltage at the open circuit.
+Thus we know the given current flows through the ideal current source as well as the resistor.
+The voltage drop on the resistor is
+\begin{equation*} R_{\rm i}= -4~\Omega \cdot 2~A \end{equation*}
+The voltage at the open circuit is
+\begin{equation*} U_{\rm S}= 2~V + 1~V + U_R \end{equation*}
+</collapse> <button size="xs" type="link" collapse="Loesung_3_3_1_2_Lösungsweg">{{icon>eye}} Final result</button><collapse id="Loesung_3_3_1_2_Lösungsweg" collapsed="true">
+The values of the substitute resistor and the currents in the branches are
+\begin{equation*}
+\text{a)} \quad R=8~\Omega \qquad I=1.25~A \\
+\text{b)} \quad R=4~\Omega \qquad U=-5~V
+\end{equation*}
+</collapse>
+</WRAP></WRAP></panel>
+{{page>task_3.1.3_with_calculation&nofooter}}
+~~PAGEBREAK~~ ~~CLEARFIX~~
+===== Common pitfalls =====
+  * **Wrong deactivation:** do **not** set an ideal voltage source to open or an ideal current source to short; the rules are: $U$-source→**short**, $I$-source→**open**.
+  * **Confusing goals:** **max power** ($R_{\rm L}=R_{\rm i}$, $\eta=50\%$) vs. **high efficiency** ($R_{\rm L}\gg R_{\rm i}$). Don’t equate them.
+  * **Ignoring ratings:** not every real source is short-circuit-proof—$I_{\rm SC}$ is a **model parameter**, not a recommended experiment.
+  * **Mixed conventions:** keep the **passive sign convention** for loads; use conventional current ($+$ to $-$).
+~~PAGEBREAK~~ ~~CLEARFIX~~
+===== Embedded resources =====
+<WRAP column half>
+DC Voltage & Current Source Theory
+{{youtube>AQK7RyecVW0}}
+</WRAP>
 <WRAP column half>
@@ Zeile 361: / Zeile 336: @@
 {{youtube>w4N9CBc_nkA}}
 </WRAP>
 <WRAP column half>
 A more complex superposition example
@@ Zeile 371: / Zeile 347: @@
   * A **linear (real) source** is fully determined by $(U_{\rm OC},I_{\rm SC})$ or equivalently $(U_0,R_{\rm i})$ / $(I_0,G_{\rm i})$; both forms are **equivalent**.
   * **Thevenin ↔ Norton**: $U_{\rm OC}=I_{\rm SC}R_{\rm i}$, $G_{\rm i}=1/R_{\rm i}$; deactivation rules let you find $R_{\rm i}$ quickly.
-  * **Efficiency vs. maximum power**: choose $R_{\rm L}\gg R_{\rm i}$ for high $\eta$, or $R_{\rm L}=R_{\rm i}$ for max $P_{\rm L}$.
   * **Two-terminal reductions** (e.g., loaded divider) simplify analysis of larger networks.