Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- electrical_engineering_and_electronics_1:block06 [2025/09/29 00:23] – angelegt mexleadmin
+++ electrical_engineering_and_electronics_1:block06 [2025/10/24 19:47] (aktuell) – mexleadmin
@@ Zeile 10: / Zeile 10: @@
 </callout>
+===== Preparation at Home =====
+And again:
+  * Please read through the following chapter.
+  * Also here, there are some clips for more clarification under 'Embedded resources'. \\ I strongly recommend to watch there the introductional video of EEVblog 1397 (at least the first 10 minutes).
+For checking your understanding please do the following exercises:
+  * E1.2
+  * 3.3.1
+~~PAGEBREAK~~ ~~CLEARFIX~~
 ===== 90-minute plan =====
   - Warm-up (8–10 min):
@@ Zeile 57: / Zeile 68: @@
 <WRAP> <imgcaption imageNo2 | Battery model with load resistor> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3AWAnC1b0DYrTAZlwBwYCsGA7AEwK7G6RIIUj7jTEjUjECmAtGGABQAdxD8wTOgjEZpUqCLECmVSEonNI0yIIDO6lQjXimYMljUQAZgEMANru6Les8OYNuLemdNU-PUOAgtg5OAE7+8i6+RoEwQqJqUa7yOokB0QFpgfLKmtqCAA6BZhb5QWo6uGRJpu6W7hCVggDm4Bq5HbhlOtAgAEI2AC5DAMoAlgC2JSAqIAA6xUyVYswzAErcuuO6QwD2YQCidpPzumBnFPNhm9u7B8enurhnkNcAant2QzYt3I9nF7nKDzV5sV6g87waEQ2G6IyQsDguFvMHEQQIMBqAhqCiQAjtSRaIKDEYTaZvXQwdiUrHQuAgqlnBHncFM9nUxR5ToqSgKUS4OpYPKlBRAA noborder}}
 </WRAP>
 </panel>
@@ Zeile 171: / Zeile 181: @@
 <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 194: @@
 <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 223: @@
 ~~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 247: @@
 ===== 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 257: @@
 #@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 282: @@
 </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>
+<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>
+<panel type="info" title="Exercise 3.3.2 Simplification by Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
+The following simulation shows four cirucits.
+. Have a look on the both circuits 1a) with U_S(1a)=10 V and 1b) U_S(1b)=5 V. Start the simulation and change the load resistors for R_L(1a) and R_L(1b) with the sliders on the right. \\ What do you see on the values for the voltage and the current on both circuits, when you choose the same resistor values? Why? \\ \\
+. Simplify the circuit 2a) with U_S(2a)=10 V by the Thevenin theorem to a linear voltage source. \\ What would be the source voltage U_S(2b) of the equivalent voltage source? What would be the resistance R_i(2b) of the inner resistor?
+<panel>
+<WRAP> <imgcaption imageNo2 | Circuits to be simplified> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxAUgpABZsKBTAWjDACgA3EJwm8YlN14g8VMbSphxY6AjYAnIX1G08I8VLgLV6nWAFRkWgO56DPPvsGQ2p4oV1Xd2NgHMl-QRZAoUfMdpOKk5OkvC2npE0aqFuINjYDk4JDr7+UBE00ZEpkTaKuSpZaipUlDZcucmJuuI0ZYYycgBG3PbxghhUKJQZdg5FJQGtuILYgjSEaj0BpjTtRdmlEQgGTsV52quCKtt5RjZzC2UGKi5HDth4fHtXfC6Kt9cUBvMOYSYvE+0bb3207TutGyQJsI0gYxqkzUCAQ-jYrSYgJqXQocIylShyMuzzqDWksm0uSBG1BIFIxHCc2yf1yfxs7jp2IoKDUsyi7TADnpgRxyioZM0DI8f28CFZjQiZ2eyWe53A3Pa3h5ii58RliveB203iBasF4S4yqVwjJVHqyEaUEJqtenIMBq0AA94ggINgsPFPVdzToAPxsF0sBI+PhI4g+ch8PzgACGAEpAz5hDGUMRplHQ+BmomXUR6OMI2nyAkJlmUDmk8H7mnuJp4uNaOWE0mwCh6GBsBGwMEaGW1QAlAD6ABkABRgFsu8WCMAWhAYbt0JuD0cTysu1hqLKzjBILL7hUgYfjlBTln0HcURaH1enjdeyyQCN9pJtlcOYcAS3Xua93M7WgJRoNVLE-IcfwrP9sHFWhCH3Vg4O3I8AFUhwAZTHKCk2wSZwFWCQkC5fsHDQzCwAfJgEA7QgICSYQwJAMiJ3PKjtwwCAIBoDA+EY5iz0TK4QAcAAxCBJAOOtjzXSdBJicgxMMTj4Ckk9fzYISPRARSJKFVS1wEjTpioHTwEklhpPvRMgA noborder}}
 </WRAP>
+</panel>
+</WRAP></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 364: @@
 {{youtube>w4N9CBc_nkA}}
 </WRAP>
 <WRAP column half>
 A more complex superposition example
@@ Zeile 371: / Zeile 375: @@
   * 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.