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electrical_engineering_and_electronics_1:block06 [2025/09/29 00:40] – angelegt mexleadminelectrical_engineering_and_electronics_1:block06 [2025/10/24 19:47] (aktuell) – mexleadmin
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 </callout> </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 ===== ===== 90-minute plan =====
   - Warm-up (8–10 min):     - Warm-up (8–10 min):  
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 <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> <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> </WRAP>
 </panel> </panel>
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 <WRAP> <imgcaption imageNob7 | Resistance of linear sources> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgDOB0YzCsICMZICYaoOyYMxgByoBsAnCZkiAgCw5UCmAtIogFABuIJRIqcP3JJh4QI1CAlFQ4rAE5ceiYUMXKIyGKwDuC3vxV6RrVIh6pU1Q7wsH1265b6qz+sPaXOHt1gEsQOdCsA9TUoeHtggxx8Sw9wOX8YqKSncCRYCJT9SNS3HWjY5WoiQqN8rJ5ix1djUxAqqwa4uwBzL1TzR3x8NLc2htSB7t7WACMQIkQkOFiiOixUePHUcmme6hZeTEW3AA9eElicGjgexGj6pB6AGx8AO3oAQ1kAHQBnAGMAV1lZejuAC7vN4Aex+H3orH2JSQODMeFhJEuiBu9yerze7BB1wBjxa9GBYNkENYQA noborder}} </WRAP> <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? 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?
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 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
 +
  
 ==== Longer exercises ==== ==== Longer exercises ====
- 
-Quick checks 
-#@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%> <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%>
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 </WRAP></WRAP></panel>  </WRAP></WRAP></panel> 
  
-<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}}+<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></WRAP></panel>+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). 
  
-<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%>+<imgcaption BildNr3_0 | Simplification by Norton / Thevenin theorem> </imgcaption> {{drawio>BildNr3_0.svg}} </WRAP>
  
-{{youtube>UU_RJJ6ne4I}}+<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*} 
  
-</WRAP></WRAP></panel>+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*} 
  
-<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%>+The current in question is the sum of both the other branches  
 +\begin{equation*} I_S= I_R + 1~A \end{equation*} 
  
-{{youtube>In3NF8f-mzg}}+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> </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 3.3.2 Simplification by Thevenin theorem"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <WRAP>
  
-<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%>+The following simulation shows four cirucits.
  
-{{youtube>ZqohGL-40a4}}+1. 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? \\ \\ 
 +2. 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?
  
-</WRAP></WRAP></panel>+<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>
  
-<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%>+</WRAP></WRAP></WRAP> 
 +</panel>
  
-{{youtube>vVDNsztDmAk}} 
- 
-</WRAP></WRAP></panel> 
  
 {{page>task_3.1.3_with_calculation&nofooter}} {{page>task_3.1.3_with_calculation&nofooter}}
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 ===== Embedded resources ===== ===== Embedded resources =====
  
-<WRAP> DC Voltage & Current Source Theory +<WRAP column half> 
 +DC Voltage & Current Source Theory
 {{youtube>AQK7RyecVW0}} {{youtube>AQK7RyecVW0}}
- 
 </WRAP> </WRAP>
- 
- 
  
 <WRAP column half> <WRAP column half>
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 {{youtube>w4N9CBc_nkA}} {{youtube>w4N9CBc_nkA}}
 </WRAP> </WRAP>
 +
 <WRAP column half> <WRAP column half>
 A more complex superposition example   A more complex superposition example  
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   * 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**.    * 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.    * **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.    * **Two-terminal reductions** (e.g., loaded divider) simplify analysis of larger networks.