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electrical_engineering_and_electronics_1:block06 [2025/09/29 00:40] – angelegt mexleadminelectrical_engineering_and_electronics_1:block06 [2025/09/29 00:57] (aktuell) – mexleadmin
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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*} 
  
-</WRAP></WRAP></panel>+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*} 
  
-<panel type="info" title="Exercise 3.2.4 Interesting circuit tasks"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> +The voltage at the open circuit is  
- +\begin{equation*} U_{\rm S}= 2~V + 1~V + U_R \end{equation*}
-{{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}}+
  
 +</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>
  
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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.