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electrical_engineering_1:aufgabe_2.7.7_mit_rechnung [2021/10/08 22:26] – [Bearbeiten - Panel] tfischerelectrical_engineering_1:aufgabe_2.7.7_mit_rechnung [2023/07/17 13:36] (aktuell) – [Bearbeiten - Panel] mexleadmin
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-{{elektrotechnik_1:schaltung_klws2020_2_1_1.jpg?400}}+{{elektrotechnik_1:schaltung_klws2020_2_1_1.jpg?300}}
 </WRAP> </WRAP>
  
 Given is the adjoining circuit with \\ Given is the adjoining circuit with \\
-$R_1=10 \Omega$\\ +$R_1=10 ~\Omega$\\ 
-$R_2=20 \Omega$\\ +$R_2=20 ~\Omega$\\ 
-$R_3=5 \Omega$\\+$R_3=5 ~\Omega$\\
 and the switch $S$. and the switch $S$.
  
-1. determine the total resistance $R_{ges}$ between A and B by summing the resistances with the switch $S$ open.+1. Determine the total resistance $R_{\rm eq}$ between A and B by summing the resistances with the switch $S$ open.
  
 <button size="xs" type="link" collapse="Solution_2_7_7_1_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_7_1_Tips" collapsed="true"> <button size="xs" type="link" collapse="Solution_2_7_7_1_Tips">{{icon>eye}} Tips for solving</button><collapse id="Solution_2_7_7_1_Tips" collapsed="true">
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 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Solution_2_7_7_1_SolutionPath">{{icon>eye}} Solution path</button><collapse id="Solution_2_7_7_1_SolutionPath" collapsed="true">+<button size="xs" type="link" collapse="Solution_2_7_7_1_SolutionPath">{{icon>eye}} Solution</button><collapse id="Solution_2_7_7_1_SolutionPath" collapsed="true">
  
 First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\ First of all, it is a good idea to reshape the circuit so that the actual structure becomes visible. \\
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 Thus $R_3$ and $R_3$ can be combined to $R_{33} = 2 \cdot R_3 = R_1$, yielding a left and a right voltage divider. \\ Thus $R_3$ and $R_3$ can be combined to $R_{33} = 2 \cdot R_3 = R_1$, yielding a left and a right voltage divider. \\
-Now it is visible that in the left and right voltage divider the same potential is at the respective branch, or at the node K1 (green) and K2 (pink).+Now it is visible that in the left and right voltage divider, the same potential is at the respective branch, or at the node K1 (green) and K2 (pink).
  
-Thus, the total resistance can be calculated as $R_{ges} = (2 \cdot R_1)||(2 \cdot R_1)$. \\ +Thus, the total resistance can be calculated as $R_{\rm eq} = (2 \cdot R_1)||(2 \cdot R_1)$. \\ 
-However, by symmetry, nodes K1 and K2 can also be short-circuited. Thus, $R_{ges} = 2 \cdot \left( R_1||R_1 \right)$ also holds.+However, by symmetry, nodes K1 and K2 can also be short-circuited. Thus, $R_{\rm eq} = 2 \cdot \left( R_1||R_1 \right)$ also holds.
  
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-<button size="xs" type="link" collapse="Solution_2_7_7_1_Final result">{{icon>eye}} Final result</button><collapse id="Solution_2_7_7_1_Final result" collapsed="true">+<button size="xs" type="link" collapse="Solution_2_7_7_1_Finalresult">{{icon>eye}} Final result</button><collapse id="Solution_2_7_7_1_Finalresult" collapsed="true">
 \begin{align*} \begin{align*}
-R_{ges} &= 2 \cdot \left( 10 \Omega || 10 \Omega \right) = 10 \Omega+R_{\rm eq} &= 2 \cdot \left( 10 ~\Omega || 10 ~\Omega \right) = 10 ~\Omega
 \end{align*} \end{align*}
  \\  \\
 </collapse> </collapse>
  
-2. what is the total resistance when switch $S$ is closed?+2. What is the total resistance when switch $S$ is closed?
  
-<button size="xs" type="link" collapse="Solution_5_1_3_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_2_Final result" collapsed="true">+<button size="xs" type="link" collapse="Solution_5_1_3_2_FinalResult">{{icon>eye}} Final result</button><collapse id="Solution_5_1_3_2_FinalResult" collapsed="true">
 Due to symmetry, the potentials at K1 and K2 are equal. Thus, no current flows across resistor $R_2$ even when the switch is closed. \\ Due to symmetry, the potentials at K1 and K2 are equal. Thus, no current flows across resistor $R_2$ even when the switch is closed. \\
 So the resistance remains the same. \\ \\ So the resistance remains the same. \\ \\