Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

Link zu dieser Vergleichsansicht

Nächste Überarbeitung		 Vorhergehende Überarbeitung
electrical_engineering_and_electronics_1:aufgabe_4.5.2_mit_rechnung [2023/03/23 14:17] – angelegt - Externe Bearbeitung 127.0.0.1electrical_engineering_and_electronics_1:aufgabe_4.5.2_mit_rechnung [2026/01/20 10:08] (aktuell) – mexleadmin
Zeile 1: Zeile 1:
 <panel type="info" title="Aufgabe 4.5.2: open circuit voltage via superposition (exam task, approx. 12 % of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Aufgabe 4.5.2: open circuit voltage via superposition (exam task, approx. 12 % of a 60-minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-<WRAP right> +<WRAP right> {{:elektrotechnik_1:schaltung_klws2020_2_3_1.jpg?400|schaltung_klws2020_2_3_1.jpg}}</WRAP>
-{{elektrotechnik_1:schaltung_klws2020_2_3_1.jpg?400}} +
-</WRAP>+
  
 A circuit is given with the following parameters\\ A circuit is given with the following parameters\\
Zeile 13: Zeile 11:
 $R_4=10 ~\Omega$ $R_4=10 ~\Omega$
  
-Determine the open circuit voltage between A and B using the principle of superposition.\\+Determine the open circuit voltage between A and B using the principle of superposition.
  
 <button size="xs" type="link" collapse="Loesung_4_5_2_1_Tipps">{{icon>eye}} Tips </button><collapse id="Loesung_4_5_2_1_Tipps" collapsed="true"> <button size="xs" type="link" collapse="Loesung_4_5_2_1_Tipps">{{icon>eye}} Tips </button><collapse id="Loesung_4_5_2_1_Tipps" collapsed="true">
 +
   * What do the individual circuits look like, by which the effects of the individual sources can be calculated? \\ Which equivalent resistor must be used to replace a current or voltage source when calculating the individual effects?   * What do the individual circuits look like, by which the effects of the individual sources can be calculated? \\ Which equivalent resistor must be used to replace a current or voltage source when calculating the individual effects?
   * Where are the open-circuit voltages applied when looking at the individual components?   * Where are the open-circuit voltages applied when looking at the individual components?
 +
 </collapse> </collapse>
  
 <button size="xs" type="link" collapse="Loesung_4_5_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_4_5_2_1_Lösungsweg" collapsed="true"> <button size="xs" type="link" collapse="Loesung_4_5_2_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_4_5_2_1_Lösungsweg" collapsed="true">
  
-First, the individual circuits must be created, from which the effect of the individual sources between points A and B can be determined. +First, the individual circuits must be created, from which the effect of the individual sources between points A and B can be determined. \\  \\ **(Voltage) source $U_1$**
-\\ \\ +
-**(Voltage) source $U_1$**+
  
   * substitute the current source $I_2$ with a short-circuit   * substitute the current source $I_2$ with a short-circuit
   * substitute the voltage source $U_3$ with an open circuit   * substitute the voltage source $U_3$ with an open circuit
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1.jpg?400}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q1.jpg?400|schaltung_klws2020_2_3_1_q1.jpg}}
  
 The components can be moved in order to understand the circuit s bit better. The components can be moved in order to understand the circuit s bit better.
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q1_1.jpg?300}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q1_1.jpg?300|schaltung_klws2020_2_3_1_q1_1.jpg}}
  
 For the open circuit, no current is flowing through any resistor. Therefore, the effect is: $U_{AB,1} = U_1$ For the open circuit, no current is flowing through any resistor. Therefore, the effect is: $U_{AB,1} = U_1$
Zeile 42: Zeile 40:
   * substitute the voltage source $U_3$ with an open circuit   * substitute the voltage source $U_3$ with an open circuit
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2.jpg?400}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q2.jpg?400|schaltung_klws2020_2_3_1_q2.jpg}}
  
 Also here, the components can be shifted for a better understanding: Also here, the components can be shifted for a better understanding:
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q2_1.jpg?300}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q2_1.jpg?300|schaltung_klws2020_2_3_1_q2_1.jpg}}
  
 Here, the current source $I_2$ creates a voltage drop $U_{AB_2}$ on the resistor $R_2$ : $U_{\rm AB,2} = - R_1 \cdot I_2$ Here, the current source $I_2$ creates a voltage drop $U_{AB_2}$ on the resistor $R_2$ : $U_{\rm AB,2} = - R_1 \cdot I_2$
Zeile 55: Zeile 53:
   * substitute the current source $I_2$ with a short-circuit   * substitute the current source $I_2$ with a short-circuit
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3.jpg?400}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q3.jpg?400|schaltung_klws2020_2_3_1_q3.jpg}}
  
-Again, rearranging the circuit might help for an understanding: +Again, rearranging the circuit might help for an understanding:
  
-{{elektrotechnik_1:schaltung_klws2020_2_3_1_q3_1.jpg?300}}+{{:elektrotechnik_1:schaltung_klws2020_2_3_1_q3_1.jpg?300|schaltung_klws2020_2_3_1_q3_1.jpg}}
  
-In this case, between the unloaded outputs $\rm A$ and $\rm B$ there will be an unloaded voltage divider given by $R_3$ and $R_4$.  +In this case, between the unloaded outputs $\rm A$ and $\rm B$ there will be an unloaded voltage divider given by $R_3$ and $R_4$. On $R_1$ there is no voltage drop since there is no current flow out of the unloaded outputs. \\ Therefore:
-On $R_1$ there is no voltage drop since there is no current flow out of the unloaded outputs. \\ +
-Therefore:+
  
-\begin{align*} +\begin{align*} U_{\rm AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3 \end{align*}
-U_{\rm AB,3} = \frac{R_4}{R_3 + R_4} \cdot U_3 +
-\end{align*}+
  
-\\ \\ + \\  \\ **resulting voltage**
-**resulting voltage**+
  
-\begin{align*} +\begin{align*} U_{\rm AB} &= U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3 \\ \end{align*}
-U_{\rm AB} &=  U_1 - R_1 \cdot I_2 + \frac{R_4}{R_3 + R_4} \cdot U_3   \\ +
-\end{align*}+
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Loesung_4_5_2_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_2_1_Endergebnis" collapsed="true"> +<button size="xs" type="link" collapse="Loesung_4_5_2_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_2_1_Endergebnis" collapsed="true"> \begin{align*} U_{\rm AB} &= 2 ~{\rm V} - 5 ~\Omega \cdot 1 ~{\rm A} + \frac{10 ~\Omega}{20 ~\Omega + 10 ~\Omega} \cdot 8 ~{\rm V} \\ \\ U_{\rm AB} & = - 0.333... ~{\rm V} \rightarrow - 0.3 ~{\rm V} \\ \end{align*} \\ </collapse>
-\begin{align*} +
-U_{\rm AB} &=  2 ~{\rm V} - 5 ~\Omega \cdot 1 ~{\rm A} + \frac{10 ~\Omega}{20 ~\Omega + 10 ~\Omega} \cdot 8 ~{\rm V} \\ \\ +
-U_{\rm AB} & = 0.333... ~{\rm V} \rightarrow 0.3 ~{\rm V} \\ +
-\end{align*} +
- \\ +
-</collapse>+
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>