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electrical_engineering_1:aufgabe_4.5.3 [2023/03/09 13:19] – [Bearbeiten - Panel] mexleadminelectrical_engineering_1:aufgabe_4.5.3 [2023/03/23 14:16] (aktuell) – mexleadmin
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-<panel type="info" title="Exercise 4.5.3 -Variation: 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="Exercise 4.5.3 -Variation: 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> {{:elektrotechnik_1:schaltung_klws2020_2_3_2.jpg?400|schaltung_klws2020_2_3_2.jpg}}</WRAP> <WRAP right> {{:elektrotechnik_1:schaltung_klws2020_2_3_2.jpg?400|schaltung_klws2020_2_3_2.jpg}}</WRAP>
 A circuit is given with the following parameters\\ A circuit is given with the following parameters\\
 $R_1=5 ~\Omega$\\ $R_1=5 ~\Omega$\\
-$U_1=2 ~V$\\ +$U_1=2 ~{\rm V}$\\ 
-$I_2=1 ~A$\\+$I_2=1 ~{\rm A}$\\
 $R_3=20 ~\Omega$\\ $R_3=20 ~\Omega$\\
-$U_3=8 ~V$\\+$U_3=8 ~{\rm V}$\\
 $R_4=10 ~\Omega$ $R_4=10 ~\Omega$
  
 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_3_1_Endergebnis">{{icon>eye}} Solution</button><collapse id="Loesung_4_5_3_1_Endergebnis" collapsed="true"> <button size="xs" type="link" collapse="Loesung_4_5_3_1_Endergebnis">{{icon>eye}} Solution</button><collapse id="Loesung_4_5_3_1_Endergebnis" collapsed="true">
-Case 1: For this case is $I_2 = 0~A$ and $U_3 = 0~V$. The voltage is at $R_4$.+Case 1: For this case is $I_2 = 0~{\rm A}$ and $U_3 = 0~{\rm V}$. The voltage is at $R_4$.
 <WRAP> {{drawio>Sloution4531.svg}} </WRAP> <WRAP> {{drawio>Sloution4531.svg}} </WRAP>
 \begin{align*} \begin{align*}
-U_{AB,1} = \frac{R_4}{R_1+R_4} U_1 = \frac{10~\Omega}{5~\Omega+10~\Omega} \cdot 2~V = 1.33~V+U_{\rm AB,1} = \frac{R_4}{R_1+R_4} U_1 = \frac{10~\Omega}{5~\Omega+10~\Omega} \cdot 2~{\rm V} = 1.33~{\rm V}
 \end{align*} \end{align*}
-Case 2: For this case is $U_1 = 0~V$ and $U_3 = 0~V$. The voltage is at $R_3$.+Case 2: For this case is $U_1 = 0~{\rm V}$ and $U_3 = 0~{\rm V}$. The voltage is at $R_3$.
 <WRAP> {{drawio>Sloution4532}} </WRAP> <WRAP> {{drawio>Sloution4532}} </WRAP>
 \begin{align*} \begin{align*}
-U_{AB,2} = R_3 I_2 = 20~\Omega \cdot 1~A = 20~V+U_{\rm AB,2} = R_3 I_2 = 20~\Omega \cdot 1~{\rm A} = 20~{\rm V}
 \end{align*} \end{align*}
-Case 3: For this case is $U_1 = 0~V$ and $I_2 = 0~A$. The voltage comes from the source $U_3$.+Case 3: For this case is $U_1 = 0~{\rm V}$ and $I_2 = 0~{\rm A}$. The voltage comes from the source $U_3$.
 <WRAP> {{drawio>Sloution4533}} </WRAP> <WRAP> {{drawio>Sloution4533}} </WRAP>
 \begin{align*} \begin{align*}
-U_{AB,3} = 8~V+U_{\rm AB,3} = 8~{\rm V}
 \end{align*} \end{align*}
 Superposition means adding the voltages of all three cases. Superposition means adding the voltages of all three cases.
 \begin{align*} \begin{align*}
-U_{AB} = U_{AB,1} + U_{AB,2} + U_{AB,3} = 1.33~V + 20~V + 8~V+U_{\rm AB} = U_{\rm AB,1} + U_{\rm AB,2} + U_{\rm AB,3} = 1.33~{\rm V} + 20~{\rm V} + 8~{\rm V}
 \end{align*} \end{align*}
 </collapse> </collapse>
 <button size="xs" type="link" collapse="Loesung_4_5_3_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_3_2_Endergebnis" collapsed="true">  <button size="xs" type="link" collapse="Loesung_4_5_3_2_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_3_2_Endergebnis" collapsed="true"> 
 \begin{align*}  \begin{align*} 
-U_{AB} = 29.333... ~V \rightarrow 29.3 ~V \\ +U_{AB} = 29.333... ~{\rm V} \rightarrow 29.3 ~{\rm V} \\ 
 \end{align*}\\ \end{align*}\\
 </collapse> </collapse>