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--- electrical_engineering_1:aufgabe_4.5.3 [2021/10/17 03:10]
tfischer [Bearbeiten - Panel]
+++ electrical_engineering_1:aufgabe_4.5.3 [2023/03/23 14:16] (aktuell)
mexleadmin
@@ Zeile 1: / Zeile 1: @@
-<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>
 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.
-<button size="xs" type="link" collapse="Loesung_4_5_3_1_Endergebnis">{{icon>eye}} Final value</button><collapse id="Loesung_4_5_3_1_Endergebnis" collapsed="true"> \begin{align*} U_{AB} = 29.333... V \rightarrow 29.3 V \\ \end{align*}\\
+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">
+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>
+\begin{align*}
+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*}
+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>
+\begin{align*}
+U_{\rm AB,2} = R_3 I_2 = 20~\Omega \cdot 1~{\rm A} = 20~{\rm V}
+\end{align*}
+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>
+\begin{align*}
+U_{\rm AB,3} = 8~{\rm V}
+\end{align*}
+Superposition means adding the voltages of all three cases.
+\begin{align*}
+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*}
+</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">
+\begin{align*}
+U_{AB} = 29.333... ~{\rm V} \rightarrow 29.3 ~{\rm V} \\
+\end{align*}\\
 </collapse>