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--- electrical_engineering_2:task_1.9.3_with_calculation [2023/03/10 11:33] – mexleadmin
+++ electrical_engineering_2:task_1.9.3_with_calculation [Unbekanntes Datum] (aktuell) – gelöscht - Externe Bearbeitung (Unbekanntes Datum) 127.0.0.1
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-<panel type="info" title="Task 1.9.3: layered plate capacitor (exam task, ca 6% of a 60 minute exam, WS2020)"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
-<WRAP right>
-{{elektrotechnik_1:schaltung_klws2020_3_1_1.jpg?200}}
-</WRAP>
-Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data:
-  * rectangular electrodes with edge length of $6 ~cm$ and $8 ~cm$.
-  * distance between the plates: $2 ~mm$
-  * dielectric A:
-    * $\varepsilon_{r,A} = 1 \:\:(air)$
-    * thickness $d_A = 1.5 ~mm$
-  * Dielectric B:
-    * $\varepsilon_{r,B} = 100 \:\:(ice)$
-    * thickness $d_B = 0.5 ~mm$
-$\varepsilon_{0} = 8.854 \cdot 10^{-12}  ~F/m$
-<button size="xs" type="link" collapse="Loesung_5_9_3_1_Tipps">{{icon>eye}} Tips for the solution</button><collapse id="Loesung_5_9_3_1_Tipps" collapsed="true">
-  * Which circuit can be used to replace a layered structure with different dielectrics?
-</collapse>
-<button size="xs" type="link" collapse="Loesung_5_9_3_1_Lösungsweg">{{icon>eye}} Solution</button><collapse id="Loesung_5_9_3_1_Lösungsweg" collapsed="true">
-The total capacitance $C$ can be divided into a partial capacitance $C_A$ and a $C_B$. These are connected in series. \\
-This results in: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\
-The partial capacitance $C_A$ can be calculated by
-\begin{align*}
-C_A &= \varepsilon_{0} \varepsilon_{r,A} \cdot \frac{A}{d_A} && | \text{with } A = 3 ~cm \cdot 5cm = 6 \cdot 10^{-2} \cdot 8 \cdot 10^{-2} ~m^2 = 48 \cdot 10^{-4} ~m^2\\
-C_A &= 8.854 \cdot 10^{-12}  ~F/m \cdot \frac{48 \cdot 10^{-4} m^2}{1.5 \cdot 10^{-3} ~m} \\
-C_A &= 28.33 \cdot 10^{-12} ~F \\
-\end{align*}
-The partial capacitance $C_B$ can be calculated by
-\begin{align*}
-C_B &= \varepsilon_{0} \varepsilon_{r,B} \cdot \frac{B}{d_B} \\
-C_B &= 100 \cdot 8.854 \cdot 10^{-12}  ~F/m \cdot \frac{48 \cdot 10^{-4} ~m^2}{0,5 \cdot 10^{-3} ~m} \\
-C_B &= 8.500 \cdot 10^{-9} ~F \\
-\end{align*}
-</collapse>
-<button size="xs" type="link" collapse="Loesung_5_9_3_1_Endergebnis">{{icon>eye}} Result</button><collapse id="Loesung_5_9_3_1_Endergebnis" collapsed="true">
-\begin{align*}
-C = 28.24 \cdot 10^{-12} ~F \rightarrow 28~pF
-\end{align*}
- \\
-</collapse>
-</WRAP></WRAP></panel>