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electrical_engineering_2:task_1.9.3_with_calculation [2023/03/24 12:05] – mexleadminelectrical_engineering_2:task_1.9.3_with_calculation [Unknown date] (current) – removed - external edit (Unknown date) 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%> 
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-{{drawio>electrical_engineering_2:schaltung_klws2020_3_1_1.svg}} 
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-Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data:  
-  * rectangular electrodes with edge length of $6 ~{\rm cm}$ and $8 ~{\rm cm}$. 
-  * distance between the plates: $2 ~{\rm mm}$ 
-  * dielectric ${\rm A}$:  
-    * $\varepsilon_{\rm r,A} = 1   \:\:\rm (air)$ 
-    * thickness $d_{\rm A}   = 1.5 ~{\rm mm}$ 
-  * Dielectric ${\rm B}$:  
-    * $\varepsilon_{\rm r,B} = 100 \:\: \rm (ice)$ 
-    * thickness $d_{\rm B}   = 0.5 ~{\rm mm}$ 
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-$\varepsilon_{0} = 8.854 \cdot 10^{-12}  ~{\rm F/m}$ 
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-<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> 
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-<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"> 
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-The total capacitance $C$ can be divided into a partial capacitance $C_{\rm A}$ and a $C_{\rm B}$. These are connected in series. \\  
-This results in: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\ 
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-The partial capacitance $C_A$ can be calculated by  
-\begin{align*} 
-C_{\rm A} &= \varepsilon_{0} \varepsilon_{r,A} \cdot \frac{A}{d_A} && | \text{with } A = 3 ~{\rm cm} \cdot 5~{\rm cm} = 6 \cdot 10^{-2} \cdot 8 \cdot 10^{-2} ~{\rm m}^2 = 48 \cdot 10^{-4} ~{\rm m}^2\\ 
-          &= 8.854 \cdot 10^{-12} ~{\rm F/m}   \cdot \frac{48 \cdot 10^{-4} ~{\rm m}^2}{1.5 \cdot 10^{-3} ~{\rm m}} \\ 
-          &= 28.33 \cdot 10^{-12} ~{\rm F} \\ 
-\end{align*} 
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-The partial capacitance $C_B$ can be calculated by  
-\begin{align*} 
-C_{\rm B} &= \varepsilon_{0} \varepsilon_{r,B}          \cdot \frac{B}{d_B} \\ 
-          &= 100 \cdot 8.854 \cdot 10^{-12}  ~{\rm F/m} \cdot \frac{48 \cdot 10^{-4} ~{\rm m}^2}{0,5 \cdot 10^{-3} ~{\rm m}} \\ 
-          &= 8.500 \cdot 10^{-9} ~{\rm F} \\ 
-\end{align*} 
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-</collapse> 
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-<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} ~{\rm F} \rightarrow 28~{\rm pF} 
-\end{align*} 
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