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electrical_engineering_2:task_1.9.3_with_calculation [2022/03/10 12:31]
tfischer 		electrical_engineering_2:task_1.9.3_with_calculation [2023/11/30 00:03]
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-<panel type="info" title="Aufgabe 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%>+<panel type="info" title="Task 1.9.6: 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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 Determine the capacitance $C$ for the plate capacitor drawn on the right with the following data:  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 $8 cm$. +  * rectangular electrodes with edge length of $6 ~{\rm cm}$ and $8 ~{\rm cm}$. 
-  * distance between the plates: $2 mm$ +  * distance between the plates: $2 ~{\rm mm}$ 
-  * dielectric A:  +  * dielectric ${\rm A}$:  
-    * $\varepsilon_{r,A} = 1 \:\:(air)$ +    * $\varepsilon_{\rm r,A} = 1   \:\:\rm (air)$ 
-    * thickness $d_A = 1.5 mm$ +    * thickness $d_{\rm A}   = 1.5 ~{\rm mm}$ 
-  * Dielectric B:  +  * Dielectric ${\rm B}$:  
-    * $\varepsilon_{r,B} = 100 \:\:(ice)$ +    * $\varepsilon_{\rm r,B} = 100 \:\: \rm (ice)$ 
-    * thickness $d_B = 0.5 mm$+    * thickness $d_{\rm B}   = 0.5 ~{\rm mm}$
  
-$\varepsilon_{0} = 8.854 \cdot 10^{-12}  F/m$+$\varepsilon_{0} = 8.854 \cdot 10^{-12}  ~{\rm F/m}$
  
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-The total capacitance $C$ can be divided into a partial capacitance $C_A$ and a $C_B$. These are connected in series. \\ +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: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\ This results in: $C = \frac{C_A \cdot C_B}{C_A + C_B}$ \\ \\
  
 The partial capacitance $C_A$ can be calculated by  The partial capacitance $C_A$ can be calculated by 
 \begin{align*} \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_{\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\\ 
-C_A &= 8.854 \cdot 10^{-12}  F/m \cdot \frac{48 \cdot 10^{-4} m^2}{1.5 \cdot 10^{-3} m} \\ +          &= 8.854 \cdot 10^{-12} ~{\rm F/m}   \cdot \frac{48 \cdot 10^{-4} ~{\rm m}^2}{1.5 \cdot 10^{-3} ~{\rm m}} \\ 
-C_A &= 28.33 \cdot 10^{-12} F \\+          &= 28.33 \cdot 10^{-12} ~{\rm F} \\
 \end{align*} \end{align*}
  
 The partial capacitance $C_B$ can be calculated by  The partial capacitance $C_B$ can be calculated by 
 \begin{align*} \begin{align*}
-C_B &= \varepsilon_{0} \varepsilon_{r,B} \cdot \frac{B}{d_B} \\ +C_{\rm 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} \\ +          &= 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}} \\ 
-C_B &= 8.500 \cdot 10^{-9} F \\+          &= 8.500 \cdot 10^{-9} ~{\rm F} \\
 \end{align*} \end{align*}
  
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 \begin{align*} \begin{align*}
-C = 28.24 \cdot 10^{-12} F \rightarrow 28pF+C = 28.24 \cdot 10^{-12} ~{\rm F} \rightarrow 28~{\rm pF}
 \end{align*} \end{align*}
  \\  \\