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electrical_engineering_2:task_1.9.3_with_calculation [2022/03/10 12:31] tfischer ↷ Seitename wurde von electrical_engineering_2:task_5.9.3_with_calculation auf electrical_engineering_2:task_1.9.3_with_calculation geändert |
electrical_engineering_2:task_1.9.3_with_calculation [2023/11/30 00:03] mexleadmin |
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- | <panel type=" | + | <panel type=" |
<WRAP right> | <WRAP right> | ||
- | {{elektrotechnik_1: | + | {{drawio> |
</ | </ | ||
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 |
- | * $\varepsilon_{r, | + | * $\varepsilon_{\rm r,A} = 1 |
- | * thickness $d_A = 1.5 mm$ | + | * thickness $d_{\rm A} = 1.5 ~{\rm mm}$ |
- | * Dielectric B: | + | * Dielectric |
- | * $\varepsilon_{r, | + | * $\varepsilon_{\rm r,B} = 100 \:\: \rm (ice)$ |
- | * thickness $d_B = 0.5 mm$ | + | * thickness $d_{\rm B} = 0.5 ~{\rm mm}$ |
- | $\varepsilon_{0} = 8.854 \cdot 10^{-12} | + | $\varepsilon_{0} = 8.854 \cdot 10^{-12} |
<button size=" | <button size=" | ||
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<button size=" | <button size=" | ||
- | 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, | + | C_{\rm A} &= \varepsilon_{0} \varepsilon_{r, |
- | C_A &= 8.854 \cdot 10^{-12} | + | &= 8.854 \cdot 10^{-12} |
- | C_A &= 28.33 \cdot 10^{-12} F \\ | + | &= 28.33 \cdot 10^{-12} |
\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, | + | C_{\rm B} &= \varepsilon_{0} \varepsilon_{r, |
- | C_B &= 100 \cdot 8.854 \cdot 10^{-12} | + | &= 100 \cdot 8.854 \cdot 10^{-12} |
- | C_B &= 8.500 \cdot 10^{-9} F \\ | + | &= 8.500 \cdot 10^{-9} |
\end{align*} | \end{align*} | ||
Zeile 44: | Zeile 44: | ||
<button size=" | <button size=" | ||
\begin{align*} | \begin{align*} | ||
- | C = 28.24 \cdot 10^{-12} F \rightarrow | + | C = 28.24 \cdot 10^{-12} |
\end{align*} | \end{align*} | ||
\\ | \\ |