91 lines
4.3 KiB
TeX
91 lines
4.3 KiB
TeX
\section{Dualitätskonstante}
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Berechnen Sie zu der gegebenen Schaltung die duale Schaltung mit der Dualitätskonstanten.\\[\baselineskip]
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$R^2_D=10000\,\ohm^2$\\
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$R_1=80\,\ohm$\\
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$L_1=50\,\milli\henry$\\
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$C_1=10\,\micro\,\farad$
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\begin{align*}
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\begin{tikzpicture}[scale=2]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Kondensator -
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.5cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm]%Spule -
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\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_1$};
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\fill (.3,-0.0667)rectangle(.7,0.0667);
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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\draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
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\fill(0,0)circle(.05) (3,0)circle(.05);
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\ifthenelse{\equal{\toPrint}{Lösung}}{%
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\begin{align}
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\intertext{Formeln:}
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\uline{Z}_1(\omega)&=R^2_D \cdot \uline{Y}_2(\omega)
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\end{align}
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Berechnung:\\[\baselineskip]
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Parallel $\Leftrightarrow$ Serie\\
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Leitwert $\Leftrightarrow$ Widerstand\\
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Kapazität $\Leftrightarrow$ Induktivität\\
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Parallelschaltung $R_1||L_1$ in Serienschaltung $R_2+C_2$
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Serienschaltung $C_1+(R_2+C_2)$ in Parallelschaltung $L_2 ||(R_2+C_2)$
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\begin{align*}
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\begin{tikzpicture}[scale=2]
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\draw node at (0,0){\phantom{.}};
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\draw node at (0,1.5){Zwischenschritt:};
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0.5cm]%Kondensator -
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0.5cm]%Kondensator -
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_2$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0.5cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0.5cm]
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% \draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
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% \draw (0,.5)--(.5,.5) (.5,.5)--(.5,.5)--(.5,-.5)--(2,-.5)
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% (2.5,.5)--(2.5,-.5)--(2,-.5)(2.5,.5)--(3,.5);
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\fill(0,0)circle(.05) (3,0)circle(.05);
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\end{scope}
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\end{tikzpicture}
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\begin{tikzpicture}[scale=2]
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\draw node at (0,1.0){Duale Schaltung:};
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\begin{scope}[>=latex,very thick,xshift=.5cm,yshift=.5cm]%Kondensator -
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_2$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=.5cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm]%Spule -
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\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_2$};
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\fill (.3,-0.0667)rectangle(.7,0.0667);
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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% \draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
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\draw (0,.5)--(.5,.5) (.5,.5)--(.5,.5)--(.5,-.5)--(2,-.5)
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(2.5,.5)--(2.5,-.5)--(2,-.5)(2.5,.5)--(3,.5);
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\fill(0,.5)circle(.05) (3,.5)circle(.05);
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\begin{align*}
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R_1\cdot R_2&=R^2_D\Rightarrow R_2=\frac{R^2_D}{R_1}=\frac{10000\,\ohm^2}{80\,\ohm}=\uuline{125\,\ohm}\\
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\text{(C zu L) aus: }\frac{L_2}{C_1}&=R^2_D\Rightarrow L_2=R^2_D\cdot C_1
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=\power{10}{4}\,\ohm^{{\cancel{2}}}
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\cdot \power{10}{-5}\,\frac{\second}{\cancel{\ohm}}
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=0{,}1\,\frac{\volt\second}{\ampere}
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=\uuline{100\,\milli\henry}\\
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\text{(L zu C) aus: }\frac{L_1}{C_2}&=R^2_D\Rightarrow C_2=\frac{L_1}{R^2_D}
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=\frac{50\,\milli\henry}{10000\,\ohm^2}
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=5\cdot \power{10}{-6}\,\frac{\cancel{\volt}\second}{\cancel{\ampere}}
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\cdot \frac{\ampere^{\cancel{2}}}{\volt^{\cancel{2}}}
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=\uuline{5\,\micro\farad}
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\end{align*}
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\clearpage
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}{}%
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