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