\section{Leitwert}
Der Widerstand des abgebildeten Netzwerkes soll $\uline{Z}=1\,\kilo\ohm\cdot e^{j60\,\degree}$ sein.\\[\baselineskip]
Wie groß müssen $R$ und $B_C$ sein, wenn $B_L=-3{,}33\,\milli\siemens$ ist?
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\text{...noch einfügen...}
%\end{align}
\begin{align*}
	\begin{tikzpicture}[scale=3]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%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] {$R$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator | 			
			\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$B_C$};
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spule | 			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$B_L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);	
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. 
			\draw (.8,1)--(2,1)--(2,.9) (0,0)--(2,0)--(2,.2);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%End Knoten 
			\fill (0,0)circle(.02); 
			\fill (0,1)circle(.02);					
			\draw node at (0,.5)[left] {$\uline{Z}\Rightarrow$};
		\end{scope}									
	\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Berechnung:}
\uline{Z}&\stackrel{!}{=}\power{10}{3}\,\ohm\cdot e^{j60\,\degree}\quad\,=(500+j866{,}25)\,\ohm\\
\uline{Z}&=R+\frac{1}{j(B_C+B_L)}=\,\,\, R\,\,\, -j\frac{1}{(B_C+B_L)}\\[\baselineskip]
\intertext{$\Re$} 
R&=\uuline{500\,\ohm}\\
\intertext{$\Im$}
B_C+B_L&=\frac{-1}{866{,}25}\,\siemens=-1{,}1547\,\milli\siemens\\
B_C&=-1{,}1547\,\milli\siemens -B_L=-1{,}1547\,\milli\siemens +3{,}33\,\milli\siemens  =\uuline{2{,}175\,\milli\siemens}\\
\end{align*}
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