\section{Zeigerdiagramm Netzwerk}
Zeichnen Sie für das abgebildete Netzwerk ein maßstäbliches Zeigerdiagramm aller Spannungen und Ströme.\\
Wie groß muß der Widerstand $R_1$ sein damit der Strom $I$ der Spannung $U$ um $30\,\degree$ nacheilt?\\[\baselineskip] 
$\varphi_u - \varphi_i = 30\,\degree$\\
$R_2 = |X_L| = |X_C| = 10\,\kilo\ohm$\\
$\uline{I}_C = 1\,\milli\ampere\cdot e^{j90\,\degree}$\\[\baselineskip]
\uline{Maßstäbe:}\\
$1\,\volt \,\widehat{=}\,0{,}8\centi\metre$\\
$1\,\milli\ampere \,\widehat{=}\,5\centi\metre$
\begin{align*}
	\begin{tikzpicture}[very thick,scale=3]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule -			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\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] {$C$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick]%Knotenpunkte
			\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(2,0)--(2,.2) (.8,1)--(2,1)--(2,.8);
			\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$};	
			\fill (-.5,0)circle(.025) (-.5,1)circle(.025);
			\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$};	
			\draw [->,red] (1.1,.9)--(1.1,.6) node at (1.1,.75)[right]{$\underline{I}_C$};
			\draw [->,black!50!] (1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$\underline{U}_{R_2}$};
			\draw [->,black!50!] (1.4,1)--(1.6,1) node at (1.5,1)[above]{$\underline{I}_2$};
			\draw [->,black!50!] (.3,.9)--(.7,.9) node at (.5,.9)[below]{$\underline{U}_L$};
			\draw [->,black!50!] (.75,1)--(.95,1) node at (.85,1)[above]{$\underline{I}_L$};				
			\draw [->,black!50!] (0,.95)--(0,.75) node at (0,.85)[left]{$\underline{I}_{R_1}$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align*}
\intertext{Berechnung:  (Platz in $x=\pm10\,\centi\metre$ und $x=10\,\centi\metre$}
R_2=|X_C|&=10\,\kilo\ohm\text{ (Stomteiler, mit gleichem Betrag des Stroms)}\\	
\text{mit }|\uline{I}_{R_2}| &=|\uline{I}_C|=1\,\milli\ampere\\
\uline{I}_C &= 1\,\milli\ampere\cdot e^{j90\,\degree}\Rightarrow\uline{I}_{R_2}=1\,\milli\ampere\cdot e^{j0\,\degree}\text{ $(\uline{I}_C$ eilt vor)}\\
\uline{U}_{R_2}&=R_2\cdot \uline{I}_{R_2}=10\,\cancel{\kilo}\ohm\cdot 1\,\cancel{\milli}\ampere\cdot e^{j0\,\degree}=10\,\volt\cdot e^{j0\,\degree}\,\widehat{=}\,8\centi\metre\\
\uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C =1\,\milli\ampere\cdot e^{j0\,\degree}+1\,\milli\ampere\cdot e^{j90\,\degree}=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\\
\uline{U}_L&=\uline{I}_L\cdot j\cdot X_L=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\cdot j10\,\kilo\ohm =14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\,11{,}3\centi\metre\text{ $(\uline{U}_L$ voreilend)}\\
\uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}=14{,}1\,\volt\cdot e^{j135\,\degree}+10\,\volt\cdot e^{j0\,\degree}
=(-10+j10+10)\,\volt=10\,\volt\cdot e^{j90\,\degree}\\
\text{Zeichnen: }\varphi_u-\varphi_i&=30\,\degree\text{ deshalb $30\,\degree$, Linie zeichnen, Schnittpunkt mit } \uline{I}_L+\uline{I}_{R_1}\Rightarrow\uline{I}=2\,\milli\ampere\cdot e^{j30\,\degree}\\
\text{Ablesen: }\qquad\,\,\uline{I}_{R_1}&=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\
R_1&=\frac{\uline{U}}{\uline{I}_{R_1}}=\frac{10\,\volt\cancel{\cdot e^{j90\,\degree}}}{0{,}72\,\milli\ampere\cancel{\cdot e^{j90\,\degree}}}=\uuline{13{,}89\,\kilo\ohm}\\
\end{align*}	
\begin{align*}	
	\begin{tikzpicture}[very thick,scale=.7]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
		  \draw [black!15!,very thin](-10,0)grid(10,10);
			\draw [->,black!25!](-10,0)--(10.5,0);
			\draw [->,black!25!](0,0)--(0,10.5);
			\draw node at (10.5,0)[right]{$\Re$};
			\draw node at (0,10.5)[above]{$\Im$};	
		  \draw [black!75!,thick](0:0)--(60:11)node [right]{$30\,\degree$ Linie};
			\draw [->,blue] (0,0)--(90:8)node [below right]{$\underline{U}$};
			\draw [->,red] (0,0)--(90:5)node [below right]{$\underline{I}_C$};
			\draw [->,red] (0,0)--(0:5)node [above left]{$\underline{I}_R$};
			\draw [->,red] (0,0)--(45:7.07)node [below right]{$\underline{I}_L$};
			\draw [->,red!50!blue] (5,5)--(5,8.65)node [below right]{$\underline{I}_{R_1}=0{,}72\,\milli\ampere\,\widehat{=}\,3{,}6\centi\metre$};
		  \draw [->,red!50!blue](0:0)--(60:10)node at (3,7.8) [above]{$\uline{I}=2\,\milli\ampere$};							
			\draw [->,blue] (0,0)--(135:11.28)node [below left]{$\underline{U}_L$};
			\draw [->,blue] (-8,8)--(0,8)node [above left]{$\underline{U}_{R_2}$};
			\draw [->,black!75!](0:0)+(90:2)arc(90:60:2)node at(.6,2)[above]{$30\,\degree$};					
		\end{scope}				
	\end{tikzpicture}
\end{align*}
\begin{align*}
\uline{I}_{R_2}&=1\,\milli\ampere\cdot e^{j0\,\degree} \,\widehat{=}\ 5\,\centi\metre\\
\uline{I}_{C}&=1\,\milli\ampere\cdot e^{j90\,\degree} \,\widehat{=}\ 5\,\centi\metre\\
\uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C \,\widehat{=}\ 7{,}05\,\centi\metre\\
\uline{U}_{L}&=14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\ 11{,}3\centi\metre \quad \uline{U}_{L}\bot \uline{I}_{L}\\
\uline{R}_{R_2}&=10\,\volt\cdot e^{j0\,\degree} \,\widehat{=}\ 8\centi\metre\\
\uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}\\
&\text{Gerade für I, $30\degree $ nacheilend}\\
\uline{I}_{L}&+\uline{I}_{R_1}=\uline{I};\qquad \uline{I}_{R_1}||\uline{U}\\
\rightarrow\ &\text{ ablesen } 3{,}6\,\centi\metre\ \widehat{=}\ \uline{I}_{R_1}=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\
\end{align*}
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