\section{Zeigerdiagramm}
Gegeben ist die Ausgangsspannung $U_a = 5\,\volt \cdot e^{j0\degree}$ und $R_1=R_2=\frac{1}{\omega C_1}=\frac{1}{\omega C_2}=1\,\kilo\ohm$\\
Zeichnen Sie ein maßstäbliches Zeigerdiagramm aller Spannungen und aller Ströme!\\
Maßstäbe: $1\,\centi\metre\,\widehat{=}\, 1\,\volt \text{;}\quad 1\,\centi\metre\,\widehat{=}\, 1\,\milli\ampere$
Entnehmen Sie dem Zeigerdiagramm Betrag und Phasenwinkel der Spannung $U_e$ !
\begin{align*}
	\begin{tikzpicture}[very thick,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] {$R_1$};
			\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$};
		\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] {$C_1$};
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R_2$};
			\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,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] {$C_2$};
		\end{scope}								
		\begin{scope}[>=latex,very thick]%Knotenpunkte
			\draw (0,0)--(2.5,0) (2,1)--(2.5,1);		
			\fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025); 
			\draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$};	
			\draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$};					
		\end{scope}			
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
e^{j\varphi}&=\cos\varphi+j\sin\varphi \quad\text{Eulersche Formel}
\end{align}
\footnotesize{R-C Ketten sind u.a. ein Ersatzbild für Leitungen (Kapazität pro Längeneinheit)}\\
Berechnung:
\begin{align*}
	\begin{tikzpicture}[very thick,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] {$R_1$};
			\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$};
		\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] {$C_1$};
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R_2$};
			\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,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] {$C_2$};
		\end{scope}								
		\begin{scope}[>=latex,very thick]%Knotenpunkte
			\draw (0,0)--(2.5,0) (2,1)--(2.5,1);		
			\fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025); 
			\draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$};	
			\draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$};					
			\draw [->,red] (.05,1.1)--(.25,1.1) node at (.1,1.1)[above]{$\underline{I}_{e}$};			
			\draw [->,red] (2.2,1.1)--(2.4,1.1) node at (2.3,1.1)[above]{$\underline{I}_{a}=0$};
			\draw [->,red] (1.1,.9)--(1.1,.7) node at (1.1,.8)[right]{$\underline{I}_{1}$};	
			\draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{2}$};					
		\end{scope}
		\begin{scope}[>=latex,thick,xshift=.4cm,yshift=.5cm]%Masche			
		   \draw [->,red!50!blue] (270:.15)arc(270:-60:.15) node at (0,0){$M_1$};
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}
\underline{U}_a&=5\,\volt\cdot e^{j0\degree}\\
\underline{I}_a&=0\text{, da kein Lastwiderstand angeschlossen ist!}\\
\underline{I}_2&=\frac{\underline{U}_a}{\underline{jX}_2}
=\frac{5\,\volt\cdot e^{j0\degree}}{1\kilo\ohm\cdot e^{-j90\degree}}
=5\,\milli\ampere\cdot e^{j90\degree}=j5\,\milli\ampere
\quad\text{(Strom eilt vor)}\quad 5\,\milli\ampere\;\angle +90\degree\\
\underline{U}_{R_2}&=R_2\cdot \underline{I}_2=1\,\kilo\ohm\cdot e^{-j90\degree}=1\,\kilo\ohm\cdot 5\,\milli\ampere\;\angle +90\degree=5\,\volt\;\angle +90\,\degree\\
\underline{U}_1&=\underline{U}_a+\underline{U}_{R_2}=5\,\volt+j5\,\volt\quad\text{(Vektoren addieren)}\quad =\sqrt{2}\cdot 5\,\volt\;\angle +45\degree \sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}\\\underline{I}_1&=\frac{\underline{U}_1}{\underline{jX}_1}=\frac{\sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}}{ 1\kilo\ohm\cdot e^{-j90\degree}}=\sqrt{2}\cdot 5\,\milli\ampere\cdot e^{j135\degree}=5\cdot (-1+j)\,\milli\ampere\quad\text{(Strom eilt $90\degree$ vor) } 5\,\milli\ampere\;\angle +135\degree\\
\underline{I}_e&=\underline{I}_1+\underline{I}_2=\sqrt{2}\cdot 5\,\milli\ampere\;\angle +135\degree+5\,\milli\ampere\;\angle +90\degree\quad\text{(Vektoren addieren)}\\[\baselineskip]
\end{align*}
\begin{align*}
\intertext{Jetzt zeichnen oder rechnerisch: (jedoch aufwendiger)}
\underline{I}_e&=I_1\cdot (\cos\varphi+j\sin\varphi)+I_2\cdot (\cos\varphi+ j\sin\varphi)\\
&=I_1\cdot (\cos135\degree+j\sin 135\degree)+I_2\cdot (\cos 90\degree+ j\sin 90\degree)\\
&=|\underline{I}_1|\cdot (-\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})+|\underline{I}_2|\cdot (0+j)\\ 
&=[\cancel{\sqrt{2}}\cdot 5\cdot (-\frac{1}{\cancel{\sqrt{2}}}+j\frac{1}{\cancel{\sqrt{2}}})+j\cdot 5]\,\milli\ampere=(-5+j10\,\milli\ampere)\\
&\quad |\underline{I}_e|= \sqrt{10^2+5^2}\,\milli\ampere=11{,}18\,\milli\ampere \\
&\quad\tan\varphi=\frac{\Im}{\Re}=\tan\frac{10}{-5}=\tan-2\Rightarrow\varphi=\arctan\frac{-2}=-1{,}107\,\rad\,\widehat{=}\,-63{,}435\degree\\
&\quad\text{(Definitionsbereich $\tan\varphi [-\pi/2\cdots\pi/2]$ beachten!)}\\
&=11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere\\[\baselineskip]
\underline{U}_{R_1}&=R_1\cdot \underline{I}_e\\
&\quad\text{(Nur zur Vollständigkeit) }\underline{U}_{R_1}=1\,\kilo\ohm\cdot 11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere=11{,}18\cdot e^{j116{,}565\degree}\,\volt\\
\underline{U}_e&=\underline{U}_{R_1}+\underline{U}_1=\uuline{15\,\volt\cdot e^{+j90\degree}}\quad\text{(Vektoren addieren)}
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=.5]
		\begin{scope}[>=latex] 
			\draw [very thin,black!50!](-5,0)grid(5,15);						
		\end{scope}
		\begin{scope}[>=latex,very thick] 			
			\draw [->,blue] (0:0)--(90:15) node at (90:7.5)[right]{$\underline{U}_e$};	
			\draw [->,blue] (0:0)--(0:5) node at (0:2.5)[below]{$\underline{U}_a$};							
			\draw [->,blue] (0:0)--(45:7.07) node at (45:3.54)[right]{$\underline{U}_1$};
			\draw [->,blue,ultra thick] (0:0)--(-5,10) node at (-2.5,5)[right]{$\underline{U}_{R_1}$};
			\draw [->,blue] (-5,10)--(0,15) node at (-2.5,12.5)[left]{$\underline{U}_{1}$};
			\draw [->,blue,ultra thick] (0:0)--(90:5) node at (90:3.5)[right]{$\underline{U}_{R_2}$};	
			\draw [->] (5,0)--(5,5) node at (5,3.5)[right]{$\underline{U}_{R_2}$};											
			\draw [->,red] (0:0)--(135:7.07) node at (135:3.54)[left]{$\underline{I}_1$};
			\draw [->,red,thick] (0:0)--(90:5) node at (90:3.5)[left]{$\underline{I}_2$};
			\draw [->] (135:7.07)--+(90:5) node at (-5,7.5)[right]{$\underline{I}_2$};
			\draw [->,red,thick] (0:0)--(-5,10) node at (-2.5,5)[left]{$\underline{I}_{e}$};								
		\end{scope}
			\draw node at (8,10)[right]{Reihenfolge$U_a\, I_2, U_{R_2},U_1, I_1, I_e, U_{R_1})$};
			\draw node at (8,9)[right]{$U_a\widehat{=}\,5\,\centi\metre\angle 0\,\degree\quad(5+j0)$};
			\draw node at (8,8)[right]{$U_{R_2}\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$};	
			\draw node at (8,7)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(5+j5)$};
			\draw node at (8,6)[right]{$I_1\widehat{=}\,7{,}07\,\centi\metre\angle 135\,\degree\quad(-5+j5)$};
			\draw node at (8,5)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$};
			\draw node at (8,4)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(-5+j5)\rightarrow(0+j5)$ addiert zu $I_1$};
			\draw node at (8,3)[right]{$I_e\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree(-5+j10)$};
			\draw node at (8,2)[right]{$U_{R_1}\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree\quad(-5+j10)$};
			\draw node at (8,1)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(-5+j10)\rightarrow(5+j5)$ addiert zu $U_{R_1}$};		
	\end{tikzpicture}
\end{align*}
\begin{align*}
\sum M_1=0=U_{R_1}+U_1-U_e\\
U_e=15\,\volt \cdot e^{j90\degree}
\end{align*}
\clearpage
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