\section {Gleichungen in Matrizenschreibweise}
Die Spannungen am gegebenen Netzwerk
sollen mit Hilfe des Knotenpotenzialverfahrens
berechnet werden. Stellen Sie die Gleichungen
in Matrizenschreibweise auf.\\
$R=2\,\kilo\ohm;\quad C=5\,\nano\farad \quad L=2{,}5\,\milli\henry; $\\
$f=63{,}662\,\kilo\hertz;$\\
$\uline{I}_1=7\,\milli\ampere;\quad \uline{U}_2=3\,\volt\cdot  e^{j90\,\degree} \quad \uline{I}_3= 5\milli\ampere \cdot e^{-j90\,\degree}$
\begin{align*}
	\begin{tikzpicture}[scale=1.5]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spule | 			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);	
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};		
        \end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=2cm,rotate=90]%Spannungsquelle | 		
			\draw (0,0)--(1,0);
			\draw (.5,0)circle(.133);
			\draw [->,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_{2}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=3cm,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=0cm,yshift=2cm]%Widerstand
			 \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=2cm]%Kondensator -		
			\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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=3cm,yshift=0.5cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=2.5cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=.5cm,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=4cm,yshift=2.5cm,rotate=90]%Stromquelle 			
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0);
			\draw (.5,0)circle(.133);
			\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_3$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw  (0,.2)--(0,0)--(4,0)--(4,.7) (0,3.8)--(0,4)--(4,4)--(4,3.4)(4,1.2)--(4,2.7)
(3,0)--(3,.7)(3,1.2)--(3,2.7)(3,2)--(4,2)(3,3.2)--(3,4)(1,0)--(1,-.2)(.9,-.2)--(1.1,-.2);
        \fill(0,2)circle(0.05cm)(3,2)circle(0.05cm)(4,2)circle(0.05cm)(3,0)circle(0.05cm)
        (4,0)circle(0.05cm)(3,4)circle(0.05cm)(4,4)circle(0.05cm)(0,4)circle(0.05cm)(1,0)circle(0.05cm);
        \draw(0,4)--+(135:.5cm)(4,0)--+(225:.5cm);
        \draw[<-,red](0,4)--+(135:.25cm)node[above right]{$\uline{I}_1$};
        \draw[->,red](4,0)--+(225:.25cm)node[below right]{$\uline{I}_0$};
		\end{scope}	
        \draw[magenta](0,2)node[left]{$1$};
        \draw[magenta](4,2)node[right]{$3$};
        \draw[magenta](3,4)node[above]{$2$};
        \draw[magenta](3,0)node[below]{$0$};
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:
\begin{align*}
	\begin{tikzpicture}[scale=1.5]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.5cm,rotate=90]%Stromquelle 			
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0);
			\draw (.5,0)circle(.133);
			\draw [->,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Y}_{10}$};		
        \end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm,rotate=90]%Stromquelle 			
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0);
			\draw (.5,0)circle(.133);
			\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Y}_{12}$};		
        \end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Y}_{23}$};		
        \end{scope}
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm,rotate=90]%Stromquelle 			
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0);
			\draw (.5,0)circle(.133);
			\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_3$};
		\end{scope}                	
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Widerstand
			 \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [below] {$\uline{Y}_{13}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Y}_{30}$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw  (0,.7)--(0,0)--(4,0)--(4,.2) (0,1.3)--(0,2)--(4,2)--(4,1.8)(1,1)--(2,1)
(3,1)--(4,1);
        \draw[magenta](1,1)node[left]{$1$};
        \draw[magenta](4,1)node[right]{$3$};
        \draw[magenta](2.5,2)node[above]{$2$};
        \draw[magenta](2.5,0)node[below]{$0$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\begin{align*}
\left[\begin{array}{c}\uline{Y}\\ \end{array} \right]\cdot
\left[\begin{array}{c}\uline{U}\\ \end{array}\right]
&=\left[\begin{array}{c} -\sum{\uline{I}_q}\\ \end{array}\right]\\
\end{align*}
$\uline{I}_0\stackrel{!}{=}\uline{I}_1$ wegen Gleichgewicht. Abfließende Quellströme positiv.
\clearpage
\begin{align*}
G&=\frac{1}{R}=0{,}5\,\milli\siemens\\
\omega&=2\pi\cdot f=0{,}4\cdot \power{10}{-6}\cdot \frac{1}{\second}\\
X_C&=\frac{-1}{\omega C}=-0{,}5\,\kilo\ohm\\
\Rightarrow\quad B_C&=\omega C=2{,}0\,\milli\siemens\\
X_L&=\omega L=1{,}0\,\kilo\ohm\\
\Rightarrow\quad B_L&=\frac{-1}{\omega L}=-1{,}0\,\milli\siemens\\[\baselineskip]
\uline{Y}_{10}&=\frac{1}{R+jX_L}=\frac{1}{(2+j1)\,\kilo\ohm}&=(0{,}4-j0{,}2)\,\milli\siemens\\
\uline{Y}_{30}&=G+jB_C&=(0{,}5+j2{,}0)\,\milli\siemens\\
\uline{Y}_{13}&=\frac{1}{R+j(X_L+X_C)}=\frac{1}{[2+j(1-0{,}5)]\,\kilo\ohm}&=(0{,}4706-j0{,}1176)\,\milli\siemens\\
\uline{Y}_{12}&=jB_C=2\,\milli\siemens\cdot e^{j90\,\degree}&=(0+j2{,}0)\,\milli\siemens\\
\uline{Y}_{23}&=G&=(0{,}5+j0)\,\milli\siemens\\[\baselineskip]
\uline{I}_1&=7\,\milli\ampere \cdot e^{j0\,\degree}&=(7-j0)\,\milli\ampere\\
\uline{I}_2&=\uline{Y}_{12}\cdot \uline{U}_2=2\,\milli\siemens\cdot e^{j90\,\degree}\cdot 3\,\volt\cdot  e^{j90\,\degree}=6\,\milli\ampere \cdot e^{j180\,\degree}&=(-6+j0)\,\milli\ampere\\
\uline{I}_3&=5\,\milli\ampere \cdot e^{-j90\,\degree}&=(0-j5)\,\milli\ampere\\
\end{align*}
%\scriptsize
\begin{align*}
&\left(
  \begin{array}{ccc}
    \uline{Y}_{10}+\uline{Y}_{12}+\uline{Y}_{13} & -\uline{Y}_{12} & -\uline{Y}_{13} \\
    -\uline{Y}_{12} & \uline{Y}_{12}+\uline{Y}_{23} & -\uline{Y}_{23} \\
    -\uline{Y}_{13} & -\uline{Y}_{23} & \uline{Y}_{30}+\uline{Y}_{13}+\uline{Y}_{23} \\
  \end{array}
\right)
\left(
  \begin{array}{c}
    \uline{U}_{10} \\
    \uline{U}_{20} \\
    \uline{U}_{30} \\
  \end{array}
\right)=
\left(
  \begin{array}{c}
    \uline{I}_2\\
    \uline{I}_1-\uline{I}_2-\uline{I}_3\\
    \uline{I}_3\\
  \end{array}
\right)\\
\end{align*}
\scriptsize
\begin{align*}
&\left(
  \begin{array}{ccc}
    (0{,}8706+j1{,}6824) & (0-j2) & (-0{,}4706+j0{,}1176) \\
    (0-j2) & (0{,}5+j2{,}0) & (0{,}5+j0) \\
    (-0{,}4706-j0{,}1176) & (0{,}5+j0) & (1{,}4706+j1{,}8823) \\
  \end{array}
\right)\,\milli\siemens
\left(
  \begin{array}{c}
    \uline{U}_{10} \\
    \uline{U}_{20} \\
    \uline{U}_{30} \\
  \end{array}
\right)=
\left(
  \begin{array}{c}
    (6-j0)\\
    (13+j5)\\
    (0-j5)\\
  \end{array}
\right)\,\milli\ampere
\end{align*}
\normalsize
\clearpage
}{}%
