\section {Impedanzmatrix}
%\enlargethispage{3cm}
Geben Sie die Impedanzmatrix $(Z)$ des Vierpols an.\\
$R_1=10\,\ohm$, $X_1=100\,\ohm$, $R_2=20\,\ohm$, $X_2=200\,\ohm$, $X_C=-200\,\ohm$, $X_M=120\,\ohm$\\
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
			\draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
			\draw (.5,0)circle(.133);	
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_1$};			
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
			\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spule | 			 
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,.0667) [left] {$X_1$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);	
            \fill(.825,.075)circle(.033);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule | 			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_2$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);	
            \fill(.825,-.075)circle(.033);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
			\draw [->,red] (1.5,.1)--(1.25,.1)node at(1.375,.1)[above]{\footnotesize$\uline{I}_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2.5cm,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) [left] {$X_C$};
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(1,0)--(1,.2) (3,0)--(1.5,0)--(1.5,.2) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9)(3,1)--(2.5,1);
            \fill (2.5,0)circle(0.025cm)(2.5,1)circle(0.025cm)(3,1)circle(0.025cm)(3,0)circle(0.025cm);
			\draw [->,blue] (3,.8)--(3,.2) node at (3,.5)[right]{$\uline{U}_2$};
        \draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$X_M$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\
Umwandlung in T-Ersatzschaltbild\\
%\begin{enumerate}
%\item Offener Schalter\\
%\begin{align*}
%\uline{I}_2&=0\quad\text{Wegen Wicklungssinn $\uline{U}_2$ negativ (bzw. $X_M$ negativ)}\\
%\uline{U}_2&=jX_M\cdot \uline{I}_1=jX_M\cdot \frac{\uline{U}_1}{(R+jX_1)}=\frac{(-j40)\,\ohm\cdot 1\,\volt}{(10+j100)\,\ohm}=(0{,}396-j0{,}04)\,\volt\\
%&=\uuline{0{,}398\,\volt\cdot e^{-j174{,}3\,\degree}}\\
%\end{align*}
%\clearpage
%\item Ersatzschaltbild\\
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
			\draw node at(.5,-.2){\footnotesize$10$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Spule -			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_1-X_M$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw node at(.5,-.2){\footnotesize$-j20$};				
        \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) [left] {$X_M$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);	
			\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$U_M$};
			\draw node at(.25,.25){\footnotesize$j120$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_2-X_M$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw node at(.5,-.2){\footnotesize$+j80$};				
        \end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
			\draw node at(.5,-.2){\footnotesize$20$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=4cm,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] {$X_C$};
 			\draw node at(.25,-.2){\footnotesize$-j200$};
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(4,0)--(4,.2) (3.8,1)--(4,1)--(4,.8)(4,0)--(4.5,0)(4,1)--(4.5,1);
            \fill (0,0)circle(0.025cm)(0,1)circle(0.025cm)(4.5,0)circle(0.025cm)(4.5,1)circle(0.025cm);
			\draw [->,blue] (0,.7)--(0,.3) node at (0,.5)[left]{$\uline{U}_1$};	
			\draw [->,blue] (4.5,.7)--(4.5,.3) node at (4.5,.5)[right]{$\uline{U}_2$};	
%		    \draw [->,blue] (3.5,.7)--(2.5,.7) node at (3,.7)[below]{$\uline{U}'_2$};
%            \draw node at (0,.5)[left]{$\uline{Z}_{ges}\Rightarrow$};
            \draw node at (1,1.5)[above]{$\uline{Z}_1$};
            \draw node at (3,1.5)[above]{$\uline{Z}_2$};
            \draw node at (1,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
            \draw node at (3,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
            \draw node at (3.25,-.5)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxx}}_{}$};
            \draw node at (3.25,-.25)[below]{$\uline{Z}'=(20-j120)\,\ohm$};
            \draw node at (3,-1)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxxxxxxx}}_{}$};
            \draw node at (3.25,-.75)[below]{$\uline{Z}_{||}=X_M||\uline{Z}'=(720+j120)\,\ohm$};
            \end{scope}	
	\end{tikzpicture}
\end{align*}
\begin{align*}
\uline{Z}_{1}&=(10-j20)\,\ohm &\uline{Z}_{2}&=(20+j80)\,\ohm \\
\uline{Z}'&=(20-j120)\,\ohm &\uline{Z}_{||}&=X_M||\uline{Z}'=\frac{j120\cdot (20-j120)}{\cancel{j120}+20-\cancel{j120}}\,\ohm=(720+j120)\,\ohm \\
\end{align*}
\clearpage
Vierpol:
\begin{align*}
\begin{bmatrix}
\uline{U}_{1}  \\
\uline{U}_{2}  
\end{bmatrix}=
\begin{bmatrix}
\uline{Z}_{11} & \uline{Z}_{12} \\
\uline{Z}_{21} & \uline{Z}_{22}  
\end{bmatrix}\cdot 
\begin{bmatrix}
\uline{I}_{1}  \\
\uline{I}_{2}  
\end{bmatrix}
\end{align*}

\begin{align*}
\uline{U}_1&=\uline{Z}_{11}\cdot \uline{I}_1+\uline{Z}_{12}\cdot \uline{I}_2\\
\uline{U}_2&=\uline{Z}_{21}\cdot \uline{I}_1+\uline{Z}_{22}\cdot \uline{I}_2\\
\uline{Z}_{11}&=\frac{\uline{U}_1}{\uline{I}_1}\Big|_{\uline{I}_2=0}\\
&=\uline{Z}_1+\uline{Z}_{||}=\uuline{(730+j100)\,\ohm}\\
\uline{Z}_{12}&=\frac{\uline{U}_1}{\uline{I}_2}\Big|_{\uline{I}_1=0}\\[\baselineskip]
\uline{I}_1&=0 \Rightarrow \text{ Leerlauf an Primärseite; Spannung an }\uline{U}_M=\uline{U}_1\\[\baselineskip]
&\underline{I}_{X_M}=\uline{I}_2\cdot \frac{jX_C}{R_2+j(X_2-\cancel{X_M})+\cancel{jX_M}+jX_C}\quad\text{( Stromteiler)}\\
\uline{U}_1&=jX_M\cdot \uline{I}_{X_M}=\uline{I}_2\cdot \underbrace{jX_M\cdot \frac{jX_C}{R_2+j(X_2+X_C)}}_{\uline{Z}_{12}}\\
\uline{Z}_{12}&=\frac{-X_M\cdot X_C}{R_2+j(X_2+X_C)}=\frac{-120\cdot (-200)}{20-j0}\,\ohm=\uuline{+1200\,\ohm}\\
\uline{Z}_{21}&=\uuline{\uline{Z}_{12}}\\
\uline{Z}_{22}&=\frac{\uline{U}_2}{\uline{I}_2}\Big|_{\uline{I}_1=0}\\
&=jX_C||(\uline{Z}_2)+jX_M)=\frac{-j200\cdot (20+j(80+120))}{20+\underbrace{j(80+120-200)}_{=0}}\,\ohm=\uuline{(2000-j200)\,\ohm}
\end{align*}
Impedanzmatrix:
\begin{align*}
\uuline{Z}&=
\left[
  \begin{array}{cc}
    730+j100 & 1200 \\
    1200 & 2000-j200 \\
  \end{array}
\right]\,\ohm
\end{align*}
\clearpage
}{}%