\section{Spannung Vierpol}
%\renewcommand{\theequation}{0}
Berechnung Sie $\uline{U}_0$
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\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 [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$\uline{U}_R=R_2\cdot \uline{I}_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick, xshift=1.5cm, yshift=0cm]% Vierpol 			 
			\draw (0,-.2)rectangle(2.0,1.2)node at(.75,.5)[right]{$(\uline{Z})$};
			\draw (0,0)--(-.5,0)(0,1)--(-.5,1)(2,0)--(2.5,0)(2,1)--(2.5,1);				 
		\end{scope}				
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0] 			
			\draw (0,0)--(1,0);
			\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
			\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);	
			\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05)	 (4,0)circle(.05) (4,1)circle(.05);					
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=.75cm,yshift=1cm]% Strompfeil		
			\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_0$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=1cm]% Strompfeil		
			\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_2$};			
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_1$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_0$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_2$};			
		\end{scope}											
	\end{tikzpicture}
\end{align*}
\begin{align*}
R_1&=300\,\ohm\\
R_2&=600\,\ohm\\
\uline{Z}_{11}&=100\,\kilo\ohm\text{ reell}\\
\uline{Z}_{21}&=-2\,\mega\ohm\text{ reell}\\
\uline{U}_1&=1{,}5\,\volt
%\end{align*}
%\begin{align*}
\intertext{Die $\uline{Z}$-Matrix des Vierpols ist gegeben:}
\begin{bmatrix}
	\uline{U}_0\\
	\uline{U}_2
\end{bmatrix}
&=
\begin{bmatrix}
	\uline{Z}_{11}&0\\
	\uline{Z}_{21}&0
\end{bmatrix}
\cdot
\begin{bmatrix}
	\uline{I}_0\\
	\uline{I}_2
\end{bmatrix}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\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]{\footnotesize$R_1\cdot (\uline{I}_2+\uline{I}_0)$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$R_2\cdot \uline{I}_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0] 			
			\draw (0,0)--(1,0);
			\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
			\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);	
			\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05)	 (4,0)circle(.05) (4,1)circle(.05);					
		\end{scope}
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0] 			
			\draw (0,0)--(1,0);
			\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
			\draw (1,0)--(1,-.5)--(4,-.5)--(4,0)(1,1)--(1.25,1)(4,1)--(3.75,1);	
			\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05)	 (4,0)circle(.05) (4,1)circle(.05);					
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1.2cm]% Strompfeil		
			\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_0$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1.2cm]% Strompfeil		
			\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_2$};			
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=-.2cm,yshift=1cm]% Strompfeil		
			\draw [->,red] (0,0)--(.2,0)node at(.1,0)[above]{\footnotesize$\uline{I}_1$};			
		\end{scope}			
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(-.2,.5)[left]{\footnotesize$\uline{U}_1$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_0=\uline{Z}_{11}\cdot \uline{I}_0$};	
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_2=\uline{Z}_{21}\cdot \uline{I}_0$};	
		\end{scope}											
	\end{tikzpicture}
\end{align*}
Aus $\uline{Z}$-Matrix folgt:
\begin{align*}
\uline{U}_0&=\uline{Z}_{11}\cdot \uline{I}_0 \tag{1}\\
\uline{U}_2&=\uline{Z}_{21}\cdot \uline{I}_0 \tag{2}
\intertext{Knoten 1:}
\uline{I}_1-\uline{I}_2-\uline{I}_0&=0 \tag{3}
\intertext{Masche 1:}
\uline{U}_1&=R_1\cdot \uline{I}_1+ \uline{U}_0\tag{4}
\intertext{Masche 2:}
\uline{U}_0&=R_2\cdot \uline{I}_2+\uline{U}_2 \tag{5}
\intertext{Zwei Gleichungen für zwei unbekante Ströme $\uline{I}_0$ und $\uline{I}_2$:}
\uline{Z}_{11}\cdot \uline{I}_0&=R_2\cdot \uline{I}_2+\uline{Z}_{21}\cdot \uline{I}_0  \tag{1\&2 in 5}\\
\Rightarrow \uline{I}_2&=\frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0  \tag{6}\\
\uline{U}_1&=\uline{I}_0 \cdot\uline{Z}_{11}+R_1(\uline{I}_0 +\uline{I}_2 ) \tag{7 1\&3 in 4}\\
\uline{U}_1&=\uline{I}_0\cdot \uline{Z}_{11}+R_1\cdot \uline{I}_0+R_1\cdot  \frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0 \tag{6 in 7}\\
&=\left[\uline{Z}_{11}+R_1+\frac{R_1}{R_2}\cdot(\uline{Z}_{11}-\uline{Z}_{21})\right]\cdot \uline{I}_0\\
&=\left[100\,\kilo\ohm+300\,\ohm+\frac{300\,\ohm}{600\,\ohm}\cdot \Big(100\,\kilo\ohm-(-2\,\mega\ohm)\Big)\right]\cdot \uline{I}_0\\
&=0{,}103+0{,}5\cdot (100+2000)\,\kilo\ohm \cdot  \uline{I}_0=1{,}15\,\mega\ohm\cdot  \uline{I}_0\\
\Rightarrow\uline{I}_0&=\frac{\uline{U}_1}{1{,}15\,\mega\ohm}=\frac{1{,}5\,\volt}{1{,}15\,\mega\ohm}=1{,}3\,\micro\ampere\\
\uline{U}_0&=\uline{Z}_{11}\cdot \uline{I}_0=100\,\kilo\ohm\cdot 1{,}3\,\micro\ampere=\uuline{130{,}4\,\milli\volt}
\intertext{Nicht gefragt:}
\uline{I}_2&=\frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0=\frac{1}{600\,\ohm}\cdot 2{,}1\mega\ohm\cdot 1{,}3\,\micro\ampere=4{,}55\,\milli\ampere >> \uline{I}_0\\
\text{Probe }\\
%&=\frac{1}{R_1}\cdot (\uline{I}_2+\uline{I}_0)\cdot \uline{U}_0\\
\uline{U}_1&=1{,}5\,\volt=R_1\cdot (\uline{I}_0+\uline{I}_2)+\uline{I}_0\cdot Z_{11}\\
&=300\,\ohm\cdot (0{,}0013+4{,}55)\,\milli\ampere+1{,}3\,\micro\ampere\cdot 100\,\kilo\ohm=1{,}498\,\volt
\end{align*}
\clearpage
\enlargethispage{1cm}
Ist Ihnen aufgefallen, daß $\uline{Z}_{21}$ einen negativen Wert hat?\\[\baselineskip]
Ein Widerstand nimmt elektrische Leistung auf, also ein Verbraucher.\\
Dann muß also ein negativer Widerstand elektrische Leistung abgeben!
Gibt es das in der Realität?\\[\baselineskip]
Was steckt in dem Vierpol?\\[\baselineskip]
Passiver Vierpol:\\
Zum Beispiel mit 3 Widerständen in $T$ oder $\Pi$ Schaltung, ergibt jedoch keine Lösung!\\
Aus der $\uline{Z}$ Matrix ist ersichtlich, daß $\uline{Z}_{21}=0$ und $\uline{Z}_{22}=0$ sind,
somit existiert keine Kopplung vom Ausgang zum Eingang, sondern nur in Vorwärtsrichtung.\\[\baselineskip]
Beispiel: Elektrodynamischer Lautsprecher der auf ein Kondensatormikrofon einwirkt. Kopplung nur vom Lautsprecher zum Mikrofon, nicht umgekehrt.\\[\baselineskip]
Aktiver Vierpol: (Für Fortgeschrittene)\\
Es muß sich um einen invertierenden Trennverstärker mit einem Eingangswiderstand von
$100\,\kilo\ohm$ und einem Ausgangswiderstand von $2\,\mega\ohm$ handeln. Die Verstärkung ist zu berechnen.
\begin{align*}
\uline{U}_2&=\uline{U}_0-R_2\cdot \uline{I}_2=130{,}4\,\milli\volt-600\,\ohm\cdot 4{,}55\,\milli\ampere=-2{,}6\,\volt\\
\uline{U}_q&=\uline{U}_2-\uline{Z}_{22}\cdot \uline{I}_2=-2{,}6\,\volt-2\,\mega\ohm\cdot 4{,}55\,\milli\ampere=-9102{,}6\,\volt\\
V&=\frac{\uline{U}_q}{\uline{U}_0}=\frac{-9102{,}6\,\volt}{130{,}4\,\milli\volt}=-69805
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=1.5]
		\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$R_1\cdot (\uline{I}_2+\uline{I}_0)$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$R_2\cdot \uline{I}_2$};
		\end{scope}
				\begin{scope}[>=latex,very thick,black!25!, xshift=1.5cm, yshift=0cm]% Vierpol mit Innenleben			
					\draw (0,-.2)rectangle(2.0,1.2);%node at(.75,.5)[right]{$(\uline{Z})$};
					\draw (0,0)--(-.5,0)(0,1)--(-.5,1)(2,0)--(2.5,0)(2,1)--(2.5,1);
%				\end{skope}	
				\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90] %Z_11 			
					\draw [red!50!blue](0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667)
[left] {$\uline{Z}_{11}$};
					\draw [<-,red!50!blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$\uline{Z}_{11}\cdot \uline{I}_0$};
				\end{scope}			
				\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Z_21			 
					\draw [red!50!blue](0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_{21}$};
					\draw [->,red!50!blue] (.3,-.2)--(.7,-.2)node at(.6,-.2)[below]{\footnotesize$-\uline{Z}_{21}\cdot \uline{I}_2$};
			\end{scope}
				\begin{scope}[>=latex,very thick,red!50!blue,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle| 			
					\draw (0,0)--(1,0);%;node at(.5,-.133)[right]{$V\cdot \uline{U}_0$};
					\draw (.5,0)circle(.133);
					\draw [->] (.3,-.2)--(.7,-.2) node at (.3,-.2)[right]{\footnotesize$V\cdot \uline{U}_q$};
					\draw node at (.3,.025)[left]{\footnotesize$+$};
					\draw node at (.7,.025)[left]{\footnotesize$-$};					 
			\end{scope}
			\begin{scope}[>=latex,very thick,red!50!blue,xshift=0cm,yshift=0cm]%
					\draw (-.5,0)--(2.5,0)(-.5,1)--(0,1)(2,1)--(2.5,1);
			\end{scope}
		\end{scope}			% Ende Vierpol Innenleben
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0] 			
			\draw (0,0)--(1,0);
			\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
			\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);	
			\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05)	 (4,0)circle(.05) (4,1)circle(.05);					
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]% Strompfeil		
			\draw [->,red] (0,0)--(.4,0)node at(.2,0)[above]{\footnotesize$\uline{I}_0$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm]% Strompfeil		
			\draw [->,red] (0,0)--(-.4,0)node at(-.2,0)[above]{\footnotesize$\uline{I}_2$};			
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=-.2cm,yshift=1cm]% Strompfeil		
			\draw [->,red] (0,0)--(.2,0)node at(.1,0)[above]{\footnotesize$\uline{I}_2+\uline{I}_0$};			
		\end{scope}			
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(-.2,.5)[left]{\footnotesize$\uline{U}_1$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(0,.25)[left]{\tiny$\uline{U}_0=\uline{Z}_{11}\cdot \uline{I}_0$};	
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil		
			\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_2=\uline{Z}_{21}\cdot \uline{I}_0$};	
		\end{scope}											
	\end{tikzpicture}
\end{align*}
\begin{minipage}[t]{1\textwidth}
\centering
\includegraphics[width =0.6\textwidth]{b16a8a}
\end{minipage}
\clearpage
}{}%
