\section{Energieübertragung}
Die Skizze zeigt ein System zur elektrischen Energieübertragung bestehend aus Quelle,
Leitung und Verbraucher. Das System soll mit einem parallel geschalteten Kondensator $X_C$ so
optimiert werden, dass die Leitungsverluste $P_{VRL}$ minimal werden.\\
\begin{align*}
	\begin{tikzpicture}[scale=3]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle |
			\draw (0,0)--(1,0)node at(.5,-.133)[right]{\footnotesize{$\underline{U}_q=100\,\volt\cdot e^{j0}$}};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2)node at (.5,.2)[left]{\footnotesize$\underline{U}_q$};
		\end{scope}
		\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] {\footnotesize{$\underline{Z}_i=(1+2j)\,\ohm$}};
		\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] {\footnotesize{$R_L=1\,\ohm$}};
		\end{scope}						
		\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=0cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [above right] {\footnotesize{$\underline{Z}_V=(10+j5)\,\ohm$}};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=3.5cm,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) [right] {\footnotesize{$jX_C$}};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]		
			\draw [dashed](2.5,0)--(3.5,0)--(3.5,.2) (2.5,1)--(3.5,1)--(3.5,.8);
			\draw [dashed](0,0)--(2.5,0) (2.5,1)--(3.5,1)--(3.5,.8);
			\draw (0,0)--(2.5,0) (2,1)--(2.5,1);			
			\draw (0,.2)--(0,0)--(0,.2) (0,.8)--(0,1)--(.2,1);
			\draw  node at(.5,0) [below] {\footnotesize Quelle};
			\draw  node at(1.5,0) [below] {\footnotesize Leitung};
			\draw  node at(2.5,0) [below] {\footnotesize Verbraucher};
			\draw [very thin, dashed] (1,-.2)--(1,1.2)(2,-.2)--(2,1.2);
			\draw [->,red](2.125,1.125)--(2.375,1.125) node [right] {\footnotesize{$\underline{I}$}};
		\end{scope}						
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte		
			\draw (0,0)circle(.035); 
			\fill [white](0,0)circle(.025); 
			\draw (1,0)circle(.035);
			\fill [white](1,0)circle(.025);
			\fill (1.5,0)circle(.025);				
			\fill (2.5,0)circle(.025);			 
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte		
			\draw (0,0)circle(.035); 
			\fill [white](0,0)circle(.025); 
			\draw (1,0)circle(.035);
			\fill [white](1,0)circle(.025);
			\fill (1.5,0)circle(.025);			
			\fill (2.5,0)circle(.025);				 
		\end{scope}				
	\end{tikzpicture}
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Bestimmen Sie $X_C$ so, dass der Blindleistungsbedarf des Verbrauchers verschwindet.
\item  Berechnen Sie die Verlustleistung $P_{VRL}$ der Leitung und die Wirkleistung $P_W$ im Verbraucher.
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\[\baselineskip]
a) Verbraucher $\uline{Z}_V\,||\,X_C$, daher Ersatzschaltbild für $\uline{Z}_V$ (ESB) in Parallelform erforderlich
\begin{align*}
\underline{Z}_V&=R_V+jX_V=(10+j5)\,\ohm\,\,\quad\text{ Scheinwiderstand, entspricht einer Reihenschaltung}\\[\baselineskip]
Z^2_V&=R_V\cdot R_p=R^2_V+X^2_V\,\qquad\qquad\text{Umwandlung in Parallel-ESB}\\
R_p&=R_V+\frac{X^2_V}{R_V}=(10+\frac{25}{10})\,\ohm=12{,}5\,\ohm\\[\baselineskip]
Z^2_V&=X_V\cdot X_{L_p}=R^2_V+X^2_V\,\,\quad\qquad\text{Umwandlung in Parallel-ESB}\\
X_{L_p}&=X_V+\frac{R^2_V}{X_V}=(5+\frac{100}{5})\,\ohm=25\,\ohm
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\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] {$R_p$};
		\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) [right] {$jX_{L_p}$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=3cm,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] {$jX_C$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]		
			\draw (0,0)--(3,0)--(3,.2) (0,1)--(3,1)--(3,.8);			
		\end{scope}						
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte		
			\fill (0,0)circle(.025); 
			\fill (1,0)circle(.025);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte		
			\fill (0,0)circle(.025); 
			\fill (1,0)circle(.025);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Knotenpunkte		
			\draw (0,0)circle(.05); 
			\fill (0,0)circle(.025); 
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Knotenpunkte		
			\draw (0,0)circle(.05);
			\fill (0,0)circle(.025);			 
		\end{scope}
		\begin{scope}[>=latex,very thick]		
			\draw node at (0,0.5) [left] {$\underline{Z}'\Rightarrow$};	
		\end{scope}
		\draw node at(1.5,0)[below]{Verbraucher $Z_V$};
%	\end{tikzpicture}
%		\begin{tikzpicture}[scale=.5,xshift=15cm,yshift=-2cm]
		\begin{scope}[>=latex,very thick,scale=.25,xshift=18cm,yshift=2cm] 			
			\draw [->](0,0)--(1,0)node [right]{$R_p$};
			\draw [->](1,0)--(1,2)node [above]{$jX_{Lp}$};
			\draw [->](1,0)--(1,-2)node[below]{$jX_C$};	
			\draw [->,red!50!blue](0,0)--(1,2)node at (.5,1)[left]{$Z_V$};								
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}
\underline{Z}'&=\underline{Z}_V || jX_C\\
\frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\frac{1}{jX_{Lp}}+\frac{1}{jX_C}\\
\intertext{Leitungsverluste sind minimal, wenn die Blindleistung $=0$ wird (Kompensation)}
\frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\cancel{\frac{1}{jX_{Lp}}}+\cancel{\frac{1}{jX_C}}\qquad\Rightarrow \underline{Z}'=R_p\\
\Im(\underline{Z}')&=0  \quad\text{oder}\quad |X_C| \stackrel{!}{=} |X_{L_p}|\text{ also}\\
%\Re(\underline{Z}')&=R_p\\
X_C&=-X_{L_p}=\uuline{-25\,\ohm}\\
\underline{Z}_{ges}&=\underline{Z}_i+R_L+\underline{Z}'\\
\underline{Z}_{ges}&=\underline{Z}_i+R_L+R_p=(1+j2+1+12{,}5)\,\ohm=(14{,}5+j2)\,\ohm\\
|\underline{Z}_{ges}|&=\sqrt{14{,}5^2+2^2}\,\ohm=14{,}64\,\ohm\\
I&=\frac{U}{|Z_{ges}|}=\frac{100\,\volt}{14{,}64\,\ohm}=6{,}83\,\ampere\\
\intertext{b) Verlust- und Wirkleistung}
P_{VR_L}&=I^2\cdot R_L=(6{,}83\,\ampere)^2\cdot 1\,\ohm=\uuline{46{,}7\,\watt}\\
P_W&=I^2\cdot R_p=(6{,}83\,\ampere)^2\cdot 12{,}5\,\ohm=\uuline{583\,\watt}
\end{align*}
\clearpage
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