\section{Wirkleistung Spannungsquelle}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie die Wirkleistung der Spannungsquelle $\uline{U}_2$.
\item Wird Wirkleistung aufgenommen oder abgegeben?
\end{enumerate}
$R=200\,\ohm\quad L=80\,\micro\henry\quad C=500\,\pico\farad \quad f=1\,\mega\hertz$\\
$\uline{I}_1=10\,\milli\ampere\cdot e^{j60\,\degree}\quad \uline{U}_2=3\,\volt\cdot e^{-j30\,\degree}$
\begin{align*}
	\begin{tikzpicture}[scale=3]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Stromquelle 			 
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$\uline{I}_1$};
			\draw (.5,0)circle(.133);
			\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize $\uline{I}_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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=1cm,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=1cm,yshift=0cm]%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] {$R$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle | 			
			\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_2$};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_2$};
		\end{scope}								
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
			\draw[very thick](0,.9)--(0,1)--(2,1)--(2,.9)(0,.1)--(0,0)--(.1,0)(2,.1)--(2,0)--(1.9,0);
			\draw [->,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\[\baselineskip]
\begin{align*}
\intertext{$C$ spielt keine Rolle, da in Reihe zu Stromquelle.}
\text{Gesucht: } P&=\Re(\uline{U}_2\cdot \uline{I}^*)
\end{align*}
Lösung mit Überlagerungsverfahren:\\[\baselineskip]
Nur Stromquelle:
\begin{align*}
	\begin{tikzpicture}[scale=3]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Stromquelle 			 
			\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$\uline{I}_1$};
			\draw (.5,0)circle(.133);
			\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\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) [right] {$L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%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] {$R$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
			\draw[very thick](0,.9)--(0,1)--(2,1)--(2,.9)(0,.1)--(0,0)--(1,0)(1.9,1)--(2,1)--(2,0)--(1.9,0);
			\draw [->,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}'$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\clearpage
Stromteiler:\\
\begin{align*}
X_L&= \omega \cdot L = 2\cdot \pi\cdot 1\,\mega\hertz\cdot 80\,\micro\henry= 503\,\ohm\\
\uline{I}'&=\uline{I}_1\cdot \frac{jX_L}{R+jX_L}
=10\,\milli\ampere\cdot e^{j60\,\degree}\cdot \frac{j503}{200+j503}\\
&=10\,\milli\ampere\cdot e^{j60\,\degree}\cdot \underbrace{\frac{503\cdot e^{j90\,\degree}}{541{,}3\cdot e^{j68{,}3\,\degree}}}_{0{,}929\,\milli\ampere\cdot e^{j21{,}7\,\degree}}\\
&=9{,}29\,\milli\ampere\cdot e^{j81{,}7\,\degree} = (1{,}34 + j9{,}19)\,\milli\ampere\\
\end{align*} 
Nur Spannungquelle:
\begin{align*}
	\begin{tikzpicture}[scale=3]
		\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) [right] {$L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%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] {$R$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle | 			
			\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_2$};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_2$};
		\end{scope}								
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
			\draw[very thick](1,.9)--(1,1)--(2,1)--(2,.9)(1,.1)--(1,0)(2,.1)--(2,0)--(1.9,0);
			\draw [<-,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}''$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\begin{align*}
\uline{I}''&=\frac{\uline{U}_2}{R+jX_L}
=\frac{3\,\volt\cdot  e^{-j30\,\degree}}{(200+j503)\,\ohm}
=\frac{3\,\volt\cdot e^{-j30\,\degree}}{541{,}3\,\ohm \cdot  e^{j68{,}32\,\degree}
}\\
&=5{,}54\,\milli\ampere\cdot e^{-j98{,}32\,\degree} = (-0{,}80-j5{,}48)\,\milli\ampere\\
\intertext{Überlagerung - vorzeichenrichtg:}
\uline{I}&=\uline{I}'-\uline{I}''= (1{,}34 + j9{,}19+0{,}80+j5{,}48)\,\milli\ampere \\
&=(2{,}14+j14{,}67)\,\milli\ampere=\uline{14{,}83\,\milli\ampere\cdot e^{j81{,}7\,\degree}}\\
S&=\uline{U}_2\cdot \uline{I}^*=3\,\volt\cdot e^{-j30\,\degree}
\cdot 14{,}83\,\milli\ampere\cdot  e^{-j81{,}7\,\degree}
=44{,}49\,\milli\volt\ampere\cdot e^{-j111{,}7\,\degree}\\
&=(\underbrace{-16{,}31}_{P}-j\underbrace{40{,}97}_{Q})\,\milli\volt\ampere\Rightarrow\\
P&=\uuline{-16{,}31\,\milli\watt}
\intertext{Verbraucher-Zählpfeilsystem $\Rightarrow$ \uuline{Spannungsquelle gibt Leistung ab.}}
\end{align*}
\clearpage
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