\section{Abgebbare Wirkleistung}
Um wieviel Prozent weicht die in dem passiven Zweipol umgesetzte Wirkleistung von der in dem aktiven Zweipol maximal abgebbaren Wirkleistung ab?\\
$C_1=2\,\nano\farad\quad
C_2=3\,\nano\farad\quad
L_1=2{,}5\,\micro\henry\quad
R_1=20\,\ohm\quad
L_2=3\,\micro\henry\quad
R_1=15\,\ohm\quad
f=3\,\mega\hertz$\\
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\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)node at(.5,-.133)[right]{$\uline{I}$};
			\draw (.5,0)circle(.133);
			\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,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] {$C_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,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] {$C_2$};
		\end{scope}			
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm,rotate=90]%Spule | 			 
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_1$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,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] {$R_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm,rotate=90]%Spule | 			 
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_2$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,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] {$R_2$};
		\end{scope}
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0cm]
			\draw(0,1.5)--(0,2)--(.1,2)(1,2)--(3,2)--(3,1.9)(1,1.5)--(1,2);
			\draw(0,.5)--(0,0)--(3,0)--(3,.1)(1,0)--(1,.5);	
			\fill (2.5,2)circle(.05);
			\fill (2.5,0)circle(.05);
			\draw [thin,dashed](-.45,-.2)rectangle(2.4,2.5);
			\draw [thin,dashed](2.6,-.2)rectangle(3.45,2.5);			
			\draw node at(1,-.25)[below]{Aktiver Zweipol};	
			\draw node at(3,-.25)[below]{Passiver};
			\draw node at(3,-.5)[below]{Zweipol};											 
		\end{scope}								
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%%\begin{align}
%%\intertext{Formeln:}
%%\end{align}
Berechnung:\\[\baselineskip]
$C_1$  spielt für die Berechnung der Wirkleistung keine Rolle, da in Reihe zu Stromquelle.\\[\baselineskip]
ESB:
\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]{$U_q$};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{q}$};
		\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] {$\uline{Z}_i$};
		\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] {$\uline{Z}_v$};
		\end{scope}
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
			\draw(0,.9)--(0,1)--(.1,1)(0,.1)--(0,0)--(1,0)--(1,.1);
			\fill (1,1)circle(.05);
			\fill (1,0)circle(.05);
		\end{scope}						
	\end{tikzpicture}
\end{align*}
Als ESB ist eine Stromquelle $I_q$ mit parallelem $\underline{Z}_i$ und dazu parallelem $\underline{Z}_v$ möglich.
\clearpage
\begin{align*}
Z_{i}&=jX_C || (R_1 + jX_{L1})\\
\omega&=2\pi\cdot f=2\pi\cdot 3\,\mega\hertz=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\\
X_{C_2}&=\frac{-1}{\omega\cdot C_2}=\frac{-1}{18{,}85\cdot \power{10}{6}\,\cdot \frac{1}{\second}\cdot 3\,\nano\farad}=-17{,}68\,\ohm\\
X_{L_1}&=\omega\cdot L_1=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\cdot 2{,}5\,\micro\henry=47{,}12\,\ohm\\
X_{L_2}&=\omega\cdot L_2=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\cdot 3\,\micro\henry=56{,}55\,\ohm\\
\uline{Z}_i&=jX_{C_2}||(R_1+jX_{L_1})
=\frac{-j17{,}68\,\ohm\cdot (20\,\ohm+j47{,}12\,\ohm)}{-j17{,}68\,\ohm + 20\,\ohm+j47{,}12\,\ohm}
=\frac{-j17{,}68\cdot (20+j47{,}12)}{20+j(47{,}12-17{,}68)}\\
&=\frac{833{,}08-j353{,}6}{20+j29{,}44}\,\ohm=\frac{905{,}02\cdot e^{-j23{,}0\,\degree}}{35{,}59\,\ohm\cdot e^{j55{},81\,\degree}}=25{,}428\,\ohm\cdot e^{-j78{,}81\,\degree}=(4{,}935-j24{,}94)\,\ohm
\end{align*}
Verbraucherwiderstand:
\begin{align*}
\uline{Z}_v&=(15+j56{,}55)\,\ohm\\
\uline{Z}_{ges}&=\uline{Z}_i+\uline{Z}_v=(4{,}935-j24{,}94)\,\ohm+(15+j56{,}55)\,\ohm=(19{,}94+j31{,}61)\,\ohm\\
\intertext{Anmerkung: $\uline{U}_q$ ist unbekannt, kürzt sich später heraus.}
P_{v,max}&=\frac{U^2_q}{4\cdot R_i}=\frac{U^2_q}{4\cdot 4{,}935\,\ohm}=\frac{U^2_q}{19{,}94\,\ohm}\\
P_v&=I^2\cdot R_v\\
I&=\frac{U_q}{Z_{ges}}\qquad \text{Anmerkung: $I$ und $U_q$ Effektivwert; $Z$ Betrag}\\
P_v&=\left(\frac{U_q}{Z_{ges}}\right)^2 \cdot R_v =\frac{U^2_q}{(19{,}94^2+31{,}61^2)\,\ohm^2}\cdot 15\,\ohm=U^2_q\cdot \frac{15}{1397\,\ohm}\\
F_ \% &=100\,\%\cdot \frac{P_v-P_{v,max}}{P_{v,max}}=100\,\%\cdot \left(\frac{P_v}{P_{v,max}}-1\right)\\
&=100\, \% \cdot \left(\frac{\cancel{U^2_q}\cdot 15}{1397\,\ohm}\cdot \frac{19{,}94\,\ohm}{\cancel{U^2_q}}-1\right)\\
&=100\, \% \cdot (0{,}212-1)=\uuline{-78{,}8\, \%}
\end{align*}
\clearpage
}{}%
