ET2_Uebung_BEI/ET2_L_B16_A2.tex

\section{Wirkleistung}
Welche Werte müssen $R$ und $C$ annehmen, damit im Verbraucher die maximale Wirkleistung umgesetzt wird?\\
Wie groß ist diese Wirkleistung ?\\[\baselineskip]
$R_1=20\,\ohm\quad C_1=3{,}18\,\micro\farad\quad L=0{,}6\,\milli\henry\quad \uline{U}=1\,\volt\cdot e^{j20\,\degree}\quad f=1000\,\hertz$\\[\baselineskip]
\begin{align*}
	\begin{tikzpicture}[scale=2]
			\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm,rotate=90]%Spannungsquelle | 			
			\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}$};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}$};
		\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] {$R_1$};
%			\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.5cm]%Kondensator -		 
			\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
%			\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C}$};
		\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] {$R$};
%			\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$U_{R}$}; 		
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=-1cm,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);
%			\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$U_{L}$};			
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1cm,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] {$C$};
%			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{C}$};
		\end{scope}	
		\begin{scope}[>=latex,very thick]
		\draw(0,0)--(0,1)--(0.2,1) (.9,.5)--(1,.5)--(1,1) (1,1)--(2,1)--(2,.9) (1.5,-.1)--(1.5,0)--(2,0)(0,-.9)--(0,-1)--(2,-1)--(2,-.9);
		\draw node at(.5,-1)[below]{Quelle};
		\draw node at(2,-1)[below]{Verbraucher};		
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1.25cm,yshift=-1cm,]%Knoten | 			
			\filldraw (0,0)circle(.05); 
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1.25cm,yshift=1cm,]%Knoten | 			
			\filldraw (0,0)circle(.05); 
		\end{scope}
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\[\baselineskip]
Phase von $\uline{U}$ ohne Bedeutung! (Berechnung über Impedanzen)\\
Maximale Wirkleistung bei Anpassung! $\uline{Z}^*_i \stackrel{!}{=} \uline{Z}_v$
\begin{align*}
\omega&=2\pi\cdot f=2\pi \cdot 1000\,\hertz=6283\,\frac{1}{\second}\\
B_1&=\omega\cdot C_1=6283\,\frac{1}{\second}\cdot 3{,}18\,\micro\farad=0{,}02\,\siemens\\
X_L&=\omega\cdot L=6283\,\frac{1}{\second}\cdot 0{,}6\,\milli\henry=3{,}77\,\ohm\\
\intertext{Quelle:}
\uline{Y}_i&=G_1+jB_1=\frac{1}{R_1}+jB_1=(0{,}05+j0{,}02)\,\siemens\\
\uline{Z}_i&=\frac{1}{\uline{Y}_i}=(17{,}24-j6{,}897)\,\ohm\\
R_i&=17{,}24\,\ohm\qquad X_i=-6{,}897\,\ohm\\
\intertext{Anpassung, wenn $\uline{Z}_v=\uline{Z}^*_i$}
\uline{Z}_v&=R_v+jX_v \stackrel{!}{=} (17{,}24+j6{,}897)\,\ohm\quad\Rightarrow\\ 
R_v&=R=\uuline{17{,}24\,\ohm}\\
X_v&=6{,}897\,\ohm\\
B_v&=B_L+B_C=\frac{-1}{X_v}=\frac{-1}{6{,}897\,\ohm}=-0{,}145\,\siemens\\
B_L&=\frac{-1}{X_L}=\frac{-1}{3{,}77\,\ohm}=-0{,}2653\,\siemens\\
B_C&=B_v-B_L=(-0{,}145+0{,}2653)\,\siemens=0{,}1203\,\siemens\\
C&=\frac{B_C}{\omega}=\frac{0{,}1203\,\siemens}{6283\,\frac{1}{\second}}=\uuline{19{,}14\,\micro\farad}\\[\baselineskip]
P_{v\text{,}max}&=\frac{U^2}{4\cdot R_v}=\frac{1\,\volt^2}{4\cdot 17{,}24\,\ohm}=\uuline{14{,}5\,\milli\watt}
\end{align*}
\clearpage
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