ET2_Uebung_BEI/ET2_L_B15_A1.tex

\section{Komplexe Wechselstromrechnung Netzwerk Strom}
Berechnen sie den Strom $I_2$\\
$I_0=3\,\milli\ampere \cdot e^{-j30\,\degree}\qquad  C_1= 1{,}2\,\nano\farad \qquad L=100\,\micro\henry$\\
$R=9,3\,\ohm \qquad C_2=820\,\pico\farad \qquad f=570\,\kilo\hertz$\\
Hinweis: Rechnung nur mit komplexen Größen\\
Kann $I_2$ größer als $I_0$ sein? Wenn ja, warum?\\
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
\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]{$I_0$};		
			\draw (.5,0)circle(.133);
			\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_{0}$};
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%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=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=-1cm,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 [->,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{$\uline{I}_{1}$}; 		
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,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] {$C_2$};
			\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}_{2}$};
			\draw [->,red] (-.2,.2)--(.2,.2) node at (0,.2)[left]{$\uline{I}_{2'}$};			
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.25cm]%Fehlstellen Eckverbindungen. 
			\draw (0,.75)--(0,1.25)--(.2,1.25) (2,.75)--(2,1.25)--(.8,1.25) (0,0)--(0,-1)-- (2,-1)--(2,0) (1,-1)--(1,-.5);
		\end{scope}							
	\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Berechnung:}
C_1&\text{ spielt keine Rolle, da in Reihe mit Stromquelle.}\\
\uline{I_2}&=-\uline{I}_2'\quad\text{, wegen Knotenpunkt $\uline{I}_0-\uline{I}_1'-\uline{I}_2'=0$}\\
\uline{I_2}&=-\uline{I}_2'=-\frac{R+jX_L}{R+j(X_L+X_{C_2})}\cdot \uline{I}_0\\
\text{mit }X_{C_2}&=\frac{-1}{\omega C_2}=-\frac{1}{2\pi\cdot 570\cdot \power{10}{3}\cdot 820\cdot \power{10}{-12}}\,\ohm=-340{,}51\,\ohm\\
X_L&=\omega \cdot L=2\pi\cdot 570\cdot \power{10}{3}\cdot \power{10}{-4}\,\ohm= +358{,}14\,\ohm\\
X_{C_2}+X_L&=(-340{,}51+358{,}14)\,\ohm=17{,}63\,\ohm\\
\uline{I_2}&=-\frac{9{,}3\,\ohm +j 358{,}14\,\ohm}{9{,}3\,\ohm +j 17{,}63\,\ohm}\cdot 3\,\milli\ampere \cdot  e^{-j30\,\degree}=(-16{,}11+j7{,}97)\cdot 3\,\milli\ampere\cdot  e^{-j30\,\degree}\\
&=17{,}97\cdot e^{-j153{,}7\,\degree}\cdot 3\,\milli\ampere\cdot  e^{-j30\,\degree}=\uuline{53{,}92\,\milli\ampere\cdot e^{j176{,}32\,\degree}}\\
\uline{I}_2&>\uline{I}_0\text{, da sehr nahe an Resonanz $(X_L\approx |X_C|$)}
\end{align*}
\clearpage
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