\section {Resonanzfrequenz Zweipol}
Berechnen Sie die Resonanzfrequenz des abgebildeten Zweipols
$L=12\,\milli\henry$, $C_1=2\,\micro\farad$, $R=160\,\ohm$
\vspace{-.5cm}
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule -			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,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] {$C$};
		\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] {$R$};
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(2,0)--(2,.2) (1,1)--(2,1)--(2,.9);
            \fill (0,0)circle(0.05cm)(0,1)circle(0.05cm);
%			\draw [->,blue] (2.5,.8)--(2.5,.2) node at (2.5,.5)[right]{$\uline{U}_2$};
%        \draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$M$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\
\begin{align*}
\text{Falsch ist: } f_{res}&=\frac{1}{2\pi\sqrt{L\cdot C}}=\frac{1}{2\pi\sqrt{12\cdot \power{10}{-3}\,\ohm\second\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\ohm}}}=\uline{1027\,\frac{1}{\second}}\\
&\text{Gilt nur für } R\rightarrow\infty \text{, reine Reihen- oder Parallelschaltung.}\\[.5\baselineskip]
\text{Bei Resonanz: }\Im\, (\uline{Z})&=0 \\
\uline{Z}=jX_L+(R||jX_C)&=j\omega L+\frac{R\cdot \frac{-j}{\omega C}}{R-j\cdot \frac{1}{\omega C}}\underbrace{\cdot \frac{R+j\frac{1}{\omega C}}{R+j\frac{1}{\omega C}}}_{\text{konj. komplex erweitern}}\hspace{-.75cm}=j\omega L+\frac{R^2\cdot \frac{-j}{\omega C}+R\frac{1}{(\omega C)^2}}{R^2+\frac{1}{(\omega C)^2}}\\
&=\underbrace{\frac{\frac{R}{(\omega C)^2}}{R^2+\frac{1}{(\omega C)^2}}}_{\Re\,= \text{ Widerstand bei Resonanz, }\omega_{res}}+j\underbrace{\Big(\omega L-\frac{R^2}{\omega C\big(R^2+(\frac{1}{\omega C})^2\big)}\Big)}_{\Im =0\text{ bei }\omega_{res}}\\
&\omega L-\frac{R^2}{\omega C\left(R^2+\frac{1}{(\omega C)^2}\right)}=0\\
\Im\quad\Rightarrow\, &\, \omega_{res}\cdot \left[L(\omega_{res}\cdot C\cdot \left(R^2+\frac{1}{(\omega_{res}\cdot C)^2}\right)\right]=R^2\\
&\omega^2_{res}\cdot L\cdot C\cdot R^2+\frac{L}{C}=R^2\\
\Rightarrow\, &\omega^2_{res}=\frac{R^2-\frac{L}{C}}{L\cdot C\cdot R^2}=\frac{1}{L\cdot C}-\frac{1}{(R\cdot C)^2}=\frac{1}{12\,\milli\henry\cdot 2\,\micro\farad}-\frac{1}{(160\,\ohm\cdot 2\,\micro\farad)^2}\\
&=\frac{1}{12\cdot \power{10}{-3}\,\cancel{\ohm}\second\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\cancel{\ohm}}}-\frac{1}{160\,\cancel{\ohm}\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\cancel{\ohm}}}\\
&=4{,}17\cdot \power{10}{7}\power{\second}{-2}-9{,}77\cdot \power{10}{6} \power{\second}{-2}
=3{,}19\cdot \power{10}{7}\cdot \power{\second}{-2}\\
\omega_{res}&=\sqrt{\omega^2_{res}}=5648\cdot \power{\second}{-1}\\
f_{res}&=\frac{\omega_{res}}{2\pi}=\uuline{899\,\hertz}\\
\end{align*}
\begin{align*}
\intertext{Nicht gefragt Resonanzwiderstand: }\\
Z&=\frac{R}{1+(\omega \cdot C\cdot R)^2}=\frac{160\,\ohm}{(5648\cdot \cancel{\power{\second}{-1}}\cdot 2\cdot \power{10}{-6}\,\cancel{\frac{\second}{\ohm}}\cdot 160\,\cancel{\ohm})^2+1}=\uuline{37{,}5\,\ohm}
\end{align*}
\clearpage
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