\section{Strom L-R-C}
Berechnen Sie den Strom $\uline{I}$
\begin{align*}
\uline{U}&=15\,\volt \cdot e^{j20\,\degree}\quad f=1\,\kilo\hertz\\
C_1&=9\,\micro\farad\quad C_2=4\,\micro\farad\quad R=20\,\ohm\quad L=2\,\milli\henry 
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\text{...noch einfügen...}
%\end{align}
\begin{align*}
	\begin{tikzpicture}[scale=2.5]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,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]{$\uline{U}$};
		\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] {$C_1$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,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=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=3cm,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] {$C_2$};
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}$};
			\draw [<-,red] (.75,0)--(.95,0) node at (.5,.2)[left]{$\uline{I}$};			
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. 
			\draw (0,.8)--(0,1)--(2,1) (.8,1)--(3,1)--(3,.9) (0,.2)--(0,0)--(3,0)--(3,.2);
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Berechnung:}
C_1& \text{ unwirksam, da parallel zur Spannungsquelle.}\\
\omega&=2\pi f=6283\,\frac{1}{\,\second}\\
\uline{Z}_{RLC_2}&=X_L+\uline{Z}_{||}\\
\uline{Y}_{||}&=G+jB_{C_2}\\
G&=\frac{1}{R}=50\,\milli\siemens\\
B_{C_2}&=\omega C_2=6283\frac{1}{\,\second}\cdot 4\,\micro\farad=25{,}13\,\milli\siemens\Rightarrow\\
X_{C_2}&=-\frac{1}{B_{C_2}}=-39{,}79\,\ohm\\
\uline{Y}_{||}&=G+jB_{C_2}=(50+j25{,}13)\,\milli\siemens=55{,}96\,\milli\siemens\cdot e^{j26{,}7\,\degree}\\
\uline{Z}_{||}&=\frac{1}{\uline{Y}_{||}}=17{,}87\cdot e^{-j26{,}7\,\degree}=(15{,}97-j8{,}025)\,\ohm\\
X_L&=\omega \cdot L=6283\frac{1}{\,\second}\cdot 2\,\milli\henry=12{,}57\,\ohm\\
\uline{Z}_{LRC_2}&=\uline{Z}_{||}+jX_L=[15{,}97+j(-8{,}025+12{,}57)]\,\ohm
=(15{,}97+j4{,}541)\,\ohm=16{,}6\,\ohm\cdot e^{j15{,}9\,\degree}\\
\uline{U}_{C_2}&=\uline{U}\cdot \frac{\uline{Z}_{||}}{\uline{Z}_{LRC_2}}=15\,\volt \cdot e^{j20\,\degree}\cdot \frac{17{,}87\cdot e^{-j26{,}7\,\degree}}{16{,}6\,\ohm\cdot e^{j15{,}9\,\degree}}=16{,}15\,\volt\cdot e^{-j22{,}6\,\degree}=(14{,}91-j6{,}21)\,\volt\\
\uline{I}&=\uline{U}_{C_2}\cdot jB_{C_2}=16{,}15\,\volt\cdot e^{-j22{,}6\,\degree}\cdot 25{,}13\,\milli\siemens\cdot e^{j90\,\degree}\\
&=\uuline{0{,}4058\,\ampere\cdot e^{j67{,}4\,\degree}}=\uuline{(0{,}156+j0{,}375)\,\ampere}\\
\end{align*}
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