\section{Wechselstrombrücke}
Gegeben ist die dargestellte Wechselstrombrücke, die zum Messen der Größe von $R_2$
und $L_2$ dient. Dabei ist $R_1=1\,\kilo\ohm$, $R_3=R_4=2\,\kilo\ohm$ und $L_1=1\,\milli\henry$.\\
Die Brücke ist bei einer Kreisfrequenz von $\omega=\power{10}{6}\,\power{\second}{-1}$ und $C_1=2\,\nano\farad$ abgeglichen.\\[\baselineskip]
Berechnen Sie $R_2$ und $L_2$!
\begin{align*}
	\begin{tikzpicture}[very thick,scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
		\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] {$L_1$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,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] {$C_1$};
        \draw[->](.3,-.2)--(.7,.2);
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_3$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.25cm]%Widerstand
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.25cm]%Spule -			 
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_2$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%Widerstand
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_4$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%V-meter
            \draw (0,0)--(.367,0) (.633,0)--(1,0);
			\draw (.5,0)circle(.133);
            \draw[->](.4,.1)--(.6,-.1);
		\end{scope}			
		\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
			\draw (.1,1)--(0,1)--(0,-.5)(0,0)--(1,0)
                  (1.01,-.25)--(1,-.25)--(1,.25)--(1.01,.25)
                  (1.99,-.25)--(2,-.25)--(2,.25)--(1.99,.25)
                  (2,0)--(3,0)
                  (3.9,1)--(4,1)--(4,-.5);
			\draw [->,blue] (.2,-.5)--(3.8,-.5)node at(3,-.5)[above]{$\underline{U}=20\,\volt$};
			\fill (0,-.5)circle(.025) (4,-.5)circle(.025);
		\end{scope}
        \draw node at (1.5,1.25)[above]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxxxxxxx}}$};
        \draw node at (3.5,1.25)[above]{$\overbrace{\phantom{xxxxx}}$};
        \draw node at (1.5,1.5)[above]{$\uline{Z}_1$};
        \draw node at (3.5,1.5)[above]{$\uline{Z}_3$};
        \draw node at (1.5,-.5)[below]{$\underbrace{\phantom{xxxxxxxxxxx}}$};
        \draw node at (3.5,-.5)[below]{$\underbrace{\phantom{xxxxx}}$};
        \draw node at (1.5,-.75)[below]{$\uline{Z}_2$};
        \draw node at (3.5,-.75)[below]{$\uline{Z}_4$};
%		\begin{scope}[>=latex,very thick]%Variablen Pfeile
%            \draw[->] (1.3,.75)--(1.7,1.25);
%            \draw[->] (2.3,.75)--(2.7,1.25);
%            \draw[dashed] (1.3,.75)--(2.3,.75);
%        \end{scope}
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
R_P&=R_S+\frac{X^2_S}{R_S}\label{eq:rp177}\\
X_P&=X_S+\frac{R^2_S}{X_S}\label{eq:xp177}
\end{align}
Berechnung:
\begin{align*}
\uline{Z}_1&=R_1+jX_{L_1}-jX_{C_1} &\uline{Z}_3&=R_3\\
\quad\uline{Z}_2&=R_2 ||X_{L_2} &\uline{Z}_4&=R_4
\end{align*}
\begin{align*}
\intertext{Brücke abgeglichen, wenn:}
\frac{\uline{Z}_1}{\uline{Z}_2}&=\frac{\uline{Z}_3}{\uline{Z}_4}=\frac{R_3}{R_4}=1\\
\uline{Z}_2&=\uline{Z}_1=R_1+j\omega L_1 -j\frac{1}{\omega C_1}\\
&=1\,\kilo\ohm+j\Big(\underbrace{\power{10}{6} \,\cancel{\second^{-1}} \cdot \power{10}{-3}\,\ohm\cancel{\second}}_{1\,\kilo\ohm} -\underbrace{\frac{1}{\power{10}{6} \,\cancel{\second^{-1}}\cdot 2 \cdot \power{10}{-9}\,\frac{\cancel{\second}}{\ohm}}}_{\frac{1}{2}\,\kilo\ohm}\Big)=\uuline{(1+j0{,}5)\,\kilo\ohm}\\
\end{align*}
Um $R_2$ und $L_2$ zu bestimmen, die Reihenschaltung $\uline{Z}_2=R_{2S}+jX_{2S}=(1+j0{,}5)\,\kilo\ohm$ in Parallelwiderstände umrechnen.
\begin{align*}
\uline{Y_2}&=\frac{1}{\uline{Z_2}}=\frac{1}{(1+j0{,}5)\,\kilo\ohm}
=(\underbrace{0{,}8}_{G_2}-\underbrace{j0{,}4}_{B_{L_2}})\,\milli\siemens\\
R_{2}&=\frac{1}{G_2}=\frac{1}{0{,}8\,\milli\siemens}=\uuline{1{,}25\,\kilo\ohm}\\
X_{L_2}&=\frac{-1}{B_{L_2}}=\frac{-1}{-0{,}4\,\milli\siemens}=\uline{+2{,}5\,\kilo\ohm}\\
\text{aus }X_{L_2}&=\omega\cdot L \Rightarrow \\
L_2&=\frac{X_{L_2}}{\omega}=\frac{2{,}5\,\kilo\ohm}{\power{10}{6}\cdot \power{\,\second}{-1}}=\uuline{2{,}5\,\milli\henry}\\
\intertext{Alternativ mit Formeln \ref{eq:rp177} und \ref{eq:xp177}}
R_{2P}&=R_{2S}+\frac{X^2_{2S}}{R_{2S}}\\
X_{2P}&=X_{2S}+\frac{R^2_{2S}}{X_{2S}}\\
R_2&=R_{2S}+\frac{X^2_{2S}}{R_{2S}}=(1+\frac{0{,}5^2}{1})\,\kilo\ohm=1{,}25\,\kilo\ohm\\
X_{L_2}&=X_{2S}+\frac{R^2_{2S}}{X_{2S}}=(0{,}5+\frac{1^2}{0{,}5})\,\kilo\ohm=2{,}5\,\kilo\ohm
\end{align*}
\clearpage
}{}%
