\section {3-Phasen System mit unsymmetrischem Verbraucher}
Ein symmetrisches 3-Phasen System mit der Phasenlage 1-2-3 speist einen unsymmetrischen
Verbraucher. Gegeben sind:\\
$\uline{U}_{31}=380\,\volt\cdot e^{j0\,\degree}$; $R_1=60\,\ohm$; $\uline{X}_{L_1}=20\,\ohm$; $R_2=100\,\ohm$; $\uline{X}_{C_2}=-80\,\ohm$; $X_{L_3}=50\,\ohm$;\\[\baselineskip]
Es ist die Anzeige des Leistungsmessinstrumentes zu berechnen!
\begin{align*}
	\begin{tikzpicture}[very thick,scale=4]
		\begin{scope}[>=latex,xshift=0cm,yshift=1.5cm]%Wattmeter			
			\draw (0,0)--(.367,0) (.633,0)--(1,0) node at (.5,0)  {$W$};
			\draw (.5,0)circle(.133);
            \draw(.25,0)--(.25,.25)--(.5,.25)--(.5,.133)(.5,-.133)--(.5,-1);
		\end{scope}
		\begin{scope}[xshift=1cm,yshift=.5cm,rotate=60,scale=.577]
		  \begin{scope}[>=latex,xshift=0cm,yshift=0cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.1330667) [left] {$R_1$};
			\draw [<-,red] (.8,.1)--(1.2,.1)node at(1,.14)[left]{$\uline{I}_{12}$};
		  \end{scope}
		  \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Spule -			
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.1330667) [ left] {$L_1$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
          \end{scope}
		\end{scope}
		  \begin{scope}[>=latex,very thick,xshift=2.01cm,yshift=.75cm,rotate=120,scale=.577]%Spule	
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,-.1330667) [ right] {$L_3$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw [<-,red] (1.2,-.1)--(.8,-.1)node at(1,-.14)[right]{$\uline{I}_{31}$};
		  \end{scope}
		  \begin{scope}[>=latex,xshift=1.289cm,yshift=.5cm,scale=.577]%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=1.289cm,yshift=.25cm,scale=.577]%Kondensator -		
			\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.1133) [above] {$C_2$};
%			\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C2}$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(2.154,0)--(2.154,.5)(.9,.5)--(1.289,.5) (1.289,.5)--(1.289,.25)(1.867,.5)--(1.867,.25);
            \filldraw(0,0)circle(0.025cm)node [left]{$L3$};
            \filldraw(0,.5)circle(0.025cm)(1,.5)--(0,.5)node [left]{$L2$};
            \filldraw(0,1.5)circle(0.025cm)node at(0,1.5) [left]{$L1$};
            \fill(.5,.5)circle(0.025cm)(1,.5)circle(0.025cm)(1.289,.5)circle(0.025cm)(1.867,.5)circle(0.025cm)(2.154,.5)circle(0.025cm)(1.577,1.5)circle(0.025cm);
			\draw [->,blue] (.8,1.4)--(.8,0.6) node at (.8,1)[right]{$\uline{U}_{12}$};
			\draw [->,red] (.9,1.6)--(1.1,1.6) node at (1,1.6)[above]{$\uline{I}_1$};
            \draw +(1,1.5)--(1.577,1.5)--(2.154,.5cm)--(1.75,.5);
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\
%\begin{align*}
%	\begin{tikzpicture}[scale=2,rotate=-150]
%		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
%            \draw[black!25!,very thin,step=.5cm](-1,-1)grid(1,1);
%            \draw[->]++(-180:1cm)--++(30:1.732cm)node [above right]{$2$};
%            \draw[->]++(-180:1cm) ++(30:1.732cm)-- ++(270:1.732cm)node [below]{$3$};
%            \draw[->]++(-180:1cm) ++(330:1.732cm)--(-180:1cm)node [left]{$1$};
%            \draw[->](-180:1cm)--(0,0)node [right]{$N$};
%            \draw[->](60:1cm)--(0,0);
%            \draw[->](300:1cm)--(0,0);
%            \draw[blue] node at(-.35,.6){$\uline{U}_{12}$};
%            \draw[blue] node at(-.35,-.6){$\uline{U}_{31}$};
%            \draw[blue] node at(.75,0){$\uline{U}_{23}$};
%            \draw[blue] node at(-.35,.15){$\uline{U}_{1}$};
%            \draw[blue] node at(.35,.25){$\uline{U}_{2}$};
%            \draw[blue] node at(.35,-.25){$\uline{U}_{3}$};
%        \end{scope}	
%	\end{tikzpicture}
%\end{align*}
%%\clearpage
\begin{minipage}[c]{0.25\textwidth}
\begin{align*}
	\begin{tikzpicture}[very thick,scale=1]
\draw[->,blue](0:0)--(0:1cm)node [right]{$\uline{U}_{31}$};
\draw[->,blue](0:0)--(120:1cm)node [above]{$\uline{U}_{23}$};
\draw[->,blue](0:0)--(-120:1cm)node [below]{$\uline{U}_{12}$};
\draw[->,red](0:0)--(141.2:.577cm)node [above left]{$\uline{I}_1$};
	\end{tikzpicture}
\end{align*}
\end{minipage}
%\hfill
\begin{minipage}[l]{0.75\textwidth}
\begin{align*}
\uline{U}_{31}&=380\,\volt\cdot e^{j0\,\degree}\\
\uline{U}_{12}&=380\,\volt\cdot e^{-j120\,\degree}\\
\end{align*}
\end{minipage}
\begin{align*}
P&=\Re\{\uline{U}_{12}\cdot I^*_1\}\\
\uline{I}_1&=\uline{I}_{12}-\uline{I}_{31}\\
\end{align*}
\begin{align*}
\uline{I}_{12}&=\frac{\uline{U}_{12}}{R_1+jX_{L_1}}
=\frac{\overbrace{380\,\volt\cdot e^{-j120\,\degree}}^{(-190-j329{,}09)\,\volt}}{\underbrace{63,246\,\ohm\cdot ^{j18{,}435\,\degree}}_{(60+j20)\,\ohm}}=6{,}01\,\ampere\cdot e^{-j138{,}4\,\degree}=(-4{,}495-j3{,}986)\,\ampere\\
\uline{I}_{31}&=\frac{\uline{U}_{31}}{jX_{L_3}}=\frac{380\,\volt\cdot e^{j0\,\degree}}{j50\,\ohm}=-j\frac{380\,\volt}{50\,\ohm}=-j7{,}6\,\ampere \qquad\tag{$\frac{1}{j}=-j$}\\
\uline{I}_1&=[(-4{,}495-j3{,}986)-(-j7{,}6)]\,\ampere\\   &=(-4{,}495+j3{,}614)\,\ampere=5{,}77\,\ampere\cdot e^{j141{,}2\,\degree}\\[\baselineskip]
\end{align*}
\begin{align*}
	\begin{tikzpicture}[very thick,scale=2.5]
\draw[->,blue](0:0)--(0:1cm)node [right]{$\uline{U}_{31}$};
\draw[->,blue](0:0)--(120:1cm)node [above]{$\uline{U}_{23}$};
\draw[->,blue](0:0)--(-120:1cm)node [below]{$\uline{U}_{12}$};
\draw[->,red!50!blue](0:0)--(-138.4:.601cm)node [below]{$\uline{I}_{12}$};
\draw[->,red!50!blue](0:0)--(-90:.76cm)node [below]{$\uline{I}_{31}$};
\draw[<-,red!75!blue](-138.4:.601cm)-- ++(90:.76cm)node [below left]{$\uline{I}_{31}$};
\draw[->,red](0:0)--(141.2:.577cm)node [above left]{$\uline{I}_1$};
	\end{tikzpicture}
\end{align*}
\begin{align*}
\uline{S}&=\uline{U}_{12}\cdot \uline{I}^*_1=\underbrace{380\,\volt\cdot e^{-j120\,\degree}}_{(-190-j329{,}09)\,\volt}\cdot \underbrace{5{,}77\,\ampere\cdot e^{-j141{,}2\,\degree}}_{(-4{,}495-j3{,}614)\,\ampere}\\
&=2192{,}6\,\volt\ampere\cdot e^{j98{,}8\,\degree}=(\underbrace{-335{,}28}_{P}+\underbrace{j2165{,}92}_{jQ})\,\volt\ampere\\
P&=S\cdot \cos(\varphi)=2192{,}6\,\volt\ampere\cdot \underbrace{\cos(98{,}8\,\degree)}_{-0{,}153}=\uuline{-335{,}44\,\watt}\\
\end{align*}
\clearpage
}{}%
