ET2_Uebung_BEI/ET2_L_B18_A6.tex

\section {3-Phasen Spannungssystem}
Ein symmetrisches 3-Phasen Spannungssystem mit der Phasenlage 1-2-3 speist einen unsymmetrischen
Verbraucher mit den Impedanzen\\
$\uline{Z}_1=R_1$; $\quad\uline{Z}_2=R_2+jX_{L2}$; $\quad\uline{Z}_3=jX_{L3}$;\\
$U_{12}=400\,\volt\cdot e^{j30\degree}$; $\quad R_2=50\,\ohm$; $\quad X_{L2}=30\,\ohm$; $\quad X_{L3}=25\,\ohm$\\
Der Leistungsmesser zeigt $1323\,\watt$ an. Berechnen Sie den Strom $\uline{I}_N$.
\begin{align*}
	\begin{tikzpicture}[very thick,scale=2]
		\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,-1.5);
		\end{scope}
		\begin{scope}[>=latex,xshift=1cm,yshift=1.5cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_1$};
			\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
		\end{scope}
		\begin{scope}[>=latex,xshift=1cm,yshift=1.cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_2$};
			\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_2$};
		\end{scope}
		\begin{scope}[>=latex,xshift=1cm,yshift=.5cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_3$};
			\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_3$};
		\end{scope}
		\begin{scope}[>=latex,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(2,0)--(2,1.5)--(1.8,1.5);% (4.5,0)--(2.5,0)--(2.5,.2) (2,.8)--(2,1)--(1.8,1)(2.5,.8)--(2.5,1)--(2.6,1);
            \filldraw(0,0)circle(0.025cm)node [left]{$N$};
            \filldraw(0,.5)circle(0.025cm)(1,.5)--(0,.5)node [left]{$L3$};
            \filldraw(0,1)circle(0.025cm)(1,1)--(0,1)node [left]{$L2$};
            \filldraw(0,1.5)circle(0.025cm)node at(0,1.5) [left]{$L1$};
            \fill(.5,0)circle(0.025cm)(2,.5)circle(0.025cm)(2,1)circle(0.025cm);
			\draw [->,red] (1.25,.1)--(1,.1) node at (1.125,.1)[above]{$\uline{I}_N$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\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
\enlargethispage{2\baselineskip}
\begin{align*}
\uline{I}_N&=\uline{I}_1+\uline{I}_2+\uline{I}_3\qquad \text{(Um Ströme zu berechnen, Spannungen ermitteln)}\\
\uline{U}_1&=\frac{\uline{U}_{12}}{\sqrt{3}}=\frac{400\,\volt}{\sqrt{3}}=230{,}9\,\volt\cdot e^{j0\,\degree}\\
\uline{U}_2&=230{,}9\,\volt\cdot e^{-j120\,\degree}\\
\uline{U}_3&=230{,}9\,\volt\cdot e^{+j120\,\degree}\\
P_{Anzeige}&=I_1\cdot U_1\cdot \cos(\varphi_1)\qquad\text{ mit }\varphi_1=0\text{, da }\uline{Z}_1=R_1\\
I_1&=\frac{P_{Anzeige}}{U_1}=\frac{1323\,\watt}{230{,}9\,\volt}=5{,}73\,\ampere \qquad\Rightarrow\uline{I}_1=(5{,}73+j0)\,\ampere\\[.5\baselineskip]
\uline{I}_2&=\frac{\uline{U}_2}{\uline{Z}_2}=\frac{230{,}9\,\volt\cdot e^{-j120\,\degree}}{(50+j30)\,\ohm}=\frac{230{,}9\,\volt\cdot e^{-j120\,\degree}}{58{,}31\,\ohm\cdot e^{j30{,}96\,\degree}}=3{,}96\,\ampere\cdot e^{-j150{,}96\,\degree}=(-3{,}463-j1{,}923)\,\ampere\\
\uline{I}_3&=\frac{\uline{U}_3}{\uline{Z}_3}=\frac{230{,}9\,\volt\cdot e^{j120\,\degree}}{25\,\ohm\cdot e^{j90\,\degree}}=9{,}24\,\ampere\cdot e^{j30\,\degree}=(8+j4{,}62)\,\ampere\\[.5\baselineskip]
\uline{I}_N&=\uline{I}_1+\uline{I}_2+\uline{I}_3=(10{,}3+j2{,}7)\,\ampere=\uuline{10{,}6\,\ampere\cdot e^{j14{,}7\,\degree}}
\end{align*}
\clearpage
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