ET2_Uebung_BEI/ET2_L_B18_A8.tex

\section {Strangströme 3-Phasen System mit unsymmetrischem Verbraucher}
Ein symmetrisches 3-Phasen System mit der Phasenlage 1-2-3 speist einen unsymmetrischen
Verbraucher. Gegeben sind:\\[\baselineskip]
$\uline{U}_{12}=400\,\volt\cdot e^{j30\,\degree}$; $\quad \uline{X}_C=-100\,\ohm$; $\quad \uline{X}_{L_1}=125\,\ohm$;  $\quad X_{L_2}=60\,\ohm$;$\quad R=80\,\ohm$\\[\baselineskip]
Berechnen Sie die $3$ Strangströme, den Leiterstrom $\uline{I}_1$ und die Anzeige des Leistungsmessers!
\begin{align*}
	\begin{tikzpicture}[very thick,scale=2]
\draw[black!15!,very thin](0,0)grid(4,3);
		\begin{scope}[>=latex,very thick,xshift=2.268cm,yshift=1.995cm,scale=1,rotate=30]%Kondensator
			\draw (-1,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(2,0)node at (.5,.1133) [left] {$X_C$};
			\draw [<-,red] (1.15,.1)--(1.55,.1)node at(1.25,.1)[left]{$\uline{I}_C$};
		\end{scope}
		  \begin{scope}[>=latex,very thick,xshift=2.268cm,yshift=1.005cm,rotate=-30,scale=1]%Spule	
			\draw (-1,0)--(.3,0) (.7,0)--(2,0)node at (.5,-.1330667) [ left] {$X_{L_1}$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw [<-,red] (-.3,-.1)--(-.7,-.1)node at(-.5,-.14)[left]{$\uline{I}_L$};
		  \end{scope}
		  \begin{scope}[>=latex,xshift=4cm,yshift=2cm,rotate=90]%Wattmeter			
			\draw (0,0)--(.367,0) (.633,0)--(1,0) node at (.5,0)  {W};
			\draw (.5,0)circle(.133);
            \draw(.45,-.125)--(.45,-1)--(0,-1)--(-2,-1)--(-2,0)(.55,-.125)--(.55,-1)--(1,-1)--(1,0);
		  \end{scope}
		  \begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm,rotate=90]%Spule |
            \draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_{L_2}$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
			\draw [->,red] (.8,-.1)--(1.2,-.1)node at(1,-.1)[right]{\footnotesize$\uline{I}_{RL}$};
    	  \end{scope}
		  \begin{scope}[>=latex,very thick,xshift=4cm,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)--(4,0)(0,1.5)--(1.433,1.5)(0,3)--(4,3);
            \filldraw(0,0)circle(0.033cm)node [left]{L3};
            \filldraw(0,1.5)circle(0.033cm)node [left]{L2};
            \filldraw(0,3)circle(0.033cm)node [left]{L1};
           	\draw [->,blue] (0,2.8)--(0,1.7)node at(0,2.25)[right]{\footnotesize$\uline{U}_{12}$}; 		
           	\draw [->,blue] (0,1.3)--(0,.2)node at(0,.75)[right]{\footnotesize$\uline{U}_{23}$};
            \fill(1.433,1.5)circle(0.033cm)(4,3)circle(0.033cm)(4,2)circle(0.033cm)(4,1)circle(0.033cm)(4,0)circle(0.033cm);
			\draw [->,red] (.8,3.1)--(1.2,3.1)node at(1,3.1)[above]{\footnotesize$\uline{I}_1$};
            \draw [red!50!blue,very thick]node at (4,3)[above]{u};
            \draw [red!50!blue,very thick]node at (4,0)[below]{w};
            \draw [red!50!blue,very thick]node at (1.433,1.5)[above left]{v};
        \end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\
\begin{minipage}[c]{0.35\textwidth}
\begin{align*}
	\begin{tikzpicture}[very thick,scale=1.5]
\draw[thin](0,0)--(1,0);
\draw[->,blue](0:0)--(30:1cm)node [right]{$\uline{U}_{12}$};
\draw[->,blue](0:0)--(-90:1cm)node [below]{$\uline{U}_{23}$};
\draw[->,blue](0:0)--(150:1cm)node [left]{$\uline{U}_{31}$};
\draw[->,red](0:0)--(-153.2:.48cm)node [left]{$\uline{I}_1$};
\draw[thin](0:.7cm)arc(0:30:.7cm)node at(10:1cm){\footnotesize$30\,\degree$};
\draw[thin](30:.7cm)arc(30:150:.7cm)node at(90:1cm){\footnotesize$120\,\degree$};
	\end{tikzpicture}
\end{align*}
\end{minipage}
\begin{minipage}[c]{0.35\textwidth}
\begin{align*}
	\begin{tikzpicture}[very thick,scale=1.5]
\draw[thin](0,0)--(1,0);
\draw[thin](0:0cm)--+(0:1cm);%node [right]{$\uline{U}_{12}$};
\draw[thin](0:1cm)--+(-60:1cm);%node [below]{$\uline{U}_{23}$};
\draw[thin](0:1cm)--+(60:1cm);%node [left]{$\uline{U}_{31}$};
\draw[->,blue](0:0cm)--+(30:1.732cm)node [right]{$\uline{U}_{12}$};
\draw[->,blue](30:1.732cm)--+(-90:1.732cm)node [below]{$\uline{U}_{23}$};
\draw[->,blue](-30:1.732cm)--(0:0cm)node [left]{$\uline{U}_{31}$};
\draw[thin](0:.7cm)arc(0:30:.7cm)node at(15:1cm){\footnotesize$30\,\degree$};
	\end{tikzpicture}
\end{align*}
\end{minipage}
%%\hfill
\begin{minipage}[l]{0.3\textwidth}
\begin{align*}
\uline{U}_{12}&=400\,\volt\cdot e^{j30\,\degree}\\
\uline{U}_{31}&=400\,\volt\cdot e^{j150\,\degree}\\
\uline{U}_{31}&=400\,\volt\cdot e^{j270\,\degree}=400\,\volt\cdot e^{-j90\,\degree}\\
\end{align*}
\end{minipage}
Strangströme:
\begin{align*}
\uline{I}_C&=\frac{\uline{U}_{12}}{jX_C}=\frac{400\,\volt\cdot e^{j30\,\degree}}{100\,\ohm\cdot e^{-j90\,\degree}}=\uuline{4\,\ampere\cdot e^{j120\,\degree}}=\uuline{(-2+j3{,}46)\,\ampere}\\
\uline{I}_L&=\frac{\uline{U}_{23}}{jX_{L_1}}=\frac{400\,\volt\cdot e^{-j90\,\degree}}{125\,\ohm\cdot e^{j90\,\degree}}=\uuline{3{,}2\,\ampere\cdot e^{-j180\,\degree}}=-3{,}2\,\ampere\\
\uline{I}_{RL}&=\frac{\uline{U}_{31}}{R+jX_{L_2}}=\frac{400\,\volt\cdot e^{j150\,\degree}}{(80+j60)\,\ohm}=\frac{400\,\volt\cdot e^{j150\,\degree}}{100\,\ohm\cdot e^{j36{,}9\,\degree}}\\
&=\uuline{4\,\ampere\cdot e^{j113{,}1\,\degree}}=\uuline{(-1{,}57+j3{,}68)\,\ampere}\\
\intertext{Leiterströme:}
\uline{I}_1&=\uline{I}_C-\uline{I}_{RL}=(-0{,}439-j0{,}214)\,\ampere=\uuline{0{,}48\,\ampere\cdot e^{-j153\,\degree}}\\
\text{Zur Vollständigkeit}\\
\uline{I}_2&=\uline{I}_L-\uline{I}_C=(-1{,}2-j3{,}46)\,\ampere=\uuline{3{,}66\,\ampere\cdot e^{j109{,}1\,\degree}}\\
\uline{I}_3&=\uline{I}_{RL}-\uline{I}_L=(1{,}63+j3{,}68))\,\ampere=\uuline{4{,}03\,\ampere\cdot e^{j66{,}1\,\degree}}
\intertext{Anzeige der Wirkleistung:}
P&=\Re\{\uline{U}_{31}\cdot I^*_{RL}\}\\
&=\Re\{400\,\volt\cdot e^{j150\,\degree}\cdot 4\,\ampere\cdot e^{-j113{,}1\,\degree}\}\\
&=\Re\{1600\,\volt\ampere\cdot e^{j37\,\degree}\}\\
&=\Re\{(1280-j960)\,\volt\ampere\}=\uuline{1280\,\watt}\\
\intertext{oder}
P&=U_{31}\cdot I_{31}\cdot \cos(\varphi_{_{31}})=400\,\volt\cdot 4\,\ampere\cdot \cos(36{,}9\,\degree)=\uuline{1280\,\watt}\\
\intertext{oder}
P&=I^2_{RL}\cdot R=(4\,\ampere)^2\cdot 80\,\ohm=\uuline{1280\,\watt}\qquad \text{(Betrag von $\uline{I}_{RL}$!)}
\end{align*}
\clearpage
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