ET2_Uebung_BEI/ET2_L_B16_A1.tex

\section{CLR Netzwerk}
Von dem Netzwerk sind folgende Daten bekannt:\\[\baselineskip]
$R_1=1,2\,\kilo\ohm\quad R_2=470\,\ohm\quad X_{C_0}=-906\,\ohm\quad X_{L_2}=628\,\ohm$\\
$\uline{I}_2=12\,\milli\ampere\cdot e^{j20\,\degree}$\\
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.5cm,rotate=90]%Spannungsquelle | 			
			\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_{0}$};
			\draw (.5,0)circle(.133);
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_0$};
			\draw [->,red] (.8,.2)--(1.2,.2)node at(1,.2)[left]{$\uline{I}_0$};	
		\end{scope}				
		\begin{scope}[>=latex,very thick,xshift=0cm,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] {$X_{C_0}$};
			\draw [->,blue] (0.3,-.25)--(0.7,-.25)node at(.25,-.4)[right]{\footnotesize$\uline{U}_{C_0}$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm,rotate=90] 			
			\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
			\draw [<-,blue] (0,-.5)--(1,-.5)node at(.5,-.5)[right]{\footnotesize$\uline{U}_2$};
			\draw [<-,red] (.8,.15)--(1.2,.15)node at(1,.1)[left]{\footnotesize$\uline{I}_1$}; 		
		\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] {$R_2$};
			\draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{I}_2$};			
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=2cm,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);
		\end{scope}		
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm]%Fehlstellen Eckverbindungen. 
			\draw (0,1.5)--(0,2)--(.2,2) (1,2)--(2,2)--(2,1.75)(0,.5)--(0,0)--(2,0)--(2,.25) 
			(1,0)--(1,.5) (1,2)--(1,1.5);
			\filldraw [red] (1,2)circle(0.02)node [above]{\footnotesize$KP$};
		\end{scope}							
	\end{tikzpicture}
\end{align*}
Berechnen Sie Schein- Wirk- und Blindleistung des Netzwerkes!\\[\baselineskip]
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:
\begin{align*}
\uline{S}&=\uline{U}\cdot \uline{I}^*\qquad\text{mit }\uline{U}_0=\uline{U}_{C_0}+\uline{U}_2\qquad\uline{I}_0=\uline{I}_1+\uline{I}_2\\
\intertext{Spannung $U_2$:}
\uline{Z}_2&=R_2+jX_{L_2}=(470+j628)\,\ohm=784{,}4\,\ohm\cdot e^{j53{,}19\,\degree}\\
\uline{U}_2&=\uline{Z}_2\cdot \uline{I}_2
=9{,}413\,\volt\cdot e^{j73{,}19\,\degree}=(2{,}722+j9{,}011)\,\volt\\
\intertext{Strom:}
\uline{I}_1&=\frac{\uline{U}_2}{R_1}=\frac{9{,}413\,\volt\cdot e^{j73{,}19\,\degree}}{1,2\,\kilo\ohm}=7{,}844\,\milli\ampere\cdot e^{j73{,}19\,\degree}
=(2{,}268+j7{,}5088)\,\milli\ampere\\
\uline{I}_0&=\uline{I}_1+\uline{I}_2=(2{,}268+j7{,}5088+11{,}276+j4{,}104)\,\milli\ampere\\
&=(13{,}544+j11{,}613)\,\milli\ampere=17{,}841\,\milli\ampere\cdot e^{j40{,}61\,\degree}\\
\intertext{Spannung:}
\uline{U}_{C_0}&=jX_{C_0}\cdot \uline{I}_0
=-j906\,\ohm\cdot 17{,}841\,\milli\ampere\cdot e^{j40{,}61\,\degree}=16{,}164\,\volt\cdot e^{-j49{,}39\,\degree}\\
&=(10{,}521-j12{,}271)\,\volt\\
\uline{U}_0&=\uline{U}_{C_0}+\uline{U}_2=(10{,}521-j12{,}271+2{,}722+j9{,}011)\,\volt\\
&=(13{,}243-j3{,}260)\,\volt=13{,}638\,\volt\cdot e^{-j13{,}83\,\degree}\\
\end{align*}
\begin{align*}
\intertext{Scheinleistung:}
\uline{S}&=\uline{U}_0\cdot \uline{I}_0^*=13{,}638\,\volt\cdot e^{-j13{,}83\,\degree}\cdot 17{,}841\,\milli\ampere\cdot e^{-j40{,}61\,\degree}=243{,}32\,\milli\volt\ampere\cdot e^{-j54{,}44\,\degree}\\[2\baselineskip]
S&=|\uline{S}|=\uuline{243{,}32\,\milli\volt\ampere}\\
P&=S\cdot \cos\varphi=243{,}32\,\milli\volt\ampere\cdot \underbrace{\cos(-54{,}44)}_{0{,}5816}=\uuline{141{,}50\,\milli\watt}\\
Q&=S\cdot \sin\varphi=243{,}32\,\milli\volt\ampere\cdot \underbrace{\sin(-54{,}44)}_{-0{,}813}=\uuline{-197{,}93\,\milli\,\var}\\
\intertext{Zweiter Weg:}
\uline{S}&=\uline{U}_0\cdot \uline{I}^*_0=\uline{I}_0\cdot \uline{Z}\cdot \uline{I}^*_0=I^2_0\cdot \uline{Z}\\
\uline{Z}&=jX_{C_0}+\frac{R_1\cdot (R_2-jX_{L_2})}{R_1+R_2-jX_{L_2}}=(445-j622)\,\ohm=764\,\ohm\cdot e^{-j54{,}44\,\degree}\\
\uline{S}&=I^2_0\cdot \uline{Z}
=(17{,}841\,\milli\ampere)^2\cdot 767\,\ohm\cdot e^{-j54{,}44\,\degree}=243{,}3\,\milli\volt\ampere\cdot e^{-j54{,}44\,\degree}\\
S&=243{,}3\,\milli\volt\ampere\\
P&=S\cdot\cos(-54{,}44\degree)=141{,}5\,\milli\watt\\
Q&=S\cdot\sin(-54{,}44\degree)=-197{,}9\,\milli\,\var
\end{align*}
\clearpage
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