ET2_Uebung_BEI/ET2_L_B18_A1.tex

\section {Übertrager im Leerlauf}
$U_1=230\,\volt$, $R_1=5\,\ohm$, $I_1=10\,\ampere$, $U_2=100\,\volt$\\
ausgangsseitiger Leerlauf\\
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie den Eigangswiderstand $\uline{Z}_1=\frac{\uline{U}_1}{\uline{I}_1}$ nach Betrag und Phase sowie die aufgenommene Wirk- und Blindleistung.
\item Berechnen sie $\omega L_1$, $\omega L_2$ und $\omega M$ unter Annahme einer idealen Kopplung.
\end{enumerate}
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
			\draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
			\draw (.5,0)circle(.133);	
			\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_1$};	
		\end{scope}				
		\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] {$R_1$};
			\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spule |	
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_1$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
            \fill(.825,.075)circle(.033);
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule |	
			\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_2$};
			\fill (.3,-0.0667)rectangle(.7,0.0667);
            \fill(.825,-.075)circle(.033);	
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
            \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
			\draw [->,red] (1,.1)--(.75,.1)node at(.875,.1)[above right]{\footnotesize$\uline{I}_2=0$};
		\end{scope}	
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
			\draw (0,0)--(1,0)--(1,.2) (2.5,0)--(1.5,0)--(1.5,.2) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9);
            \fill (2.5,0)circle(0.025cm)(2.5,1)circle(0.025cm);
			\draw [->,blue] (2.5,.8)--(2.5,.2) node at (2.5,.5)[right]{$\uline{U}_2$};
        \draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$M$};
		\end{scope}	
	\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\
\begin{enumerate}
\item Komplexer Eingangswiderstand\\
Begriffe:\\
$\uline{Z}$ Impedanz oder komplexer Widerstand\\
$Z=|\uline{Z}|$ Scheinwiderstand \\
\begin{minipage}[c]{0.25\textwidth}
    \begin{tikzpicture}[scale=.2]
        \draw[->](0,0)--(5,0)node[right]{$R_1$};
        \draw[->](5,0)--(5,22.45)node at(5,10)[right]{$\omega L_1$};
        \draw[->](0,0)--(77.44:23cm)node at(0,10)[left]{$\uline{Z}_1$};
        \draw[->](2,0)arc(0:77.4:2cm)node at(1,2)[right]{$\varphi_1$};
	\end{tikzpicture}
\end{minipage}
\begin{minipage}[c]{0.75\textwidth}
\begin{align*}
Z_1&=\frac{U_1}{I_1}=\frac{230\,\volt}{10\,\ampere}=23\,\ohm\\
\text{mit }R_1&=Z_1\cdot \cos\varphi_1\Rightarrow\\
\varphi_1&=\arccos\frac{R_1}{Z_1}=\arccos\frac{5\,\ohm}{23\,\ohm}=77{,}44\,\degree\\
\uline{Z}_1&=Z_1\cdot e^{j\varphi_1}=\uuline{23\,\ohm\cdot e^{j77{,}44\,\degree}}\\
P&=U_1\cdot I_1\cdot \cos\varphi_1=230\,\volt\cdot 10\,\ampere\cdot \cos(77{,}44\,\degree)=\uuline{500\,\watt}\\
Q&=U_1\cdot I_1\cdot \sin\varphi_1=230\,\volt\cdot 10\,\ampere\cdot \sin(77{,}44\,\degree)=\uuline{2245\,var}\\
\text{oder  }\uline{S}&=P+jQ=\frac{\uline{U}_1^2}{\uline{Z_1}}=\frac{(230\,\volt)^2}{23\,\ohm\cdot e^{j77{,}44\,\degree}}=(\underbrace{500}_{P}-j\underbrace{2245}_{Q})\,\volt\ampere
\end{align*}
\end{minipage}
\clearpage
\item  Blindwiderstände und Kopplungswiderstand\\
Begriffe:\\
$L$ Selbstinduktivität\\
$M$ Gegeninduktivität\\
$X_L=\omega L$ Reaktanz oder Blindwiderstand\\ 
$X_M=\omega M$ Kopplungswiderstand\\
\begin{align*}
\omega L_1&=Z_1\cdot \sin\varphi_1=23\,\ohm\cdot \sin(77{,}44\,\degree)=\uuline{22{,}45\,\ohm}\\
U_2&=\omega M\cdot I_1\Rightarrow\\
\omega M&=\frac{U_2}{I_1}=\frac{100\,\volt}{10\,\ampere}=\uuline{10\,\ohm}\\
M&=\sqrt{L_1\cdot L_2}\\
\omega M&=\sqrt{\omega L_1\cdot \omega L_2}\\
\Rightarrow\omega L_2&=\frac{(\omega M)^2}{\omega L_1}=\frac{(10\,\ohm)^2}{22{,}45\,\ohm}=\uuline{4{,}45\,\ohm}
\end{align*}
\end{enumerate}
\clearpage
}{}%