ET2_Uebung_BEI/ET2_L_B13_A7.tex

\section{Werte $R_L$ und $L$ einer Spule}
Aus den drei gemessenen sinusförmigen Spannungen $U$, $U_N$, und $U_{SP}$ lassen sich die Werte $R_L$ und $L$ einer Spule bestimmen.
\begin{align*}
U=100\,\volt\\
U_N=60\,\volt\\
U_{SP}=70\,\volt\\
R_N=60\,\ohm\\
f = 50\,\hertz
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Zeichnen Sie ein qualitatives Zeigerdiagramm der Spannungen!
\item Bestimmen Sie $R_L$ und $L$!
\end{enumerate}
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand			
			\draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_N$};
			\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_N$};
			\draw (0,1)--(0,0)--(.1,0) (3,1)--(3,0)--(2.9,0);%anschuß und Füllt die Ecken der Verbindung!
			\fill (0,1)circle (0.025) (3,1)circle (0.025);
			\draw [->,blue](.2,1)--(2.8,1)node at (1.5,1)[below]{\footnotesize$U$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Widerstand			
			\draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_L$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]%Spule 			
			\draw (0,0)--(.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$L$};
			\fill (.2,-0.1)rectangle(.8,0.1);
			\draw [->,blue] (-.7,-.2)--(.7,-.2) node at (0,-.2)[below]{\footnotesize$U_{SP}$};		 
		\end{scope}					
	\end{tikzpicture}		
\end{align*}	
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
\begin{align*}
\intertext{Berechnung:}
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=.5]
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
		  \draw [black!25!,very thin](0,0)grid(8,7);
			\draw [->,red, thick](0,0)--(8,0)node [right] {$\underline{I}$};						 
			\draw [->](0,0)--(6,0)node at(3,0)[below] {$\underline{U}_N$};	
			\draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{U}_{SP}$};
			\draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{U}$};
			\draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$\underline{U}_{R_L}$};
			\draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$\underline{U}_{L}$};
			\draw [black!50!](33:10)arc(33:53:10)node [left]{$\underline{U}=100\,\volt\widehat{=}10\,\centi\metre$};
			\draw [black!50!](6,0)+(90:7)arc(90:70:7)node [right]{$\uline{U}_{SP}=70\,\volt\widehat{=}7\,\centi\metre$};
			\draw [black!50!](3,-1)node[below]{$\underline{U}_N=60\,\volt\widehat{=}6\,\centi\metre$};		
		\end{scope}
		\begin{scope}[>=latex,very thick, xshift=12cm, yshift=2cm]
		\draw node at(0,2)[right]{$I$ zeichnen};
		\draw node at(0,1)[right]{$u_N || I$};
		\draw node at(0,0)[right]{Mit Zirkel $U$ und $U_{SP}$};
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}
I&=\frac{U_N}{R_N}=\frac{60\,\volt}{60\,\ohm}=\uuline{1\,\ampere}\\
\underline{U}_{SP}&=70\,\volt=\sqrt{U^2_{RL}+U^2_L}\\
\end{align*}
\clearpage
Widerstandsoperatoren:\\

\footnotesize{Impedanzdreieck wie Spannungsdreieck}
\begin{align*}
	\begin{tikzpicture}[scale=.5]
		\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
		  \draw [black!25!,very thin](0,0)grid(8,7); 		
			\draw [->](0,0)--(6,0)node at(3,0)[below] {$R_N$};	
			\draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{Z}_{SP}$};
			\draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{Z}$};
			\draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$R_L$};
			\draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$X_L$};																 
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}
Z_{SP}&=\frac{U_{SP}}{I}=\frac{70\,\volt}{1\,\ampere}=70\,\ohm \quad \text{\footnotesize{(Nur Effektivwerte - ohne Winkel)}}\\%=\sqrt{R^2_L+X^2_L}\\
Z^2_{SP}&=R^2_L+X^2_L=(70\,\ohm)^2\\
X^2_L&=(70\,\ohm)^2-R^2_L \tag{1}\\[\baselineskip]
Z&=\frac{U}{I}=\frac{100\,\volt}{1\,\ampere}=100\,\ohm\\
Z^2&=(R_N+R_L)^2+X^2_L=(100\,\ohm)^2\\
X^2_L&=(100\,\ohm)^2-(R_N+R_L)^2\\
&=(100\,\ohm)^2-(R^2_N+2\cdot R_N\cdot R_L+R^2_L) \tag{2}\\[\baselineskip]
(70\,\ohm)^2-\cancel{R^2_L}&=(100\,\ohm)^2-R^2_N-2\cdot R_N\cdot R_L-\cancel{R^2_L}\tag{$1$ in $2$}\\
2\cdot R_N\cdot R_L&=(100\,\ohm)^2-R^2_N-(70\,\ohm)^2\\
R_L&=\frac{(100\,\ohm)^2-R^2_N-(70\,\ohm)^2}{2\cdot R_N}=\frac{(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2}{2\cdot 60\,\ohm}\\
&=\frac{1500\,\ohm^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm}\\
%(100\,\ohm)^2&=(60\,\ohm)^2+2\cdot 60\,\ohm\cdot R_L+\cancel{R^2_L}+(70\,\ohm)^2 -\cancel{R^2_L}\\
%2\cdot 60\,\ohm\cdot R_L&=(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2=1500(\,\ohm)^2\\
%R_L&=\frac{1500(\,\ohm)^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm} \tag{in $1$}\\
\text{in (1) }\qquad X_L&=\sqrt{(70\,\ohm)^2-(12{,}5\,\ohm)^2}=68{,}87\,\ohm\\
L&=\frac{X_L}{\omega}=\frac{68{,}87\,\ohm}{2\pi\cdot 50\,\frac{1}{\second}}=\uuline{0{,}219\,\henry}
\end{align*}
\clearpage
}{}%