187 lines
11 KiB
TeX
187 lines
11 KiB
TeX
\section{Stromortskurve}
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Konstruieren Sie die Stromortskurve $\uline{I}=\uline{g}(f)$ von dem abgebildeten Netzwerk!\\
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Entnehmen Sie der Stromortskurve den Strom $\uline{I}$ für $f_0=2,5\,\kilo\hertz$!\\[\baselineskip]
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Gegeben sind: $R_1=15\,\ohm$; $R_2=50\,\ohm$; $C=1{,}59\,\micro\farad$;\\ $0{,}5\,\kilo\hertz \leq f \leq 2{,}5\,\kilo\hertz$;\\
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$\uline{U}=10\,\volt$ = konstant\\[\baselineskip]
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Maßstäbe: $5\,\milli\siemens\,\widehat{=}\,1\,\centi\metre$ : $ 10\,\ohm\,\widehat{=}\, 1\,\centi\metre$ (Platzbedarf in x: $14\,\centi\metre$; in y: $12\,\centi\metre$)\\
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\begin{align*}
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\begin{tikzpicture}[very thick,scale=2]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
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% \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$};
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\end{scope}
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% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
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% \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
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% \end{scope}
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\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Kondensator |
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\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
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\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(2,0)--(2,.2) (.8,1)--(2,1)--(2,.8);
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\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$};
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\fill (-.5,0)circle(.025) (-.5,1)circle(.025);
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\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$};
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% \draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{R_C}$};
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% \draw [->,red] (3.1,.9)--(3.1,.7) node at (3.1,.8)[right]{$\underline{I}_C$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\ifthenelse{\equal{\toPrint}{Lösung}}{%
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%\begin{align}
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%\intertext{Formeln:}
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%\end{align}
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Berechnung:
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\begin{align*}
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\intertext{Parallelschaltung $\uline{Y}_P$}
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\uline{Y}_P&=G_2+jB\\
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G_2&=\frac{1}{R_2}=\frac{1}{50\,\ohm}=20\,\milli\siemens\\
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f_u&=0{,}5\,\kilo\hertz\qquad f_o=2{,}5\,\kilo\hertz \qquad\text{(Anmerkung)}\\
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B_u&=\omega_u\cdot C=2\cdot \pi\cdot f_u=2\cdot \pi\cdot 0{,}5\,\kilo\hertz\cdot 1{,}59\,\micro\farad =5\,\milli\siemens\\
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B_o&=\omega_o\cdot C=25\,\milli\siemens\qquad\text{(5-facher Wert von $B_u$)}\\
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\uline{Y}_P&=20\,\milli\siemens+j(5\ldots 25)\,\milli\siemens\\
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\end{align*}
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\vspace{-1cm}
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\begin{align*}
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\begin{aligned}
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\uline{Y}_{P,min}&=G_2 & (\text{für } & f=0) \qquad
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&\uline{Z}_{P,min}&=0 \quad & (\text{für } & f\negmedspace\rightarrow\negmedspace\infty)\\
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\uline{Y}_{P,max}&=\infty & (\text{für } & f\negmedspace\rightarrow\negmedspace\infty) \qquad
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&\uline{Z}_{P,max}&=R_2 \quad &(\text{für } & f=0)\\
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\end{aligned}
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\end{align*}\\
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Zeichnen der Leitwertgeraden $\uline{Y}_P$; Konstruktion des Halbkreises für $\uline{Z}_P$
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\begin{align*}
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(G_2=20\,\milli\siemens\,\widehat{=}\,4\,\centi\metre : R_2 = 50\,\ohm\,\widehat{=}\, 5\,\centi\metre)
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\end{align*}
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Einzeichnen der $\uline{Z}_{P_o}$, $\uline{Z}_{P_u}$ Linien (Zeiger).\\
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\clearpage
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Berechnen der Gesamtschaltung $\uline{Y}$\\
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Graphisch wird der Widerstand $R_1$ addiert, durch verschieben der $\Im$ Achse um $1{,}5\,\centi\metre$ nach links.\\
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\begin{align*}
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\uline{Y}_{min}&=\frac{1}{R_1+R_2}=\frac{1}{65\,\ohm}=15{,}4\,\milli\siemens\,\widehat{=}\,3{,}08\,\centi\metre \qquad (f=0)\\
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\uline{Y}_{max}&=\frac{1}{R_1}=\frac{1}{15\,\ohm}=66{,}7\,\milli\siemens\,\widehat{=}\,13{,}33\,\centi\metre\qquad (f\rightarrow\infty)\\
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\end{align*}
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Zeichnen des Leitwertkreises $\uline{Y}$ mit Radius $r$:
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\begin{align*}
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r&=\frac{\uline{Y}{max}-\uline{Y}_{min}}{2}\cdot \hspace{-1cm} \underbrace{\frac{1\,\centi\metre}{5\,\milli\siemens}}_{\text{Maßstabsumwandlung}}\hspace{-1cm}=5{,}12\,\centi\metre\\
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\end{align*}
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Der Mittelpunkt ergibt sich aus:
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\begin{align*}
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\uline{Y}_{min}+r=3{,}08\,\centi\metre+5{,}12\,\centi\metre=8{,}2\,\centi\metre
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\end{align*}
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zu messen ab neuer imaginärer Achse.\\
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Einzeichnen der $\uline{Z}_o$, $\uline{Z}_u$ Linien (Zeiger).\\
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Spiegeln der Zeiger $\uline{Z}_o$, $\uline{Z}_u$ an der reellen Achse liefert Schnittpunkt mit $\uline{Y}$. Diese sind entsprechend $\uline{Y}_o$, $\uline{Y}_u$.\\
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\clearpage
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$x=14\,\centi\metre; y=12\,\centi\metre$
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\begin{align*}
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\begin{tikzpicture}[very thick,scale=1]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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\draw[ultra thin,black!50!](-2,-6)grid(12,6);
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\draw[thin](5,-6)--(5,6)(10,-6)--(10,6)(-2,5)--(12,5)(-2,5)--(12,5);
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\draw[thin,->](0,0)--(12.5,0)node[right]{$\Re$};
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\draw[thin,->](0,-6.5)--(0,6.5)node[above]{$\Im$};
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\draw[blue,->,ultra thick](0,0)--(5,0)node[below left]{$R_2$};
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\draw[blue,thin](2.5,0)circle(2.5cm);
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\fill[blue](2.5,0)circle(.075cm);
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\draw[red,->](0,0)--(4,0)node[below left]{$G_2$};
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\draw[red!50!blue,thin](6.67,0)circle(5.125cm);% Y-Kreis
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\fill[red!50!blue](6.67,0)circle(.075cm);
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\draw[red!50!blue](1.525,-.1)--(1.525,.1)(1.525,0)node[below]{$\uline{Y}_{min}$};
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\draw[red!50!blue](11.807,-.1)--(11.807,.1)(11.807,0)node[below]{$\uline{Y}_{max}$};
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\draw[red,thin](4,-6)--(4,6);
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\draw[red,|-|](4,1)--(4,5)node at (4.5,5){$f_o$};
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\draw[red]node at (4.5,1){$f_u$};
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\draw[red]node at (4.5,3){$\uline{Y}_P$};
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\draw[red]node at (4.5,-3){$\uline{Y}^*_P$};
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\draw[red]node at (6.5,4.5){$\uline{Y}$};
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\draw[red,|-|](4,-1)--(4,-5)node at (4.5,-5){$f_o$};
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\draw[red]node at (4.5,-0.75){$f_u$};
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\draw[blue,thin](0,0)--(8,-2)node at (3.0,-1.75){$\uline{Z}_P$; $\uline{Z}$};
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\draw[blue,thick,->](0,0)--(-14.036:4.851cm)node at(3,-.5){$\uline{Z}_{P_u}$};
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\draw[green!50!black,->](-1.5,0)--(-14.036:4.851cm)node at(3,-1.25){$\uline{Z}_u$};
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\draw[->,green!50!black](-1.5,0)--(1.9,-2.4)node at (-.5,-1){$\uline{Z}_o$};
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\draw[blue,thin](0,0)--(4,-5);
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\draw[->,blue](0,0)--(1.9,-2.4)node at(1.2,-1){$\uline{Z}_{P_o}$};
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\draw[blue,ultra thick](0:2.5cm)+(257:2.5cm)arc(257:332:2.5cm);%Mittelpunkt+Start arc Start:End:Radius
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\draw[blue]node at (5.5,-1.25){$f_u$};
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\draw[blue]node at (2,-2.75){$f_o$};
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\draw[blue!50!red,thin](-1.5,0)--(9.5,2)node at (2.5,1.5){$\uline{Y}$}; \draw[blue!50!red,thin](-1.5,0)--+(35.2:8.5cm);
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\draw[blue!50!red,ultra thick](0:6.67cm)+(148:5.125cm)arc(148:174.5:5cm);%Mittelpunkt+Start arc Start:End:Radius
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% \draw[blue!50!red]node at (1.25,.75){$f_u$};
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\draw[blue!50!red]node at (2,2.825){$f_o$};
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\draw[blue!20!red,ultra thick,->](-1.5,0)--+(35.2:4.72cm)node at (.5,2){$\uline{Y}(f_o)$};
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\draw[blue!50!red,ultra thick,->](-1.5,0)--+(10.3:3.15cm)node at (.5,.625){$\uline{Y}(f_u)$};
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\draw[blue!50!red]node at (5.5,1.25){$f_u$};
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% \draw[blue!50!red]node at (2.5,3.25){$f_o$};
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\draw[blue!50!red,thin,->](-1.5,-6.5)--(-1.5,6.5)node[above]{$\Im$ neu};
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\draw[blue!50!red,->](-1.5,0)--(0,0)node[below left]{$R_1$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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Ablesen der $\uline{Y}$ Werte und Maßstabsumrechnung $M=5\,\milli\siemens/\centi\metre$ ergibt Bereich zwischen $f_u$ und $f_o$.
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\begin{align*}
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\text{ablesen: } \uline{Y}(f_o)&=23{,}6\,\milli\siemens\cdot e^{(j35{,}2\,\degree)}\quad\text{aus Zeichnung $4{,}72\,\centi\metre$}\\
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\uline{I}&=\uline{Y}\cdot \uline{U}=\uuline{236\,\milli\ampere\cdot e^{(j35{,}2\,\degree)}}\\
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\uline{I}&=\uuline{(193+j136)\,\milli\ampere}\\
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\uline{Y}_P \text{ Gerade } \Rightarrow \uline{Z}_P \text{ Kreis } \\
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\uline{Z}_{P_u}, \uline{Z}_{P_o} \text{ Linie }\\
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R_1 \text{ addieren }\\
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\uline{Y}_{min} \text{ Kreis } \\
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\Rightarrow \uline{Y}(f_u); \uline{Y}(f_o)
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\end{align*}
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\clearpage
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\subsubsection*{Kurzanleitung}
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(Siehe Merksätze zu Inversion im Script, hier graphisch veranschaulicht:)
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\begin{align*}
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\begin{tikzpicture}[very thick,scale=1.25]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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\draw(0,-1)--(0,1)node[above]{$\Im$};
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\draw(0,0)--(2,0);
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\draw[red](1,-1)--(1,1);
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\draw[red,->](0,0)--(1,.75)node at(1.25,.75){$\uline{Y}_p$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]
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\draw(0,-1)--(0,1)node[above]{$\Im$};
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\draw(0,0)--(2,0);
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\draw[blue](.5,0)circle(.5cm);
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\draw[blue,->](0,0)--(.7,-.5cm)node at(1.25,-.5){$\uline{Z}_p$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=6cm,yshift=0cm]
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\draw(0,-1)--(0,1);
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\draw(-.5,0)--(2,0);
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\draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
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\draw[blue](.5,0)circle(.5cm);
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\draw[green!50!black,->](-.5,0)--(.7,-.5cm)node at(1,-.5){$\uline{Z}$}; \end{scope}
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\begin{scope}[>=latex,very thick,xshift=9cm,yshift=0cm]
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\draw(0,-1)--(0,1);
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\draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
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\draw(-.5,0)--(2.5,0);
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\draw[blue](1.25,0)circle(1cm);
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\draw[blue!50!red,->](-.5,0)--(2.1,.5)node at(2.5,.5){$\uline{Y}$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\begin{enumerate}
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\item Leitwertortskuve der Parallelschaltung von $R_2$ und $C$ ergibt eine Gerade die im Abstand $G_2$ parallel zur imaginären Achse liegt.\\
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$G_2=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre$
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\item Grenzen $f_u: \uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre$, $f_o: \uline{Y}_P=25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre$ einzeichnen, auch für $\uline{Y}^*_P$
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%Grenzen eintragen für $\uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{ auch für $\uline{Y}^*_P$}$\\
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\item Inversion von $\uline{Y}_P$ liefert die Widerstandsortskurve $\uline{Z}_P$ ein Kreis durch den Ursprung.\\
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$\uline{Z}_{P_{min}}=0;\quad \uline{Z}_{P_{max}}=R_2=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\rightarrow\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}$
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\item Der Serienwiderstand $R_1$ wird durch verschieben der imaginären Achse addiert $R_1=15\,\ohm\,\widehat{=}\,1{,}5\,\centi\metre$ (Verschiebung nach links).
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\item Punktweise Inversion von $\uline{Z}$ liefert $\uline{Y}$ mit\\ $\uline{Y}_{min}=15{,}4\,\milli\siemens\,\widehat{=}\,3{,}08\,\centi\metre$ und $\uline{Y}_{max}=66{,}7\,\milli\siemens\,\widehat{=}\,13{,}3\,\centi\metre$,\\ wieder einen Kreis der jetzt nicht mehr durch den Ursprung geht.
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\item Ablesen von $\uline{Y}(f_o)=4{,}72\,\centi\metre$ und Winkel $35{,}2\,\degree \\ \Rightarrow \uline{Y}(f_o)=23{,}6\,\milli\siemens\cdot e^{(j35{,}2\,\degree)}$.
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\item Berechnen des Stroms \ldots \quad ;-)
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\end{enumerate}
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\clearpage
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}{}%
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