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- \section{Überlagerungsmethode}
- Berechnen Sie mit der Überlagerungsmethode den Strom $\uline{I}_C$
- \begin{align*}
- R&=R_L=10\,\ohm\quad L=50\,\milli\henry\quad C=100\,\micro\farad\\
- f&=50\,\hertz\quad \uline{U}_q=5\,\volt \cdot e^{j20\,\degree}\quad \uline{I}_q=2\,\ampere \cdot e^{-j60\,\degree}\\
- \end{align*}
- \begin{align*}
- \begin{tikzpicture}[scale=3]
- \begin{scope}[>=latex,very thick,xshift=0cm,yshift=.333cm,rotate=90,scale=.75]%Spule |
- \draw (.2,0)--(.3,0) (.7,0)--(0.9,0)node at(.5,-.0667) [right] {$L$};
- \fill (.3,-0.0667)rectangle(.7,0.0667);
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.083cm,rotate=90,scale=.75]
- \draw (0.1,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_L$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle |
- \draw (0,0)--(1,0) node at (.5,-.133) [right] {$\uline{U}_q$};
- \draw (.5,0)circle(.133);
- \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_q$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617
- \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Stromquelle
- \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$I_q$};
- \draw (.5,0)circle(.133);
- \draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_q$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator |
- \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
- \draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}_C$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
- \draw (0,.98)--(0,1)--(1,1) (1.8,1)--(3,1)--(3,.9) (0,.02)--(0,0)--(3,0)--(3,.2);
- \end{scope}
- \end{tikzpicture}
- \end{align*}
- \ifthenelse{\equal{\toPrint}{Lösung}}{%
- %\begin{align}
- %\intertext{Formeln:}
- %\text{...noch einfügen...}
- %\end{align}
- Berechnung:\\[\baselineskip]
- $R_L+jX_L \text{ unwirksam, da parallel zur Spannungsquelle.}$\\[\baselineskip]
- $\omega=2\pi f=2\pi\cdot 50\,\frac{1}{\second}=314{,}2\,\frac{1}{\second}$
- \begin{align*}
- \intertext{a) nur Spannungsquelle $\Rightarrow\uline{I}_q=0$}
- \begin{tikzpicture}[scale=3]
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle |
- \draw (0,0)--(1,0) node at (.5,-.133) [right] {$\uline{U}_q$};
- \draw (.5,0)circle(.133);
- \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_q$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617
- \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Kondensator |
- \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
- \draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}'_C$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
- \draw (1,.98)--(1,1)--(1.1,1) (1.8,1)--(2,1)--(2,.9) (1,.02)--(1,0)--(2,0)--(2,.2);
- \end{scope}
- \end{tikzpicture}
- \end{align*}
- \begin{align*}
- B_C&=\omega \cdot C=314{,}2\frac{1}{\,\second}\cdot 100\,\micro\farad=31{,}42\,\milli\siemens\\
- X_C&=-\frac{1}{B_C}=-\frac{1}{31{,}42\,\milli\siemens}=-31{,}831\,\ohm\\
- \uline{I}'_C&=\frac{\uline{U}_q}{R+jX_C}=\frac{4{,}698+j1{,}710}{10-j31{,}83}\,\ampere=\uline{(-6{,}693+j149{,}7)\,\milli\ampere}=149{,}8\,\milli\ampere\cdot e^{j92{,}3\,\degree}
- \end{align*}
- \begin{align*}
- \intertext{b) Nur Stromquelle $\Rightarrow\uline{U}_q=0$}
- \begin{tikzpicture}[scale=3]
- \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] {$G=\frac{1}{R}$};
- \draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}''$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Stromquelle
- \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$I_q$};
- \draw (.5,0)circle(.133);
- \draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_q$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator |
- \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
- \draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}''_C$};
- \end{scope}
- \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
- \draw (1,.98)--(1,1)--(2,1) (1.8,1)--(3,1)--(3,.9) (1,.02)--(1,0)--(3,0)--(3,.2);
- \end{scope}
- \end{tikzpicture}
- \end{align*}
- \begin{align*}
- \uline{I}''_C&=\uline{U}''\cdot jB_C\qquad \uline{U}''=\uline{I}_q\cdot \uline{Z}_{||}\\
- &\text{oder Stromteiler } \uline{I}''_C=\uline{I_q}\cdot\frac{R}{R-jX_C}\\[\baselineskip]
- \uline{I}_q&=2\,\ampere \cdot e^{-j60\,\degree}=(1-j1{,}732)\,\ampere\\
- G&=\frac{1}{R}=100\,\milli\siemens\\
- \uline{Y}_{||}&=G+jB_C=(100+j31{,}42)\,\milli\siemens=0{,}105\,\siemens \cdot e^{j17{,}44\,\degree}\\
- \uline{Z}_{||}&=\frac{1}{\uline{Y}_{||}}=9{,}54\,\ohm\cdot e^{-j17{,}44\,\degree}\\
- \uline{U}''&=\uline{I}_q\cdot \uline{Z}_{||}
- %\frac{\uline{I}_q}{Y_{||}}=\frac{\uline{I}_q}{G+jB_C}=
- %\frac{2\,\ampere \cdot e^{-j60\,\degree}}{(0{,}1+j0{,}03142)\,\siemens}\\
- %&=\frac{2\,\ampere \cdot e^{-j60\,\degree}}{0{,}105\,\siemens \cdot e^{-j17{,}44\,\degree}}
- =2\,\ampere \cdot e^{-j60\,\degree}\cdot 9{,}54\,\ohm \cdot e^{-j17{,}44\,\degree}\\
- &=19{,}08\,\volt\cdot e^{-j77{,}44\,\degree}=(4{,}15-j18{,}62)\,\volt\\[\baselineskip]
- %jB_C&=31{,}42\,\milli\siemens=31{,}42\,\milli\siemens\cdot e^{j90\,\degree}\\
- \uline{I}''_C&=\uline{U}''\cdot jB_C=19{,}08\,\volt\cdot e^{-j77{,}44\,\degree}
- \cdot 31{,}42\,\milli\siemens \cdot e^{j90\,\degree}\\
- &=600\,\milli\ampere\cdot e^{j12{,}56\,\degree}=\uline{(585{,}1+j130{,}3)\,\milli\ampere}\\
- %&=\frac{\uline{I}_q\cdot X_C}{G+jB_C}\\
- %=\frac{(1-j1{,}732)\,\ampere\cdot (-j31{,}42\cdot \power{10}{-3})\,\ampere\second\per\volt}{(0{,}1+j31{,}42\cdot \power{10}{-3})\,\ampere\second\per\volt}\\
- %&=\frac{(54{,}42+j31{,}42)\cdot \power{10}{-3}}{(0{,}1+j31{,}42\cdot \power{10}{-3})}\,\ampere
- %=\frac{(54{,}42+j31{,}42)}{(100+j31{,}42)\cdot }\,\ampere=(585{,}1+j130{,}3)\,\milli\ampere
- &\text{oder alternativ:}\\
- \uline{I}''_C&=\uline{I}_q\cdot \frac{R}{R+jX_C}\\
- \intertext{Überlagerung}
- \uline{I}_C&=\uline{I}'_C+\uline{I}''_C=(-6{,}69+j149{,}7+585{,}1+j130{,3})\,\milli\ampere\\
- &=\uuline{(578{,}41+j279{,}99)\,\milli\ampere}=\uuline{642{,}6\,\milli\ampere\cdot e^{+j25{,}38\,\degree}}\\
- \end{align*}
- \clearpage
- }{}%
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