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- \section {Ringspule}
- Durch das Zentrum einer Ringspule mit rechteckigem Querschnitt (Länge
- $l=12\,\milli\metre$, $r_i=10\,\milli\metre$, $r_a=20\,\milli\metre$;
- Windungszahl $N= 1800$) wird ein Leiter geführt, in dem ein Wechselstrom fließt.
- ($I_{eff}=40\,\ampere$; $f=50\,\hertz$).\\
- Berechnen Sie den Effektivwert $U$ der Spannung $u$.\\
- $\mu=\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
- \begin{align*}
- \begin{tikzpicture}[scale=2.5, z=0.2pt]
- \begin{scope}[>=latex, xshift=0, yshift=0]
- \foreach \z in {0,0.2,...,1}
- \fill [black!25!](0,0,\z cm) circle (0.7cm);% Zylinder außen
- \draw [black!75!](0,0,1 cm) circle (0.7cm);%Hintere äußere Ringkante
- %\foreach \z in {0,0.2,...,0.5}
- %\fill [black!50!](0,0,\z cm) circle (0.5cm);% Zylinder innen
- \fill [black!25!](0,0) circle (0.7cm);% Stirnseitenfüllung
- \fill [black!35](0,0) circle (0.5cm); %Füllung innen
- \begin{scope}%Clip Füllung
- \clip (0,0,1cm) circle (0.5cm);%Clip mit hinterer inneren Ringkante
- \fill [black!0!](0,0) circle (0.5cm);%Innere Ringkante
- \end{scope}
- \begin{scope}%Clip hintere innere Kante
- \clip (0,0) circle (0.5cm);%Clip mit hinterer inneren Ringkante
- \draw [black!75!](0,0,1cm) circle (0.5cm);%Hintere innere Ringkante
- \end{scope}
- \draw [black!75!](0,0) circle (0.5cm);% Innere Ringkante
- \draw [black!75!](0,0) circle (0.7cm);% Äußere Ringkante
- \foreach \a in {-45,-30,-15,15,30,45,60,75,90,105,120,135} %Rechte Drähte
- \draw [red!60!black, ultra thick] (\a:0.5)--(\a:0.7)--+(45:.3);
- \foreach \a in {150,165,...,300} %Linke Drähte
- \draw [red!60!black, ultra thick] (\a:0.7)--(\a:0.5)--+(45:.3);
- \draw [red!60!black, ultra thick] (0.5,0,0)--(1.5,0,0)circle(0.01);%Anschußdrähte
- \draw [red!60!black, ultra thick] (0.7,0,1cm)--(1.5,0,1cm)circle(0.01);%Anschußdrähte
- \draw [->,blue, thick] (1.6,0,1cm) -- (1.6,0,0) node at (1.6,0,.5cm)[right] {$u$};
- \draw [blue, ultra thick] (0,0,3.5cm) -- (0,0,7cm);
- \draw [->,red, thick] (0,0,7.5cm) -- (0,0,8cm)node [right] {$i$};
- \draw [blue, ultra thick] (0,0,1.75cm) -- (0,0,-5cm)node [above left]{Leiter};
- \end{scope}
- \end{tikzpicture}
- \begin{tikzpicture}[scale=1.4]%Querschnitt
- \begin{scope}[>=latex,xshift=0,yshift=0]
- \fill [black!25!] (0,0)rectangle(1,1);
- \end{scope}
- \begin{scope}[>=latex,xshift=0,yshift=3cm]
- \fill [black!25!] (0,0)rectangle(1,1);
- \end{scope}
- \begin{scope}[>=latex,xshift=0,yshift=0cm]
- \draw [very thick] (0,0)rectangle(1,4);
- \end{scope}
- \begin{scope}[>=latex,xshift=-1cm,yshift=1.9cm]
- \fill [blue!50!] (0,0)rectangle(3,.2);
- \end{scope}
- \begin{scope}[>=latex,xshift=0cm,yshift=4cm]%Hilfslinien
- \draw [very thin] (-1,0)--(0,0)--(0,.5) (1,0)--(1,.5)(-.5,-1)--(0,-1);
- \draw [very thin, dashed] (-1.5,-2)--(2.5,-2);
- \draw [->,very thin] (-.9,-2)--(-.9,0)node at(-.9,-1)[right]{$r_a$};
- \draw [->,very thin] (-.4,-2)--(-.4,-1)node at(-.4,-1.5)[right]{$r_i$};
- \draw [<->,very thin] (0,0.4)--(1,0.4)node at(.5,0.4)[above]{$l$};
- \draw node at(.5,-.5){$\mu=\mu_0$};
- \draw [->,red,thin] (1.25,-1.75)--(1.5,-1.75) node [right]{$i$};
- \end{scope}
- \end{tikzpicture}
- \end{align*}
- \ifthenelse{\equal{\toPrint}{Lösung}}{%
- \begin{align}
- \intertext{Formeln:}
- u&=N\cdot \frac{d\phi(t)}{dt}
- \end{align}
- Berechnung:
- Strom im Leiter bewirkt Fluss durch die Spule:
- \begin{align*}
- \phi&=\widehat{\phi}_0\cdot \sin(\omega t)\\
- \widehat{\phi}_0&=\int_{r_i}^{r_a}l\cdot \widehat{B}(r)\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \int_{r_i}^{r_a}\frac{1}{r}\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \ln\frac{r_a}{r_i}\\
- \text{mit }\widehat{i}&=\sqrt{2}\cdot I\\
- \Rightarrow \widehat{\phi}_0&=\frac{1{,}26\cdot \power{10}{-6}\, \frac{\volt\second}{\ampere\metre}\cdot 12\cdot \power{10}{-3}\,\metre\cdot \sqrt{2}\cdot 40\,\ampere}{2\pi}\cdot \ln(2)=9{,}44\cdot \power{10}{-8}\,\volt\second\\
- \phi_{eff}&=\frac{\widehat{\phi}_0}{\sqrt{2}}=6{,}672\cdot \power{10}{-8}\,\volt\second\\
- \intertext{Veränderlicher Fluss induziert Spannung:}
- u&=N\cdot \frac{d\phi(t)}{dt}=N\cdot \widehat{\phi}_0\cdot \omega\cdot \cos(\omega t)=\widehat{u}\cdot \cos(\omega t)\\
- \widehat{u}&=N\cdot \phi_0\cdot \omega=1800\cdot 9{,}44\cdot \power{10}{-8}\,\volt\second\cdot 2\pi\cdot 50\,\power{\second}{-1}=53{,}36\,\milli\volt\\
- U&=\frac{\widehat{u}}{\sqrt{2}}=\frac{53{,}36\,\milli\volt}{\sqrt{2}}=\uuline{37{,}73\,\milli\volt}
- \intertext{Anmerkung:}
- u_i&=2\cdot a\cdot r\cdot N\cdot B\cdot \omega\cdot \sin(\omega t)\\
- &=2\cdot a\cdot r\cdot N\cdot \frac{\Phi}{A}\cdot \omega\cdot \sin(\omega t)\\
- &\text{mit $2\cdot a\cdot r=A$}\\
- &=N\cdot \Phi \cdot \omega\cdot \sin(\omega t)\\
- &=\widehat{u}_i\cdot \sin(\omega t)\\
- \intertext{oder}
- U&=\mu_0\cdot l\cdot \ln\frac{r_a}{r_i}\cdot N\cdot f\cdot I\\
- &=1{,}26\cdot \power{10}{-6}\,\frac{\volt\cancel{\second}}{\cancel{\ampere}\cancel{\metre}}\cdot 12\cdot \power{10}{-3}\,\cancel{\metre}\cdot \ln\frac{20\,\cancel{\milli\metre}}{10\,\cancel{\milli\metre}}\cdot 1800\cdot 50\,\cancel{\frac{1}{\second}}\cdot 40\,\cancel{\ampere}=\uuline{37{,}73\,\milli\volt}
- \end{align*}
- \clearpage
- }{}%
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