91 lines
4.6 KiB
TeX
91 lines
4.6 KiB
TeX
\section{Momentan Leistung}
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In der Reihenschaltung fließt der Strom\\
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\begin{align*}
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i(t)&=
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\begin{cases}
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&0 \text{ für }t<0\\
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&I_0\cdot \sin (\omega t) \text{ für }t\geq 0\\
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\end{cases}
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\end{align*}
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Der Kondensator ist zur Zeit $t=0$ entladen.\\
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\renewcommand{\labelenumi}{\alph{enumi})}
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\begin{enumerate}
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\item Berechnen Sie die momentanen Spannungen an $R$, $L$ und $C$ zur Zeit $t_1=350\,\micro\second$.
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\item Welche Leistung nimmt die Schaltung in diesem Moment auf?
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\item Hinweis: Rechnung mit komplexen Größen wäre hier falsch. Warum?
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\end{enumerate}
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\begin{align*}
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R&=12\,\ohm\qquad L=1{,}3\,\milli\henry\qquad C=8{,}7\,\micro\farad\\
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f&=2\,\kilo\hertz\qquad I_0=10\,\milli\ampere\\
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\end{align*}
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\begin{align*}
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\begin{tikzpicture}[scale=2.5]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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] {$R$};
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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=1cm,yshift=0cm]%Spule -
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\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$};
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\fill (.3,-0.0667)rectangle(.7,0.0667);
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\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{L}$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]%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) [above] {$C$};
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\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C}$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Strompfeil
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\draw [->,red] (-.2,-.2)--(.2,-.2)node at(0,-.2)[below]{\footnotesize$I_0$};
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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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%\text{...noch einfügen...}
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%\end{align}
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Berechnung:
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\begin{align*}
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\omega&=2\pi f=1{,}257\cdot\power{10}{4}\,\frac{1}{\second}\\
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\omega t_1&=1{,}257\cdot \power{10}{4}\,\frac{1}{\second}\cdot 350\cdot \power{10}{-6}\,\second=4{,}398\,[rad]\,\widehat{=}\,252\,\degree\\
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\end{align*}
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Zur Erklärung wie die Schwingung aussieht:
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\begin{align*}
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\begin{tikzpicture}[scale=1]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
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\draw [->] (0,2)--(4.2,2)node[right] {$\omega t$}; % Draw x axis;
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\draw [->] (0,0)--(0,4)node[above] {$i$}; % Draw y axis;
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\draw [red, very thick](0,2) sin (1,4) cos (2,2) sin (3,0) cos (4,2);
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\draw node at(1.8,3)[right] {$i(t)$};
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\draw node at(2,2)[below] {$\pi$};
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\draw node at(4,2)[below] {$2\pi$};
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\draw node at(2.8,2)[below] {$\omega t_1$};
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\filldraw (2.8,2)--(2.8,.10)circle (2pt)node [below] {$t_1=350\,\micro\second$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=10cm,yshift=2cm]
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\draw [thin](-2.2,0)--(2.2,0) (0,-2.2)--(0,2.2);
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\draw (0,0) circle (2);
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\draw [->,red](0:0)--(252:2)node [below]{$252\,\degree$};
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\draw [red!70!blue](270:1.92)--(0:0)node at (270:.951)[right]{$\sin (\omega t_1)=-0{,}951$};
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\draw [red!70!blue](252:2)--(270:1.92)node at (261:2.4)[below]{$\cos (\omega t_1)=-0{,}309$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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a) Berechnung der momentanen Spannung bei $t=t_1$
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\begin{align*}
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i(t)&=I_0\cdot \sin(\omega t)=10\,\milli\ampere\cdot \sin{(\underbrace{1{,}257\cdot \power{10}{4}\,\frac{1}{\,\second}\cdot 350\,\micro\second}_{4{,}3995 rad})}\\
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&=10\,\milli\ampere\cdot (-0{,}951)=-9{,}51\,\milli\ampere
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\intertext{Für $t=t_1$:}
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u_R(T)&=R\cdot i(t)=12\,\ohm\cdot (-9{,}51\,\milli\ampere)=\uuline{-114{,}1\,\milli\volt}\\
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u_L(T)&=L\cdot \frac{di}{dt}\Bigg{|}_{t=T}=L\cdot I_0\cdot \hspace{-.9cm}\underbrace{\omega}_{\mathrm{nachdifferenzieren}}\hspace{-.9cm}\cdot \cos(\omega t)\\
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&=1{,}3\,\milli\henry\cdot 10\,\milli\ampere\cdot 1{,}257\cdot \power{10}{4}\,\frac{1}{\second}\cdot (-0{,}309)=\uuline{-50{,}5\,\milli\volt}\\
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u_C(T)&=\underbrace{U_0}_0+\frac{1}{C}\int_{t=0}^{T}{I_0\cdot \sin(\omega t)\cdot dt} =\frac{I_0}{C}\cdot \frac{1}{\omega}\cdot \hspace{-1,5cm}\underbrace{\Big{|}-\cos(\omega t)\Big{|}_{t=0}^{T}}_{-\cos(\omega T)+\cos(0)=-(-0{,}309)+1=1,309}\\
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&=\frac{0{,}01\,\ampere\cdot 1{,}309}{8{,}7\cdot \power{10}{-6}\,\farad\cdot 1{,}257\cdot \power{10}{4}\,\frac{1}{\second}}=\uuline{119{,}7\,\milli\volt}\\[\baselineskip]
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\intertext{b) Momentanleistung bei $t=t_1$}
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p(t)&=u(t)\cdot i(t)=[u_R(t)+u_L(t)+u_C(t)]\cdot i(t)
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\intertext{Für $t=T$:}
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p(T)&=u(T)\cdot i(T)=(-114{,}1-50{,}5+119{,}7)\,\milli\volt\cdot (-9{,}51)\,\milli\ampere\\
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&=(-44{,}9)\,\milli\volt\cdot (-9{,}51)\,\milli\ampere=\uuline{0{,}427\,\milli\watt}
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\end{align*}
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c) Komplexe Größen verwenden den Effektivwert der Schwingung!
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\clearpage
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}{}% |