ET2_Uebung_BEI/ET2_L_B13_A6.tex

\section{Parallelschaltung von L und C}
An der Parallelschaltung von $L$ und $C$ liegt die Spannung $u(t)$ (siehe Diagramm).\\
Bei $t=0$ ist $i_L=0$.\\
Berechnen Sie den Strom $i$ bei $t=t_2$!
\begin{align*}
U_0&=5\,\volt\\
t_1&=3\milli\second\\
t_2&=5\milli\second\\
L&=6\,\milli\henry\\
C&=100\,\micro\farad
\end{align*}
\begin{align*}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%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);
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Kondensator 			
			\draw (0,0)--(.45,0) (.45,-.2)--(.45,.2) (.55,-.2)--(.55,.2) (.55,0)--(1,0)node at(.75,.05)[above]{$C$};
		\end{scope}
		\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
			\draw (0,2)--(0,0)--(.1,0) (1,2)--(1,0)--(.9,0);
			\fill (0,2)circle(.05) (1,2)circle(.05);
			\draw [->,blue] (.3,2)--(.7,2) node at (.5,2)[below]{$u(t)$};
			\draw [->,red] (.2,1.75)--(.2,1.25) node at (.2,1.5)[right]{$i(t)$};			
		\end{scope}						
		\begin{scope}[>=latex,thick,scale=.5,xshift=4cm,yshift=.5cm]
		  \draw [very thin,step=0.5cm](0,0)grid(5,2.5);			
			\draw [->](0,0) -- (5.5,0) node [right] {$t\,[\milli\second]$};
			\draw [->](0,0) -- (0,3) node [above] {$u\,[\volt]$};
			\draw [blue,very thick](0,0)--(3,2.5)--(5,2.5)node at (1.5,1.25)[right]{$u(t)$};				 
			\foreach \x in {1,2,...,5}
				\draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$};
			\foreach \y in {0,5}
				\draw (0,\y/2) -- (-0.2,\y/2) node[anchor=east] {$\y$};
		\end{scope}
	\end{tikzpicture}		
\end{align*}	
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
i_C&=C\cdot \frac{du}{dt}\\
u_L&=L\cdot \frac{di}{dt}\\
\text{KNP: }\sum i&=0
\end{align}
Berechnung:\\
\begin{align*}
i(t)&=i_C(t)+i_L(t)\\
i_C(t)&=C\cdot \frac{du}{dt}\tag{1}\label{eq:paralleschaltung-ic}\\
u_L(t)&=L\cdot \frac{di}{dt}\tag{2}\label{eq:paralleschaltung-ul}\\
\end{align*}
\begin{align*}
\text{aus \ref{eq:paralleschaltung-ic}}\quad\, i_C(t_2)&=0 \quad \text{zum Zeitpunkt }t_2=5\,\milli\second \quad \frac{du}{dt}=0\\
&\Rightarrow i(t_2)=i_L(t_2)\\
\text{aus \ref{eq:paralleschaltung-ul}}\qquad\, di_L&=\frac{1}{L}\cdot u_L\cdot dt \qquad \Big|\int \\
\big[i_L(t)\big]_{t_a}^{t_b}&=i_L(t_b)-i_L(t_a)=\frac{1}{L}\int_{t_a}^{t_b}{u_L\cdot dt}\\[\baselineskip]
\text{für }0&\leq t \leq t_1\\
i_L(t_1)-\underbrace{i_L(t=0)}_{0}&=\frac{1}{L}\int_{0}^{t_1}{\frac{U_0}{t_1}\cdot t\cdot dt}=\frac{U_0}{L\cdot t_1}\left[\frac{t^2}{2}\right]_{0}^{t_1}=\frac{U_0\cdot t_1}{2\cdot L}\\
i_L(t_1)&=\frac{5\,\volt\cdot 3\cdot \,\milli\second}{2\cdot 6\cdot \frac{\,\milli\volt\second}{\ampere}}=1{,}25\,\ampere\\[\baselineskip]
\text{für }t_1&\leq t \leq t_2\\
i_L(t_2)-i_L(t_1)&=\frac{1}{L}\int_{t_1}^{t_2}{U_0\cdot dt}=\frac{U_0}{L}\cdot (t_2-t_1)\\
i_L(t_2)&=1{,}25\,\ampere+\frac{5\,\volt\cdot 2\cdot \power{10}{-3}\,\second}{6\cdot \power{10}{-3}\frac{\,\volt\second}{\ampere}}=1{,}25\,\ampere+1{,}67\,\ampere=\uuline{2{,}92\,\ampere}
\end{align*}
\begin{align*}
\intertext{Alternativ Graphisch: $i_L \textrm{ ist proportional zur \textit{Fläche}}\cdot \frac{1}{L}+\textrm{const.}$}
	\begin{tikzpicture}[scale=2]
		\begin{scope}[>=latex,thick,scale=.5,xshift=4cm,yshift=.5cm]
			\fill [black!10!](0,0)--(3,2.5)--(3,0)--(0,0);
			\fill [black!25!](3,2.5)--(5,2.5)--(5,0)--(3,0);			
		  \draw [very thin,step=0.5cm](0,0)grid(5,2.5);			
			\draw [->](0,0) -- (5.5,0) node [right] {$t\,[\milli\second]$};
			\draw [->](0,0) -- (0,3) node [above] {$u\,[\volt]$};
			\draw [blue,very thick](0,0)--(3,2.5)--(5,2.5)node at (1.5,1.25)[right]{$u(t)$};								 
			\foreach \x in {1,2,...,5}
				\draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$};
			\foreach \y in {0,5}
				\draw (0,\y/2) -- (-0.2,\y/2) node[anchor=east] {$\y$};
		\end{scope}
	\end{tikzpicture}
\end{align*}
\begin{align*}		
i(t_2)&=\frac{1}{L}\cdot \underbrace{\left(\frac{U_0\cdot t_1}{2}+U_0(t_2-t_1)\right)}_{\textrm{Fläche}} =\frac{1}{6\,\milli\henry}\cdot \left(\frac{5\,\volt\cdot 3\,\milli\second}{2}+	5\,\volt\cdot (5-3)\,\milli\second\right)\\
&=\frac{(7{,}5+10)\cdot \power{10}{-3}\,\volt\second}{6\cdot \power{10}{-3}\,\frac{\volt\second}{\ampere}}=	\uuline{2{,}92\,\ampere}
\end{align*}
\footnotesize{Warum ist $i_L(t_2)-i_L(t_1)\not= 0$? Strom ändert sich noch, nur Spannung ist konstant.\\
Wenn $u=konstant \rightarrow$ Strom steigt unendlich an. $u=L\frac{di}{dt} \Rightarrow i(t)=1/L\int u(t)dt$}
\clearpage
}{}%