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\pagenumbering{roman}

%\ifthenelse{\equal{\toPrint}{Lösung}}{\input{ET2_Deckblatt}}{\input{ET2_Deckblatt_A}}

\begin{titlepage}
\vspace*{3cm}
\begin{center}
{\bfseries \large {Technische Hochschule Nürnberg \\ Georg Simon Ohm}\\[\baselineskip]
Übung Grundlagen der Elektrotechnik 2}\\[\baselineskip]

% Prüfen ob Musterlösung oder nicht, wenn ja, dann mit Dozentenkommentar, ansonsten nur Aufgabenstellung für Studierende
\ifthenelse{\equal{\toPrint}{Lösung}}%
{%
{\normalsize Prof.\,Dr.\,C.\,Niebler}\\[2\baselineskip]

\textcolor{red}{Nur für Dozentengebrauch, nicht zur Weitergabe an Studenten.}\\[\baselineskip]
{\large \textbf{Dozenten Exemplar mit Lösungen für}}\\[2\baselineskip]

\textbf{\ProfName}\\[2\baselineskip]

%\copyright\ Dr. Christine Niebler\\

}% Dozentenausgabe
{}% Studentenausgabbe
\end{center}
\end{titlepage}





\thispagestyle{empty}
\cleardoublepage




\clearpage
\pagenumbering{arabic}

\enlargethispage{1cm}
%\chead{Inhaltsverzeichnis}
\footnotesize{\tableofcontents}

\clearpage

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entering extended mode
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att_A.tex

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tt_A.tex
LaTeX2e <2011/06/27>
Babel <v3.8m> and hyphenation patterns for english, afrikaans, ancientgreek, ar
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See the LaTeX manual or LaTeX Companion for explanation.
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l.2 \begin{titlepage}
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l.2 \begin{titlepage}
Your command was ignored.
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\pagenumbering{roman}
\begin{titlepage}
\vspace*{3cm}
\begin{center}
{\bfseries \Large {Technische Hochschule Nürnberg\\
Georg Simon Ohm\\[\baselineskip]
{Übung Grundlagen der Elektrotechnik 2}}\\[\baselineskip]
\normalsize{Prof.\,Dr.\,C.\,Niebler}\\[\baselineskip]
\Large\textbf{Aufgabenstellung}}\\[\baselineskip]
% SS 2013}\\[\baselineskip]
\end{center}
\end{titlepage}
\thispagestyle{empty}
\cleardoublepage



\clearpage
\pagenumbering{arabic}

\enlargethispage{1cm}
%\chead{Inhaltsverzeichnis}
%\footnotesize{
\tableofcontents
%}

\clearpage

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\section{Blitzableiter}
Eine $l=4\,\metre$ lange Verbindungsleitung, deren Leiter voneinander einen Abstand $d = 2\,\centi\metre$ haben, ist im Abstand $D=3\,\metre$ parallel zu einem Blitzableiter verlegt. (Siehe Skizze).\\
$\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Welche Spannung $u$ (Betrag !) wird induziert, wenn der
Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt?
\item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !)
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom
\draw [->] (0,4) -- (0,0) node [below] {$i$};
\end{scope}
\begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung
\draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$};
\draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$};
\end{scope}
\begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D
\draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$};
\draw [dashed] (3.1,.5)--(3.1,-.5);
\end{scope}
\begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d
\draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$};
\draw [<-] (0.2,0) -- (0.7,0);
\end{scope}
\begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l
\draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$};
\draw (0,0) -- (-0.5,0);
\draw (0,3) -- (-0.5,3);
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
Formeln:
\begin{align}
% \intertext{Formeln:}
u&=-N\cdot \frac{d\Phi}{dt}\\
\Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\
H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r}
\end{align}
Magnetischer Fluss $\Phi$; Flussdichte $B$; Feldstärke $H$;\\
Magnetische Permeabilität $\mu=\mu_0\cdot \mu_r$; In Luft $\mu_r=1$\\
\clearpage
Berechnung:
\begin{align*}
u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\qquad u=-\frac{d\Phi}{dt}\\
\Phi&=\int{\vec{B}\cdot \vec{dA}} \qquad \text{mit $B\bot A$}\qquad \Phi=B\cdot A=\underbrace{\mu\cdot H}_{B}\cdot \underbrace{l\cdot d}_{A}\\
H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\
\phantom{blablabla}bla\\
\vphantom{u=-\frac{y}{x}}..\\
u&=-\frac{d\Phi}{dt}\hspace{-.5cm}\underbrace{=}_{linearer Anstieg}\hspace{-.5cm}-\frac{\Phi}{t}=-\mu\cdot l\cdot d\cdot \frac{i}{2\cdot \pi\cdot D\cdot t}=-\frac{\mu\cdot i\cdot l\cdot d}{2\cdot \pi\cdot D\cdot t}\\
% } %end pahntom
&=-\frac{1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)\cdot 15000\,\ampere\cdot 4\,\metre\cdot 0{,}02\,\metre}{2\cdot \pi \cdot 3\,\metre\cdot 0{,}6\cdot \power{10}{-6}\,\second}=\uuline{-134\,\volt}\\[\baselineskip]
&\text{Leitung symmetrisch zur Mittellinie bei } 2{,}99\,\metre \text{ und } 3{,}01\,\metre\\
\text{mit }&\Phi=\frac{\mu\cdot i\cdot l}{2\cdot \pi}\cdot\ln\frac{D+d/2}{D-d/2}\\
u&=-\frac{\mu_0\cdot 15\cdot \power{10}{3}\,\ampere\cdot 4\,\metre}{0{,}6\cdot \power{10}{-6}\,\second\cdot 2\cdot \pi}\cdot \underbrace{\ln\frac{3{,}01\,\metre}{2{,}99\,\metre}}_{6{,}66\cdot \power{10}{-3}}=-133{,}7\,\volt\\
\text{\uline{Lentzsche Regel:}}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=2.5]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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] {$R$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
\draw [<-,blue] (.2,.35)--(.8,.35)node at(.5,.35)[above]{\footnotesize$u$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=.25cm,yshift=-.25cm]% Leiterschleife
\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=-1cm]% O.
\draw (.5,0)circle(.133)node at (1,0)[right]{$\frac{d\Phi}{dt}$};
\fill (.5,0)circle(.05);
\end{scope}
\begin{scope}[>=latex,very thick,green!50!black,xshift=0cm,yshift=-1.33cm]% Ox
\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
\draw [black]node at (1.5,-1)[below]{Wirkung $\frac{d\Phi_{ind}}{dt}$ entgegen Ursache $\frac{d\Phi}{dt}$};
\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%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] {$R$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=3cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=-0.25cm]% Leiterschleife
\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=-2.25cm]%Stromquelle
\draw [green!50!black](.25,0)--(.75,0);
\draw node at (.5,.133) [above] {$i_{}$};
\draw [green!50!black](.5,0)circle(.133);
\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$i_{ind}$};
\end{scope}
\begin{scope}[>=latex,very thick,green!50!black,xshift=3cm,yshift=-1.33cm]% Ox
\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1.25cm]%
\draw [->,red] (-2.5,.5)--(-2.5,-.5) node at(-2.5,0)[right]{$\frac{di}{dt}$};
\draw [<-,green!50!black] (.5,.5)--(.5,-.5) node at(.5,0)[left]{$i_{ind}$};
\draw [<-,red!75!black] (-1.75,.5)--(-1.75,-.5) node at(-1.75,0)[left]{$i_{ind}$};
\end{scope}
\end{tikzpicture}
\end{align*}
$u$ hat negatives Vorzeichen, da Polung von $i_{ind}$ entgegen Bezugspfeilen aus Skizze\\
($u_R = -u$)
\clearpage
}{}%

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\section{Blitzableiter}
Eine $l=4\,\metre$ lange Verbindungsleitung, deren Leiter voneinander einen Abstand $d = 2\,\centi\metre$ haben, ist im Abstand $D=3\,\metre$ parallel zu einem Blitzableiter verlegt. (Siehe Skizze).\\
$\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Welche Spannung $u$ (Betrag !) wird induziert, wenn der
Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt?
\item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !)
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom
\draw [->] (0,4) -- (0,0) node [below] {$i$};
\end{scope}
\begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung
\draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$};
\draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$};
\end{scope}
\begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D
\draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$};
\draw [dashed] (3.1,.5)--(3.1,-.5);
\end{scope}
\begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d
\draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$};
\draw [<-] (0.2,0) -- (0.7,0);
\end{scope}
\begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l
\draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$};
\draw (0,0) -- (-0.5,0);
\draw (0,3) -- (-0.5,3);
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
Formeln:
\begin{align}
% \intertext{Formeln:}
u&=-N\cdot \frac{d\Phi}{dt}\\
\Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\
H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r}
\end{align}
Magnetischer Fluss $\Phi$; Flussdichte $B$; Feldstärke $H$;\\
Magnetische Permeabilität $\mu=\mu_0\cdot \mu_r$; In Luft $\mu_r=1$\\
\clearpage
Berechnung:
\begin{align*}
u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\qquad u=-\frac{d\Phi}{dt}\\
\Phi&=\int{\vec{B}\cdot \vec{dA}} \qquad \text{mit $B\bot A$}\qquad \Phi=B\cdot A=\underbrace{\mu\cdot H}_{B}\cdot \underbrace{l\cdot d}_{A}\\
H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\
u&=-\frac{d\Phi}{dt}\hspace{-.5cm}\underbrace{=}_{linearer Anstieg}\hspace{-.5cm}-\frac{\Phi}{t}=-\mu\cdot l\cdot d\cdot \frac{i}{2\cdot \pi\cdot D\cdot t}=-\frac{\mu\cdot i\cdot l\cdot d}{2\cdot \pi\cdot D\cdot t}\\
&=-\frac{1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)\cdot 15000\,\ampere\cdot 4\,\metre\cdot 0{,}02\,\metre}{2\cdot \pi \cdot 3\,\metre\cdot 0{,}6\cdot \power{10}{-6}\,\second}=\uuline{-134\,\volt}\\[\baselineskip]
&\text{Leitung symmetrisch zur Mittellinie bei } 2{,}99\,\metre \text{ und } 3{,}01\,\metre\\
\text{mit }&\Phi=\frac{\mu\cdot i\cdot l}{2\cdot \pi}\cdot\ln\frac{D+d/2}{D-d/2}\\
u&=-\frac{\mu_0\cdot 15\cdot \power{10}{3}\,\ampere\cdot 4\,\metre}{0{,}6\cdot \power{10}{-6}\,\second\cdot 2\cdot \pi}\cdot \underbrace{\ln\frac{3{,}01\,\metre}{2{,}99\,\metre}}_{6{,}66\cdot \power{10}{-3}}=-133{,}7\,\volt\\
\text{\uline{Lentzsche Regel:}}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=2.5]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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] {$R$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
\draw [<-,blue] (.2,.35)--(.8,.35)node at(.5,.35)[above]{\footnotesize$u$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=.25cm,yshift=-.25cm]% Leiterschleife
\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=-1cm]% O.
\draw (.5,0)circle(.133)node at (1,0)[right]{$\frac{d\Phi}{dt}$};
\fill (.5,0)circle(.05);
\end{scope}
\begin{scope}[>=latex,very thick,green!50!black,xshift=0cm,yshift=-1.33cm]% Ox
\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
\draw [black]node at (1.5,-1)[below]{Wirkung $\frac{d\Phi_{ind}}{dt}$ entgegen Ursache $\frac{d\Phi}{dt}$};
\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%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] {$R$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
\end{scope}
\begin{scope}[>=latex,very thick,red,xshift=3cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=-0.25cm]% Leiterschleife
\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=-2.25cm]%Stromquelle
\draw [green!50!black](.25,0)--(.75,0);
\draw node at (.5,.133) [above] {$i_{}$};
\draw [green!50!black](.5,0)circle(.133);
\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$i_{ind}$};
\end{scope}
\begin{scope}[>=latex,very thick,green!50!black,xshift=3cm,yshift=-1.33cm]% Ox
\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1.25cm]%
\draw [->,red] (-2.5,.5)--(-2.5,-.5) node at(-2.5,0)[right]{$\frac{di}{dt}$};
\draw [<-,green!50!black] (.5,.5)--(.5,-.5) node at(.5,0)[left]{$i_{ind}$};
\draw [<-,red!75!black] (-1.75,.5)--(-1.75,-.5) node at(-1.75,0)[left]{$i_{ind}$};
\end{scope}
\end{tikzpicture}
\end{align*}
$u$ hat negatives Vorzeichen, da Polung von $i_{ind}$ entgegen Bezugspfeilen aus Skizze\\
($u_R = -u$)
\clearpage
}{}%

+ 123
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ET2_L_B12_A2.tex View File

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\section{Drahtschleife}
Eine Drahtschleife $N=1$ wird von einem zeitlich veränderlichen Fluss durchsetzt.\\
$\Phi(t)=\Phi_0\cdot (1-e^{-t/T})$ mit $\Phi_0=60\cdot \power{10}{-6
}\,\volt\second$ und $T=1\,\milli\second$.
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie die Spannung $u$ für $t=0{,}5\cdot T$!
\item Geben Sie die Polarität der Anschlussklemmen der Drahtschleife a-b für diesen Zeitpunkt an und begründen Sie ihre Angabe!
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex,red,ultra thick,xshift=0,yshift=0]
\draw (0,0) -- (0,1);
\draw [dashed] (0,1) -- (0,2);
\end{scope}
\begin{scope}[>=latex,blue,very thick,xshift=0,yshift=2cm]
\draw (0,0) ellipse (2cm and 1cm);
\end{scope}
\begin{scope}[>=latex,blue,very thick,xshift=1.9cm,yshift=1.75cm]
\filldraw [black!0!] (0,0.02) rectangle (0.2,0.48);
\draw [fill] (1pt,0) -- (1,0) circle (2pt) node [right] {b};
\draw [fill] (1pt,0.5cm) -- (1,0.5) circle (2pt) node [right] {a};
\draw node at (1.5,0.25)[right] {$u$};
\end{scope}
\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0]
\draw [->] (0,2) -- (0,4) node [above] {$\Phi(t)$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
u&=-N\cdot \frac{d\Phi}{dt}
\end{align}
\begin{align*}
\intertext{Berechnung:}
\intertext{a) Spannung $u$ für $t=0{,}5\cdot T$}
u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\\
&=-\frac{d\left(\Phi_0\cdot \left(1-e^{-\frac{t}{T}}\right)\right)}{dt}\\
&=-\Phi_0\cdot \frac{d}{dt} \left(1-e^{-\frac{t}{T}}\right)\\
&=-\Phi_0\cdot \Bigg[-e^{-\frac{t}{T}}\cdot \hspace{-.7cm}\underbrace{ \left(-\frac{1}{T}\right)}_{\mathrm{Nachdifferenzieren}}\hspace{-.5cm}\Bigg]\\
&\text{mit $t=0{,}5\cdot T\qquad T=1\,\milli\second$}\\
u&=-60\cdot \power{10}{-6}\,\volt\second\cdot \left[-e^{-0{,}5}\cdot \left(-\frac{1}{1\,\milli\second}\right)\right]=\uuline{-36{,}39\,\milli\volt}
\end{align*}
\begin{align*}
\intertext{b) Polarität der Spannung $u$ für $t=0{,}5\cdot T$}
\frac{d\Phi}{dt}&>0 \Rightarrow \text{Linke-Hand für Stromrichtung}\\
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=1]
\foreach \xs in {0} { % Enter Start value of x label
\foreach \ys in {0} { % Enter Start value of y label
\foreach \ii in {10} { % Enter Number of Decades in x
\foreach \jj in {6} { % Enter Number of Decades in y
\foreach \xe in {5} { % Enter End value of x label
\foreach \ye in {60} { % Enter End value of y label
\foreach \i in {1,2,...,\ii} {
\foreach \j in {1,2,...,\jj} {
}}% End Log Grid
\draw[black!80!] (0,0) grid (\ii,\jj); % Draw Linear grid
\draw [<->,thick] (0,\jj+.2) node (yaxis) [above] {$\Phi\,[\micro\volt\second]$} |- (\ii+.2,0) node (xaxis) [right] {$t\,[\milli\second]$}; % Draw axes
\draw [thick] (-1.25,\jj+.2) node (yaxis) [above] {$u\,[\milli\volt]$} ; % Draw axes
\foreach \x in {\xs,.5,...,\xe}% x Axis Label:
\node [black,anchor=north] at(\x*2-\xs,0){$\x$};
\foreach \y in {\ys,10,...,\ye}% y Axis Label:
\node [red,anchor=east] at(0,\y/10-\ys){$\y$};
\foreach \y in {\ys,10,...,\ye}% y Axis Label:
\node [blue,anchor=east] at(-1,\y/10-\ys){$-\y$};
}}}}}}
\draw [black,thick,dashed] (0,0) -- (2,6) -- (2,0.3);
\draw [blue,thick,dashed] (1,3.64) -- (-.75,3.64) node [left]{$-36{,}39\,\milli\volt$};
\draw[color=red, very thick,domain=0:10] plot[id=tau] function{6*(1-exp(-x/2))};
\draw[color=blue, very thick,domain=0:10] plot[id=tau] function{6*(exp(-x/2))};
\draw [black] node at (2,0)[above] {$T$};
\draw [red] node at (2,3.5) [right]{$\Phi=\Phi_0\cdot \left(1-e^{-t/T}\right)$};
\draw [blue] node at (2,2.5) [right]{$u=-\Phi_0\cdot \left(\frac{e^{-t/T}}{T}\right)$};
\draw node at (0,-1) [right]{Induzierte Spannung bei zunehmendem Fluss};
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{\uline{Lenzsche Regel:}}
\text{Klemme a $\Rightarrow$ $+$}\\
\text{Klemme b $\Rightarrow$ $-$}\\
%F_{\sqcup\sqcap}\\ %-------------------Versuch mit Symbolen--------------
%\ding{43}\\
%\oplus\ominus\odot\\
%\circlearrowright\\
%\lightning\\
%\curvearrowright\\
%\leftthumbsup\\
%\SmallSquare\\
%\SmallTriangleUp\\ %-------------------Versuch mit Symbolen--------------
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex,red,ultra thick,xshift=0,yshift=0]
\draw (0,0) -- (0,1);
\draw [dashed] (0,1) -- (0,2);
\end{scope}
\begin{scope}[>=latex,blue,very thick,xshift=0,yshift=2cm]
\draw (0,0) ellipse (2cm and 1cm);
\end{scope}
\begin{scope}[>=latex,blue,very thick,xshift=1.9cm,yshift=1.75cm]
\filldraw [black!0!] (0,0.02) rectangle (0.2,0.48);
\draw [fill] (1pt,0) -- (1,0) circle (2pt) node [below] {b $-$};
\draw [fill] (1pt,0.5cm) -- (1,0.5) circle (2pt) node [above] {a $+$};
\draw node at (1.5,0.25)[right] {$u$};
\end{scope}
\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0]
\draw [->] (0,2) -- (0,4) node [above] {$\Phi(t)$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=5cm,yshift=1.5cm,rotate=90]
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$U_{R}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=2cm]
\draw [dashed] (0,.25)--(2,.5) (0,-.25)--(2,-.5);
\draw [->,red] (0,.25) +(7.2:.75) -- +(7.2:1.25) node [above left]{$i$};
\end{scope}
\end{tikzpicture}
\end{align*}
\clearpage
}{}%

+ 125
- 0
ET2_L_B12_A3.tex View File

@@ -0,0 +1,125 @@
\section{Metallstab}
Ein Metallstab $M$ rotiert um die Achse $A$ mit der Winkelgeschwindigkeit $\omega$.\\
Der Metallstab schleift auf dem Metallring $S$. Senkrecht zu dem Metallstab und Metallring wirkt eine homogene magnetische Flussdichte $\vec{B}=B_z\cdot \vec{e}_z$\\
Zwischen Schleifring und Achse wird eine Gleichspannung $U_{SA}$ gemessen.\\[\baselineskip]
Berechnen Sie die Flussdichte $B_z$.\\[\baselineskip]
$R=2\,\centi\metre$; $\omega =100\cdot \pi\cdot\frac{1}{\second}$ ; $U_{SA}=25\,\milli\volt$.\\
\begin{align*}
\begin{tikzpicture}[scale=1.5]
\begin{scope}[>=latex,thick,xshift=0,yshift=0]
\draw [->] (0,0) -- (0,5) node [above] {$y$};
\draw [->] (0,0) -- (5,0) node [right] {$x$};
\draw [draw=black,fill=white,very thick](0,0) circle (0.1) node at (0,-.25) [below]{$z$};
\fill [fill=black] (0,0) circle (0.04);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=2.5cm]
\draw [red,ultra thick](-2,-2) -- (2,2) node [below right]{M};
\draw [->](0,0) -- (0,-2) node at (0,-1)[right]{$R$};
\draw [->](-1.414,1.414) -- (0,0)node at (-0.707,0.707) [right]{$U_{SA}$};
\draw [blue](0,0) circle (2)node at (0,2)[above] {S};
\draw [->,red](0,0) -- (45:1.414) arc (45:30:1.5cm) node [below] {$\omega$};
\fill [blue](0,0) circle (0.1) node [below right] {A};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3.5cm,yshift=2.5cm]
\draw [red] (0,0) circle (0.1) node [below]{$\vec{B}$};
\fill [red] (0,0) circle (0.04);
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
u_{ind}&=-N\cdot \Bigg(\underbrace{\frac{dB(t)}{dt}\cdot A(t)}_{\mathrm{Ruheinduktion}}+\underbrace {\frac{B(t)}{dt}\cdot dA(t)}_{\mathrm{Bewegungsinduktion}}\Bigg)
\end{align}
\begin{align*}
\intertext{Berechnung:}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex,thick,xshift=0,yshift=0]
\draw [->] (0,0) -- (0,5) node [above] {$y$};
\draw [->] (0,0) -- (5,0) node [right] {$x$};
\draw [draw=black,fill=white,very thick](0,0) circle (0.1) node at (0,-.25) [below]{$z$};
\fill [fill=black] (0,0) circle (0.04);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=2.5cm]
\fill [black!15!](0,0) -- (45:2) arc (45:30:2cm) -- (0,0);
\draw node at (30:2)[above right]{$dt$};
\draw [red,ultra thick](-2,-2) -- (2,2) node [below right]{M};
\draw [->](0,0) -- (0,-2) node at (0,-1)[right]{$R$};
\draw [->](-1.414,1.414) -- (-0.05,0.05)node at (-0.707,0.707) [right]{$U_{SA}$};
\draw [blue](0,0) circle (2)node at (0,2)[above] {S};
\draw [->,red](0,0) -- (45:1.414) arc (45:30:1.5cm) node [below] {$\omega$};
\fill [blue](0,0) circle (0.1) node [below right] {A};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3.5cm,yshift=2.5cm]
\draw [red] (0,0) circle (0.1) node [below]{$\vec{B}$};
\fill [red] (0,0) circle (0.04);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=2.5cm]% Voltmeter
\draw [red!50!blue](-.5,0)--(.25,0) (.75,0)--(1.5,0) node at (.5,.25) [above] {$V$};
\draw [red!50!blue](.5,0)circle(.25);
\draw [->,red!50!blue](.25,-.5)--(.75,.5)node at (.2,0) [above] {$+$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Bewegungsinduktion! $N=1$, homogenes zeitlich unverändertes Feld $B$ $\bot$ zu $\omega$}
u_{ind}&=-\frac{d\Phi}{dt}=-B\cdot \frac{dA(t)}{dt}\\
\intertext{Der Metallstab überstreicht im Zeitintervall $dt$ den vom Fluß $\Phi$ durchsetzten Kreissektor mit der Fläche}
dA(t)&=\underbrace{R^2\cdot \pi}_{\text{Kreisfläche}}\cdot \underbrace{\frac{\omega\cdot dt}{2\pi}}_{\text{Segment}}\\
dA(t)&=\frac{1}{2}\cdot R^2\cdot \omega\cdot dt\\
\frac{dA(t)}{dt}&=\frac{1}{2}\cdot R^2\cdot \omega\\
\frac{d\Phi}{dt}&=B\cdot \frac{1}{2}\cdot R^2\cdot \omega \\
u_{ind}&=-\frac{d\Phi}{dt}=-B\cdot \frac{1}{2}\cdot R^2\cdot \omega\\
|u_{ind}|&=|B_z|\cdot \frac{1}{2}\cdot R^2\cdot \omega\\
|B_z|&=\frac{2\cdot U_{SA}}{R^2\cdot \omega}=\frac{2\cdot 25\,\milli\volt}{(2\,\centi\metre)^2\cdot 100\cdot \pi\cdot \frac{1}{\second}}=\uuline{0{,}398\,\frac{\volt\second}{\square\metre}}
\end{align*}
\clearpage
\textbf{Alternativ:}\\
Bewegungsinduktion mit $v(r)=\omega\cdot r$ (radiusabhängig)
\begin{align*}
% \intertext{Bewegungsinduktion mit $v(r)=\omega\cdot r$ (radiusabhängig)}
|u_{ind}|&=\int{(\vec{v} \times \vec{B})\cdot \vec{dl}}=\int{v\cdot |B_z|\cdot dr}=\omega\cdot |B_z|\int_{r=0}^{R}{r\cdot dr}=\omega\cdot |B_z|\left[\frac{r^2}{2}\right]_{0}^{R}=\frac{1}{2}\cdot \omega\cdot |B_z|\cdot R^2\\
|B_z|&=\uuline{0{,}398\,\frac{\volt\second}{\square\metre}}
% \intertext{Richtung der Lorenzkraft $\vec{F}_L$ wirkt so, daß positive Ladungsträger $q$ zum Zentrum $(A)$ gedrückt werden (entspricht der technischen Stromrichtung $I$).}
\end{align*}
Richtung der Lorenzkraft $\vec{F}_L$ wirkt so, daß positive Ladungsträger $q$ zum Zentrum $(A)$ gedrückt werden (entspricht der technischen Stromrichtung $I$).\\

$\vec{F}_L=q(\vec{v}\times \vec{B});\qquad (F_L=q\cdot v\cdot B\text{, wenn }v\bot B)$\\

Rechte Hand Regel:\\
Der \textbf{Daumen} zeigt in Richtung der Ursache:\\
a.) Bewegter Leiter im Magnetfeld: Die Relativbewegung $\vec{v}$ des Leiters im Magnetfeld\\
b.) Strom durch Leiter im Magnetfeld: Die technische Stromrichtung $I$ bzw. Bewegungsrichtung der positiven Ladung $q$\\
Der \textbf{Zeigefinger} zeigt senkrecht zum Daumen in Richtung der magnetischen Feldlinien, also der Vermittlung (auch Verknüpfung), also dem Magnetfeld $\vec{B}$\\
Der \textbf{Mittelfinger} zeigt senkrecht zu Daumen und Zeigefinger in Richtung der Wirkung, der Lorentzkraft $\vec{F}_L$ \\

a.) Bewegter Leiter im Magnetfeld
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex, xshift=0, yshift=0]
\draw [->](0,0)--(1,0,0) node [right]{$\vec{v}$\,\text{(Daumen)}};
\draw [->](0,0)--(0,1,0) node [above]{$\vec{B}$\,\text{(Zeigefinger)}};
\draw [->](0,0)--(0,0,1.41) node [below]{$\vec{F}_L$ \,\text{(Mittelfinger) Kraft auf pos. Ladung} $q\Rightarrow I$};
% \draw node at (0,2.25)[left]{Rechte Hand Regel};
\end{scope}
\end{tikzpicture}\\
% \hspace{1cm}
\end{align*}
\begin{align*}
&\text{Da $U_{SA}$ positiv, muß $B_z$ negativ sein. }
\Rightarrow B_z=\uuline{-0{,}398\,\frac{\volt\second}{\square\metre}}\\
\end{align*}

b.) Strom durch Leiter im Magnetfeld:
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex, xshift=0, yshift=1cm]
\draw [->](0,0)--(1,0,0) node [right]{$\vec{F}_L$ \,\text{(Mittelfinger)}};
\draw [->](0,0)--(0,-1,0) node [below]{$\vec{B}$ \,\text{(Zeigefinger)}};
\draw [->](0,0)--(0,0,1.41) node [left]{$i=\frac{dq}{dt}$\,\text{(Daumen)}}; %\ \sim\ \vec{v}
\end{scope}
\end{tikzpicture}
\end{align*}

\clearpage
}{}%

+ 50
- 0
ET2_L_B12_A4.tex View File

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\section{Spannungsverlauf}
Gegeben ist die dargestellte Spannung:
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex,thick]
\draw [->](0,0) -- (6,0) node [right] {$t\,[\milli\second]$};
\draw [->](0,-1.25) -- (0,1.25) node [above] {$u\,[\volt]$};
\draw [red,very thick](0,0)--(1,1)--(1,0)--(1.5,0)--(1.5,-1)
--(2.5,-1)--(2.5,0)--(3.5,1)--(3.5,0)--(4,0)--(4,-1)--(5,-1)--(5,0)--(6,1);
\foreach \x in {10,20,...,50}
\draw (\x/10,0) -- (\x/10,-0.2) node[anchor=north] {$\x$};
\foreach \y in {-10,0,10}
\draw (0,\y/10) -- (-0.2,\y/10) node[anchor=east] {$\y$};
\end{scope}
\end{tikzpicture}
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Ermitteln Sie die Frequenz der Grundschwingung!
\item Berechnen Sie den Gleichrichtwert der Spannung!
\item Berechnen Sie den Effektivwert der Spannung!
\item Berechnen Sie den Formfaktor der Spannung!
\item Nun wird die dargestellte Spannung an einen Ohmschen Widerstand von $100\,\ohm$ angelegt. Welche Verlustleistung tritt im Widerstand auf?\\
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
\overline{|u|}&=\frac{1}{T}\cdot \int_{t=0}^{T}{|u(t)|\cdot dt}&\text{Gleichrichtwert}\\
U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&\\
F&=\frac{U}{\overline{|u|}}=\frac{\text{Effektivwert}}{\text{Gleichrichtwert}}
\end{align}
\begin{align*}
\intertext{Berechnung:}
\intertext{a) Grundschwingung mit $T=25\,\milli\second$:}
f&=\frac{1}{T}=40\,\hertz\\
\intertext{b) Gleichrichtwert der Spannung:}
\overline{|u|}&=\frac{1}{T}\cdot (F_{\triangle} +F_{\sqcup\hspace{-.2cm}\sqcap})=\frac{1}{25\,\milli\second}\cdot (\frac{1}{2}\cdot 10\,\volt\cdot 10\,\milli\second+10\,\volt\cdot 10\,\milli\second)=\frac{150\,\volt\cdot \milli\second}{25\milli\second}=\uuline{6\,\volt}\\
\intertext{c) Effektivwert der Spannung:}
U&=\sqrt{\frac{1}{T}\int_{0}^{T}{u^2}\cdot dt}\\
U^2&=\frac{1}{T}\left(\int_{0}^{10\,\milli\second}{\left(\frac{10\,\volt}{\power{10}{-2}\,\second}\cdot t\right)^2\cdot dt}+\int_{15\,\milli\second}^{25\,\milli\second}{(-10\,\volt)^2\cdot dt}\right)\\
&=\frac{1}{25\,\milli\second}\left(\frac{100\,\square\volt}{\power{10}{-4}\,\square\second}\cdot \left[\frac{t^3}{3}\right]_{0}^{10\,\milli\second}+100\,\square\volt\cdot \big[t\big]_{15\,\milli\second}^{25\,\milli\second}\right)\\
&=\frac{100\,\square\volt}{25\,\milli\second}\left(\power{10}{4}\frac{1}{\,\square\second}\cdot \frac{1}{3}\cdot \power{10}{-6}\,\cubic\second+10\,\milli\second\right)=53{,}33\,\square\volt\\
U&=\sqrt{53{,}33}\,\volt=\uuline{7{,}30\,\volt}
\intertext{d) Formfaktor der Spannung:}
F&=\frac{U}{\overline{|u|}}=\frac{7{,}30\,\volt}{6\,\volt}=\uuline{1{,}22}
\intertext{e) Verlustleistung im Widerstand:}
P&=\frac{U^2}{R}=\frac{53{,}33\,\square\volt}{100\,\ohm}=\uuline{0{,}533\,\watt}
\end{align*}
\clearpage
}{}%

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ET2_L_B12_A5.tex View File

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\section{Phasenanschnitt}
Berechnen Sie den Effektivwert dieser sinusförmigen Spannung mit Phasenanschnitt.\\
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex,thick]
\draw [ultra thin] (0,-2)grid(6,2);
\draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$};
\draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$};
\draw[color=red,very thick,domain=0.6:2,smooth,samples=100](0,0)--(0.6,0)--(0.6,1.618)
plot[id=sina]function{2*sin(.5*3.14*x)};
\draw[color=red,very thick,domain=0.6:2,smooth,samples=100]plot[id=sinb]function{2*sin(0.5*3.14*x)};
\draw[color=red,very thick,domain=2.6:4,smooth,samples=100] (2,0)--(2.6,0)-- (2.6,-1.618) plot[id=sinc] function{2*(sin(0.5*3.14*x))};
\draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618)plot[id=sind]function{2*sin(0.5*3.14*x)};
\foreach \x in {10,20,30}
\draw(\x/5,0)--(\x/5,-0.2) node[anchor=north]{$\x$};
\draw (3/5,0)--(3/5,-0.2)node[anchor=north]{$3$};
\foreach \y in {-400,-200,...,400}
\draw (0,\y/200)--(-0.2,\y/200)node[anchor=east]{$\y$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&\\
\sin^2\alpha&=\frac{1}{2}(1-\cos 2\alpha)
\end{align}
\begin{align*}
\intertext{Berechnung:}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=1]
\begin{scope}[>=latex,thick]
\draw [ultra thin] (0,-2)grid(6,2);
\draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$};
\draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$};
\draw[color=red,very thick,domain=0.6:2,smooth,samples=100] (0,0)--(0.6,0)--(0.6,1.618) plot[id=sin12a53] function{2*sin(0.5*3.14*x)};
\draw[color=red,very thick,domain=2.6:4,smooth,samples=100] (2,0)--(2.6,0)--(2.6,-1.618) plot[id=sin12a54] function{2*(sin(0.5*3.14*x))};
\draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618) plot[id=sin12a55] function{2*sin(0.5*3.14*x)};
\draw[color=blue,very thick,dashed, domain=0:4,smooth,samples=100] plot[id=sin12a56] function{2*sin(0.5*3.14*x)};
\foreach \x in {10,20,30}
\draw (\x/5,0) -- (\x/5,-0.2) node[anchor=north] {$\x$};
\draw (3/5,0) -- (3/5,-0.2)node[anchor=north] {$3$};
\foreach \y in {-400,-200,0,200,400}
\draw (0,\y/200) -- (-0.2,\y/200) node[anchor=east] {$\y$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{Periodendauer $T=20\,\milli\second$, da Symmetrie in einer Periode. Betrachtung nur einer Sinus-Halbwelle mit $\frac{1}{2}\cdot T=10\,\milli\second$}
\omega&=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{20\,\milli\second}=314\,\frac{1}{\second}\\
u(t)&=
\begin{cases}
0&\text{ für }t=0\ldots 3\,\milli\second\\
400\,\volt\cdot \sin(\omega t)=400\,\volt\cdot \sin(314\,\frac{1}{\second}\cdot t) &\text{ für }t=3\,\milli\second\ldots 10\,\milli\second\\
\end{cases}\\[\baselineskip]
U^2&=\frac{1}{T/2}\int_{3\,\milli\second}^{10\,\milli\second}{(400\,\volt\cdot \sin(\omega t))^2\cdot dt}\\
&=\frac{1}{T/2}\cdot (400\,\volt)^2\int_{3\,\milli\second}^{10\,\milli\second}{\sin^2(\omega t)\cdot dt}\\[\baselineskip]
&\text{mit }\sin^2\alpha=\frac{1}{2}(1-\cos 2\alpha)\\[\baselineskip]
U^2&=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot\frac{1}{2}\cdot \left(\int_{3\,\milli\second}^{10\,\milli\second}{1\, dt} -\int_{3\,\milli\second}^{10\,\milli\second}{\cos(2 \omega t)\cdot dt}\right)\\
&=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot \frac{1}{2}\cdot \left(\Big[t\Big]_{3\,\milli\second}^{10\,\milli\second}-\left[\sin(2 \omega t)\cdot \frac{1}{2\omega} \right]_{3\,\milli\second}^{10\,\milli\second}\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{1}{2\cdot \omega}\cdot \underbrace{\left(sin(\cancel{2}\cdot \frac{2\pi}{\cancel{20\,\milli\second}}\cdot \cancel{10\,\milli\second}\right)}_{sin(2\pi)=0}-sin\left(2\cdot \frac{\pi}{20\,\milli\second}\cdot 3\,\milli\second\right)\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{20\,\milli\second}{4\pi}\cdot \underbrace{(-sin(0{,}6\pi))}_{-0{,}951}\right)\\
&=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot 8{,}51\,\milli\second=68109\,\square\volt\\[\baselineskip]
U&=400\,\volt\cdot \sqrt{\frac{8{,}51\,\milli\second}{20\,\milli\second}}=\uuline{261\,\volt}\\
\intertext{Zum Vergleich: Sinus ohne Phasenanschnitt hätte einen Effektivwert von}
U_{(sin)}&=\frac{400\,\volt}{\sqrt{2}}=283\,\volt
\end{align*}
\clearpage
}{}%

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ET2_L_B12_A6.tex View File

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\section{Rechteckspannung}
Gegeben ist eine periodische Rechteckspannung mit der Periodendauer von $10\,\milli\second$.\\
Berechnen Sie den Effektivwert, wenn der arithmetische Mittelwert gleich Null ist.\\
\begin{align*}
\begin{tikzpicture}[scale=0.5]
\begin{scope}[>=latex,thick]
\draw [->](0,-2) -- (10.75,-2) node [right] {$t\,[\milli\second]$};
\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};
\draw [<->,blue, very thick] (-.5,0)--(-.5,5) node at (0,2.5)[right]{$5\,\volt$};
\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)
--(10,5)--(10.5,5);
\foreach \x in {0,1,...,10}
\draw (\x,-2) -- (\x,-2.2) node[anchor=north] {$\x$};
\foreach \y in {0,1,...,5}
\draw (0,\y) -- (-0.2,\y);% node[anchor=east] {$\y$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
\overline{u}&=\frac{1}{T}\cdot \int_{t=0}^{T}{u(t)\cdot dt}&\text{Arithmetischer Mittelwert}\\
U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&
\end{align}
\begin{align*}
\intertext{Berechnung:}
\intertext{a) Arithmetischer Mittelwert $\overline{u}=0$}
&\text{$\Rightarrow$ Fläche ober- und unterhalb der Nulllinie muß gleich sein!}\\
&\text{$\Rightarrow$ Wo ist die Nulllinie?}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=0.5]
\begin{scope}[>=latex,thick]
\draw [->](0,0) -- (10.75,0) node [right] {$t\,[\milli\second]$};
\draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$};
\fill [black!15!] (0,3.5)rectangle(7,5) (7,3.5)rectangle(10,0);
\draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0)
--(10,5)--(10.5,5);
\draw [blue,very thick] (0,3.5)--(10,3.5) node [right]{NULL};
\draw [->,blue,very thick] (3.5,3.5)--(3.5,5) node at(3.5,4.25)[right]{$u_1$};
\draw [<-,blue,very thick] (8.5,0)--(8.5,3.5) node at(8.5,1.75)[right]{$u_2$};
\foreach \x in {0,1,...,10}
\draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$};
\foreach \y in {-3.5,-2.5,...,1.5}
\draw (0,\y+3.5) -- (-0.2,\y+3.5) node[anchor=east] {$\y$};
\end{scope}
\end{tikzpicture}
\end{align*}
\enlargethispage{1cm}
\begin{align*}
&u_1+(-u_2)=5\,\volt\rightarrow u_2=-(5\,\volt-u_1)\\
&\text{Fläche:}\\
&\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\
&\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\
&u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\
&u_1\cdot 7\,\milli\second-(5\,\volt-U_1)\cdot 3\,\milli\second=0\\
&(7\,\milli\second+3\,\milli\second)\cdot u_1=15\,\volt\cdot \milli\second\\
&u_1=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uuline{1{,}5\,\volt}\\
&u_2=-(5\,\volt-1{,}5\,\volt)=\uuline{-3{,}5\,\volt}
\intertext{Alternativ mit Beträgen}
&|u_1|+|u_2|=5\,\volt\rightarrow |u_2|=5\,\volt-|u_1|\\
&\text{Fläche:}\\
&\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\
&\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\
&u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\
&u_1\cdot 7\,\milli\second-(5\,\volt-|u_1|)\cdot 3\,\milli\second=0\\
&(7\,\milli\second+3\,\milli\second)\cdot |u_1|=15\,\volt\cdot \milli\second\\
&|u_1|=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uline{1{,}5\,\volt}\qquad
|u_2|=5\,\volt-1{,}5\,\volt)=\uline{3{,}5\,\volt}\\
&\text{Da $u_1$ positives Vorzeichen in der Skizze hat, muss $u_2$ ein negatives Vorzeichen erhalten.}\\
\Rightarrow& u_1=\uuline{1{,}5\,\volt} \qquad u_2=\uuline{-3{,}5\,\volt}
\intertext{b) Effektivwert}
U^2&=U^2_{\textrm{eff}}=\frac{1}{T}\int{\left(u(t)\right)^2\cdot dt}\\
&=\frac{1}{10\,\milli\second}\left(\int_{0}^{7\,\milli\second}{(1{,}5\,\volt)^2\cdot dt}+\int_{7\,\milli\second}^{10\,\milli\second}{(-3{,}5\,\volt)^2\cdot dt}\right)\\
&=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot \big[t\big]_{0}^{7\,\milli\second}+12{,}25\,\volt^2\cdot \big[t\big]_{7\,\milli\second}^{10\,\milli\second}\right)\\
&=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot 7\,\milli\second+12{,}25\,\volt^2\cdot (10\,\milli\second-7\,\milli\second)\right)=5{,}25\,\volt^2\\
U&=\sqrt{5{,}25\,\volt^2}=\uuline{2{,}29\,\volt}
\end{align*}
\clearpage
}{}%

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ET2_L_B13_A1.tex View File

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\section{Scheinersatzwiderstände}
Der Eingangswiderstand eines linearen Zweipols beträgt bei der Frequenz $f=800\,\hertz$\\
$Z=600\,\ohm$, sein Phasenwinkel ist $\varphi=30\,\degree$ induktiv.
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie die Schaltungselemente $R_r$ und $L_r$ der gleichwertigen Reihenersatzschaltung!
\item Berechnen Sie die Schaltungselemente $R_p$ und $L_p$ der gleichwertigen Parallelersatzschaltung!
\item Wie ändern sich die Scheinersatzwiderstände (Betrag und Phase) beider Ersatzschaltungen, wenn die Frequenz $f'= 600\,\hertz$ beträgt?
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
\underline{Z}& &\text{Scheinwiderstand (Impedanz)}\\
Z&=|\underline{Z}| &\text{Betrag des Scheinwiderstandes}\\
X&=\omega\cdot L &\text{Blindwiderstand (Reaktdanz)}\\
B&=-\frac{1}{\omega\cdot L} &\text{Blindleitwert (Suszepdanz)}
\end{align}
\begin{align*}
\intertext{Berechnung:}
\intertext{a) Widerstandsebene:}
R_r&=Z\cdot \cos(\varphi_r)=600\,\ohm\cdot \cos(30\degree)=\uuline{520\,\ohm}\\
X_r&=Z\cdot \sin(\varphi_r)=600\,\ohm\cdot \sin(30\degree)=300\,\ohm\\
L_r&=\frac{X_r}{\omega}=\frac{X_r}{2\pi f}=\frac{300\,\ohm}{2\pi\cdot 800\,\frac{1}{\second}}=\uuline{60\,\milli\henry}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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] {$R_r$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R_r}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Spule -
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_r$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{L_r}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.025) node [above left] {\footnotesize$1$};
\fill (2,0)circle(.025) node [above right] {\footnotesize$2$};
\end{scope}
\begin{scope}[>=latex,very thick, xshift=3cm, yshift=-.25cm]
\draw [->,thin](0,0)--(1,0)node[right]{$R$};
\draw [->](0,0)--(.52,0)node at (.26,0)[below]{$R_r$};
\draw [->,thin](0,0)--(0,.5)node[above]{$X$};
\draw [->](.52,0)--(.52,.3)node at (.52,.15)[right]{$X_r$};
\draw [->](0:0)--(30:.6)node at (30:.3)[above left]{$Z$};
\draw [->,red,thin] (0:.26) arc (0:30:.26cm) node at (15:.26) [right] {$\varphi_r$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{b) Leitwertebene:}
Y&=\frac{1}{Z}=\frac{1}{600\,\ohm}=1{,}667\,\milli\siemens \text{; }\qquad\varphi_p=-\varphi_r=-30\degree\\
G_p&=Y\cdot \cos(\varphi_p)=1{,}667\,\milli\siemens\cdot \cos(-30\degree)
=1{,}443\,\milli\siemens\Rightarrow R_p=\frac{1}{G_p}=\uuline{693\,\ohm}\\
B_p&=Y\cdot \sin(\varphi_p)=1{,}667\,\milli\siemens\cdot \sin(-30\degree)
=-0{,}833\,\milli\siemens\Rightarrow X_p=-\frac{1}{B_p}=1200\,\ohm\\
%&\text{mit }B_p=-\frac{1}{\omega\cdot L_p}\Rightarrow \\
L_p&=-\frac{1}{\omega\cdot B_p}=-\frac{1}{2\pi f\cdot B_p}=\frac{-1}{2\pi\cdot 800\,\frac{1}{\second}\cdot (-0{,}833\cdot \power{10}{-3}\,\frac{1}{\ohm})}=\uuline{239\,\milli\henry}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,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] {$G_p$};
\draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$i_{G_p}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spule |
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$B_p$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$i_{B_p}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=-1cm,yshift=0cm]%End Knoten
\draw (2,.5)--(2,0)--(0,0);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=-1cm,yshift=1cm]%End Knoten
\draw (2,-.5)--(2,0)--(0,0);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=-1cm,yshift=0cm]%Knotenpunkte
\fill (0,1)circle(.025) node [above left] {\footnotesize$1$};
\fill (0,0)circle(.025) node [above left] {\footnotesize$2$};
\end{scope}
\begin{scope}[>=latex,very thick, xshift=2cm, yshift=.5cm]
\draw [->,thin](0,0)--(1,0)node[right]{$G$};
\draw [->](0,0)--(.52,0)node at (.26,0)[above]{$G_p$};
\draw [->,thin](0,-.5)--(0,.5)node[above]{$jB$};
\draw [->](.52,0)--(.52,-.3)node at (.52,-.15)[right]{$B_p$};
\draw [->](0:0)--(-30:.6)node at (-30:.3)[below left]{$Y$};
\draw [->,red,thin] (0:.26) arc (0:-30:.26cm) node at (-15:.26) [right] {$\varphi_p$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{c) Frequenz $f'$ \newline Reihenschaltung:}
R'_r&\stackrel{!}{=} R_r=520\,\ohm\\
X'_r&=\omega'\cdot L_r=2\pi \cdot 600\,\frac{1}{\second}\cdot 0{,}06\,\ohm\second=226\,\ohm\\
Z'_r&=\sqrt{R'^2_r+X'^2_r}=\sqrt{520^2+226^2}\,\ohm=\uuline{567\,\ohm}\\
%\varphi'&=\arctan\frac{\Im\mathfrak m}{\Re\mathfrak e}=\arctan\frac{X'_r}{R'_r}=\arctan\frac{226\,\ohm}{520\,\ohm}=\uuline{23{,}5\,\degree}
\varphi'_r&=\arctan\frac{\Im}{\Re}=\arctan\frac{X'_r}{R'_r}=\arctan\frac{226\,\ohm}{520\,\ohm}=\uuline{23{,}5\,\degree}
\intertext{Parallelschaltung:}
G'_p&\stackrel{!}{=} G_p=1{,}443\,\milli\siemens\\
B'_p&=\frac{-1}{\omega'\cdot L_p}=\frac{-1}{2\pi \cdot 600\,\frac{1}{\second}\cdot 0{,}239\,\ohm\second}=\frac{-1}{901\,\ohm}=-1{,}11\,\milli\siemens\\
Y'_p&=\sqrt{G'^2_r+B'^2_r}=\sqrt{1{,}443^2+(-1{,}11)^2}\,\milli\siemens=1{,}82\,\milli\siemens\\
Z'_p&=\frac{1}{Y'_p}=\uuline{549\,\ohm}\\
\varphi'_p&=\arctan\frac{-1{,}11\,\milli\siemens}{1{,}443\,\milli\siemens}=\uuline{-37{,}6\,\degree}
\end{align*}
\clearpage
}{}%

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ET2_L_B13_A2.tex View File

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\section{Verbraucherleistung}
An einem Verbraucher liegt die Spannung $u(t)=310\,\volt \cdot \sin(\omega t+55\,\degree)$ an, er nimmt einen Strom von $i(t)=8,5\,\ampere\cdot \cos (\omega t)$ auf.
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Berechnen Sie den zeitlichen Verlauf des Momentanwertes der Verbraucherleistung!
\item Berechnen Sie die Schein-, Wirk- und Blindleistung!
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Merksatz:\\
Kondensat\textbf{o}r, Strom eilt v\textbf{o}r\\
Induktivit\textbf{ä}t, Strom ist zu sp\textbf{ä}t\\[\baselineskip]
Berechnung:\\
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex, xshift=0cm, yshift=0]
\foreach \ii in {5} { % Enter Number of Decades in x
\foreach \jj in {2} { % Enter Number of Decades in y
\foreach \i in {1,2,...,\ii} {
\foreach \j in {1,2,...,\jj} {
\draw[black!50!, step=0.5] (0,0) grid (\ii,\jj); % Draw Sub Linear grid
}}% End Log Grid
\draw[black!80!] (0,0) grid (\ii,\jj); % Draw Linear grid
\draw [->,blue,thick] (5,0)--(5,\jj+.25) node (yaxis) [above] {$u\,[\volt]$};
\draw [->,red,thick] (0,0)--(0,\jj+.25) node (yaxis) [above] {$i\,[\ampere]$};
\draw [->,thick] (0,0)--(\ii+.25,0) node (xaxis) [right] {$\omega t\,[\degree]$}; % Draw axes
\foreach \x in {-90,0,90,180,270,360}% x Axis Label:
\node [blue,anchor=north] at(\x/90+1,0){$\x$};
\foreach \y in {0,10}% y Axis Label:
\node [red,anchor=east] at(0,\y/10+1){$\y$};
\foreach \y in {0,500}% y Axis Label:
\node [blue,anchor=west] at(5,\y/500+1){$\y$};
}}
\draw[very thick](1,0)--(1,2);
\draw[->,very thick](1,1)--(.389,1) node [below right]{ $-55\degree$};
\draw[<-,very thick](2,1)--(2.389,1) node at (2,1.2)[above right]{$-35\degree$ $i$ vor $u \Rightarrow$ kapazitiv};
\end{scope}
\begin{scope}[>=latex, xshift=0cm, yshift=1cm]
\draw[color=red,thick,domain=0:5,smooth,samples=100] plot[id=i] function{.85*cos(.5*3.14*x-1.57)};
\draw[color=blue,thick,domain=0:5,smooth,samples=100] plot[id=u] function{.62*sin(.5*3.14*x+.96-1.57)};
%\draw[->,blue, very thick] (1.5,1) -- (2.5,1);
\draw[red] node at (1.5,1.25) {{\footnotesize $i(t)=8,5\,\ampere\cdot \cos (\omega t)$}};
\draw[blue] node at (3.5,1.25) {{\footnotesize $u(t)=310\,\volt \cdot \sin(\omega t+55\,\degree)$}};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\intertext{a) Leistungsverlauf}
p(t)&=u(t)\cdot i(t) \qquad\text{Momentane Leistung}\\
p(t)&=310\,\volt\cdot 8{,}5\,\ampere\cdot \sin x\cdot \cos y \\
&\text{mit $x=\omega t+55\degree=\omega t+0{,}96\,\text{rad} \qquad y=\omega t$}\\
&\text{und } \sin x\cdot \cos y=\frac{1}{2}[\sin(x-y)+\sin(x+y)]\Rightarrow\\
p(t)&=\widehat{u}\cdot \widehat{i}\cdot \frac{1}{2}\cdot [\sin(x-y)+\sin(x+y)]\\
&=310\,\volt\cdot 8{,}5\,\ampere\cdot \frac{1}{2}\cdot \big[\sin(\cancel{\omega t}+0{,}96-\cancel{\omega t})+\sin(\omega t+0{,}96+\omega t)\big]\\
&=\underbrace{310\,\volt\cdot 8{,}5\,\ampere\cdot \frac{1}{2}\vphantom{\frac{1}{2}}}_{S=1318\,\volt\ampere}\cdot \big[\underbrace{\sin(0{,}96)\vphantom{\frac{1}{1}}}_{0{,}819}+\sin(2\omega t+0{,}96)\big]\\
p(t)&=\uuline{1079\,\watt+1318\,\volt\ampere\cdot \sin(2\omega t+0{,}96)}
\intertext{b) $S$ Schein-, $P$ Wirk- und $Q$ Blindleistung}
\cos(\omega t)&=\sin(\omega t+90\degree)\\
\varphi_i&=+90\degree\quad\varphi_u=+55\degree\\
\varphi_u-\varphi_i&=+55\degree-90\degree=-35\degree\\
S&=U\cdot I=\frac{\widehat{u}}{\sqrt{2}}\cdot \frac{\widehat{i}}{\sqrt{2}}=\frac{1}{2}\cdot \widehat{u}\cdot \widehat{i}=\frac{2635}{2}\,\volt\ampere=\uuline{1318\,\volt\ampere}\\
P&=S\cdot \cos(-35\degree)=S\cdot 0{,}819=\uuline{1079\,\watt}\\
Q&=S\cdot \sin(-35\degree)=S\cdot (-0{,}576)=\uuline{-756\,\mathrm{var}}\qquad\text{Lies: Volt-Ampere-reaktiv}\\
\end{align*}
\clearpage
}{}%

+ 74
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ET2_L_B13_A3.tex View File

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\section{Blindleistungskompensation}
Die Daten der beiden Verbraucher am Einphasen-Wechselstromnetz sind:\\
Heizwiderstand $R_H$:
Aufgenommene Leistung $P_H=1,5\,\kilo\watt$\\
Motor $M$:
Aufgenommene Leistung $P_{auf}=2,5\,\kilo\watt$\\
Leistungsfaktor $\cos\varphi=0,7$\\
$U_N=230\,\volt$; $f=50\,\hertz$
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Wie groß ist der dem Netz entnommene Strom $I_N$?
\item Welche Phasenverschiebung hat der Strom $\uline{I}_N$ zu der Spannung $\uline{U}_N$?
\item Welche Kapazität muss ein Kondensator, parallel zu den Verbrauchern geschaltet, haben, damit der Blindstrom voll kompensiert wird?
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[scale=.75]
\begin{scope}[>=latex,very thick]
\draw (0,0) circle (1) node [above]{$M$};
\draw node at (0,0) [below]{$1 \approx$}; % F E H L E R
\draw (180:1)--(-2,0)--(-2,5)circle(.025) (0:1)--(2,0)--(2,5)circle(.025) ;
\draw (-1,2.5)--(-2,2.5)circle(.025) (1,2.5)--(2,2.5)circle(.025) (-1,2.2 )rectangle (1,2.8);
\draw (0,2.5)node{$R_H$};
\draw [->,blue](-1.8,5)--(1.8,5) node at (0,5) [above]{$\underline{U}_N$;$f$};
\draw [->,red](-1.8,4.8)--(-1.8,2.7) node at (-1.8,3.75) [right]{$\underline{I}_N$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
\begin{align*}
\intertext{Berechnung:}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=1.5]
\begin{scope}[>=latex,very thick]
\draw [->,red!50!blue](0:0)--(45.6:3.571)node [above left]{$\underline{S}_M$};
\draw [->,red!50!blue](0:0)--(0:2.5)node at (1.25,0)[above]{$P_M$};
\draw [->](0:0)--(32.52:4.744)node [right]{$\underline{S}_{gesamt}$};
\draw [->](45.6:3.571)--+(0:1.5)node at (3.25,2.551)[above]{$P_H$};
\draw [->,red!50!blue](2.5,0)--+(90:2.551)node at (2.5,1.25)[right]{$Q_M$};
\draw [red!50!blue]node at (1.5,.75)[below]{Motor};
\draw [->](2.5,0)--(4,0)node at (3.25,0)[above]{$P_H$};
\draw [->,red!50!blue](4,0)--+(90:2.551)node at (4,1.25)[right]{$Q_M$};
\end{scope}
\begin{scope}[>=latex, xshift=0, yshift=0]
\draw [black!50!] (0,0)grid(4,3);
\foreach \x in {0,1,...,4}% x Axis Label:
\node [anchor=north] at(\x,0){$\x$};
\foreach \y in {0,1,...,3}% y Axis Label:
\node [anchor=east] at(0,\y){$\y$};
\draw [->,thick](0,0)--(4.25,0) node [right]{$P\,[\kilo\watt]$};
\draw [->,thick](0,0)--(0,3.25) node [above]{$Q\,[\kilo\var]$};
\end{scope}
\end{tikzpicture}\\
\centering \underline{S}_{gesamt}=\sqrt{(P_M+P_H)^2+Q_M^2}
\end{align*}
\begin{align*}
\intertext{a) Nennstrom}
\varphi&=\arccos(0{,}7)=45{,}6\degree\\
S_M&=\frac{P_{auf}}{0{,}7}=\uline{3571\,\volt\ampere} &\text{Motor Scheinleistung}\\
Q_M&=S_M\cdot \sin\varphi=3571\,\volt\ampere\cdot 0{,}7141=\uline{2551\,\var} &\text{Motor Blindleistung}\\
P&=P_{auf}+P_H=\uline{4\,\kilo\watt} &\text{Gesamte Wirkleistung}\\
S&=\sqrt{P^2+Q^2_M}=\uline{4744\,\volt\ampere} &\text{Gesamte Scheinleistung}\\
I_N&=\frac{S}{U_N}=\frac{4744\,\volt\ampere}{230\,\volt}=\uuline{20{,}63\,\ampere}
\intertext{b) Phasenverschiebung}
\varphi_N&=\arccos\frac{P}{S}=\arccos\frac{4000\,\watt}{4744\,\volt\ampere}=\uuline{32,52\degree}
\intertext{c) Kompensation}
|Q_C|&=Q_M=2551\,\var=U^2_N\cdot \omega\cdot C\\
C&=\frac{2551\,\var}{2\pi\cdot 50\,\frac{1}{\second}\cdot (230\,\volt)^2}=\uuline{153{,}6\,\micro\farad}
\end{align*}
\clearpage
}{}%

+ 133
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ET2_L_B13_A4.tex View File

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\section{Energieübertragung}
Die Skizze zeigt ein System zur elektrischen Energieübertragung bestehend aus Quelle,
Leitung und Verbraucher. Das System soll mit einem parallel geschalteten Kondensator $X_C$ so
optimiert werden, dass die Leitungsverluste $P_{VRL}$ minimal werden.\\
\begin{align*}
\begin{tikzpicture}[scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle |
\draw (0,0)--(1,0)node at(.5,-.133)[right]{\footnotesize{$\underline{U}_q=100\,\volt\cdot e^{j0}$}};
\draw (.5,0)circle(.133);
\draw [<-,blue] (.3,.2)--(.7,.2)node at (.5,.2)[left]{\footnotesize$\underline{U}_q$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,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] {\footnotesize{$\underline{Z}_i=(1+2j)\,\ohm$}};
\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] {\footnotesize{$R_L=1\,\ohm$}};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=0cm,rotate=90]
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [above right] {\footnotesize{$\underline{Z}_V=(10+j5)\,\ohm$}};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=3.5cm,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] {\footnotesize{$jX_C$}};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [dashed](2.5,0)--(3.5,0)--(3.5,.2) (2.5,1)--(3.5,1)--(3.5,.8);
\draw [dashed](0,0)--(2.5,0) (2.5,1)--(3.5,1)--(3.5,.8);
\draw (0,0)--(2.5,0) (2,1)--(2.5,1);
\draw (0,.2)--(0,0)--(0,.2) (0,.8)--(0,1)--(.2,1);
\draw node at(.5,0) [below] {\footnotesize Quelle};
\draw node at(1.5,0) [below] {\footnotesize Leitung};
\draw node at(2.5,0) [below] {\footnotesize Verbraucher};
\draw [very thin, dashed] (1,-.2)--(1,1.2)(2,-.2)--(2,1.2);
\draw [->,red](2.125,1.125)--(2.375,1.125) node [right] {\footnotesize{$\underline{I}$}};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte
\draw (0,0)circle(.035);
\fill [white](0,0)circle(.025);
\draw (1,0)circle(.035);
\fill [white](1,0)circle(.025);
\fill (1.5,0)circle(.025);
\fill (2.5,0)circle(.025);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte
\draw (0,0)circle(.035);
\fill [white](0,0)circle(.025);
\draw (1,0)circle(.035);
\fill [white](1,0)circle(.025);
\fill (1.5,0)circle(.025);
\fill (2.5,0)circle(.025);
\end{scope}
\end{tikzpicture}
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Bestimmen Sie $X_C$ so, dass der Blindleistungsbedarf des Verbrauchers verschwindet.
\item Berechnen Sie die Verlustleistung $P_{VRL}$ der Leitung und die Wirkleistung $P_W$ im Verbraucher.
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:\\[\baselineskip]
a) Verbraucher $\uline{Z}_V\,||\,X_C$, daher Ersatzschaltbild für $\uline{Z}_V$ (ESB) in Parallelform erforderlich
\begin{align*}
\underline{Z}_V&=R_V+jX_V=(10+j5)\,\ohm\,\,\quad\text{ Scheinwiderstand, entspricht einer Reihenschaltung}\\[\baselineskip]
Z^2_V&=R_V\cdot R_p=R^2_V+X^2_V\,\qquad\qquad\text{Umwandlung in Parallel-ESB}\\
R_p&=R_V+\frac{X^2_V}{R_V}=(10+\frac{25}{10})\,\ohm=12{,}5\,\ohm\\[\baselineskip]
Z^2_V&=X_V\cdot X_{L_p}=R^2_V+X^2_V\,\,\quad\qquad\text{Umwandlung in Parallel-ESB}\\
X_{L_p}&=X_V+\frac{R^2_V}{X_V}=(5+\frac{100}{5})\,\ohm=25\,\ohm
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R_p$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spule |
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$jX_{L_p}$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\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] {$jX_C$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw (0,0)--(3,0)--(3,.2) (0,1)--(3,1)--(3,.8);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte
\fill (0,0)circle(.025);
\fill (1,0)circle(.025);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte
\fill (0,0)circle(.025);
\fill (1,0)circle(.025);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Knotenpunkte
\draw (0,0)circle(.05);
\fill (0,0)circle(.025);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Knotenpunkte
\draw (0,0)circle(.05);
\fill (0,0)circle(.025);
\end{scope}
\begin{scope}[>=latex,very thick]
\draw node at (0,0.5) [left] {$\underline{Z}'\Rightarrow$};
\end{scope}
\draw node at(1.5,0)[below]{Verbraucher $Z_V$};
% \end{tikzpicture}
% \begin{tikzpicture}[scale=.5,xshift=15cm,yshift=-2cm]
\begin{scope}[>=latex,very thick,scale=.25,xshift=18cm,yshift=2cm]
\draw [->](0,0)--(1,0)node [right]{$R_p$};
\draw [->](1,0)--(1,2)node [above]{$jX_{Lp}$};
\draw [->](1,0)--(1,-2)node[below]{$jX_C$};
\draw [->,red!50!blue](0,0)--(1,2)node at (.5,1)[left]{$Z_V$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\underline{Z}'&=\underline{Z}_V || jX_C\\
\frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\frac{1}{jX_{Lp}}+\frac{1}{jX_C}\\
\intertext{Leitungsverluste sind minimal, wenn die Blindleistung $=0$ wird (Kompensation)}
\frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\cancel{\frac{1}{jX_{Lp}}}+\cancel{\frac{1}{jX_C}}\qquad\Rightarrow \underline{Z}'=R_p\\
\Im(\underline{Z}')&=0 \quad\text{oder}\quad |X_C| \stackrel{!}{=} |X_{L_p}|\text{ also}\\
%\Re(\underline{Z}')&=R_p\\
X_C&=-X_{L_p}=\uuline{-25\,\ohm}\\
\underline{Z}_{ges}&=\underline{Z}_i+R_L+\underline{Z}'\\
\underline{Z}_{ges}&=\underline{Z}_i+R_L+R_p=(1+j2+1+12{,}5)\,\ohm=(14{,}5+j2)\,\ohm\\
|\underline{Z}_{ges}|&=\sqrt{14{,}5^2+2^2}\,\ohm=14{,}64\,\ohm\\
I&=\frac{U}{|Z_{ges}|}=\frac{100\,\volt}{14{,}64\,\ohm}=6{,}83\,\ampere\\
\intertext{b) Verlust- und Wirkleistung}
P_{VR_L}&=I^2\cdot R_L=(6{,}83\,\ampere)^2\cdot 1\,\ohm=\uuline{46{,}7\,\watt}\\
P_W&=I^2\cdot R_p=(6{,}83\,\ampere)^2\cdot 12{,}5\,\ohm=\uuline{583\,\watt}
\end{align*}
\clearpage
}{}%

+ 59
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ET2_L_B13_A5.tex View File

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\section{Wechselstrommotor}
Ein Einphasen- Wechselstrommotor liegt an einer Spannung von $230\,\volt - 50\,\hertz$ und gibt eine Leistung von $2\,\kilo\watt$ ab, wobei sein Wirkungsgrad $\eta= 80\%$ und sein Leistungsfaktor $\cos\varphi=0,7$ beträgt.
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Wie groß ist die Stromaufnahme des Motors?
\item Welche Kapazität muss parallelgeschaltet werden, um eine Blindstromkompensation auf
$cos\varphi=0{,}9$ zu erreichen?
\item Wie groß ist der dem Netz bei $cos\varphi=0{,}9$ entnommene Strom?
\end{enumerate}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
S&=U\cdot I\qquad Scheinleistung\\
P_{el}&=S\cdot\cos(\varphi)\qquad Wirkleistung\\
P_{ab}&=\eta\cdot P_{el}\\
\cos\varphi&=\frac{P}{S}
\end{align}
\begin{align*}
\intertext{Berechnung:}
\intertext{a) Stromaufnahme:}
P_{el}&=\frac{P_{ab}}{\eta}=\frac{2\,\kilo\watt}{0{,}8}=2{,}5\,\kilo\watt\\
S&=\frac{P_{el}}{\cos(\varphi)}=\frac{2{,}5\,\kilo\watt}{0{,}7}=3{,}571\,\kilo\volt\ampere\\
I&=\frac{S}{U}=\frac{3{,}571\,\kilo\volt\ampere}{230\,\volt}=\uuline{15{,}5\,\ampere}
\end{align*}
\clearpage
b) Kapazität:
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex, xshift=0, yshift=0]
\draw [black!50!,very thin](0,0)grid(2,2);
\draw [->,red,very thick] (2.525,0)--(2.525,2.55)node [right]{$Q$};
\draw [->,thick](0:0)--(0:2.5)node [right]{$P$};
\draw [->,thick] (0:0)--(90:2.55)node [above]{$Q$};
\draw [->,very thick](0:0)--(25.8:2.778) node [above left] {$\underline{S}'$};
\draw [->,very thick] (0,0)--(45.6:3.571) node [above left] {$\underline{S}$};
\draw [->,blue,ultra thick] (45.6:3.571)--(25.8:2.778)node [above right] {$Q_C$};
\draw [->,black!50!green, very thick] (2.5,0)--(2.5,1.209)node [below right]{$Q'$};
\draw (0:1)arc(0:45.6:1)node [right]{$45{,}6\degree$};
\draw (0:1.75)arc(0:25.8:1.75)node [below right]{$25{,}8\degree$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\varphi&=\arccos(0{,}7)=45{,}6\degree\\
{\color{red}Q}&=S\cdot \sin(45{,}6\degree)=3{,}571\,\kilo\volt\ampere \cdot 0{,}714=2{,}55\,\kilo \var\\
\varphi'&=\arccos(0{,}9)=25{,}86\degree\\
S'&=\frac{P}{\cos \varphi'}=\frac{2{,}5\,\kilo\watt}{0{,}9}=2{,}778\,\kilo\volt\ampere\\
{\color{black!50!green}Q'}&=S'\cdot \sin \varphi '=2{,}778\,\kilo\volt\ampere \cdot 0{,}435=1{,}209\,\kilo \var\\
\intertext{für Kompensation muß gelten:}
Q+Q_C-Q'&=0 \\
\Rightarrow Q_C=Q'-Q&=1{,}209\,\kilo \var -2{,}55\,\kilo \var=-1{,}341\,\kilo \var\\
{\color{blue}|Q_C|}&=\frac{U^2}{|X_C|}\quad\Rightarrow |X_C|=\frac{U^2}{|Q_C|}=\frac{(230\,\volt)^2}{1341\,\var}=39{,}4\,\ohm=\frac{1}{\omega C}\Rightarrow\\
C&=\frac{1}{\omega |X_C|}=\frac{1}{2\pi\cdot 50\frac{1}{\,\second}\cdot 39{,}4\,\frac{\,\volt}{\ampere}}=8{,}06\cdot\power{10}{-5}\,\frac{\ampere\second}{\volt}=\uuline{80{,}6\,\micro\farad}\\
\intertext{c) Stromaufnahme bei $\cos\varphi=0{,}9$:}
S'&=U\cdot I'\Rightarrow I'=\frac{S'}{U}=\frac{2778\,\volt\ampere}{230\,\volt}=\uuline{12{,}1\,\ampere}
\end{align*}
Nicht auf $cos(\varphi)=1$ kompensieren, da dann Schwingkreis !
\clearpage
}{}%

+ 88
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ET2_L_B13_A6.tex View File

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\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
}{}%

+ 101
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ET2_L_B13_A7.tex View File

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\section{Werte $R_L$ und $L$ einer Spule}
Aus den drei gemessenen sinusförmigen Spannungen $U$, $U_N$, und $U_{SP}$ lassen sich die Werte $R_L$ und $L$ einer Spule bestimmen.
\begin{align*}
U=100\,\volt\\
U_N=60\,\volt\\
U_{SP}=70\,\volt\\
R_N=60\,\ohm\\
f = 50\,\hertz
\end{align*}
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Zeichnen Sie ein qualitatives Zeigerdiagramm der Spannungen!
\item Bestimmen Sie $R_L$ und $L$!
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[scale=2]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand
\draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_N$};
\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_N$};
\draw (0,1)--(0,0)--(.1,0) (3,1)--(3,0)--(2.9,0);%anschuß und Füllt die Ecken der Verbindung!
\fill (0,1)circle (0.025) (3,1)circle (0.025);
\draw [->,blue](.2,1)--(2.8,1)node at (1.5,1)[below]{\footnotesize$U$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Widerstand
\draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_L$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]%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);
\draw [->,blue] (-.7,-.2)--(.7,-.2) node at (0,-.2)[below]{\footnotesize$U_{SP}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%\begin{align}
%\intertext{Formeln:}
%\end{align}
\begin{align*}
\intertext{Berechnung:}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=.5]
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
\draw [black!25!,very thin](0,0)grid(8,7);
\draw [->,red, thick](0,0)--(8,0)node [right] {$\underline{I}$};
\draw [->](0,0)--(6,0)node at(3,0)[below] {$\underline{U}_N$};
\draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{U}_{SP}$};
\draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{U}$};
\draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$\underline{U}_{R_L}$};
\draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$\underline{U}_{L}$};
\draw [black!50!](33:10)arc(33:53:10)node [left]{$\underline{U}=100\,\volt\widehat{=}10\,\centi\metre$};
\draw [black!50!](6,0)+(90:7)arc(90:70:7)node [right]{$\uline{U}_{SP}=70\,\volt\widehat{=}7\,\centi\metre$};
\draw [black!50!](3,-1)node[below]{$\underline{U}_N=60\,\volt\widehat{=}6\,\centi\metre$};
\end{scope}
\begin{scope}[>=latex,very thick, xshift=12cm, yshift=2cm]
\draw node at(0,2)[right]{$I$ zeichnen};
\draw node at(0,1)[right]{$u_N || I$};
\draw node at(0,0)[right]{Mit Zirkel $U$ und $U_{SP}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
I&=\frac{U_N}{R_N}=\frac{60\,\volt}{60\,\ohm}=\uuline{1\,\ampere}\\
\underline{U}_{SP}&=70\,\volt=\sqrt{U^2_{RL}+U^2_L}\\
\end{align*}
\clearpage
Widerstandsoperatoren:\\

\footnotesize{Impedanzdreieck wie Spannungsdreieck}
\begin{align*}
\begin{tikzpicture}[scale=.5]
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
\draw [black!25!,very thin](0,0)grid(8,7);
\draw [->](0,0)--(6,0)node at(3,0)[below] {$R_N$};
\draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{Z}_{SP}$};
\draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{Z}$};
\draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$R_L$};
\draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$X_L$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
Z_{SP}&=\frac{U_{SP}}{I}=\frac{70\,\volt}{1\,\ampere}=70\,\ohm \quad \text{\footnotesize{(Nur Effektivwerte - ohne Winkel)}}\\%=\sqrt{R^2_L+X^2_L}\\
Z^2_{SP}&=R^2_L+X^2_L=(70\,\ohm)^2\\
X^2_L&=(70\,\ohm)^2-R^2_L \tag{1}\\[\baselineskip]
Z&=\frac{U}{I}=\frac{100\,\volt}{1\,\ampere}=100\,\ohm\\
Z^2&=(R_N+R_L)^2+X^2_L=(100\,\ohm)^2\\
X^2_L&=(100\,\ohm)^2-(R_N+R_L)^2\\
&=(100\,\ohm)^2-(R^2_N+2\cdot R_N\cdot R_L+R^2_L) \tag{2}\\[\baselineskip]
(70\,\ohm)^2-\cancel{R^2_L}&=(100\,\ohm)^2-R^2_N-2\cdot R_N\cdot R_L-\cancel{R^2_L}\tag{$1$ in $2$}\\
2\cdot R_N\cdot R_L&=(100\,\ohm)^2-R^2_N-(70\,\ohm)^2\\
R_L&=\frac{(100\,\ohm)^2-R^2_N-(70\,\ohm)^2}{2\cdot R_N}=\frac{(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2}{2\cdot 60\,\ohm}\\
&=\frac{1500\,\ohm^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm}\\
%(100\,\ohm)^2&=(60\,\ohm)^2+2\cdot 60\,\ohm\cdot R_L+\cancel{R^2_L}+(70\,\ohm)^2 -\cancel{R^2_L}\\
%2\cdot 60\,\ohm\cdot R_L&=(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2=1500(\,\ohm)^2\\
%R_L&=\frac{1500(\,\ohm)^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm} \tag{in $1$}\\
\text{in (1) }\qquad X_L&=\sqrt{(70\,\ohm)^2-(12{,}5\,\ohm)^2}=68{,}87\,\ohm\\
L&=\frac{X_L}{\omega}=\frac{68{,}87\,\ohm}{2\pi\cdot 50\,\frac{1}{\second}}=\uuline{0{,}219\,\henry}
\end{align*}
\clearpage
}{}%

+ 130
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ET2_L_B14_A1.tex View File

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\section{Zeigerdiagramm}
Gegeben ist die Ausgangsspannung $U_a = 5\,\volt \cdot e^{j0\degree}$ und $R_1=R_2=\frac{1}{\omega C_1}=\frac{1}{\omega C_2}=1\,\kilo\ohm$\\
Zeichnen Sie ein maßstäbliches Zeigerdiagramm aller Spannungen und aller Ströme!\\
Maßstäbe: $1\,\centi\metre\,\widehat{=}\, 1\,\volt \text{;}\quad 1\,\centi\metre\,\widehat{=}\, 1\,\milli\ampere$
Entnehmen Sie dem Zeigerdiagramm Betrag und Phasenwinkel der Spannung $U_e$ !
\begin{align*}
\begin{tikzpicture}[very thick,scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,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] {$R_1$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$C_1$};
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$};
\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] {$R_2$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$};
\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] {$C_2$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (0,0)--(2.5,0) (2,1)--(2.5,1);
\fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025);
\draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$};
\draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
e^{j\varphi}&=\cos\varphi+j\sin\varphi \quad\text{Eulersche Formel}
\end{align}
\footnotesize{R-C Ketten sind u.a. ein Ersatzbild für Leitungen (Kapazität pro Längeneinheit)}\\
Berechnung:
\begin{align*}
\begin{tikzpicture}[very thick,scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,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] {$R_1$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$C_1$};
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$};
\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] {$R_2$};
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$};
\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] {$C_2$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (0,0)--(2.5,0) (2,1)--(2.5,1);
\fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025);
\draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$};
\draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$};
\draw [->,red] (.05,1.1)--(.25,1.1) node at (.1,1.1)[above]{$\underline{I}_{e}$};
\draw [->,red] (2.2,1.1)--(2.4,1.1) node at (2.3,1.1)[above]{$\underline{I}_{a}=0$};
\draw [->,red] (1.1,.9)--(1.1,.7) node at (1.1,.8)[right]{$\underline{I}_{1}$};
\draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{2}$};
\end{scope}
\begin{scope}[>=latex,thick,xshift=.4cm,yshift=.5cm]%Masche
\draw [->,red!50!blue] (270:.15)arc(270:-60:.15) node at (0,0){$M_1$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\underline{U}_a&=5\,\volt\cdot e^{j0\degree}\\
\underline{I}_a&=0\text{, da kein Lastwiderstand angeschlossen ist!}\\
\underline{I}_2&=\frac{\underline{U}_a}{\underline{jX}_2}
=\frac{5\,\volt\cdot e^{j0\degree}}{1\kilo\ohm\cdot e^{-j90\degree}}
=5\,\milli\ampere\cdot e^{j90\degree}=j5\,\milli\ampere
\quad\text{(Strom eilt vor)}\quad 5\,\milli\ampere\;\angle +90\degree\\
\underline{U}_{R_2}&=R_2\cdot \underline{I}_2=1\,\kilo\ohm\cdot e^{-j90\degree}=1\,\kilo\ohm\cdot 5\,\milli\ampere\;\angle +90\degree=5\,\volt\;\angle +90\,\degree\\
\underline{U}_1&=\underline{U}_a+\underline{U}_{R_2}=5\,\volt+j5\,\volt\quad\text{(Vektoren addieren)}\quad =\sqrt{2}\cdot 5\,\volt\;\angle +45\degree \sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}\\\underline{I}_1&=\frac{\underline{U}_1}{\underline{jX}_1}=\frac{\sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}}{ 1\kilo\ohm\cdot e^{-j90\degree}}=\sqrt{2}\cdot 5\,\milli\ampere\cdot e^{j135\degree}=5\cdot (-1+j)\,\milli\ampere\quad\text{(Strom eilt $90\degree$ vor) } 5\,\milli\ampere\;\angle +135\degree\\
\underline{I}_e&=\underline{I}_1+\underline{I}_2=\sqrt{2}\cdot 5\,\milli\ampere\;\angle +135\degree+5\,\milli\ampere\;\angle +90\degree\quad\text{(Vektoren addieren)}\\[\baselineskip]
\end{align*}
\begin{align*}
\intertext{Jetzt zeichnen oder rechnerisch: (jedoch aufwendiger)}
\underline{I}_e&=I_1\cdot (\cos\varphi+j\sin\varphi)+I_2\cdot (\cos\varphi+ j\sin\varphi)\\
&=I_1\cdot (\cos135\degree+j\sin 135\degree)+I_2\cdot (\cos 90\degree+ j\sin 90\degree)\\
&=|\underline{I}_1|\cdot (-\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})+|\underline{I}_2|\cdot (0+j)\\
&=[\cancel{\sqrt{2}}\cdot 5\cdot (-\frac{1}{\cancel{\sqrt{2}}}+j\frac{1}{\cancel{\sqrt{2}}})+j\cdot 5]\,\milli\ampere=(-5+j10\,\milli\ampere)\\
&\quad |\underline{I}_e|= \sqrt{10^2+5^2}\,\milli\ampere=11{,}18\,\milli\ampere \\
&\quad\tan\varphi=\frac{\Im}{\Re}=\tan\frac{10}{-5}=\tan-2\Rightarrow\varphi=\arctan\frac{-2}=-1{,}107\,\rad\,\widehat{=}\,-63{,}435\degree\\
&\quad\text{(Definitionsbereich $\tan\varphi [-\pi/2\cdots\pi/2]$ beachten!)}\\
&=11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere\\[\baselineskip]
\underline{U}_{R_1}&=R_1\cdot \underline{I}_e\\
&\quad\text{(Nur zur Vollständigkeit) }\underline{U}_{R_1}=1\,\kilo\ohm\cdot 11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere=11{,}18\cdot e^{j116{,}565\degree}\,\volt\\
\underline{U}_e&=\underline{U}_{R_1}+\underline{U}_1=\uuline{15\,\volt\cdot e^{+j90\degree}}\quad\text{(Vektoren addieren)}
\end{align*}
\begin{align*}
\begin{tikzpicture}[scale=.5]
\begin{scope}[>=latex]
\draw [very thin,black!50!](-5,0)grid(5,15);
\end{scope}
\begin{scope}[>=latex,very thick]
\draw [->,blue] (0:0)--(90:15) node at (90:7.5)[right]{$\underline{U}_e$};
\draw [->,blue] (0:0)--(0:5) node at (0:2.5)[below]{$\underline{U}_a$};
\draw [->,blue] (0:0)--(45:7.07) node at (45:3.54)[right]{$\underline{U}_1$};
\draw [->,blue,ultra thick] (0:0)--(-5,10) node at (-2.5,5)[right]{$\underline{U}_{R_1}$};
\draw [->,blue] (-5,10)--(0,15) node at (-2.5,12.5)[left]{$\underline{U}_{1}$};
\draw [->,blue,ultra thick] (0:0)--(90:5) node at (90:3.5)[right]{$\underline{U}_{R_2}$};
\draw [->] (5,0)--(5,5) node at (5,3.5)[right]{$\underline{U}_{R_2}$};
\draw [->,red] (0:0)--(135:7.07) node at (135:3.54)[left]{$\underline{I}_1$};
\draw [->,red,thick] (0:0)--(90:5) node at (90:3.5)[left]{$\underline{I}_2$};
\draw [->] (135:7.07)--+(90:5) node at (-5,7.5)[right]{$\underline{I}_2$};
\draw [->,red,thick] (0:0)--(-5,10) node at (-2.5,5)[left]{$\underline{I}_{e}$};
\end{scope}
\draw node at (8,10)[right]{Reihenfolge$U_a\, I_2, U_{R_2},U_1, I_1, I_e, U_{R_1})$};
\draw node at (8,9)[right]{$U_a\widehat{=}\,5\,\centi\metre\angle 0\,\degree\quad(5+j0)$};
\draw node at (8,8)[right]{$U_{R_2}\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$};
\draw node at (8,7)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(5+j5)$};
\draw node at (8,6)[right]{$I_1\widehat{=}\,7{,}07\,\centi\metre\angle 135\,\degree\quad(-5+j5)$};
\draw node at (8,5)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$};
\draw node at (8,4)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(-5+j5)\rightarrow(0+j5)$ addiert zu $I_1$};
\draw node at (8,3)[right]{$I_e\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree(-5+j10)$};
\draw node at (8,2)[right]{$U_{R_1}\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree\quad(-5+j10)$};
\draw node at (8,1)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(-5+j10)\rightarrow(5+j5)$ addiert zu $U_{R_1}$};
\end{tikzpicture}
\end{align*}
\begin{align*}
\sum M_1=0=U_{R_1}+U_1-U_e\\
U_e=15\,\volt \cdot e^{j90\degree}
\end{align*}
\clearpage
}{}%

+ 143
- 0
ET2_L_B14_A2.tex View File

@@ -0,0 +1,143 @@
\section{Gesamtwiderstand}
Von der Schaltung (Bild 1) sind die Zeiger $\uline{U}_0$ und $\uline{I}_0$ gegeben (Bild 2).
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Ist der Gesamtwiderstand $\uline{Z}$ induktiv, ohmsch oder kapazitiv? (Stichwortartige Begründung !)
\item Vervollständigen Sie Bild 2 zu einem qualitativen Zeigerdiagramm \uline{aller} Ströme und Spannungen.\\
(Rechte Winkel oder Parallelen sind zu kennzeichnen. Alle Ströme und Spannungen müssen im Schaltbild (Bild 1) und im Zeigerbild unmissverständlich benannt werden.)\\
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[very thick,scale=2]
\begin{scope}[>=latex,very thick,xshift=0.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_C$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=-0.5cm,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] {$C$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-2cm,rotate=90]%Spule |
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (-.5,.2)--(-0.5,0)--(.5,0)--(.5,.2) (-1,1)--(.5,1)--(.5,.8) (-1,-2)--(0,-2)--(0,-1.8);
\draw [->,blue] (-1,.9)--(-1,-1.9)node at(-1,-.5)[right]{$\underline{U}_{0}$};
\draw node at(-1,-.5)[left]{$\underline{Z}\Rightarrow$};
\fill (-1,1)circle(.025) (-1,-2)circle(.025);
\draw [->,red] (-.9,1.1)--(-.6,1.1) node at (-.75,1.1)[above]{$\underline{I}_{0}$};
\draw node at (0,-2.2)[below]{Bild 1};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]
\draw [->,blue] (0,0)--(3,0)node [right]{$\underline{U}_{0}$};
\draw [->,red] (0,0)--(-33:1.5)node [right]{$\underline{I}_{0}$};
\draw node at (1.5,-2.2)[below]{Bild 2};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Merksätze:}
&\text{Ohm'scher Widerstand: Strom und Spannung in Phase}\\
&\text{Kondensator: Strom eilt $90\degree$ vor}\\
&\text{Induktivität: Spannung eilt $90\degree$ vor}
\end{align}
Berechnung:\\[\baselineskip]
a) Induktiv, da nacheilender Strom.\\
\begin{align*}
\intertext{b) Schaltbild mit Strom- und Spannungspfeilen ergänzen, Zeigerdiagramm erstellen.}
\begin{tikzpicture}[very thick,scale=2]
\begin{scope}[>=latex,very thick,xshift=0.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_C$};
\draw [<-,blue] (.3,.5)--(.7,.5)node at(.5,.5)[right]{\footnotesize$\uline{U}_{C}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=-0.5cm,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 [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$\uline{U}_{C}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{R}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-2cm,rotate=90]%Spule |
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{L}$};
\draw [<-,blue] (.5,-.5)--(1.5,-.5)node at(1,-.5)[right]{\footnotesize$\uline{U}_{RL}$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (-.5,.2)--(-0.5,0)--(.5,0)--(.5,.2) (-1,1)--(.5,1)--(.5,.8) (-1,-2)--(0,-2)--(0,-1.8);
\draw [->,blue] (-1,.9)--(-1,-1.9)node at(-1,-.5)[right]{$\underline{U}_{0}$};
\draw node at(-1,-.5)[left]{$\underline{Z}\Rightarrow$};
\fill (-1,1)circle(.025) (-1,-2)circle(.025);
\draw [->,red] (-.9,1.1)--(-.6,1.1) node at (-.75,1.1)[above]{$\underline{I}_{0}$};
\draw [<-,red] (-.7,.3)--(-.7,.7) node at (-.7,.5)[left]{$\underline{I}_{C}$};
\draw [<-,red] (1.,.3)--(1.,.7) node at (1.,.5)[right]{$\underline{I}_{RC}$};
% \draw node at (0,-2.2)[below]{Bild 1};
\end{scope}
\end{tikzpicture}
\end{align*}
%\begin{align*}
%\uline{I}_0&=\uline{I}_{RC}+\uline{I}_C\\
%\uline{I}_C\,& \bot \,\uline{I}_{RC}\\
%\uline{I}_C &\text{ eilt vor } [\text{ für }\uline{I}_{RC}\text{ und }\uline{I}_{C}]
%\end{align*}
\begin{align*}
\begin{tikzpicture}[very thick,scale=2.5]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [->,blue] (0,0)--(3,0)node [above left]{$\underline{U}_{0}$};
\draw [->,red] (0,0)--(-33:1.5)node at (-33:1)[above right]{$\underline{I}_{0}$};
\fill [black!50!](-33:.75)circle(.02);
\draw [very thin](-33:.0)arc(147:327:.75);
\draw (-66:1.258)+(24:.15)arc(24:114:.15) ; %Rechter Winkel
\fill (-66:1.258)+(69:.075)circle(.02) ; %Punkt Rechter Winkel
\draw [->,red] (0,0)--(-66:1.258)node at(-68:.8)[above left]{$\underline{I}_{RC}$};
\draw [->,red] (-66:1.258)--(-33:1.5)node at (-55:1.2)[above right]{$\underline{I}_{C}$};
\draw (4,-.3)node [right] {$\uline{I}_0=\uline{I}_{RC}+\uline{I}_C$ und};
\draw (4,-.6)node [right] {$\uline{I}_C\, \bot \,\uline{I}_{RC}$ (Thaleskreis) und};
\draw (4,-.9)node [right] {$\uline{I}_C$ eilt $\uline{I}_{RC}$ vor.};
% \draw node at (1.5,-2.2)[below]{Bild 2};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\begin{tikzpicture}[very thick,scale=2.5]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [->,blue] (0,0)--(3,0)node at(2,0)[above left]{$\underline{U}_{0}$};
\draw [->,red] (0,0)--(-33:1.5)node at (-33:1)[above right]{$\underline{I}_{0}$};
\fill [black!50!](-33:.75)circle(.02);
\draw [very thin](-33:.0)arc(147:327:.75);
\draw [->,red] (0,0)--(-66:1.258)node at(-68:.8)[above left]{$\underline{I}_{RC}$};
\draw [->,red] (-66:1.258)--(-33:1.5)node at (-55:1.2)[above right]{$\underline{I}_{C}$};
% \draw node at (1.5,-2.2)[below]{Bild 2};
\draw [->,blue] (0,0)--(-66:2)node at(-68:1.5)[above left]{$\underline{U}_{C}$};
\draw [->,blue] (-66:2)--(3,0)node at (2.5,-.2)[below left]{$\underline{U}_{RL}$};
% \draw [red,very thin,dashed](-66:.5)arc(114:-66:1.5) (0:3)+(114:1.5)arc(114:294:1.5);%Seitenhalbierende
\draw [very thin,dashed](-66:2.0)+(70:1.5)arc(70:10:1.5) (0:3)+(190:1.5)arc(190:250:1.5);%Seitenhalbierende Teilkreise
\fill [black!60!](-25.6:2.115)circle(.02); %Mittelpunkt Thaleskreis 2
\draw [very thin](0:3)arc(40:-140:1.425); %Tahleskreis 2
\draw [very thin](-14.3:1.5)--(-31.9:2.8); %Seitenhalbierende
\draw (-66:1.258)+(24:.15)arc(24:114:.15) ; %Rechter Winkel
\fill (-66:1.258)+(69:.075)circle(.02) ; %Punkt Rechter Winkel
\draw [->,blue!50!red](-66:2)--+(-30:1)node at(1.2,-2.3)[below right]{$\underline{U}_{R}$};
\draw [->,blue!50!red](-66:2)--+(-30:1)--(0:3)node at(2.6,-.7)[below right]{$\underline{U}_{L}$};
\draw (-66:2)++(-30:1)++(60:0.15)arc(60:150:.15) ; %Rechter Winkel
\fill (-66:2)++(-30:1)++(106:.075)circle(.02) ; %Punkt Rechter Winkel
\draw (4,-.6)node [right] {$\uline{U}_C\, ||\, \uline{I}_{RC}$};
\draw (4,-.9)node [right] {$\uline{U}_{0}=\uline{U}_C+\uline{U}_{RL}$};
\draw (4,-1.5)node [right] {$\uline{U}_R\, ||\, \uline{I}_{0}$};
\draw (4,-1.8)node [right] {$\uline{U}_{RL}=\uline{U}_{R}+\uline{U}_{L}$};
\draw (4,-2.1)node [right] {$\uline{U}_L\, \bot \,\uline{U}_R$ (oder $\uline{U}_L\, \bot \,\uline{I}_0$)};
\draw (4,-2.4)node [right] {$\uline{U}_L\,$ eilt $\uline{U}_R\,$ vor};
\end{scope}
\begin{scope}[>=latex,xshift=.86cm,yshift=-.552cm]%Parallelen
\draw [double](0:-.1)--+(57:.1);
\end{scope}
\begin{scope}[>=latex,xshift=1.4cm,yshift=-2.167cm]%Parallelen
\draw [double](0:-.1)--+(57:.1);
\end{scope}
\end{tikzpicture}
\end{align*}
\clearpage
}{}%

+ 133
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ET2_L_B14_A3.tex View File

@@ -0,0 +1,133 @@
\section{Brückenschaltung}
%\enlargethispage{0cm}
Zeichnen Sie zu der abgebildeten Brückenschaltung ein \underline{maßstäbliches} Zeigerdiagramm aller Ströme und Spannungen.\\
Entnehmen Sie dem Zeigerdiagramm die Spannung $U_a$ und geben Sie von dieser Spannung Betrag und Phasenwinkel an.\\[\baselineskip]
\uline{Maßstäbe: }$ 10\,\volt\,\widehat{=}\,1\centi\metre\qquad 0,2\,\ampere\,\widehat{=}\,1\,\centi\metre $ (Platzbedarf in x und y $15\,\centi\metre$)\\
$R_1=100\,\ohm \qquad R_2=80\,\ohm \qquad X_L=200\,\ohm \qquad X_C=-120\,\ohm\qquad\uline{U}=150\,\volt\cdot e^{j0\degree}$
\begin{align*}
\begin{tikzpicture}[very thick,scale=2]
\begin{scope}[>=latex,very thick,xshift=0.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
\draw [<-,blue] (.3,.5)--(.7,.5)node at(.5,.5)[right]{\footnotesize$\uline{U}_{R_1}$};
\draw [<-,red] (.05,.0)--(.25,.0)node at(.15,.0)[right]{\footnotesize$\uline{I}_{1}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$};
\draw [<-,blue] (.3,.5)--(.7,.5)node at(.5,.5)[right]{\footnotesize$\uline{U}_{R_2}$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=.5cm,yshift=-1cm,rotate=90]%Spule |
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_L$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=-1cm,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] {$X_C$};
\draw [->,red] (.35,0)--(.15,0) node at (.25,0)[right]{\footnotesize$\uline{I}_2$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (1.5,.8)--(1.5,1)--(-.5,1) (1.5,-.8)--(1.5,-1)--(-.5,-1);
\fill (-.5,1)circle(.025) (-.5,-1)circle(.025);
\draw [->,blue] (.6,0)--(1.4,0)node at(1,0)[below]{$\underline{U}_{a}$};
% \draw node at(-1,-.5)[left]{$\underline{Z}\Rightarrow$};
\fill (.5,0)circle(.025) node at (.5,0)[left]{A};
\fill (1.5,0)circle(.025)node at (1.5,0)[right]{B};
\draw [->,blue] (-.5,.8)--(-.5,-.8) node at (-.5,0)[left]{$\underline{U}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
%%begin{align}
%\intertext{Formeln:}
%\end{align}
Berechnung:
\begin{align*}
Z_1&=\sqrt{R^2_1+X^2_L}=\sqrt{100^2+200^2}\,\ohm=223{,}6\,\ohm\\
Z_2&=\sqrt{R^2_2+X^2_C}=\sqrt{80^2+(-120)^2}\,\ohm=144{,}2\,\ohm\\
\varphi_1&=\arctan\frac{\Im}{\Re}=\arctan\frac{200\,\ohm}{100\,\ohm}=\arctan 2=63{,}4\,\degree\\
\varphi_2&=\arctan\frac{-120\,\ohm}{80\,\ohm}=\arctan (-1{,}5)=-56{,}3\,\degree\\
\uline{Z}_1&=Z_1\cdot e^{j\varphi_1}=223{,}6\,\ohm\cdot e^{j63{,}4\,\degree}\\
\uline{Z}_2&=Z_2\cdot e^{j\varphi_2}=144{,}2\,\ohm\cdot e^{-j56{,}3\,\degree}\\
\uline{I}_1&=\frac{U}{Z_1}=\frac{150\,\volt\cdot e^{j0\degree}}{223{,}6\,\ohm\cdot e^{j63{,}4\,\degree}}=0{,}67\,\ampere\cdot e^{-j63{,}4\,\degree} \approx 3{,}4\,\centi\metre\\
\uline{I}_2&=\frac{U}{Z_2}=\frac{150\,\volt\cdot e^{j0\degree}}{144{,}2\,\ohm\cdot e^{-j56{,}3\,\degree}}=1{,}04\,\ampere\cdot e^{j56{,}3\,\degree} \approx 5{,}2\,\centi\metre\\
\uline{I}&=\uline{I_1}+\uline{I_2}\\
\uline{U}_{R_1}&=I_1\cdot R_1= 0{,}671\,\ampere\cdot e^{j63{,}4\,\degree}\cdot 100\,\ohm=67{,}1\,\volt\cdot e^{-j63{,}4\,\degree}\\
\uline{U}_{R_2}&=I_2\cdot R_2=1{,}04\,\ampere\cdot 80\,\ohm=83{,}2\,\volt\cdot e^{j56{,}3\,\degree}\\
\uline{U}_a&=\uline{U}_{R_2}-\uline{U}_{R_1}\\
\uline{U}_L&=X_L\cdot \uline{I}_1=200\,\ohm\cdot e^{j90\degree}0{,}67\,\ampere\cdot e^{-j63{,}4\,\degree}=134\,\volt\cdot e^{j26{,}6\,\degree}\\
\uline{U}_C&=X_C\cdot \uline{I}_2=120\,\ohm\cdot e^{-j90\degree}\cdot 1{,}04\,\ampere\cdot e^{j56{,}3\,\degree}=124{,}8\,\volt\cdot e^{-j33{,}7\,\degree}
\end{align*}
\clearpage
Zeigerdiagramm Teil 1: (Maßstäblich)
\enlargethispage{2\baselineskip}
\begin{align*}
\begin{tikzpicture}[very thick,scale=.6]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [black!25!,very thin](0,-8)grid(15,8);
\draw [black!50!,thin,dashed](16,0)--(18,0);
\draw [black!50!,thin,dashed](3,-6)--(5,-6);
\draw [black!50!,thin](7.5,0)circle(7.5);
\draw [->,blue](0,0)--(15,0)node [right]{$\uline{U}$};
\draw [->,blue](0,0)--(-63.4:6.71)node [left]{$\uline{U}_{R_1}$};
\draw [->,blue](0,0)--(56.3:8.32)node [left]{$\uline{U}_{R_2}$};
\draw [->,blue!50!red](0,0)+(-63.4:6.71)--(15,0)node at(15,-1.5)[below]{$\uline{U}_L=134\,\volt\cdot e^{j26{,}63{,}4\,\degree}$};
\draw [->,blue!50!red](0,0)+(56.3:8.32)--(15,0)node at(15,2)[above]{$\uline{U}_C=124{,}8\,\volt\cdot e^{-j33{,}73\,\degree}$};
\draw [->,blue](0,0)+(-63.4:6.71)--(56.3:8.32)node at (4,2)[right]{$\uline{U}_{AB}=\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$};
\draw node at (3,-6)[below]{$A$};
\draw node at (4.8,7)[above]{$B$};
\draw node at (16,0)[above right]{$\Re (A)$};
\draw [black!50!](0,0)++(-63.4:6.71)+(26.6:1)arc(26.6:116.6:1);%Rechter Winkel bei UR1
\fill [black!50!](0,0)++(-63.4:6.71)+(71.6:.5)circle(0.1);
\draw [black!50!](0,0)++(56.3:8.32)+(-33.7:1)arc(-33.7:-123.7:1);%Rechter Winkel bei UR2
\fill [black!50!](0,0)++(56.3:8.32)+(-78.7:.5)circle(0.1);
\draw [<->,black!50!](0,0)+(-63.4:1)arc(-63.4:53.3:1);%Winkel UR1 UR2
\draw node at (0.8,0)[below right]{$\varphi_1=-63{,}4\,\degree$};
\draw node at (0.8,0)[above right]{$\varphi_2=53{,}3\,\degree$};
\draw [->,black!75!](3,-6)+(0:2)arc(0:83:2)node at (5,-5.5)[right]{$\varphi_{\uline{U}} =83\,\degree$};%U - UAB
\draw node at (19,5) [right]{$y=15\,\centi\metre$};
\draw node at (19,4) [right]{Kreis da};
\draw node at (19,3) [right]{$\uline{U}_{R_2}+\uline{U}_{C}=\uline{U}$};
\draw node at (19,2) [right]{$\uline{U}_{R_1}+\uline{U}_{L}=\uline{U}$};
\draw node at (19,1) [right]{$\uline{U}_{R_2}\bot \uline{U}_{C}$};
\draw node at (19,0) [right]{$\uline{U}_{R_1}\bot \uline{U}_{L}$};
\draw node at (19,-1) [right]{A und B einzeichnen};
\draw node at (19,-2) [right]{Messen $\uline{U}_{AB}=\uline{U}_a$};
\end{scope}
\end{tikzpicture}
\end{align*}
Zeigerdiagramm vollständig:
\begin{align*}
\begin{tikzpicture}[very thick,scale=.6]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [black!25!,very thin](0,-8)grid(15,8);
\draw [black!50!,thin,dashed](16,0)--(18,0);
\draw [black!50!,thin,dashed](3,-6)--(5,-6);
\draw [black!50!, thin](7.5,0)circle(7.5);
\draw [->,blue](0,0)--(15,0)node [right]{$\uline{U}$};
\draw [->,blue](0,0)--(-63.4:6.71)node [left]{$\uline{U}_{R_1}$};
\draw [->,blue](0,0)--(56.3:8.32)node [left]{$\uline{U}_{R_2}$};
\draw [->,blue!50!red](0,0)+(-63.4:6.71)--(15,0)node at(12,-1.5)[below]{$\uline{U}_L$};
\draw [->,blue!50!red](0,0)+(56.3:8.32)--(15,0)node at(12,2)[above]{$\uline{U}_C$};
\draw [->,blue](0,0)+(-63.4:6.71)--(56.3:8.32)node at (4,2)[right]{$\uline{U}_{AB}=\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$};
\draw [black!50!](0,0)++(-63.4:6.71)+(26.6:1)arc(26.6:116.6:1);%Rechter Winkel bei UR1
\fill [black!50!](0,0)++(-63.4:6.71)+(71.6:.5)circle(0.1);
\draw [black!50!](0,0)++(56.3:8.32)+(-33.7:1)arc(-33.7:-123.7:1);%Rechter Winkel bei UR2
\fill [black!50!](0,0)++(56.3:8.32)+(-78.7:.5)circle(0.1);
\draw [->,red](0,0)--(-63.4:3.34)node [left]{$\uline{I}_1$};
\draw [->,red](0,0)--(56.3:5.2)node [left]{$\uline{I}_2$};
\draw [<-,red,thin](0:0)++(56.3:5.2)++(-63.4:3.34)--(-63.4:3.34)node at(3,-.5) [left]{$\uline{I}_2$};
\draw [<-,red,ultra thick](0:0)++(56.3:5.2)++(-63.4:3.34)--(0:0)node at (6,.6)[above]{$\uline{I}=\uline{I}_1+\uline{I}_2$};
\draw [<-,black](0,0)+(17.0:3)arc(17.0:0:3)node at (1.0,1.5)[right]{$\varphi_{\uline{I}}=17\,\degree$};%Winkel I
\draw node at (3,-6)[below]{$A$};
\draw node at (4.8,7)[above]{$B$};
\draw node at (16,0)[above right]{$\Re (A)$};
\draw [<->,black!50!](0,0)+(-63.4:1)arc(-63.4:53.3:1);%Winkel UR1 UR2
\draw node at (.7,-.35)[below right]{$\varphi_1$};
\draw node at (.7,.35)[above right]{$\varphi_2$};
\draw [->,black!75!](3,-6)+(0:2)arc(0:83:2)node at (5,-5.5)[right]{$\varphi_{\uline{U}} =83\,\degree$};%U - UAB
\draw node at (20.8,5) [right]{$\uline{I}_1 \approx 3{,}4\,\centi\metre$};
\draw node at (20.8,4) [right]{$\uline{I}_2 \approx 5{,}2\,\centi\metre$};
\draw node at (20.8,3) [right]{$\uline{I}=\uline{I_1}+\uline{I_2}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\clearpage
}{}%

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ET2_L_B14_A4.tex View File

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\section{Zeigerdiagramm Netzwerk}
Zeichnen Sie für das abgebildete Netzwerk ein maßstäbliches Zeigerdiagramm aller Spannungen und Ströme.\\
Wie groß muß der Widerstand $R_1$ sein damit der Strom $I$ der Spannung $U$ um $30\,\degree$ nacheilt?\\[\baselineskip]
$\varphi_u - \varphi_i = 30\,\degree$\\
$R_2 = |X_L| = |X_C| = 10\,\kilo\ohm$\\
$\uline{I}_C = 1\,\milli\ampere\cdot e^{j90\,\degree}$\\[\baselineskip]
\uline{Maßstäbe:}\\
$1\,\volt \,\widehat{=}\,0{,}8\centi\metre$\\
$1\,\milli\ampere \,\widehat{=}\,5\centi\metre$
\begin{align*}
\begin{tikzpicture}[very thick,scale=3]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule -
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$C$};
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(2,0)--(2,.2) (.8,1)--(2,1)--(2,.8);
\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$};
\fill (-.5,0)circle(.025) (-.5,1)circle(.025);
\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$};
\draw [->,red] (1.1,.9)--(1.1,.6) node at (1.1,.75)[right]{$\underline{I}_C$};
\draw [->,black!50!] (1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$\underline{U}_{R_2}$};
\draw [->,black!50!] (1.4,1)--(1.6,1) node at (1.5,1)[above]{$\underline{I}_2$};
\draw [->,black!50!] (.3,.9)--(.7,.9) node at (.5,.9)[below]{$\underline{U}_L$};
\draw [->,black!50!] (.75,1)--(.95,1) node at (.85,1)[above]{$\underline{I}_L$};
\draw [->,black!50!] (0,.95)--(0,.75) node at (0,.85)[left]{$\underline{I}_{R_1}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align*}
\intertext{Berechnung: (Platz in $x=\pm10\,\centi\metre$ und $x=10\,\centi\metre$}
R_2=|X_C|&=10\,\kilo\ohm\text{ (Stomteiler, mit gleichem Betrag des Stroms)}\\
\text{mit }|\uline{I}_{R_2}| &=|\uline{I}_C|=1\,\milli\ampere\\
\uline{I}_C &= 1\,\milli\ampere\cdot e^{j90\,\degree}\Rightarrow\uline{I}_{R_2}=1\,\milli\ampere\cdot e^{j0\,\degree}\text{ $(\uline{I}_C$ eilt vor)}\\
\uline{U}_{R_2}&=R_2\cdot \uline{I}_{R_2}=10\,\cancel{\kilo}\ohm\cdot 1\,\cancel{\milli}\ampere\cdot e^{j0\,\degree}=10\,\volt\cdot e^{j0\,\degree}\,\widehat{=}\,8\centi\metre\\
\uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C =1\,\milli\ampere\cdot e^{j0\,\degree}+1\,\milli\ampere\cdot e^{j90\,\degree}=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\\
\uline{U}_L&=\uline{I}_L\cdot j\cdot X_L=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\cdot j10\,\kilo\ohm =14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\,11{,}3\centi\metre\text{ $(\uline{U}_L$ voreilend)}\\
\uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}=14{,}1\,\volt\cdot e^{j135\,\degree}+10\,\volt\cdot e^{j0\,\degree}
=(-10+j10+10)\,\volt=10\,\volt\cdot e^{j90\,\degree}\\
\text{Zeichnen: }\varphi_u-\varphi_i&=30\,\degree\text{ deshalb $30\,\degree$, Linie zeichnen, Schnittpunkt mit } \uline{I}_L+\uline{I}_{R_1}\Rightarrow\uline{I}=2\,\milli\ampere\cdot e^{j30\,\degree}\\
\text{Ablesen: }\qquad\,\,\uline{I}_{R_1}&=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\
R_1&=\frac{\uline{U}}{\uline{I}_{R_1}}=\frac{10\,\volt\cancel{\cdot e^{j90\,\degree}}}{0{,}72\,\milli\ampere\cancel{\cdot e^{j90\,\degree}}}=\uuline{13{,}89\,\kilo\ohm}\\
\end{align*}
\begin{align*}
\begin{tikzpicture}[very thick,scale=.7]
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
\draw [black!15!,very thin](-10,0)grid(10,10);
\draw [->,black!25!](-10,0)--(10.5,0);
\draw [->,black!25!](0,0)--(0,10.5);
\draw node at (10.5,0)[right]{$\Re$};
\draw node at (0,10.5)[above]{$\Im$};
\draw [black!75!,thick](0:0)--(60:11)node [right]{$30\,\degree$ Linie};
\draw [->,blue] (0,0)--(90:8)node [below right]{$\underline{U}$};
\draw [->,red] (0,0)--(90:5)node [below right]{$\underline{I}_C$};
\draw [->,red] (0,0)--(0:5)node [above left]{$\underline{I}_R$};
\draw [->,red] (0,0)--(45:7.07)node [below right]{$\underline{I}_L$};
\draw [->,red!50!blue] (5,5)--(5,8.65)node [below right]{$\underline{I}_{R_1}=0{,}72\,\milli\ampere\,\widehat{=}\,3{,}6\centi\metre$};
\draw [->,red!50!blue](0:0)--(60:10)node at (3,7.8) [above]{$\uline{I}=2\,\milli\ampere$};
\draw [->,blue] (0,0)--(135:11.28)node [below left]{$\underline{U}_L$};
\draw [->,blue] (-8,8)--(0,8)node [above left]{$\underline{U}_{R_2}$};
\draw [->,black!75!](0:0)+(90:2)arc(90:60:2)node at(.6,2)[above]{$30\,\degree$};
\end{scope}
\end{tikzpicture}
\end{align*}
\begin{align*}
\uline{I}_{R_2}&=1\,\milli\ampere\cdot e^{j0\,\degree} \,\widehat{=}\ 5\,\centi\metre\\
\uline{I}_{C}&=1\,\milli\ampere\cdot e^{j90\,\degree} \,\widehat{=}\ 5\,\centi\metre\\
\uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C \,\widehat{=}\ 7{,}05\,\centi\metre\\
\uline{U}_{L}&=14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\ 11{,}3\centi\metre \quad \uline{U}_{L}\bot \uline{I}_{L}\\
\uline{R}_{R_2}&=10\,\volt\cdot e^{j0\,\degree} \,\widehat{=}\ 8\centi\metre\\
\uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}\\
&\text{Gerade für I, $30\degree $ nacheilend}\\
\uline{I}_{L}&+\uline{I}_{R_1}=\uline{I};\qquad \uline{I}_{R_1}||\uline{U}\\
\rightarrow\ &\text{ ablesen } 3{,}6\,\centi\metre\ \widehat{=}\ \uline{I}_{R_1}=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\
\end{align*}
\clearpage
}{}%

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ET2_L_B14_A5.tex View File

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\section{Blind- Wirk- und Scheinleistung}
Von dem untenstehenden Schaltbild ist gegeben:\\
$R_L = X_L = 20\,\ohm \quad R_C=200\,\ohm \quad X_C=-100\,\ohm\quad\uline{U}=230\,\volt\cdot e^{j0\,\degree}$\\
\renewcommand{\labelenumi}{\alph{enumi})}
\begin{enumerate}
\item Der Eingangswiderstand Z der Schaltung nach Betrag und Phasenwinkel
\item Sämtliche Ströme und Spannungen nach Betrag und Phasenwinkel
\item Wirk- Blind- und Scheinleistungsaufnahme der Schaltung
\item Qualitatives Zeigerdiagramm aller Ströme und Spannungen unter der Annahme, daß sich
die Gesamtschaltung induktiv verhält.
\end{enumerate}
\begin{align*}
\begin{tikzpicture}[very thick,scale=2]
\begin{scope}[>=latex,very thick,xshift=0cm,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] {$R_L$};
% \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$};
\end{scope}
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
% \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
% \end{scope}
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Spule -
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_L$};
\fill (.3,-0.0667)rectangle(.7,0.0667);
\end{scope}
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_C$};
\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] {$X_C$};
\end{scope}
\begin{scope}[>=latex,very thick]%Knotenpunkte
\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(3,0)--(3,.2) (.8,1)--(3,1)--(3,.8);
\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$};
\fill (-.5,0)circle(.025) (-.5,1)circle(.025);
\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$};
\draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{R_C}$};
\draw [->,red] (3.1,.9)--(3.1,.7) node at (3.1,.8)[right]{$\underline{I}_C$};
\draw [->,black!50!] (.3,.8)--(.7,.8) node at (.5,.8)[below]{$\underline{U}_{RL}$};
\draw [->,black!50!] (1.3,.8)--(1.7,.8) node at (1.5,.8)[below]{$\underline{U}_L$};
\draw [->,black!50!] node at (-.5,.5)[left]{$\uline{Z} =>$};
\draw [|-|,black!50!] (2.,-.2)--(3,-.2) node at (2.5,-.2)[below]{$\uline{Z}_{||}$};
\end{scope}
\end{tikzpicture}
\end{align*}
\ifthenelse{\equal{\toPrint}{Lösung}}{%
\begin{align}
\intertext{Formeln:}
e^{j\varphi}&=\cos\varphi+j\sin\varphi \quad\text{Eulersche Formel}\\
\uline{Z}&=Z\cdot e^{\pm j\varphi}=R\pm jX\\
\cos\varphi&=\frac{R}{Z}\\
\sin\varphi&=\frac{X}{Z}
\end{align}