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ET2_Deckblatt.tex
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ET2_Deckblatt.tex
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\pagenumbering{roman}
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%\ifthenelse{\equal{\toPrint}{Lösung}}{\input{ET2_Deckblatt}}{\input{ET2_Deckblatt_A}}
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\begin{titlepage}
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\vspace*{3cm}
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\begin{center}
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{\bfseries \large {Technische Hochschule Nürnberg \\ Georg Simon Ohm}\\[\baselineskip]
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Übung Grundlagen der Elektrotechnik 2}\\[\baselineskip]
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% Prüfen ob Musterlösung oder nicht, wenn ja, dann mit Dozentenkommentar, ansonsten nur Aufgabenstellung für Studierende
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\ifthenelse{\equal{\toPrint}{Lösung}}%
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{%
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{\normalsize Prof.\,Dr.\,C.\,Niebler}\\[2\baselineskip]
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\textcolor{red}{Nur für Dozentengebrauch, nicht zur Weitergabe an Studenten.}\\[\baselineskip]
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{\large \textbf{Dozenten Exemplar mit Lösungen für}}\\[2\baselineskip]
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\textbf{\ProfName}\\[2\baselineskip]
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%\copyright\ Dr. Christine Niebler\\
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}% Dozentenausgabe
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{}% Studentenausgabbe
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\end{center}
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\end{titlepage}
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\thispagestyle{empty}
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\cleardoublepage
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\clearpage
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\pagenumbering{arabic}
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\enlargethispage{1cm}
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%\chead{Inhaltsverzeichnis}
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\footnotesize{\tableofcontents}
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\clearpage
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ET2_Deckblatt_A.log
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ET2_Deckblatt_A.log
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This is pdfTeX, Version 3.1415926-2.5-1.40.14 (MiKTeX 2.9 64-bit) (preloaded format=pdflatex 2014.2.26) 17 MAR 2014 17:43
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entering extended mode
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**C:/Users/Niebler/Documents/Vorlesungen/ETechnik/TeX/UeET2_20130923/ET2_Deckbl
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att_A.tex
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(C:/Users/Niebler/Documents/Vorlesungen/ETechnik/TeX/UeET2_20130923/ET2_Deckbla
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tt_A.tex
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LaTeX2e <2011/06/27>
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Babel <v3.8m> and hyphenation patterns for english, afrikaans, ancientgreek, ar
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lician, german, german-x-2013-05-26, greek, gujarati, hindi, hungarian, iceland
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ic, indonesian, interlingua, irish, italian, kannada, kurmanji, latin, latvian,
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german-x-2013-05-26, nynorsk, oriya, panjabi, pinyin, polish, portuguese, roman
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ian, russian, sanskrit, serbian, slovak, slovenian, spanish, swedish, swissgerm
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an, tamil, telugu, turkish, turkmen, ukenglish, ukrainian, uppersorbian, usengl
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ishmax, welsh, loaded.
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! LaTeX Error: Environment titlepage undefined.
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See the LaTeX manual or LaTeX Companion for explanation.
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Type H <return> for immediate help.
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...
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l.2 \begin{titlepage}
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! Emergency stop.
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l.2 \begin{titlepage}
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Your command was ignored.
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Type I <command> <return> to replace it with another command,
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or <return> to continue without it.
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Here is how much of TeX's memory you used:
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9 strings out of 493921
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477 string characters out of 3147276
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49031 words of memory out of 3000000
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3401 multiletter control sequences out of 15000+200000
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3640 words of font info for 14 fonts, out of 3000000 for 9000
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841 hyphenation exceptions out of 8191
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5i,0n,4p,116b,16s stack positions out of 5000i,500n,10000p,200000b,50000s
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! ==> Fatal error occurred, no output PDF file produced!
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ET2_Deckblatt_A.tex
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ET2_Deckblatt_A.tex
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\pagenumbering{roman}
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\begin{titlepage}
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\vspace*{3cm}
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\begin{center}
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{\bfseries \Large {Technische Hochschule Nürnberg\\
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Georg Simon Ohm\\[\baselineskip]
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{Übung Grundlagen der Elektrotechnik 2}}\\[\baselineskip]
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\normalsize{Prof.\,Dr.\,C.\,Niebler}\\[\baselineskip]
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\Large\textbf{Aufgabenstellung}}\\[\baselineskip]
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% SS 2013}\\[\baselineskip]
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\end{center}
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\end{titlepage}
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\thispagestyle{empty}
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\cleardoublepage
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\clearpage
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\pagenumbering{arabic}
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\enlargethispage{1cm}
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%\chead{Inhaltsverzeichnis}
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%\footnotesize{
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\tableofcontents
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%}
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\clearpage
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116
ET2_LS_B12_A1.tex
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ET2_LS_B12_A1.tex
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\section{Blitzableiter}
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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).\\
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$\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
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\renewcommand{\labelenumi}{\alph{enumi})}
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\begin{enumerate}
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\item Welche Spannung $u$ (Betrag !) wird induziert, wenn der
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Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt?
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\item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !)
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\end{enumerate}
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\begin{align*}
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\begin{tikzpicture}[scale=2]
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\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom
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\draw [->] (0,4) -- (0,0) node [below] {$i$};
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\end{scope}
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\begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung
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\draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$};
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\draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$};
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\end{scope}
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\begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D
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\draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$};
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\draw [dashed] (3.1,.5)--(3.1,-.5);
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\end{scope}
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\begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d
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\draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$};
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\draw [<-] (0.2,0) -- (0.7,0);
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\end{scope}
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\begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l
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\draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$};
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\draw (0,0) -- (-0.5,0);
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\draw (0,3) -- (-0.5,3);
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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\ifthenelse{\equal{\toPrint}{Lösung}}{%
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Formeln:
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\begin{align}
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% \intertext{Formeln:}
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u&=-N\cdot \frac{d\Phi}{dt}\\
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\Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\
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H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r}
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\end{align}
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Magnetischer Fluss $\Phi$; Flussdichte $B$; Feldstärke $H$;\\
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Magnetische Permeabilität $\mu=\mu_0\cdot \mu_r$; In Luft $\mu_r=1$\\
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\clearpage
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Berechnung:
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\begin{align*}
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u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\qquad u=-\frac{d\Phi}{dt}\\
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\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}\\
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H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\
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\phantom{blablabla}bla\\
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\vphantom{u=-\frac{y}{x}}..\\
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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}\\
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% } %end pahntom
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&=-\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]
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&\text{Leitung symmetrisch zur Mittellinie bei } 2{,}99\,\metre \text{ und } 3{,}01\,\metre\\
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\text{mit }&\Phi=\frac{\mu\cdot i\cdot l}{2\cdot \pi}\cdot\ln\frac{D+d/2}{D-d/2}\\
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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\\
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\text{\uline{Lentzsche Regel:}}
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\end{align*}
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\begin{align*}
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\begin{tikzpicture}[scale=2.5]
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\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
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\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
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\draw [<-,blue] (.2,.35)--(.8,.35)node at(.5,.35)[above]{\footnotesize$u$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=.25cm,yshift=-.25cm]% Leiterschleife
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\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
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\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
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\end{scope}
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\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=0cm]%Knotenpunkte
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\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
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\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
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\end{scope}
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\begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=-1cm]% O.
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\draw (.5,0)circle(.133)node at (1,0)[right]{$\frac{d\Phi}{dt}$};
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\fill (.5,0)circle(.05);
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\end{scope}
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\begin{scope}[>=latex,very thick,green!50!black,xshift=0cm,yshift=-1.33cm]% Ox
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\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
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\draw [black]node at (1.5,-1)[below]{Wirkung $\frac{d\Phi_{ind}}{dt}$ entgegen Ursache $\frac{d\Phi}{dt}$};
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\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%Widerstand - nach EN 60617
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\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
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\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$};
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\end{scope}
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\begin{scope}[>=latex,very thick,red,xshift=3cm,yshift=0cm]%Knotenpunkte
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\fill (0,0)circle(.05) node [above right] {\footnotesize$+$};
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\fill (1,0)circle(.05) node [above left] {\footnotesize$-$};
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=-0.25cm]% Leiterschleife
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\draw (0,0)--(0,-2)--(.5,-2)--(.5,0);
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\draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25);
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=3cm,yshift=-2.25cm]%Stromquelle
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\draw [green!50!black](.25,0)--(.75,0);
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\draw node at (.5,.133) [above] {$i_{}$};
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\draw [green!50!black](.5,0)circle(.133);
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\draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$i_{ind}$};
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\end{scope}
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\begin{scope}[>=latex,very thick,green!50!black,xshift=3cm,yshift=-1.33cm]% Ox
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\draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$};
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\draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133);
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\end{scope}
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\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1.25cm]%
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\draw [->,red] (-2.5,.5)--(-2.5,-.5) node at(-2.5,0)[right]{$\frac{di}{dt}$};
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\draw [<-,green!50!black] (.5,.5)--(.5,-.5) node at(.5,0)[left]{$i_{ind}$};
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\draw [<-,red!75!black] (-1.75,.5)--(-1.75,-.5) node at(-1.75,0)[left]{$i_{ind}$};
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\end{scope}
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\end{tikzpicture}
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\end{align*}
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$u$ hat negatives Vorzeichen, da Polung von $i_{ind}$ entgegen Bezugspfeilen aus Skizze\\
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($u_R = -u$)
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\clearpage
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}{}%
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113
ET2_L_B12_A1.tex
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113
ET2_L_B12_A1.tex
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@ -0,0 +1,113 @@
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\section{Blitzableiter}
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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).\\
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$\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
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\renewcommand{\labelenumi}{\alph{enumi})}
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\begin{enumerate}
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\item Welche Spannung $u$ (Betrag !) wird induziert, wenn der
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Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt?
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\item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !)
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\end{enumerate}
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\begin{align*}
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\begin{tikzpicture}[scale=2]
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\begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom
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\draw [->] (0,4) -- (0,0) node [below] {$i$};
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\end{scope}
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\begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung
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||||
\draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$};
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\draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$};
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\end{scope}
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\begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D
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\draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$};
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\draw [dashed] (3.1,.5)--(3.1,-.5);
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\end{scope}
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||||
\begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d
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\draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$};
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\draw [<-] (0.2,0) -- (0.7,0);
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\end{scope}
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||||
\begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l
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\draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$};
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\draw (0,0) -- (-0.5,0);
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\draw (0,3) -- (-0.5,3);
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||||
\end{scope}
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||||
\end{tikzpicture}
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||||
\end{align*}
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\ifthenelse{\equal{\toPrint}{Lösung}}{%
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Formeln:
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||||
\begin{align}
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||||
% \intertext{Formeln:}
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||||
u&=-N\cdot \frac{d\Phi}{dt}\\
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||||
\Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\
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H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r}
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||||
\end{align}
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||||
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}\\
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||||
H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\
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||||
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}\\
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||||
&=-\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\\
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||||
\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
ET2_L_B12_A2.tex
Normal file
123
ET2_L_B12_A2.tex
Normal file
@ -0,0 +1,123 @@
|
||||
\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
ET2_L_B12_A3.tex
Normal file
125
ET2_L_B12_A3.tex
Normal 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
ET2_L_B12_A4.tex
Normal file
50
ET2_L_B12_A4.tex
Normal file
@ -0,0 +1,50 @@
|
||||
\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
|
||||
}{}%
|
||||
70
ET2_L_B12_A5.tex
Normal file
70
ET2_L_B12_A5.tex
Normal file
@ -0,0 +1,70 @@
|
||||
\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
|
||||
}{}%
|
||||
80
ET2_L_B12_A6.tex
Normal file
80
ET2_L_B12_A6.tex
Normal file
@ -0,0 +1,80 @@
|
||||
\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
|
||||
}{}%
|
||||
106
ET2_L_B13_A1.tex
Normal file
106
ET2_L_B13_A1.tex
Normal file
@ -0,0 +1,106 @@
|
||||
\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
|
||||
}{}%
|
||||
68
ET2_L_B13_A2.tex
Normal file
68
ET2_L_B13_A2.tex
Normal file
@ -0,0 +1,68 @@
|
||||
\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
ET2_L_B13_A3.tex
Normal file
74
ET2_L_B13_A3.tex
Normal file
@ -0,0 +1,74 @@
|
||||
\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
ET2_L_B13_A4.tex
Normal file
133
ET2_L_B13_A4.tex
Normal file
@ -0,0 +1,133 @@
|
||||
\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
ET2_L_B13_A5.tex
Normal file
59
ET2_L_B13_A5.tex
Normal file
@ -0,0 +1,59 @@
|
||||
\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
ET2_L_B13_A6.tex
Normal file
88
ET2_L_B13_A6.tex
Normal file
@ -0,0 +1,88 @@
|
||||
\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
ET2_L_B13_A7.tex
Normal file
101
ET2_L_B13_A7.tex
Normal file
@ -0,0 +1,101 @@
|
||||
\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
ET2_L_B14_A1.tex
Normal file
130
ET2_L_B14_A1.tex
Normal file
@ -0,0 +1,130 @@
|
||||
\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
ET2_L_B14_A2.tex
Normal file
143
ET2_L_B14_A2.tex
Normal 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
ET2_L_B14_A3.tex
Normal file
133
ET2_L_B14_A3.tex
Normal 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
|
||||
}{}%
|
||||
87
ET2_L_B14_A4.tex
Normal file
87
ET2_L_B14_A4.tex
Normal file
@ -0,0 +1,87 @@
|
||||
\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
|
||||
}{}%
|
||||
180
ET2_L_B14_A5.tex
Normal file
180
ET2_L_B14_A5.tex
Normal file
@ -0,0 +1,180 @@
|
||||
\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}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\intertext{a) Eingangswiderstand (ist ohmisch-kapazitiv)}
|
||||
\uline{Z}_{||}&=\frac{R_C\cdot jX_C}{R_C+jX_C}=\frac{200\cdot (-j100)}{200-j100}\,\ohm=(40-j80)\,\ohm\\
|
||||
\uline{Z}&=\uline{Z}_{||}+R_L+X_L=[20+40+j(20-80)]\,\ohm=\uuline{(60-j60)\,\ohm}=\uuline{84{,}85\,\ohm\cdot e^{-j45\,\degree}}\\
|
||||
&(\Rightarrow \varphi=-45\,\degree)
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
\intertext{b) Ströme}
|
||||
\uline{I}&=\frac{\uline{U}}{\uline{Z}}=\frac{230\,\volt\cdot e^{j0\,\degree}}{84{,}5\,\ohm\cdot e^{-j45\,\degree}}=\uuline{2{,}71\,\ampere\cdot e^{j45\,\degree}}=\uuline{(1{,}916+j1{,}916)\,\ampere}
|
||||
% \end{align*}
|
||||
% \clearpage
|
||||
% \begin{align*}
|
||||
\intertext{Stromteiler}
|
||||
\uline{I}_{R_C}&=\uline{I}\cdot \frac{jX_C}{R_C+jX_C}=\uline{I}\cdot \frac{-j100}{200-j100}
|
||||
=\uline{I}\cdot \frac{-j}{2-j}\cdot \frac{2+j}{2+j}=\uline{I}\cdot \frac{1-j2}{4+1}=\uline{I}\cdot (0{,}2-j0{,}4)\\
|
||||
&=2{,}71\,\ampere\cdot e^{j45\,\degree}\cdot 0{,}447\cdot e^{-j63{,}4\,\degree}=\uuline{1{,}21\,\ampere\cdot e^{-j18{,}4\,\degree}}=\uuline{(1{,}150-j0{,}383)\,\ampere}\\%Ende IRC
|
||||
\uline{I}_{C}&=\uline{I}-\uline{I}_{R_C}=(1{,}916+j1{,}916)\,\ampere -(1{,}150-j0{,}383)\,\ampere=\uuline{(0{,}766+j2{,}30)\,\ampere}\\
|
||||
&=\uuline{2{,}42\,\ampere\cdot e^{+j71{,}6\,\degree}} \\
|
||||
\text{ alternativ}&\\
|
||||
\uline{I}_{C}&=\uline{I}\cdot \frac{R_C}{R_C+jX_C}=\uline{I}\cdot \frac{200}{200-j100}=\uline{I}\cdot \frac{1}{1-j0{,}5}\quad\text{ konjugiert komplex erweitern}\\
|
||||
&=\uline{I}\cdot \frac{1}{1-j0{,}5}\cdot \frac{1+j0{,}5}{1+j0{,}5}
|
||||
=\uline{I}\cdot \frac{1+j0{,}5}{1+0{,}5^2}
|
||||
=\uline{I}\cdot (0{,}8+j0{,}4)=\uline{I}\cdot 0{,}894\cdot e^{0{,}5j}\\
|
||||
&=2{,}71\,\ampere\cdot e^{j45\,\degree}\cdot 0{,}894\cdot e^{j26{,}6\,\degree}=\uuline{2{,}42\,\ampere\cdot e^{+j71{,}6\,\degree}}=\uuline{(0{,}766+j2{,}30)\,\ampere}
|
||||
\intertext{Spannungen}
|
||||
\uline{U}_{R_C}&=R_C\cdot \uline{I}_{R_C}=200\,\ohm\cdot 1{,}21\,\ampere\cdot e^{-j18{,}4\,\degree}=\uuline{242\,\volt\cdot e^{-j18{,}4\,\degree}}\\
|
||||
\uline{U}_{R_L}&=R_L\cdot \uline{I}=20\,\ohm\cdot 2{,}71\,\ampere \cdot e^{j45\,\degree}=\uuline{54{,}2\,\volt\cdot e^{j45\,\degree}}\\
|
||||
\uline{U}_L&=X_L\cdot \uline{I}=20\,\ohm\cdot e^{j90\,\degree} \cdot 2{,}71\,\ampere \cdot e^{j45\,\degree}=\uuline{54.2\,\volt\cdot e^{j135\,\degree}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
\intertext{Zeigerdiagramm: Beginne mit $\uline{U}$ und $\uline{I}\qquad (50\,\volt\per\centi\metre$; $1\,\ampere\per\centi\metre$)}
|
||||
&\uline{I} \text{ um Winkel $\varphi=45\,\degree$ voreilend, kapazitives Gesamtverhalten.}\\
|
||||
&\uline{I}=\uline{I}_{R_C}+\uline{I}_C \qquad(\uline{I}_C \,\bot\, \uline{I}_{R_C}) \qquad (\uline{I}_C \text{ voreilend})\\
|
||||
&\uline{U}_{R_L}\,||\,\uline{I}\\
|
||||
&\uline{U}_L\,\bot\, \uline{I} \qquad (\uline{U}_L\text{ voreilend})\\
|
||||
&\uline{U}_{R_C}\,||\,\uline{I}_{R_C}\\
|
||||
&\uline{U}_{R_L}=\uline{U}_L+\uline{U}_{R_C}=\uline{U}\ \widehat{=}\ 4{,}6\,\centi\metre
|
||||
\end{align*}
|
||||
%Trennzeile
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2.]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw [black!25!,very thin](0,-1)grid(5,2);
|
||||
\draw [->](0,0)--(5.2,0)node[right]{$\Re$};
|
||||
\draw [->](0,-1)--(0,2.2)node[above]{$\Im$};
|
||||
\draw [->,blue] (0,0)--(0:4.6)node [below left]{$\underline{U}$};
|
||||
\draw [->,red] (0,0)--(45:2.71)node at(45:2.3) [above left]{$\underline{I}$};
|
||||
\draw [->,red] (0,0)--(-18.4:1.21)node at(1.5,.5)[right]{$\underline{I}_C$};
|
||||
\draw [->,red] (0,0)++(-18.4:1.21)--+(71.6:2.42)node at(.6,-.3)[below]{$\underline{I}_{R_C}$};
|
||||
\draw [black!50!] (0,0)++(-18.4:1.21)+(71.6:.25)arc(71.6:161.6:.25);%Rechter Winkel
|
||||
\fill [black!50!] (0,0)++(-18.4:1.21)+(116.6:.125)circle(.025);%Rechter Winkel
|
||||
\draw [->,blue] (0,0)--(45:1.084)node at(0,.6)[right]{$\underline{U}_{R_L}$};
|
||||
\draw [black!50!] (0,0)++(45:1.084)+(225:.25)arc(225:135:.25);%Rechter Winkel
|
||||
\fill [black!50!] (0,0)++(45:1.084)+(180:.125)circle(.025);%Rechter Winkel
|
||||
\draw [->,blue] (0,0)++(45:1.084)--+(135:1.084)node at(3,.5)[above right]{$\underline{U}_{R_C}$};
|
||||
\draw [<-,blue] (0,0)++(0:4.6)--+(161.6:4.84)node at(0,1)[right]{$\underline{U}_L$};
|
||||
\draw [->,black!75!] (0,0)+(0:.5)arc(0:45:.5)node at(22.5:.5) [right]{$\varphi =45\,\degree$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=0.55cm,yshift=-.1cm]%Parallelen
|
||||
\draw [double](0:-.1)--+(-108.4:.1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=2.45cm,yshift=.8cm]%Parallelen
|
||||
\draw [double](0:-.1)--+(-108.4:.1);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
%Trennzeile
|
||||
\begin{align*}
|
||||
\intertext{c) Scheinleistung}
|
||||
S&=U\cdot I=230\,\volt\cdot 2{,}71\,\ampere=\uuline{623\,\volt\ampere} &\text{Scheinleistung}\\
|
||||
P&=S\cdot \cos \varphi=623\,\volt\ampere\cdot \cos (-45\,\degree)=\uuline{447\,\watt} &\text{Wirkleistung}\\
|
||||
Q&=S\cdot \sin \varphi=623\,\volt\ampere\cdot \sin (-45\,\degree)=\uuline{-447\,\volt\ampere r} &\text{Blindleistung}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
d) Annahme, daß sich die Gesamtschaltung induktiv verhält.
|
||||
\begin{align*}
|
||||
% \intertext{d) Annahme, daß sich die Gesamtschaltung induktiv verhält.}
|
||||
&\text{Reihenfolge: $\qquad (25\,\volt\per\centi\metre$; $1\,\ampere\per\centi\metre$)}\\
|
||||
&\text{Beginne mit $\uline{U}=230\,\volt\cdot e^{j0\,\degree}$ und $\uline{I}=2{,}71\,\ampere\cdot e^{-j45\,\degree}$}\\
|
||||
&\uline{I} \text{ um Winkel $\varphi=-45\,\degree$ nacheilend, da induktives Gesamtverhalten.}\\
|
||||
&\uline{I}=\uline{I}_{R_C}+\uline{I}_C \qquad(\uline{I}_C \,\bot\, \uline{I}_{R_C}) \qquad (\uline{I}_C \text{ voreilend}) \qquad [\text{Thaleskreis}]\\
|
||||
&\uline{U}_{R_C}\,||\,\uline{I}_{R_C} \hspace{14em} [\text{Gerade von der Spitze }\underline{U}]\\
|
||||
&\uline{U}_{R_L}\,||\,\uline{I} \hspace{15em}\, [\text{ Gerade } \bot\ \underline{U}_{RL}]\\
|
||||
&\uline{U}_L\,\bot\, \uline{I} \qquad (\uline{U}_L\text{ voreilend})\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1.5]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw [black!25!,very thin,scale=.5](0,-6)grid(10,4);
|
||||
\draw [->](0,0)--(5.2,0)node[right]{$\Re$};
|
||||
\draw [->](0,-3)--(0,2.2)node[above]{$\Im$};
|
||||
\draw [->,blue] (0,0)--(0:4.6)node [below right]{$\underline{U}$};
|
||||
\draw [->,red] (0,0)--(-45:2.71)node [above right]{$\underline{I}$};
|
||||
\draw [->,red] (0,0)++(-71.6:2.42)--+(18.4:1.21)node at(1.5,-2.2)[below]{$\underline{I}_{C}$};
|
||||
\draw [black!50!] (0,0)++(-71.6:2.42)+(18.4:.25)arc(18.4:108.4:.25);%Rechter Winkel
|
||||
\fill [black!50!] (0,0)++(-71.6:2.42)+(63.4:.125)circle(.02);%Rechter Winkel
|
||||
\draw [->,red] (0,0)--(-71.6:2.42)node at(.5,-1.5)[left]{$\underline{I}_{R_C}$};
|
||||
\draw [->,blue] (0,0)--(-45:2.168)node at(1,-1)[above right]{$\underline{U}_{R_L}$};
|
||||
\draw [black!50!] (0,0)++(-45:2.418)arc(-45:45:.25);%Rechter Winkel
|
||||
\fill [black!50!] (0,0)++(-45:2.168)+(0:.125)circle(.02);%Rechter Winkel
|
||||
\draw [->,blue] (0,0)++(-45:2.168)--+(45:3.8)node at(3.5,.5)[left]{$\underline{U}_L$};
|
||||
\draw [<-,blue] (0,0)++(0:4.6)--+(-251.6:1.21)node at(4.5,.5)[right]{$\underline{U}_{R_C}$};
|
||||
\draw [->,black!75!] (0,0)+(0:.5)arc(0:-45:.5)node at(-22.5:.5) [right]{$\varphi =-45\,\degree$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=.38cm,yshift=-1cm]%Parallelen
|
||||
\draw [double](0:-.1)--+(18.4:.1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=4.38cm,yshift=.8cm]%Parallelen
|
||||
\draw [double](0:-.1)--+(18.4:.1);
|
||||
\end{scope}
|
||||
\fill [black!60!](-45:1.355)circle(.05); %Mittelpunkt Thaleskreis 2
|
||||
\draw [very thin](-45:2.71)arc(-45:-225:1.355); %Tahleskreis 2
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{U}&=\uuline{230\,\volt \cdot e^{j0\,\degree}} \ \widehat{=}\ 9{,}2\,\centi\metre\\
|
||||
\uline{I}&=\uuline{2{,}71\,\ampere\cdot e^{-j45\,\degree}}\ \widehat{=}\ 2{,}71\,\centi\metre\\
|
||||
\uline{I}_{R_C}&=\uuline{2{,}42\,\ampere\cdot e^{-j71{,}6\,\degree}}\ \widehat{=}\ 2{,}42\,\centi\metre\\
|
||||
\uline{I}_C&=\uuline{1{,}21\,\ampere\cdot e^{j18{,}4\,\degree}}\ \widehat{=}\ 1{,}21\,\centi\metre\\
|
||||
\uline{U}_{R_L}&=\uuline{108{,}4\,\volt \cdot e^{-j45\,\degree}}\ \widehat{=}\ 4{,}3\,\centi\metre\\
|
||||
\uline{U}_{R_C}&=\uuline{60{,}4\,\volt \cdot e^{-j71{,}6\,\degree}}\ \widehat{=}\ 2{,}4\,\centi\metre\\
|
||||
\uline{U}_L&=\uuline{190\,\volt \cdot e^{j45\,\degree}}\ \widehat{=}\ 7{,}6\,\centi\metre
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
52
ET2_L_B15_A1.tex
Normal file
52
ET2_L_B15_A1.tex
Normal file
@ -0,0 +1,52 @@
|
||||
\section{Komplexe Wechselstromrechnung Netzwerk Strom}
|
||||
Berechnen sie den Strom $I_2$\\
|
||||
$I_0=3\,\milli\ampere \cdot e^{-j30\,\degree}\qquad C_1= 1{,}2\,\nano\farad \qquad L=100\,\micro\henry$\\
|
||||
$R=9,3\,\ohm \qquad C_2=820\,\pico\farad \qquad f=570\,\kilo\hertz$\\
|
||||
Hinweis: Rechnung nur mit komplexen Größen\\
|
||||
Kann $I_2$ größer als $I_0$ sein? Wenn ja, warum?\\
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.5cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$I_0$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_{0}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-1cm,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 [->,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{$\uline{I}_{1}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-.5cm,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] {$C_2$};
|
||||
\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}_{2}$};
|
||||
\draw [->,red] (-.2,.2)--(.2,.2) node at (0,.2)[left]{$\uline{I}_{2'}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.25cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,.75)--(0,1.25)--(.2,1.25) (2,.75)--(2,1.25)--(.8,1.25) (0,0)--(0,-1)-- (2,-1)--(2,0) (1,-1)--(1,-.5);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Berechnung:}
|
||||
C_1&\text{ spielt keine Rolle, da in Reihe mit Stromquelle.}\\
|
||||
\uline{I_2}&=-\uline{I}_2'\quad\text{, wegen Knotenpunkt $\uline{I}_0-\uline{I}_1'-\uline{I}_2'=0$}\\
|
||||
\uline{I_2}&=-\uline{I}_2'=-\frac{R+jX_L}{R+j(X_L+X_{C_2})}\cdot \uline{I}_0\\
|
||||
\text{mit }X_{C_2}&=\frac{-1}{\omega C_2}=-\frac{1}{2\pi\cdot 570\cdot \power{10}{3}\cdot 820\cdot \power{10}{-12}}\,\ohm=-340{,}51\,\ohm\\
|
||||
X_L&=\omega \cdot L=2\pi\cdot 570\cdot \power{10}{3}\cdot \power{10}{-4}\,\ohm= +358{,}14\,\ohm\\
|
||||
X_{C_2}+X_L&=(-340{,}51+358{,}14)\,\ohm=17{,}63\,\ohm\\
|
||||
\uline{I_2}&=-\frac{9{,}3\,\ohm +j 358{,}14\,\ohm}{9{,}3\,\ohm +j 17{,}63\,\ohm}\cdot 3\,\milli\ampere \cdot e^{-j30\,\degree}=(-16{,}11+j7{,}97)\cdot 3\,\milli\ampere\cdot e^{-j30\,\degree}\\
|
||||
&=17{,}97\cdot e^{-j153{,}7\,\degree}\cdot 3\,\milli\ampere\cdot e^{-j30\,\degree}=\uuline{53{,}92\,\milli\ampere\cdot e^{j176{,}32\,\degree}}\\
|
||||
\uline{I}_2&>\uline{I}_0\text{, da sehr nahe an Resonanz $(X_L\approx |X_C|$)}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
73
ET2_L_B15_A2.tex
Normal file
73
ET2_L_B15_A2.tex
Normal file
@ -0,0 +1,73 @@
|
||||
\section{Übergang Zeitabhängige zu Komplexen Größen}
|
||||
Bestimmen Sie den momentanen Strom $i_{L}(t)$ zum Zeitpunkt $t=T$\\[\baselineskip]
|
||||
$u(t)=U_0+\widehat{U}_1\cdot cos(\omega t+\varphi)$\\[\baselineskip]
|
||||
$U_0 = 2\,\volt \qquad \widehat{U}_1=3\,\volt \qquad f=20\,\kilo\hertz \qquad \varphi= 50\,\degree$\\[\baselineskip]
|
||||
$C = 130\,\nano\farad \qquad R = 60\,\ohm \qquad L = 480\,\micro\henry \qquad T= 26\,\micro\second$\\[\baselineskip]
|
||||
\uline{Hinweise:} An dieser Aufgabe sollen Sie den Übergang von der realen, zeitabhängigen Größe
|
||||
$u(t)$ in komplexe Größen $\uline{U}$, $\uline{I}$ und $\uline{I}_L$ und wieder zurück in die reale Größe $i_L(t)$ lernen. (Sehr grundsätzliche und wichtige Übung !)\\
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Wie wirkt sich der Gleichanteil der Spannung $U_0$ aus?
|
||||
\item Übergang vom Zeitbereich in komplexen \glqq Bild\grqq -Bereich $u(t)\Rightarrow \uline{U}$
|
||||
\item Berechnen Sie als Zwischenschritt den komplexen Strom $\uline{I}_L$!
|
||||
\item Übergang von komplexen Bereich in Zeitbereich $\uline{I}_L\Rightarrow i_L(t)$
|
||||
\item Achten Sie auf die Darstellung des Phasenwinkels $\omega t+\varphi$. (in Grad oder rad?!!)
|
||||
\end{enumerate}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\text{...noch einfügen...}
|
||||
%\end{align}
|
||||
\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] {$u(t)$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$u(t)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\end{scope}
|
||||
\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] {$R$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$u_{1}$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (.75,.0)--(.95,.0)node at(.85,.0)[left]{$i_L(t)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,.9)--(0,1)--(.2,1) (.8,1)--(2,1)--(2,.9) (0,.2)--(0,0)--(2,0)--(2,.2);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Berechnung:}
|
||||
&\text{a) $U_0$ spielt wegen $C$ keine Rolle - kein Gleichstrom!}\\
|
||||
\intertext{Transformation in komplexen \glqq Bildbereich\grqq $u(t)\rightarrow\uline{U}$}
|
||||
\uline{U}&=\frac{\widehat{U_1}}{\sqrt{2}}\cdot e^{j\varphi}=\frac{3\,\volt}{\sqrt{2}}\cdot e^{j50\,\degree}=2{,}12\,\volt\cdot e^{j50\,\degree}\\
|
||||
\uline{Z}_C&=\frac{-1}{\omega C}=\frac{-j1}{2\pi\cdot 20\cdot \power{10}{3}\frac{1}{\,\second}\cdot 130\cdot \power{10}{-9}\,\frac{\ampere\second}{\volt}}=-j61{,}21\,\ohm\\
|
||||
\uline{Z}_L&=j\omega L=j2\pi\cdot 20\cdot \power{10}{3}\frac{1}{\,\second}\cdot 480\cdot \power{10}{-6}\,\frac{\volt\second}{\ampere}=+j60{,}32\,\ohm\\
|
||||
\uline{Y}_{||}&=\frac{1}{R}+\frac{1}{j\omega L}=(\frac{1}{60}+\frac{1}{j60{,}32})\,\siemens=(16{,}67-j16{,}58)\,\milli\siemens\\
|
||||
\uline{Z}_{||}&=\frac{1}{Y}_{||}=(30{,}16+j30)\,\ohm\\
|
||||
\uline{U}_1&=\uline{U}_L=\uline{U}\cdot\frac{\uline{Z}_{||}}{\uline{Z}_C+\uline{Z}_{||}}
|
||||
=2{,}12\,\volt\cdot e^{j50\,\degree}\cdot \frac{(30{,}16+j30)\,\ohm}{-j61{,}21\,\ohm+(30{,}16+j30)\,\ohm}\\
|
||||
&=2{,}12\,\volt\cdot e^{j50\,\degree}\cdot \frac{(30{,}16+j30)\,\ohm}{(30{,}16-j31{,}21)\,\ohm}\\
|
||||
&=2{,}12\,\volt\cdot e^{j50\,\degree}\cdot (-0{,}0142+j0{,}98)=2{,}12\,\volt\cdot e^{j50\,\degree}\cdot 0{,}98\cdot e^{j90{,}83\,\degree}\\
|
||||
&=2{,}078\,\volt\cdot e^{j140{,}83\,\degree}
|
||||
\intertext{c) komplexer Strom}
|
||||
\uline{I}_L&=\frac{\uline{U}_{||}}{\uline{Z}_L}=\frac{2{,}078\,\volt\cdot e^{j140{,}83\,\degree}}{60{,}32\,\ohm\cdot e^{j90\,\degree}}=\uuline{34{,}45\,\milli\ampere\cdot e^{j50{,}83\,\degree}}
|
||||
\intertext{d) Übergang in den Zeitbereich}
|
||||
i_L(t)&=\sqrt{2}\cdot |\uline{I}_L|\cdot \cos(\omega T+\varphi_{i_L})
|
||||
=\sqrt{2}\cdot 34{,}45\,\milli\ampere\cdot \cos(2\pi \cdot 20\cdot \power{10}{3}\frac{1}{\,\second}\cdot t+50{,}83\,\degree)\\
|
||||
&=48{,}72\,\milli\ampere\cdot \cos(4 \cdot \power{10}{4}\frac{1}{\,\second}+50{,}83\,\degree)
|
||||
\intertext{e)}
|
||||
&\text{mit } T=26\,\micro\second\\
|
||||
i_L(T)&=48{,}72\,\milli\ampere\cdot \cos(\underbrace{3{,}267}_{\rad}+50{,}83\,\degree)\qquad\text{Achtung! }\\
|
||||
&=48{,}72\,\milli\ampere\cdot \cos(3{,}267\cdot \frac{360\,\degree}{2\pi}+50{,}83\,\degree)\\
|
||||
&=48{,}72\,\milli\ampere\cdot \cos(187{,}2\,\degree+50{,}83\,\degree)=48{,}72\,\milli\ampere\cdot \cos(238\,\degree)\\
|
||||
i_L(T)&=48{,}72\,\milli\ampere\cdot \cos(238\,\degree)=48{,}72\,\milli\ampere\cdot (-0{,}529)=\uuline{-25{,}79\,\milli\ampere}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
71
ET2_L_B15_A2rekap.tex
Normal file
71
ET2_L_B15_A2rekap.tex
Normal file
@ -0,0 +1,71 @@
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\section*{Rekapitulieren}
|
||||
\textbf{Zusammenfassung} An Tafel rekapitulieren\\[\baselineskip]
|
||||
\uline{1. Transformation:}$\quad u(t) \Rightarrow \uline{U}$\\[\baselineskip]
|
||||
\uline{2. Lösung komplexer algebraischer Gleichungen:} Rechnung mit komplexem Effektivwert.\\[\baselineskip]
|
||||
\uline{3. Rücktransformation:}$\quad\uline{I}_L\Rightarrow i_L(t)$\\[\baselineskip]
|
||||
%&\uline{U}\qquad \text{komplexer Effektivwert}\\
|
||||
%&U\qquad \text{Betrag $=$ Effektivwert $(U=\frac{\widehat{u}}{\sqrt{2}})$}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1.25]
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=3.5cm]
|
||||
\draw node at (7,.5)[right]{$\uline{U}$ komplexer Effektivwert};
|
||||
\draw node at (7,0)[right]{$U=|\uline{U}|=\frac{\widehat{u}}{\sqrt{2}}$ Effektivwert};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=-7cm]
|
||||
\draw node at (0,0)[right]{$\widehat{i}_L\qquad$ Scheitelwert $(\,\widehat{i}_L={\sqrt{2}}\cdot I_L)$};
|
||||
\draw node at (0,-.5)[right]{$\uline{I_L}\qquad$ komplexer Effektivwert};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=0cm]
|
||||
\draw (0,0)rectangle(4,2)node at(2,2)[above] {Zeitbereich};
|
||||
\draw node at (2,1){$u(t)=\widehat{u}\cdot \cos(\omega t+\varphi_u)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=7cm, yshift=0cm]
|
||||
\draw (0,0)rectangle(4,2)node at(2,2)[above] {Komplexer Bildbereich};
|
||||
\draw node at (2,1.25){$\uline{U}=U\cdot e^{j\varphi_u}$};
|
||||
\draw node at (2,.75){mit $U=\frac{\widehat{u}}{\sqrt{2}}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0, yshift=-3cm]
|
||||
\draw (0,0)rectangle(4,2);
|
||||
\draw node at (2,1.5){Lösen von};
|
||||
\draw node at (2,1){Differentialgleichungen};
|
||||
\draw node at (2,.5){ist schwierig};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=7cm, yshift=-3cm]
|
||||
\draw (0,0)rectangle(4,2);
|
||||
\draw node at (2,1.5){Lösen von};
|
||||
\draw node at (2,1){komplexen algebraischen};
|
||||
\draw node at (2,.5){Gleichungen};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0, yshift=-6cm]
|
||||
\draw (0,0)rectangle(4,2);
|
||||
\draw node at (2,1.5){Beachte $\omega t$ [rad] $\varphi$ [\degree]};
|
||||
\draw node at (2,1){$i_L(t)=\widehat{i}_L\cdot \cos(\omega t+\varphi_{I_L})$};
|
||||
\draw node at (2,.5){mit $\widehat{i}_L=\sqrt{2}\cdot \uline{I}_L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=7cm, yshift=-6cm]
|
||||
\draw (0,0)rectangle(4,2);
|
||||
\draw node at (2,1){$\uline{I}_L=I_L\cdot e^{j\varphi_{I_L}}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0cm]
|
||||
\draw [->,red,dashed](2,0)--(2,-1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0cm]
|
||||
\draw [->](4,1)--(7,1)node at (5.5,1)[above]{Transformation};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=-3cm]
|
||||
\draw [->,red,dashed](2,0)--(2,-1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=-6cm]
|
||||
\draw [<-](4,1)--(7,1)node at (5.5,1)[above]{Rücktransformation};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=7cm, yshift=0cm]
|
||||
\draw [->](2,0)--(2,-1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=7cm, yshift=-3cm]
|
||||
\draw [->](2,0)--(2,-1);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
42
ET2_L_B15_A3.tex
Normal file
42
ET2_L_B15_A3.tex
Normal file
@ -0,0 +1,42 @@
|
||||
\section{Leitwert}
|
||||
Der Widerstand des abgebildeten Netzwerkes soll $\uline{Z}=1\,\kilo\ohm\cdot e^{j60\,\degree}$ sein.\\[\baselineskip]
|
||||
Wie groß müssen $R$ und $B_C$ sein, wenn $B_L=-3{,}33\,\milli\siemens$ ist?
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\text{...noch einfügen...}
|
||||
%\end{align}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[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] {$R$};
|
||||
\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] {$B_C$};
|
||||
\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] {$B_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (.8,1)--(2,1)--(2,.9) (0,0)--(2,0)--(2,.2);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%End Knoten
|
||||
\fill (0,0)circle(.02);
|
||||
\fill (0,1)circle(.02);
|
||||
\draw node at (0,.5)[left] {$\uline{Z}\Rightarrow$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Berechnung:}
|
||||
\uline{Z}&\stackrel{!}{=}\power{10}{3}\,\ohm\cdot e^{j60\,\degree}\quad\,=(500+j866{,}25)\,\ohm\\
|
||||
\uline{Z}&=R+\frac{1}{j(B_C+B_L)}=\,\,\, R\,\,\, -j\frac{1}{(B_C+B_L)}\\[\baselineskip]
|
||||
\intertext{$\Re$}
|
||||
R&=\uuline{500\,\ohm}\\
|
||||
\intertext{$\Im$}
|
||||
B_C+B_L&=\frac{-1}{866{,}25}\,\siemens=-1{,}1547\,\milli\siemens\\
|
||||
B_C&=-1{,}1547\,\milli\siemens -B_L=-1{,}1547\,\milli\siemens +3{,}33\,\milli\siemens =\uuline{2{,}175\,\milli\siemens}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
58
ET2_L_B15_A4.tex
Normal file
58
ET2_L_B15_A4.tex
Normal file
@ -0,0 +1,58 @@
|
||||
\section{Strom L-R-C}
|
||||
Berechnen Sie den Strom $\uline{I}$
|
||||
\begin{align*}
|
||||
\uline{U}&=15\,\volt \cdot e^{j20\,\degree}\quad f=1\,\kilo\hertz\\
|
||||
C_1&=9\,\micro\farad\quad C_2=4\,\micro\farad\quad R=20\,\ohm\quad L=2\,\milli\henry
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\text{...noch einfügen...}
|
||||
%\end{align}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2.5]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$\uline{U}$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}$};
|
||||
\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] {$C_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator |
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C_2$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}$};
|
||||
\draw [<-,red] (.75,0)--(.95,0) node at (.5,.2)[left]{$\uline{I}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,.8)--(0,1)--(2,1) (.8,1)--(3,1)--(3,.9) (0,.2)--(0,0)--(3,0)--(3,.2);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Berechnung:}
|
||||
C_1& \text{ unwirksam, da parallel zur Spannungsquelle.}\\
|
||||
\omega&=2\pi f=6283\,\frac{1}{\,\second}\\
|
||||
\uline{Z}_{RLC_2}&=X_L+\uline{Z}_{||}\\
|
||||
\uline{Y}_{||}&=G+jB_{C_2}\\
|
||||
G&=\frac{1}{R}=50\,\milli\siemens\\
|
||||
B_{C_2}&=\omega C_2=6283\frac{1}{\,\second}\cdot 4\,\micro\farad=25{,}13\,\milli\siemens\Rightarrow\\
|
||||
X_{C_2}&=-\frac{1}{B_{C_2}}=-39{,}79\,\ohm\\
|
||||
\uline{Y}_{||}&=G+jB_{C_2}=(50+j25{,}13)\,\milli\siemens=55{,}96\,\milli\siemens\cdot e^{j26{,}7\,\degree}\\
|
||||
\uline{Z}_{||}&=\frac{1}{\uline{Y}_{||}}=17{,}87\cdot e^{-j26{,}7\,\degree}=(15{,}97-j8{,}025)\,\ohm\\
|
||||
X_L&=\omega \cdot L=6283\frac{1}{\,\second}\cdot 2\,\milli\henry=12{,}57\,\ohm\\
|
||||
\uline{Z}_{LRC_2}&=\uline{Z}_{||}+jX_L=[15{,}97+j(-8{,}025+12{,}57)]\,\ohm
|
||||
=(15{,}97+j4{,}541)\,\ohm=16{,}6\,\ohm\cdot e^{j15{,}9\,\degree}\\
|
||||
\uline{U}_{C_2}&=\uline{U}\cdot \frac{\uline{Z}_{||}}{\uline{Z}_{LRC_2}}=15\,\volt \cdot e^{j20\,\degree}\cdot \frac{17{,}87\cdot e^{-j26{,}7\,\degree}}{16{,}6\,\ohm\cdot e^{j15{,}9\,\degree}}=16{,}15\,\volt\cdot e^{-j22{,}6\,\degree}=(14{,}91-j6{,}21)\,\volt\\
|
||||
\uline{I}&=\uline{U}_{C_2}\cdot jB_{C_2}=16{,}15\,\volt\cdot e^{-j22{,}6\,\degree}\cdot 25{,}13\,\milli\siemens\cdot e^{j90\,\degree}\\
|
||||
&=\uuline{0{,}4058\,\ampere\cdot e^{j67{,}4\,\degree}}=\uuline{(0{,}156+j0{,}375)\,\ampere}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
120
ET2_L_B15_A5.tex
Normal file
120
ET2_L_B15_A5.tex
Normal file
@ -0,0 +1,120 @@
|
||||
\section{Überlagerungsmethode}
|
||||
Berechnen Sie mit der Überlagerungsmethode den Strom $\uline{I}_C$
|
||||
\begin{align*}
|
||||
R&=R_L=10\,\ohm\quad L=50\,\milli\henry\quad C=100\,\micro\farad\\
|
||||
f&=50\,\hertz\quad \uline{U}_q=5\,\volt \cdot e^{j20\,\degree}\quad \uline{I}_q=2\,\ampere \cdot e^{-j60\,\degree}\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.333cm,rotate=90,scale=.75]%Spule |
|
||||
\draw (.2,0)--(.3,0) (.7,0)--(0.9,0)node at(.5,-.0667) [right] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.083cm,rotate=90,scale=.75]
|
||||
\draw (0.1,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$\uline{U}_q$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_q$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$I_q$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_q$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator |
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
|
||||
\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}_C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,.98)--(0,1)--(1,1) (1.8,1)--(3,1)--(3,.9) (0,.02)--(0,0)--(3,0)--(3,.2);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\text{...noch einfügen...}
|
||||
%\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
$R_L+jX_L \text{ unwirksam, da parallel zur Spannungsquelle.}$\\[\baselineskip]
|
||||
$\omega=2\pi f=2\pi\cdot 50\,\frac{1}{\second}=314{,}2\,\frac{1}{\second}$
|
||||
\begin{align*}
|
||||
\intertext{a) nur Spannungsquelle $\Rightarrow\uline{I}_q=0$}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$\uline{U}_q$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_q$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Kondensator |
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
|
||||
\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}'_C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (1,.98)--(1,1)--(1.1,1) (1.8,1)--(2,1)--(2,.9) (1,.02)--(1,0)--(2,0)--(2,.2);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
B_C&=\omega \cdot C=314{,}2\frac{1}{\,\second}\cdot 100\,\micro\farad=31{,}42\,\milli\siemens\\
|
||||
X_C&=-\frac{1}{B_C}=-\frac{1}{31{,}42\,\milli\siemens}=-31{,}831\,\ohm\\
|
||||
\uline{I}'_C&=\frac{\uline{U}_q}{R+jX_C}=\frac{4{,}698+j1{,}710}{10-j31{,}83}\,\ampere=\uline{(-6{,}693+j149{,}7)\,\milli\ampere}=149{,}8\,\milli\ampere\cdot e^{j92{,}3\,\degree}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{b) Nur Stromquelle $\Rightarrow\uline{U}_q=0$}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$G=\frac{1}{R}$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}''$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$I_q$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_q$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator |
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$};
|
||||
\draw [<-,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{I}''_C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (1,.98)--(1,1)--(2,1) (1.8,1)--(3,1)--(3,.9) (1,.02)--(1,0)--(3,0)--(3,.2);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{I}''_C&=\uline{U}''\cdot jB_C\qquad \uline{U}''=\uline{I}_q\cdot \uline{Z}_{||}\\
|
||||
&\text{oder Stromteiler } \uline{I}''_C=\uline{I_q}\cdot\frac{R}{R-jX_C}\\[\baselineskip]
|
||||
\uline{I}_q&=2\,\ampere \cdot e^{-j60\,\degree}=(1-j1{,}732)\,\ampere\\
|
||||
G&=\frac{1}{R}=100\,\milli\siemens\\
|
||||
\uline{Y}_{||}&=G+jB_C=(100+j31{,}42)\,\milli\siemens=0{,}105\,\siemens \cdot e^{j17{,}44\,\degree}\\
|
||||
\uline{Z}_{||}&=\frac{1}{\uline{Y}_{||}}=9{,}54\,\ohm\cdot e^{-j17{,}44\,\degree}\\
|
||||
\uline{U}''&=\uline{I}_q\cdot \uline{Z}_{||}
|
||||
%\frac{\uline{I}_q}{Y_{||}}=\frac{\uline{I}_q}{G+jB_C}=
|
||||
%\frac{2\,\ampere \cdot e^{-j60\,\degree}}{(0{,}1+j0{,}03142)\,\siemens}\\
|
||||
%&=\frac{2\,\ampere \cdot e^{-j60\,\degree}}{0{,}105\,\siemens \cdot e^{-j17{,}44\,\degree}}
|
||||
=2\,\ampere \cdot e^{-j60\,\degree}\cdot 9{,}54\,\ohm \cdot e^{-j17{,}44\,\degree}\\
|
||||
&=19{,}08\,\volt\cdot e^{-j77{,}44\,\degree}=(4{,}15-j18{,}62)\,\volt\\[\baselineskip]
|
||||
%jB_C&=31{,}42\,\milli\siemens=31{,}42\,\milli\siemens\cdot e^{j90\,\degree}\\
|
||||
\uline{I}''_C&=\uline{U}''\cdot jB_C=19{,}08\,\volt\cdot e^{-j77{,}44\,\degree}
|
||||
\cdot 31{,}42\,\milli\siemens \cdot e^{j90\,\degree}\\
|
||||
&=600\,\milli\ampere\cdot e^{j12{,}56\,\degree}=\uline{(585{,}1+j130{,}3)\,\milli\ampere}\\
|
||||
%&=\frac{\uline{I}_q\cdot X_C}{G+jB_C}\\
|
||||
%=\frac{(1-j1{,}732)\,\ampere\cdot (-j31{,}42\cdot \power{10}{-3})\,\ampere\second\per\volt}{(0{,}1+j31{,}42\cdot \power{10}{-3})\,\ampere\second\per\volt}\\
|
||||
%&=\frac{(54{,}42+j31{,}42)\cdot \power{10}{-3}}{(0{,}1+j31{,}42\cdot \power{10}{-3})}\,\ampere
|
||||
%=\frac{(54{,}42+j31{,}42)}{(100+j31{,}42)\cdot }\,\ampere=(585{,}1+j130{,}3)\,\milli\ampere
|
||||
&\text{oder alternativ:}\\
|
||||
\uline{I}''_C&=\uline{I}_q\cdot \frac{R}{R+jX_C}\\
|
||||
\intertext{Überlagerung}
|
||||
\uline{I}_C&=\uline{I}'_C+\uline{I}''_C=(-6{,}69+j149{,}7+585{,}1+j130{,3})\,\milli\ampere\\
|
||||
&=\uuline{(578{,}41+j279{,}99)\,\milli\ampere}=\uuline{642{,}6\,\milli\ampere\cdot e^{+j25{,}38\,\degree}}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
91
ET2_L_B15_A6.tex
Normal file
91
ET2_L_B15_A6.tex
Normal file
@ -0,0 +1,91 @@
|
||||
\section{Momentan Leistung}
|
||||
In der Reihenschaltung fließt der Strom\\
|
||||
\begin{align*}
|
||||
i(t)&=
|
||||
\begin{cases}
|
||||
&0 \text{ für }t<0\\
|
||||
&I_0\cdot \sin (\omega t) \text{ für }t\geq 0\\
|
||||
\end{cases}
|
||||
\end{align*}
|
||||
Der Kondensator ist zur Zeit $t=0$ entladen.\\
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie die momentanen Spannungen an $R$, $L$ und $C$ zur Zeit $t_1=350\,\micro\second$.
|
||||
\item Welche Leistung nimmt die Schaltung in diesem Moment auf?
|
||||
\item Hinweis: Rechnung mit komplexen Größen wäre hier falsch. Warum?
|
||||
\end{enumerate}
|
||||
\begin{align*}
|
||||
R&=12\,\ohm\qquad L=1{,}3\,\milli\henry\qquad C=8{,}7\,\micro\farad\\
|
||||
f&=2\,\kilo\hertz\qquad I_0=10\,\milli\ampere\\
|
||||
\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$U_{R}$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{L}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
|
||||
\draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Strompfeil
|
||||
\draw [->,red] (-.2,-.2)--(.2,-.2)node at(0,-.2)[below]{\footnotesize$I_0$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\text{...noch einfügen...}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\omega&=2\pi f=1{,}257\cdot\power{10}{4}\,\frac{1}{\second}\\
|
||||
\omega t_1&=1{,}257\cdot \power{10}{4}\,\frac{1}{\second}\cdot 350\cdot \power{10}{-6}\,\second=4{,}398\,[rad]\,\widehat{=}\,252\,\degree\\
|
||||
\end{align*}
|
||||
Zur Erklärung wie die Schwingung aussieht:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw [->] (0,2)--(4.2,2)node[right] {$\omega t$}; % Draw x axis;
|
||||
\draw [->] (0,0)--(0,4)node[above] {$i$}; % Draw y axis;
|
||||
\draw [red, very thick](0,2) sin (1,4) cos (2,2) sin (3,0) cos (4,2);
|
||||
\draw node at(1.8,3)[right] {$i(t)$};
|
||||
\draw node at(2,2)[below] {$\pi$};
|
||||
\draw node at(4,2)[below] {$2\pi$};
|
||||
\draw node at(2.8,2)[below] {$\omega t_1$};
|
||||
\filldraw (2.8,2)--(2.8,.10)circle (2pt)node [below] {$t_1=350\,\micro\second$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=10cm,yshift=2cm]
|
||||
\draw [thin](-2.2,0)--(2.2,0) (0,-2.2)--(0,2.2);
|
||||
\draw (0,0) circle (2);
|
||||
\draw [->,red](0:0)--(252:2)node [below]{$252\,\degree$};
|
||||
\draw [red!70!blue](270:1.92)--(0:0)node at (270:.951)[right]{$\sin (\omega t_1)=-0{,}951$};
|
||||
\draw [red!70!blue](252:2)--(270:1.92)node at (261:2.4)[below]{$\cos (\omega t_1)=-0{,}309$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
a) Berechnung der momentanen Spannung bei $t=t_1$
|
||||
\begin{align*}
|
||||
i(t)&=I_0\cdot \sin(\omega t)=10\,\milli\ampere\cdot \sin{(\underbrace{1{,}257\cdot \power{10}{4}\,\frac{1}{\,\second}\cdot 350\,\micro\second}_{4{,}3995 rad})}\\
|
||||
&=10\,\milli\ampere\cdot (-0{,}951)=-9{,}51\,\milli\ampere
|
||||
\intertext{Für $t=t_1$:}
|
||||
u_R(T)&=R\cdot i(t)=12\,\ohm\cdot (-9{,}51\,\milli\ampere)=\uuline{-114{,}1\,\milli\volt}\\
|
||||
u_L(T)&=L\cdot \frac{di}{dt}\Bigg{|}_{t=T}=L\cdot I_0\cdot \hspace{-.9cm}\underbrace{\omega}_{\mathrm{nachdifferenzieren}}\hspace{-.9cm}\cdot \cos(\omega t)\\
|
||||
&=1{,}3\,\milli\henry\cdot 10\,\milli\ampere\cdot 1{,}257\cdot \power{10}{4}\,\frac{1}{\second}\cdot (-0{,}309)=\uuline{-50{,}5\,\milli\volt}\\
|
||||
u_C(T)&=\underbrace{U_0}_0+\frac{1}{C}\int_{t=0}^{T}{I_0\cdot \sin(\omega t)\cdot dt} =\frac{I_0}{C}\cdot \frac{1}{\omega}\cdot \hspace{-1,5cm}\underbrace{\Big{|}-\cos(\omega t)\Big{|}_{t=0}^{T}}_{-\cos(\omega T)+\cos(0)=-(-0{,}309)+1=1,309}\\
|
||||
&=\frac{0{,}01\,\ampere\cdot 1{,}309}{8{,}7\cdot \power{10}{-6}\,\farad\cdot 1{,}257\cdot \power{10}{4}\,\frac{1}{\second}}=\uuline{119{,}7\,\milli\volt}\\[\baselineskip]
|
||||
\intertext{b) Momentanleistung bei $t=t_1$}
|
||||
p(t)&=u(t)\cdot i(t)=[u_R(t)+u_L(t)+u_C(t)]\cdot i(t)
|
||||
\intertext{Für $t=T$:}
|
||||
p(T)&=u(T)\cdot i(T)=(-114{,}1-50{,}5+119{,}7)\,\milli\volt\cdot (-9{,}51)\,\milli\ampere\\
|
||||
&=(-44{,}9)\,\milli\volt\cdot (-9{,}51)\,\milli\ampere=\uuline{0{,}427\,\milli\watt}
|
||||
\end{align*}
|
||||
c) Komplexe Größen verwenden den Effektivwert der Schwingung!
|
||||
\clearpage
|
||||
}{}%
|
||||
77
ET2_L_B16_A1.tex
Normal file
77
ET2_L_B16_A1.tex
Normal file
@ -0,0 +1,77 @@
|
||||
\section{CLR Netzwerk}
|
||||
Von dem Netzwerk sind folgende Daten bekannt:\\[\baselineskip]
|
||||
$R_1=1,2\,\kilo\ohm\quad R_2=470\,\ohm\quad X_{C_0}=-906\,\ohm\quad X_{L_2}=628\,\ohm$\\
|
||||
$\uline{I}_2=12\,\milli\ampere\cdot e^{j20\,\degree}$\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.5cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_{0}$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_0$};
|
||||
\draw [->,red] (.8,.2)--(1.2,.2)node at(1,.2)[left]{$\uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$X_{C_0}$};
|
||||
\draw [->,blue] (0.3,-.25)--(0.7,-.25)node at(.25,-.4)[right]{\footnotesize$\uline{U}_{C_0}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
|
||||
\draw [<-,blue] (0,-.5)--(1,-.5)node at(.5,-.5)[right]{\footnotesize$\uline{U}_2$};
|
||||
\draw [<-,red] (.8,.15)--(1.2,.15)node at(1,.1)[left]{\footnotesize$\uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$};
|
||||
\draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_{L_2}$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,1.5)--(0,2)--(.2,2) (1,2)--(2,2)--(2,1.75)(0,.5)--(0,0)--(2,0)--(2,.25)
|
||||
(1,0)--(1,.5) (1,2)--(1,1.5);
|
||||
\filldraw [red] (1,2)circle(0.02)node [above]{\footnotesize$KP$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Berechnen Sie Schein- Wirk- und Blindleistung des Netzwerkes!\\[\baselineskip]
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\uline{S}&=\uline{U}\cdot \uline{I}^*\qquad\text{mit }\uline{U}_0=\uline{U}_{C_0}+\uline{U}_2\qquad\uline{I}_0=\uline{I}_1+\uline{I}_2\\
|
||||
\intertext{Spannung $U_2$:}
|
||||
\uline{Z}_2&=R_2+jX_{L_2}=(470+j628)\,\ohm=784{,}4\,\ohm\cdot e^{j53{,}19\,\degree}\\
|
||||
\uline{U}_2&=\uline{Z}_2\cdot \uline{I}_2
|
||||
=9{,}413\,\volt\cdot e^{j73{,}19\,\degree}=(2{,}722+j9{,}011)\,\volt\\
|
||||
\intertext{Strom:}
|
||||
\uline{I}_1&=\frac{\uline{U}_2}{R_1}=\frac{9{,}413\,\volt\cdot e^{j73{,}19\,\degree}}{1,2\,\kilo\ohm}=7{,}844\,\milli\ampere\cdot e^{j73{,}19\,\degree}
|
||||
=(2{,}268+j7{,}5088)\,\milli\ampere\\
|
||||
\uline{I}_0&=\uline{I}_1+\uline{I}_2=(2{,}268+j7{,}5088+11{,}276+j4{,}104)\,\milli\ampere\\
|
||||
&=(13{,}544+j11{,}613)\,\milli\ampere=17{,}841\,\milli\ampere\cdot e^{j40{,}61\,\degree}\\
|
||||
\intertext{Spannung:}
|
||||
\uline{U}_{C_0}&=jX_{C_0}\cdot \uline{I}_0
|
||||
=-j906\,\ohm\cdot 17{,}841\,\milli\ampere\cdot e^{j40{,}61\,\degree}=16{,}164\,\volt\cdot e^{-j49{,}39\,\degree}\\
|
||||
&=(10{,}521-j12{,}271)\,\volt\\
|
||||
\uline{U}_0&=\uline{U}_{C_0}+\uline{U}_2=(10{,}521-j12{,}271+2{,}722+j9{,}011)\,\volt\\
|
||||
&=(13{,}243-j3{,}260)\,\volt=13{,}638\,\volt\cdot e^{-j13{,}83\,\degree}\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Scheinleistung:}
|
||||
\uline{S}&=\uline{U}_0\cdot \uline{I}_0^*=13{,}638\,\volt\cdot e^{-j13{,}83\,\degree}\cdot 17{,}841\,\milli\ampere\cdot e^{-j40{,}61\,\degree}=243{,}32\,\milli\volt\ampere\cdot e^{-j54{,}44\,\degree}\\[2\baselineskip]
|
||||
S&=|\uline{S}|=\uuline{243{,}32\,\milli\volt\ampere}\\
|
||||
P&=S\cdot \cos\varphi=243{,}32\,\milli\volt\ampere\cdot \underbrace{\cos(-54{,}44)}_{0{,}5816}=\uuline{141{,}50\,\milli\watt}\\
|
||||
Q&=S\cdot \sin\varphi=243{,}32\,\milli\volt\ampere\cdot \underbrace{\sin(-54{,}44)}_{-0{,}813}=\uuline{-197{,}93\,\milli\,\var}\\
|
||||
\intertext{Zweiter Weg:}
|
||||
\uline{S}&=\uline{U}_0\cdot \uline{I}^*_0=\uline{I}_0\cdot \uline{Z}\cdot \uline{I}^*_0=I^2_0\cdot \uline{Z}\\
|
||||
\uline{Z}&=jX_{C_0}+\frac{R_1\cdot (R_2-jX_{L_2})}{R_1+R_2-jX_{L_2}}=(445-j622)\,\ohm=764\,\ohm\cdot e^{-j54{,}44\,\degree}\\
|
||||
\uline{S}&=I^2_0\cdot \uline{Z}
|
||||
=(17{,}841\,\milli\ampere)^2\cdot 767\,\ohm\cdot e^{-j54{,}44\,\degree}=243{,}3\,\milli\volt\ampere\cdot e^{-j54{,}44\,\degree}\\
|
||||
S&=243{,}3\,\milli\volt\ampere\\
|
||||
P&=S\cdot\cos(-54{,}44\degree)=141{,}5\,\milli\watt\\
|
||||
Q&=S\cdot\sin(-54{,}44\degree)=-197{,}9\,\milli\,\var
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
14
ET2_L_B16_A1a.tex
Normal file
14
ET2_L_B16_A1a.tex
Normal file
@ -0,0 +1,14 @@
|
||||
\section{Wirkleistung vs. Blindleistung}
|
||||
Was sind die Unterschiede von \\
|
||||
\textbf{Wirkleistungsanpassung} und \textbf{Blindleistungskompensation}\\[\baselineskip]
|
||||
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\textbf{Wirkleistungsanpassung}\\[\baselineskip]
|
||||
|
||||
Maximale Wirkleistung bei Anpassung! $\uline{Z}^*_i \stackrel{!}{=} \uline{Z}_v$\\
|
||||
|
||||
\textbf{Blindleistungskompensation}\\[\baselineskip]
|
||||
|
||||
Maximale Blindleistungskompensation bei $\cos(\varphi=0)$
|
||||
\clearpage
|
||||
}{}%
|
||||
72
ET2_L_B16_A2.tex
Normal file
72
ET2_L_B16_A2.tex
Normal file
@ -0,0 +1,72 @@
|
||||
\section{Wirkleistung}
|
||||
Welche Werte müssen $R$ und $C$ annehmen, damit im Verbraucher die maximale Wirkleistung umgesetzt wird?\\
|
||||
Wie groß ist diese Wirkleistung ?\\[\baselineskip]
|
||||
$R_1=20\,\ohm\quad C_1=3{,}18\,\micro\farad\quad L=0{,}6\,\milli\henry\quad \uline{U}=1\,\volt\cdot e^{j20\,\degree}\quad f=1000\,\hertz$\\[\baselineskip]
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-1cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}$};
|
||||
\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] {$R_1$};
|
||||
% \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.5cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
% \draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$};
|
||||
% \draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$U_{R}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=-1cm,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$U_{L}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,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] {$C$};
|
||||
% \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{C}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]
|
||||
\draw(0,0)--(0,1)--(0.2,1) (.9,.5)--(1,.5)--(1,1) (1,1)--(2,1)--(2,.9) (1.5,-.1)--(1.5,0)--(2,0)(0,-.9)--(0,-1)--(2,-1)--(2,-.9);
|
||||
\draw node at(.5,-1)[below]{Quelle};
|
||||
\draw node at(2,-1)[below]{Verbraucher};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.25cm,yshift=-1cm,]%Knoten |
|
||||
\filldraw (0,0)circle(.05);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.25cm,yshift=1cm,]%Knoten |
|
||||
\filldraw (0,0)circle(.05);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
Phase von $\uline{U}$ ohne Bedeutung! (Berechnung über Impedanzen)\\
|
||||
Maximale Wirkleistung bei Anpassung! $\uline{Z}^*_i \stackrel{!}{=} \uline{Z}_v$
|
||||
\begin{align*}
|
||||
\omega&=2\pi\cdot f=2\pi \cdot 1000\,\hertz=6283\,\frac{1}{\second}\\
|
||||
B_1&=\omega\cdot C_1=6283\,\frac{1}{\second}\cdot 3{,}18\,\micro\farad=0{,}02\,\siemens\\
|
||||
X_L&=\omega\cdot L=6283\,\frac{1}{\second}\cdot 0{,}6\,\milli\henry=3{,}77\,\ohm\\
|
||||
\intertext{Quelle:}
|
||||
\uline{Y}_i&=G_1+jB_1=\frac{1}{R_1}+jB_1=(0{,}05+j0{,}02)\,\siemens\\
|
||||
\uline{Z}_i&=\frac{1}{\uline{Y}_i}=(17{,}24-j6{,}897)\,\ohm\\
|
||||
R_i&=17{,}24\,\ohm\qquad X_i=-6{,}897\,\ohm\\
|
||||
\intertext{Anpassung, wenn $\uline{Z}_v=\uline{Z}^*_i$}
|
||||
\uline{Z}_v&=R_v+jX_v \stackrel{!}{=} (17{,}24+j6{,}897)\,\ohm\quad\Rightarrow\\
|
||||
R_v&=R=\uuline{17{,}24\,\ohm}\\
|
||||
X_v&=6{,}897\,\ohm\\
|
||||
B_v&=B_L+B_C=\frac{-1}{X_v}=\frac{-1}{6{,}897\,\ohm}=-0{,}145\,\siemens\\
|
||||
B_L&=\frac{-1}{X_L}=\frac{-1}{3{,}77\,\ohm}=-0{,}2653\,\siemens\\
|
||||
B_C&=B_v-B_L=(-0{,}145+0{,}2653)\,\siemens=0{,}1203\,\siemens\\
|
||||
C&=\frac{B_C}{\omega}=\frac{0{,}1203\,\siemens}{6283\,\frac{1}{\second}}=\uuline{19{,}14\,\micro\farad}\\[\baselineskip]
|
||||
P_{v\text{,}max}&=\frac{U^2}{4\cdot R_v}=\frac{1\,\volt^2}{4\cdot 17{,}24\,\ohm}=\uuline{14{,}5\,\milli\watt}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
104
ET2_L_B16_A3.tex
Normal file
104
ET2_L_B16_A3.tex
Normal file
@ -0,0 +1,104 @@
|
||||
\section{Abgebbare Wirkleistung}
|
||||
Um wieviel Prozent weicht die in dem passiven Zweipol umgesetzte Wirkleistung von der in dem aktiven Zweipol maximal abgebbaren Wirkleistung ab?\\
|
||||
$C_1=2\,\nano\farad\quad
|
||||
C_2=3\,\nano\farad\quad
|
||||
L_1=2{,}5\,\micro\henry\quad
|
||||
R_1=20\,\ohm\quad
|
||||
L_2=3\,\micro\henry\quad
|
||||
R_1=15\,\ohm\quad
|
||||
f=3\,\mega\hertz$\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.5cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$\uline{I}$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=2cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.5cm,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,xshift=2cm,yshift=1cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,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] {$R_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0cm]
|
||||
\draw(0,1.5)--(0,2)--(.1,2)(1,2)--(3,2)--(3,1.9)(1,1.5)--(1,2);
|
||||
\draw(0,.5)--(0,0)--(3,0)--(3,.1)(1,0)--(1,.5);
|
||||
\fill (2.5,2)circle(.05);
|
||||
\fill (2.5,0)circle(.05);
|
||||
\draw [thin,dashed](-.45,-.2)rectangle(2.4,2.5);
|
||||
\draw [thin,dashed](2.6,-.2)rectangle(3.45,2.5);
|
||||
\draw node at(1,-.25)[below]{Aktiver Zweipol};
|
||||
\draw node at(3,-.25)[below]{Passiver};
|
||||
\draw node at(3,-.5)[below]{Zweipol};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%%\begin{align}
|
||||
%%\intertext{Formeln:}
|
||||
%%\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
$C_1$ spielt für die Berechnung der Wirkleistung keine Rolle, da in Reihe zu Stromquelle.\\[\baselineskip]
|
||||
ESB:
|
||||
\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]{$U_q$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{q}$};
|
||||
\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] {$\uline{Z}_i$};
|
||||
\end{scope}
|
||||
\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] {$\uline{Z}_v$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
|
||||
\draw(0,.9)--(0,1)--(.1,1)(0,.1)--(0,0)--(1,0)--(1,.1);
|
||||
\fill (1,1)circle(.05);
|
||||
\fill (1,0)circle(.05);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Als ESB ist eine Stromquelle $I_q$ mit parallelem $\underline{Z}_i$ und dazu parallelem $\underline{Z}_v$ möglich.
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
Z_{i}&=jX_C || (R_1 + jX_{L1})\\
|
||||
\omega&=2\pi\cdot f=2\pi\cdot 3\,\mega\hertz=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\\
|
||||
X_{C_2}&=\frac{-1}{\omega\cdot C_2}=\frac{-1}{18{,}85\cdot \power{10}{6}\,\cdot \frac{1}{\second}\cdot 3\,\nano\farad}=-17{,}68\,\ohm\\
|
||||
X_{L_1}&=\omega\cdot L_1=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\cdot 2{,}5\,\micro\henry=47{,}12\,\ohm\\
|
||||
X_{L_2}&=\omega\cdot L_2=18{,}85\cdot \power{10}{6}\,\frac{1}{\second}\cdot 3\,\micro\henry=56{,}55\,\ohm\\
|
||||
\uline{Z}_i&=jX_{C_2}||(R_1+jX_{L_1})
|
||||
=\frac{-j17{,}68\,\ohm\cdot (20\,\ohm+j47{,}12\,\ohm)}{-j17{,}68\,\ohm + 20\,\ohm+j47{,}12\,\ohm}
|
||||
=\frac{-j17{,}68\cdot (20+j47{,}12)}{20+j(47{,}12-17{,}68)}\\
|
||||
&=\frac{833{,}08-j353{,}6}{20+j29{,}44}\,\ohm=\frac{905{,}02\cdot e^{-j23{,}0\,\degree}}{35{,}59\,\ohm\cdot e^{j55{},81\,\degree}}=25{,}428\,\ohm\cdot e^{-j78{,}81\,\degree}=(4{,}935-j24{,}94)\,\ohm
|
||||
\end{align*}
|
||||
Verbraucherwiderstand:
|
||||
\begin{align*}
|
||||
\uline{Z}_v&=(15+j56{,}55)\,\ohm\\
|
||||
\uline{Z}_{ges}&=\uline{Z}_i+\uline{Z}_v=(4{,}935-j24{,}94)\,\ohm+(15+j56{,}55)\,\ohm=(19{,}94+j31{,}61)\,\ohm\\
|
||||
\intertext{Anmerkung: $\uline{U}_q$ ist unbekannt, kürzt sich später heraus.}
|
||||
P_{v,max}&=\frac{U^2_q}{4\cdot R_i}=\frac{U^2_q}{4\cdot 4{,}935\,\ohm}=\frac{U^2_q}{19{,}94\,\ohm}\\
|
||||
P_v&=I^2\cdot R_v\\
|
||||
I&=\frac{U_q}{Z_{ges}}\qquad \text{Anmerkung: $I$ und $U_q$ Effektivwert; $Z$ Betrag}\\
|
||||
P_v&=\left(\frac{U_q}{Z_{ges}}\right)^2 \cdot R_v =\frac{U^2_q}{(19{,}94^2+31{,}61^2)\,\ohm^2}\cdot 15\,\ohm=U^2_q\cdot \frac{15}{1397\,\ohm}\\
|
||||
F_ \% &=100\,\%\cdot \frac{P_v-P_{v,max}}{P_{v,max}}=100\,\%\cdot \left(\frac{P_v}{P_{v,max}}-1\right)\\
|
||||
&=100\, \% \cdot \left(\frac{\cancel{U^2_q}\cdot 15}{1397\,\ohm}\cdot \frac{19{,}94\,\ohm}{\cancel{U^2_q}}-1\right)\\
|
||||
&=100\, \% \cdot (0{,}212-1)=\uuline{-78{,}8\, \%}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
115
ET2_L_B16_A4.tex
Normal file
115
ET2_L_B16_A4.tex
Normal file
@ -0,0 +1,115 @@
|
||||
\section{Wirkleistung Spannungsquelle}
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie die Wirkleistung der Spannungsquelle $\uline{U}_2$.
|
||||
\item Wird Wirkleistung aufgenommen oder abgegeben?
|
||||
\end{enumerate}
|
||||
$R=200\,\ohm\quad L=80\,\micro\henry\quad C=500\,\pico\farad \quad f=1\,\mega\hertz$\\
|
||||
$\uline{I}_1=10\,\milli\ampere\cdot e^{j60\,\degree}\quad \uline{U}_2=3\,\volt\cdot e^{-j30\,\degree}$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$\uline{I}_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize $\uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_2$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[very thick](0,.9)--(0,1)--(2,1)--(2,.9)(0,.1)--(0,0)--(.1,0)(2,.1)--(2,0)--(1.9,0);
|
||||
\draw [->,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
\begin{align*}
|
||||
\intertext{$C$ spielt keine Rolle, da in Reihe zu Stromquelle.}
|
||||
\text{Gesucht: } P&=\Re(\uline{U}_2\cdot \uline{I}^*)
|
||||
\end{align*}
|
||||
Lösung mit Überlagerungsverfahren:\\[\baselineskip]
|
||||
Nur Stromquelle:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Stromquelle
|
||||
\draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0)node at(.5,-.133)[right]{$\uline{I}_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->,red] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{I}_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[very thick](0,.9)--(0,1)--(2,1)--(2,.9)(0,.1)--(0,0)--(1,0)(1.9,1)--(2,1)--(2,0)--(1.9,0);
|
||||
\draw [->,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}'$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
Stromteiler:\\
|
||||
\begin{align*}
|
||||
X_L&= \omega \cdot L = 2\cdot \pi\cdot 1\,\mega\hertz\cdot 80\,\micro\henry= 503\,\ohm\\
|
||||
\uline{I}'&=\uline{I}_1\cdot \frac{jX_L}{R+jX_L}
|
||||
=10\,\milli\ampere\cdot e^{j60\,\degree}\cdot \frac{j503}{200+j503}\\
|
||||
&=10\,\milli\ampere\cdot e^{j60\,\degree}\cdot \underbrace{\frac{503\cdot e^{j90\,\degree}}{541{,}3\cdot e^{j68{,}3\,\degree}}}_{0{,}929\,\milli\ampere\cdot e^{j21{,}7\,\degree}}\\
|
||||
&=9{,}29\,\milli\ampere\cdot e^{j81{,}7\,\degree} = (1{,}34 + j9{,}19)\,\milli\ampere\\
|
||||
\end{align*}
|
||||
Nur Spannungquelle:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle |
|
||||
\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}_2$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[very thick](1,.9)--(1,1)--(2,1)--(2,.9)(1,.1)--(1,0)(2,.1)--(2,0)--(1.9,0);
|
||||
\draw [<-,red] (1.3,1.1)--(1.7,1.1)node at (1.5,1.1)[above]{\footnotesize$\uline{I}''$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{I}''&=\frac{\uline{U}_2}{R+jX_L}
|
||||
=\frac{3\,\volt\cdot e^{-j30\,\degree}}{(200+j503)\,\ohm}
|
||||
=\frac{3\,\volt\cdot e^{-j30\,\degree}}{541{,}3\,\ohm \cdot e^{j68{,}32\,\degree}
|
||||
}\\
|
||||
&=5{,}54\,\milli\ampere\cdot e^{-j98{,}32\,\degree} = (-0{,}80-j5{,}48)\,\milli\ampere\\
|
||||
\intertext{Überlagerung - vorzeichenrichtg:}
|
||||
\uline{I}&=\uline{I}'-\uline{I}''= (1{,}34 + j9{,}19+0{,}80+j5{,}48)\,\milli\ampere \\
|
||||
&=(2{,}14+j14{,}67)\,\milli\ampere=\uline{14{,}83\,\milli\ampere\cdot e^{j81{,}7\,\degree}}\\
|
||||
S&=\uline{U}_2\cdot \uline{I}^*=3\,\volt\cdot e^{-j30\,\degree}
|
||||
\cdot 14{,}83\,\milli\ampere\cdot e^{-j81{,}7\,\degree}
|
||||
=44{,}49\,\milli\volt\ampere\cdot e^{-j111{,}7\,\degree}\\
|
||||
&=(\underbrace{-16{,}31}_{P}-j\underbrace{40{,}97}_{Q})\,\milli\volt\ampere\Rightarrow\\
|
||||
P&=\uuline{-16{,}31\,\milli\watt}
|
||||
\intertext{Verbraucher-Zählpfeilsystem $\Rightarrow$ \uuline{Spannungsquelle gibt Leistung ab.}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
90
ET2_L_B16_A5.tex
Normal file
90
ET2_L_B16_A5.tex
Normal file
@ -0,0 +1,90 @@
|
||||
\section{Dualitätskonstante}
|
||||
Berechnen Sie zu der gegebenen Schaltung die duale Schaltung mit der Dualitätskonstanten.\\[\baselineskip]
|
||||
$R^2_D=10000\,\ohm^2$\\
|
||||
$R_1=80\,\ohm$\\
|
||||
$L_1=50\,\milli\henry$\\
|
||||
$C_1=10\,\micro\,\farad$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.5cm]%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] {$R_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
|
||||
\fill(0,0)circle(.05) (3,0)circle(.05);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
\uline{Z}_1(\omega)&=R^2_D \cdot \uline{Y}_2(\omega)
|
||||
\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
Parallel $\Leftrightarrow$ Serie\\
|
||||
Leitwert $\Leftrightarrow$ Widerstand\\
|
||||
Kapazität $\Leftrightarrow$ Induktivität\\
|
||||
Parallelschaltung $R_1||L_1$ in Serienschaltung $R_2+C_2$
|
||||
Serienschaltung $C_1+(R_2+C_2)$ in Parallelschaltung $L_2 ||(R_2+C_2)$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\draw node at (0,0){\phantom{.}};
|
||||
\draw node at (0,1.5){Zwischenschritt:};
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0.5cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0.5cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0.5cm]%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] {$R_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0.5cm]
|
||||
% \draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
|
||||
% \draw (0,.5)--(.5,.5) (.5,.5)--(.5,.5)--(.5,-.5)--(2,-.5)
|
||||
% (2.5,.5)--(2.5,-.5)--(2,-.5)(2.5,.5)--(3,.5);
|
||||
\fill(0,0)circle(.05) (3,0)circle(.05);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\draw node at (0,1.0){Duale Schaltung:};
|
||||
\begin{scope}[>=latex,very thick,xshift=.5cm,yshift=.5cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=.5cm]%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] {$R_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.5cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
% \draw (1.1,.5)--(1,.5)--(1,-.5)--(1.1,-.5)(1.9,.5)--(2,.5)--(2,-.5)--(1.9,-.5)(2,0)--(3,0);
|
||||
\draw (0,.5)--(.5,.5) (.5,.5)--(.5,.5)--(.5,-.5)--(2,-.5)
|
||||
(2.5,.5)--(2.5,-.5)--(2,-.5)(2.5,.5)--(3,.5);
|
||||
\fill(0,.5)circle(.05) (3,.5)circle(.05);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
R_1\cdot R_2&=R^2_D\Rightarrow R_2=\frac{R^2_D}{R_1}=\frac{10000\,\ohm^2}{80\,\ohm}=\uuline{125\,\ohm}\\
|
||||
\text{(C zu L) aus: }\frac{L_2}{C_1}&=R^2_D\Rightarrow L_2=R^2_D\cdot C_1
|
||||
=\power{10}{4}\,\ohm^{{\cancel{2}}}
|
||||
\cdot \power{10}{-5}\,\frac{\second}{\cancel{\ohm}}
|
||||
=0{,}1\,\frac{\volt\second}{\ampere}
|
||||
=\uuline{100\,\milli\henry}\\
|
||||
\text{(L zu C) aus: }\frac{L_1}{C_2}&=R^2_D\Rightarrow C_2=\frac{L_1}{R^2_D}
|
||||
=\frac{50\,\milli\henry}{10000\,\ohm^2}
|
||||
=5\cdot \power{10}{-6}\,\frac{\cancel{\volt}\second}{\cancel{\ampere}}
|
||||
\cdot \frac{\ampere^{\cancel{2}}}{\volt^{\cancel{2}}}
|
||||
=\uuline{5\,\micro\farad}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
81
ET2_L_B16_A6.tex
Normal file
81
ET2_L_B16_A6.tex
Normal file
@ -0,0 +1,81 @@
|
||||
\section{Dualitätskonstante verlustbehaftete Bauelemente}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.5cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,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] {$G$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=.5cm,yshift=-.5cm]%Widerstand - nach EN 60617
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=-.5cm]%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=0cm,yshift=0cm]
|
||||
\draw (0,0)--(1,0) (1.1,.5)--(1,.5)--(1,0) (.6,-.5)--(.5,-.5)--(.5,0) (1.9,.5)--(2,.5)--(2,0)(2.4,-.5)--(2.5,-.5)--(2.5,0) (2,0)--(3,0);
|
||||
\fill(.5,0)circle(.03) (1,0)circle(.03)(2,0)circle(.03) (2.5,0)circle(.03);
|
||||
\fill(0,0)circle(.05)node[left]{$a$};
|
||||
\fill(3,0)circle(.05)node[right]{$b$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
$\hspace{3cm} R=20\,\ohm\quad L=1\,\milli\henry\quad C=100\,\nano\,\farad\quad G=2\,\milli\siemens$\\
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Geben Sie das duale Schaltbild für den Zweipol $a-b$ an und berechnen Sie dessen Elemente für $R^2_D=(100\,\ohm)^2$.
|
||||
\item Welches verlustbehaftete Bauelement stellt die Reihenschaltung $R$ und $L$ dar?
|
||||
\item Welches verlustbehaftete Bauelement stellt die Parallelschaltung $G$ und $C$ dar?
|
||||
\item Interpretieren Sie das Ergebnis der Bauelementegrößen der beiden dualen Schaltungen in Bezug auf verlustbehaftete Bauelemente.
|
||||
\end{enumerate}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
%\uline{Z}_1(\omega)&=R^2_D \cdot \uline{Y}_2(\omega)
|
||||
R\cdot R_{dual}&=R^2_D \quad = \quad\text{Dualitätskonstante}\\
|
||||
R^2_D&=\frac{L}{C}=\frac{L_{dual}}{C}=\frac{L}{C_{dual}}
|
||||
\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
Duales Netzwerk:\\
|
||||
Parallel $\Leftrightarrow$ Serie\\
|
||||
Leitwert $\Leftrightarrow$ Widerstand\\
|
||||
Kapazität $\Leftrightarrow$ Induktivität\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=.25cm]%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] {$G_R$};
|
||||
\draw node at (.75,.35) [above] {\footnotesize{Verlustbehafteter Kondensator}};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=-.25cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_L$};
|
||||
\end{scope}
|
||||
\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] {$R_G$};
|
||||
\draw node at (.75,-.5) [above] {\footnotesize{Verlustbehaftete Spule}};
|
||||
\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] {$L_C$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw (2.1,.25)--(2,.25)--(2,-.25)--(2.1,-.25) (2.9,.25)--(3,.25)--(3,-.25)--(2.9,-.25) (3,0)--(4,0);
|
||||
\fill(2,0)circle(.03)(3,0)circle(.03);
|
||||
\fill(0,0)circle(.05)node[left]{$a$};
|
||||
\fill(4,0)circle(.05)node[right]{$b$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{a) Duale Bauelemente:}
|
||||
R_G&=\frac{R^2_D}{R}=R^2_D\cdot G=(100\,\ohm)^2\cdot 2\,\milli\siemens=\uuline{20\,\ohm}\quad(=R\text{ !})\\
|
||||
L_C&=C\cdot R^2_D=100\cdot \power{10}{-9}\frac{\,\ampere\second}{\volt}\cdot (100\frac{\,\volt}{\ampere})^2=\uuline{1\,\milli\henry}=\quad(=L\text{ !})\\
|
||||
G_R&=\frac{R}{R^2_D}=\frac{20\,\ohm}{(100\,\ohm)^2}=\uuline{2\,\milli\siemens} \quad(=G\text{ !)}\\
|
||||
C_L&=\frac{L}{R^2_D}=\frac{1\,\milli\henry}{(100\,\ohm)^2}=\uuline{100\,\nano\farad} \quad(=C\text{ !)}\\
|
||||
\intertext{b) Verlustbehaftete Spule.}
|
||||
\intertext{c) Verlustbehafteter Kondensator.}
|
||||
\intertext{d) Die duale Schaltung ist die Reihenschaltung der gleichen verlustbehafteten Bauteile Spule und Kondensator.}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
239
ET2_L_B16_A7.tex
Normal file
239
ET2_L_B16_A7.tex
Normal file
@ -0,0 +1,239 @@
|
||||
\section{Vierpol Y-Parameter}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=3]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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]%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=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);
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_2$};
|
||||
\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] {$2L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]
|
||||
\fill (0,0)circle(.05)(0,1)circle(.05)(1,1)circle(.05)(1,0)circle(.05)(2,0)circle(.05)(2,1)circle(.05);
|
||||
\draw [->,blue] (0,0.9)--(0,.1) node at (0,.5)[right]{$\uline{U}_{1}$};
|
||||
\draw [->,blue] (2,0.9)--(2,.1) node at (2,.5)[right]{$\uline{U}_{2}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie die $\uline{Y}$-Parameter des Vierpols in Abhängigkeit von $L$.\\
|
||||
\item Bestimmen Sie die $\uline{Z}$-Parameter
|
||||
\end{enumerate}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\intertext{a)\hspace{0.4cm} $Y$ Parameter; Einträge in Leitwertmatrix; Achtung Serienschaltung $\uline{Y}_{Serie}=\frac{\uline{Y}_1\cdot \uline{Y}_2}{\uline{Y}_1+\uline{Y}_2}!$}
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
\uline{I}_1 \\
|
||||
\uline{I}_2 \\
|
||||
\end{array}
|
||||
\right]
|
||||
&=
|
||||
\left[
|
||||
\begin{array}{cc}
|
||||
\uline{Y}_{11} & \uline{Y}_{12} \\
|
||||
\uline{Y}_{21} & \uline{Y}_{22} \\
|
||||
\end{array}
|
||||
\right]\cdot
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
\uline{U}_1 \\
|
||||
\uline{U}_2 \\
|
||||
\end{array}
|
||||
\right]\\[\baselineskip]
|
||||
\uline{I}_1&=\uline{Y}_{11}\cdot \uline{U}_1+\uline{Y}_{12}\cdot \uline{U}_2\\
|
||||
\uline{I}_2&=\uline{Y}_{21}\cdot \uline{U}_1+\uline{Y}_{22}\cdot \uline{U}_2\\
|
||||
\uline{Y}_L&=\uuline{\frac{1}{j\omega\cdot L}}=\uuline{-j\frac{1}{\omega\cdot L}}
|
||||
\end{align*}
|
||||
%\begin{center}
|
||||
%\intertext{$Y_{11}$: $U_2=0$ d.h. Kurzschluß am Ausgang}
|
||||
$Y_{11}$: $U_2=0$ d.h. Kurzschluß am Ausgang\\[\baselineskip]
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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]%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=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);
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_2$};
|
||||
\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] {$2L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]
|
||||
\fill (0,0)circle(.05)(0,1)circle(.05)(1,1)circle(.05)(1,0)circle(.05)(2,0)circle(.05)(2,1)circle(.05);
|
||||
\draw [->,blue] (0,0.9)--(0,.1) node at (0,.5)[right]{$\uline{U}_{1}$};
|
||||
% \draw [->,blue] (2,0.9)--(2,.1) node at (2,.5)[right]{$\uline{U}_{2}$};
|
||||
\draw node at (0,.5)[left]{$\uline{Y}_{11}\Rightarrow$};
|
||||
\draw(2,0)--(2,1);
|
||||
\draw[red!50!blue,thick,dashed](.8,-.33)rectangle(2.2,1.33)node at (1.5,-.33)[below]{$L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,thick,xshift=0cm]
|
||||
\draw node at (0,-1)[right]{$\uline{Y}_{11}=\frac{\uline{I}_1}{\uline{U}_1}\Big{|}_{U_2=0}=\uuline{\frac{1}{3}\cdot \uline{Y}_L}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\hspace{1cm}
|
||||
% ----------------------------------
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$Y_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\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] {$Y_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\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] {$Y_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
% \draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_1$};
|
||||
\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] {$Y_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
% \draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_2$};
|
||||
\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] {$\frac{1}{2} \cdot Y_L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]
|
||||
\fill (0,0)circle(.05)(0,1)circle(.05)(1,1)circle(.05)(1,0)circle(.05)(2,0)circle(.05)(2,1)circle(.05);
|
||||
% \draw [->,blue] (0,0.9)--(0,.1) node at (0,.5)[right]{$\uline{U}_{1}$};
|
||||
% \draw [->,blue] (2,0.9)--(2,.1) node at (2,.5)[right]{$\uline{U}_{2}$};
|
||||
\draw node at (0,.5)[left]{$\uline{Y}_{11}\Rightarrow$};
|
||||
\draw(2,0)--(2,1);
|
||||
% \draw[red!50!blue,thick,dashed](.8,-.33)rectangle(2.2,1.33)node at (1.5,-.33)[below]{$Y_L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,thick,xshift=0cm]
|
||||
\draw[white] node at (0,-1)[right]{$\uline{Y}_{L}=\frac{1}{j\omega \cdot L}=-j\frac{1}{\omega\cdot L}$};
|
||||
\end{scope}
|
||||
\begin{scope}[black!75!,>=latex,thick,xshift=0cm]
|
||||
\draw[black] node at (1.5,-.33)[above]{$\underbrace{\phantom{xxxxxxxx}}$};
|
||||
\draw node at (1.5,-.2)[below]{\footnotesize{$\frac{1}{2\cdot X_L}=\frac{1}{2}\cdot Y_L$}};
|
||||
\draw[black] node at (1.4,-.67)[above]{$\underbrace{\phantom{xxxxxxxxxxx}}$};
|
||||
\draw node at (1.4,-.54)[below]{\footnotesize{$\frac{1}{2}\cdot Y_L+\frac{1}{2}\cdot Y_L= Y_L$}};
|
||||
\draw[black] node at (1,-1)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxxx}}$};
|
||||
\draw node at (1,-.87)[below]{\footnotesize{$\frac{1}{3\cdot X_L}=\frac{1}{3}\cdot Y_L$}};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
% ----------------------------------
|
||||
%\end{center}
|
||||
\begin{align*}
|
||||
\intertext{$Y_{12}$: $U_1=0$ d.h. Kurzschluß am Eingang}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%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]%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=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);
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_1$};
|
||||
\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] {$L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{$\uline{I}_2$};
|
||||
\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] {$2L$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]
|
||||
\fill (0,0)circle(.05)(0,1)circle(.05)(1,1)circle(.05)(1,0)circle(.05)(2,0)circle(.05)(2,1)circle(.05);
|
||||
% \draw [->,blue] (0,0.9)--(0,.1) node at (0,.5)[right]{$\uline{U}_{1}$};
|
||||
\draw [->,blue] (2,0.9)--(2,.1) node at (2,.5)[left]{$\uline{U}_{2}$};
|
||||
\draw node at (2,.5)[right]{$\Leftarrow\uline{Y}_{L}$};
|
||||
\draw(0,0)--(0,1);
|
||||
\draw[red!50!blue,thick,dashed](-.2,-.33)rectangle(1.2,1.33)node at (.5,-.33)[below]{$L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,thick,xshift=.5cm,yshift=.5cm]
|
||||
\draw node at (2.5,.5)[right]{aus $\uline{I}_{1}=\uline{Y}_{11}\cdot \uline{U}_1+\uline{Y}_{12}\cdot \uline{U}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,thick,xshift=.5cm,yshift=.25cm]
|
||||
\draw node at (2.5,.5)[right]{folgt mit $\uline{U}_{1}=0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,thick,xshift=.5cm,yshift=-.25cm]
|
||||
\draw node at (2.5,.5)[right]{$\uline{Y}_{12}=\frac{\uline{I}_1}{\uline{U}_2}\big{|}_{U_1=0}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\text{Stromteiler: }
|
||||
-\uline{I}_1&=\frac{2\cdot Z_L}{4\cdot Z_L}\cdot \uline{I}_2=\frac{1}{2}\cdot \uline{I}_2\\
|
||||
%
|
||||
%
|
||||
\uline{I}_2&=\frac{\uline{U}_2}{3\cdot Z_L} = \uline{U}_2\cdot \frac{1}{3}\cdot \uline{Y}_L\\
|
||||
\uline{I}_1&=-\frac{1}{2}\cdot \uline{U}_2 \cdot \frac{1}{3}\cdot \uline{Y}_L = -\frac{1}{6}\cdot \uline{U}_2\cdot \uline{Y}_L\\
|
||||
\uline{Y}_{12}&=\frac{-\frac{1}{6}\cdot \uline{U}_2\cdot \uline{Y}_L}{\uline{U}_2}=\uuline{-\frac{1}{6}\cdot \uline{Y}_L}\\
|
||||
%
|
||||
%
|
||||
%\uline{Y}_{12}&=\uuline{-\frac{1}{6}\cdot \uline{Y}_L}\text{ ??? 1/3 oder 1/6 ???}\\
|
||||
\uline{Y}_{21}&=\uline{Y}_{12}=\uuline{-\frac{1}{6}\cdot \uline{Y}_L}\text{ da spiegelsymmetrisch}\\
|
||||
\uline{Y}_{22}&=\frac{\uline{I}_2}{\uline{U}_2}\Big{|}_{_{U_1=I_1=0}}=\text{spiegelbildlich zu $\uline{Y}_{11}$ d.h. }\\
|
||||
\uline{Y}_{22}&=\uline{Y}_{11}=\uuline{\frac{1}{3}\cdot \uline{Y}_L}\\
|
||||
\intertext{b)\hspace{0.4cm} $Z$-Parameter Leerlauf}
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
\uline{U}_1 \\
|
||||
\uline{U}_2 \\
|
||||
\end{array}
|
||||
\right]
|
||||
&=
|
||||
\left[
|
||||
\begin{array}{cc}
|
||||
\uline{Z}_{11} & \uline{Z}_{12} \\
|
||||
\uline{Z}_{21} & \uline{Z}_{22} \\
|
||||
\end{array}
|
||||
\right]\cdot
|
||||
\left[
|
||||
\begin{array}{c}
|
||||
\uline{I}_1 \\
|
||||
\uline{I}_2 \\
|
||||
\end{array}
|
||||
\right]\\[\baselineskip]
|
||||
\uline{U}_1&=\uline{Z}_{11}\cdot \uline{I}_1+\uline{Z}_{12}\cdot \uline{I}_2\\
|
||||
\uline{U}_2&=\uline{Z}_{21}\cdot \uline{I}_1+\uline{Z}_{22}\cdot \uline{I}_2\\
|
||||
\uline{Z}_{11}&=\frac{\uline{U}_{1}}{\uline{I}_{1}}\Big{|}_{_{I_2=0}}=4\cdot j\omega\cdot L=\uuline{\uline{Z}_{22}}\\[\baselineskip]
|
||||
\uline{Z}_{12}&=\frac{\uline{U}_{1}}{\uline{I}_{2}}\Big{|}_{_{I_1=0}}\\
|
||||
\uline{U}_{1}&=2\cdot X_L \cdot \uline{I}_{2}\\
|
||||
\uline{Z}_{12}&=\frac{2\cdot X_L \cdot \uline{I}_{2}}{\uline{I}_{2}}=2\cdot X_L = 2\cdot j\omega\cdot L=\uuline{\uline{Z}_{21}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
224
ET2_L_B16_A8.tex
Normal file
224
ET2_L_B16_A8.tex
Normal file
@ -0,0 +1,224 @@
|
||||
\section{Spannung Vierpol}
|
||||
%\renewcommand{\theequation}{0}
|
||||
Berechnung Sie $\uline{U}_0$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[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] {$R_1$};
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$\uline{U}_R=R_2\cdot \uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=1.5cm, yshift=0cm]% Vierpol
|
||||
\draw (0,-.2)rectangle(2.0,1.2)node at(.75,.5)[right]{$(\uline{Z})$};
|
||||
\draw (0,0)--(-.5,0)(0,1)--(-.5,1)(2,0)--(2.5,0)(2,1)--(2.5,1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
|
||||
\draw (0,0)--(1,0);
|
||||
\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
|
||||
\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);
|
||||
\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05) (4,0)circle(.05) (4,1)circle(.05);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=.75cm,yshift=1cm]% Strompfeil
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=1cm]% Strompfeil
|
||||
\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[below]{\footnotesize$\uline{U}_2$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
R_1&=300\,\ohm\\
|
||||
R_2&=600\,\ohm\\
|
||||
\uline{Z}_{11}&=100\,\kilo\ohm\text{ reell}\\
|
||||
\uline{Z}_{21}&=-2\,\mega\ohm\text{ reell}\\
|
||||
\uline{U}_1&=1{,}5\,\volt
|
||||
%\end{align*}
|
||||
%\begin{align*}
|
||||
\intertext{Die $\uline{Z}$-Matrix des Vierpols ist gegeben:}
|
||||
\begin{bmatrix}
|
||||
\uline{U}_0\\
|
||||
\uline{U}_2
|
||||
\end{bmatrix}
|
||||
&=
|
||||
\begin{bmatrix}
|
||||
\uline{Z}_{11}&0\\
|
||||
\uline{Z}_{21}&0
|
||||
\end{bmatrix}
|
||||
\cdot
|
||||
\begin{bmatrix}
|
||||
\uline{I}_0\\
|
||||
\uline{I}_2
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[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] {$R_1$};
|
||||
\draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$R_1\cdot (\uline{I}_2+\uline{I}_0)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$R_2\cdot \uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
|
||||
\draw (0,0)--(1,0);
|
||||
\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
|
||||
\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);
|
||||
\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05) (4,0)circle(.05) (4,1)circle(.05);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
|
||||
\draw (0,0)--(1,0);
|
||||
\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
|
||||
\draw (1,0)--(1,-.5)--(4,-.5)--(4,0)(1,1)--(1.25,1)(4,1)--(3.75,1);
|
||||
\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05) (4,0)circle(.05) (4,1)circle(.05);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1.2cm]% Strompfeil
|
||||
\draw [->,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1.2cm]% Strompfeil
|
||||
\draw [<-,red] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=-.2cm,yshift=1cm]% Strompfeil
|
||||
\draw [->,red] (0,0)--(.2,0)node at(.1,0)[above]{\footnotesize$\uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(-.2,.5)[left]{\footnotesize$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_0=\uline{Z}_{11}\cdot \uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_2=\uline{Z}_{21}\cdot \uline{I}_0$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Aus $\uline{Z}$-Matrix folgt:
|
||||
\begin{align*}
|
||||
\uline{U}_0&=\uline{Z}_{11}\cdot \uline{I}_0 \tag{1}\\
|
||||
\uline{U}_2&=\uline{Z}_{21}\cdot \uline{I}_0 \tag{2}
|
||||
\intertext{Knoten 1:}
|
||||
\uline{I}_1-\uline{I}_2-\uline{I}_0&=0 \tag{3}
|
||||
\intertext{Masche 1:}
|
||||
\uline{U}_1&=R_1\cdot \uline{I}_1+ \uline{U}_0\tag{4}
|
||||
\intertext{Masche 2:}
|
||||
\uline{U}_0&=R_2\cdot \uline{I}_2+\uline{U}_2 \tag{5}
|
||||
\intertext{Zwei Gleichungen für zwei unbekante Ströme $\uline{I}_0$ und $\uline{I}_2$:}
|
||||
\uline{Z}_{11}\cdot \uline{I}_0&=R_2\cdot \uline{I}_2+\uline{Z}_{21}\cdot \uline{I}_0 \tag{1\&2 in 5}\\
|
||||
\Rightarrow \uline{I}_2&=\frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0 \tag{6}\\
|
||||
\uline{U}_1&=\uline{I}_0 \cdot\uline{Z}_{11}+R_1(\uline{I}_0 +\uline{I}_2 ) \tag{7 1\&3 in 4}\\
|
||||
\uline{U}_1&=\uline{I}_0\cdot \uline{Z}_{11}+R_1\cdot \uline{I}_0+R_1\cdot \frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0 \tag{6 in 7}\\
|
||||
&=\left[\uline{Z}_{11}+R_1+\frac{R_1}{R_2}\cdot(\uline{Z}_{11}-\uline{Z}_{21})\right]\cdot \uline{I}_0\\
|
||||
&=\left[100\,\kilo\ohm+300\,\ohm+\frac{300\,\ohm}{600\,\ohm}\cdot \Big(100\,\kilo\ohm-(-2\,\mega\ohm)\Big)\right]\cdot \uline{I}_0\\
|
||||
&=0{,}103+0{,}5\cdot (100+2000)\,\kilo\ohm \cdot \uline{I}_0=1{,}15\,\mega\ohm\cdot \uline{I}_0\\
|
||||
\Rightarrow\uline{I}_0&=\frac{\uline{U}_1}{1{,}15\,\mega\ohm}=\frac{1{,}5\,\volt}{1{,}15\,\mega\ohm}=1{,}3\,\micro\ampere\\
|
||||
\uline{U}_0&=\uline{Z}_{11}\cdot \uline{I}_0=100\,\kilo\ohm\cdot 1{,}3\,\micro\ampere=\uuline{130{,}4\,\milli\volt}
|
||||
\intertext{Nicht gefragt:}
|
||||
\uline{I}_2&=\frac{1}{R_2}\cdot (\uline{Z}_{11}-\uline{Z}_{21})\cdot \uline{I}_0=\frac{1}{600\,\ohm}\cdot 2{,}1\mega\ohm\cdot 1{,}3\,\micro\ampere=4{,}55\,\milli\ampere >> \uline{I}_0\\
|
||||
\text{Probe }\\
|
||||
%&=\frac{1}{R_1}\cdot (\uline{I}_2+\uline{I}_0)\cdot \uline{U}_0\\
|
||||
\uline{U}_1&=1{,}5\,\volt=R_1\cdot (\uline{I}_0+\uline{I}_2)+\uline{I}_0\cdot Z_{11}\\
|
||||
&=300\,\ohm\cdot (0{,}0013+4{,}55)\,\milli\ampere+1{,}3\,\micro\ampere\cdot 100\,\kilo\ohm=1{,}498\,\volt
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\enlargethispage{1cm}
|
||||
Ist Ihnen aufgefallen, daß $\uline{Z}_{21}$ einen negativen Wert hat?\\[\baselineskip]
|
||||
Ein Widerstand nimmt elektrische Leistung auf, also ein Verbraucher.\\
|
||||
Dann muß also ein negativer Widerstand elektrische Leistung abgeben!
|
||||
Gibt es das in der Realität?\\[\baselineskip]
|
||||
Was steckt in dem Vierpol?\\[\baselineskip]
|
||||
Passiver Vierpol:\\
|
||||
Zum Beispiel mit 3 Widerständen in $T$ oder $\Pi$ Schaltung, ergibt jedoch keine Lösung!\\
|
||||
Aus der $\uline{Z}$ Matrix ist ersichtlich, daß $\uline{Z}_{21}=0$ und $\uline{Z}_{22}=0$ sind,
|
||||
somit existiert keine Kopplung vom Ausgang zum Eingang, sondern nur in Vorwärtsrichtung.\\[\baselineskip]
|
||||
Beispiel: Elektrodynamischer Lautsprecher der auf ein Kondensatormikrofon einwirkt. Kopplung nur vom Lautsprecher zum Mikrofon, nicht umgekehrt.\\[\baselineskip]
|
||||
Aktiver Vierpol: (Für Fortgeschrittene)\\
|
||||
Es muß sich um einen invertierenden Trennverstärker mit einem Eingangswiderstand von
|
||||
$100\,\kilo\ohm$ und einem Ausgangswiderstand von $2\,\mega\ohm$ handeln. Die Verstärkung ist zu berechnen.
|
||||
\begin{align*}
|
||||
\uline{U}_2&=\uline{U}_0-R_2\cdot \uline{I}_2=130{,}4\,\milli\volt-600\,\ohm\cdot 4{,}55\,\milli\ampere=-2{,}6\,\volt\\
|
||||
\uline{U}_q&=\uline{U}_2-\uline{Z}_{22}\cdot \uline{I}_2=-2{,}6\,\volt-2\,\mega\ohm\cdot 4{,}55\,\milli\ampere=-9102{,}6\,\volt\\
|
||||
V&=\frac{\uline{U}_q}{\uline{U}_0}=\frac{-9102{,}6\,\volt}{130{,}4\,\milli\volt}=-69805
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1.5]
|
||||
\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$R_1\cdot (\uline{I}_2+\uline{I}_0)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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]{\footnotesize$R_2\cdot \uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,black!25!, xshift=1.5cm, yshift=0cm]% Vierpol mit Innenleben
|
||||
\draw (0,-.2)rectangle(2.0,1.2);%node at(.75,.5)[right]{$(\uline{Z})$};
|
||||
\draw (0,0)--(-.5,0)(0,1)--(-.5,1)(2,0)--(2.5,0)(2,1)--(2.5,1);
|
||||
% \end{skope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90] %Z_11
|
||||
\draw [red!50!blue](0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667)
|
||||
[left] {$\uline{Z}_{11}$};
|
||||
\draw [<-,red!50!blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$\uline{Z}_{11}\cdot \uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Z_21
|
||||
\draw [red!50!blue](0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_{21}$};
|
||||
\draw [->,red!50!blue] (.3,-.2)--(.7,-.2)node at(.6,-.2)[below]{\footnotesize$-\uline{Z}_{21}\cdot \uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,red!50!blue,xshift=1cm,yshift=0cm,rotate=90]%Spannungsquelle|
|
||||
\draw (0,0)--(1,0);%;node at(.5,-.133)[right]{$V\cdot \uline{U}_0$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [->] (.3,-.2)--(.7,-.2) node at (.3,-.2)[right]{\footnotesize$V\cdot \uline{U}_q$};
|
||||
\draw node at (.3,.025)[left]{\footnotesize$+$};
|
||||
\draw node at (.7,.025)[left]{\footnotesize$-$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,red!50!blue,xshift=0cm,yshift=0cm]%
|
||||
\draw (-.5,0)--(2.5,0)(-.5,1)--(0,1)(2,1)--(2.5,1);
|
||||
\end{scope}
|
||||
\end{scope} % Ende Vierpol Innenleben
|
||||
\begin{scope}[>=latex,very thick, xshift=0, yshift=0]
|
||||
\draw (0,0)--(1,0);
|
||||
\draw (1,1)--(1,2)--(2,2) (3,2)--(4,2)--(4,1);
|
||||
\draw (1,0)--(1,-.5)--(4,-.5)--(4,0);
|
||||
\fill (0,0)circle(.05) (0,1)circle(.05) (1,0)circle(.05) (1,1)circle(.05) (4,0)circle(.05) (4,1)circle(.05);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]% Strompfeil
|
||||
\draw [->,red] (0,0)--(.4,0)node at(.2,0)[above]{\footnotesize$\uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm]% Strompfeil
|
||||
\draw [->,red] (0,0)--(-.4,0)node at(-.2,0)[above]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=-.2cm,yshift=1cm]% Strompfeil
|
||||
\draw [->,red] (0,0)--(.2,0)node at(.1,0)[above]{\footnotesize$\uline{I}_2+\uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(-.2,.5)[left]{\footnotesize$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(0,.25)[left]{\tiny$\uline{U}_0=\uline{Z}_{11}\cdot \uline{I}_0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,yshift=0cm]% Spannungspfeil
|
||||
\draw [->,blue] (0,.8)--(0,.2)node at(.2,.5)[right]{\footnotesize$\uline{U}_2=\uline{Z}_{21}\cdot \uline{I}_0$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{minipage}[t]{1\textwidth}
|
||||
\centering
|
||||
\includegraphics[width =0.6\textwidth]{b16a8a}
|
||||
\end{minipage}
|
||||
\clearpage
|
||||
}{}%
|
||||
186
ET2_L_B17_A1.tex
Normal file
186
ET2_L_B17_A1.tex
Normal file
@ -0,0 +1,186 @@
|
||||
\section{Stromortskurve}
|
||||
Konstruieren Sie die Stromortskurve $\uline{I}=\uline{g}(f)$ von dem abgebildeten Netzwerk!\\
|
||||
Entnehmen Sie der Stromortskurve den Strom $\uline{I}$ für $f_0=2,5\,\kilo\hertz$!\\[\baselineskip]
|
||||
Gegeben sind: $R_1=15\,\ohm$; $R_2=50\,\ohm$; $C=1{,}59\,\micro\farad$;\\ $0{,}5\,\kilo\hertz \leq f \leq 2{,}5\,\kilo\hertz$;\\
|
||||
$\uline{U}=10\,\volt$ = konstant\\[\baselineskip]
|
||||
Maßstäbe: $5\,\milli\siemens\,\widehat{=}\,1\,\centi\metre$ : $ 10\,\ohm\,\widehat{=}\, 1\,\centi\metre$ (Platzbedarf in x: $14\,\centi\metre$; in y: $12\,\centi\metre$)\\
|
||||
\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] {$R_1$};
|
||||
% \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=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,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] {$C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%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] (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]{$\underline{I}_C$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\intertext{Parallelschaltung $\uline{Y}_P$}
|
||||
\uline{Y}_P&=G_2+jB\\
|
||||
G_2&=\frac{1}{R_2}=\frac{1}{50\,\ohm}=20\,\milli\siemens\\
|
||||
f_u&=0{,}5\,\kilo\hertz\qquad f_o=2{,}5\,\kilo\hertz \qquad\text{(Anmerkung)}\\
|
||||
B_u&=\omega_u\cdot C=2\cdot \pi\cdot f_u=2\cdot \pi\cdot 0{,}5\,\kilo\hertz\cdot 1{,}59\,\micro\farad =5\,\milli\siemens\\
|
||||
B_o&=\omega_o\cdot C=25\,\milli\siemens\qquad\text{(5-facher Wert von $B_u$)}\\
|
||||
\uline{Y}_P&=20\,\milli\siemens+j(5\ldots 25)\,\milli\siemens\\
|
||||
\end{align*}
|
||||
\vspace{-1cm}
|
||||
\begin{align*}
|
||||
\begin{aligned}
|
||||
\uline{Y}_{P,min}&=G_2 & (\text{für } & f=0) \qquad
|
||||
&\uline{Z}_{P,min}&=0 \quad & (\text{für } & f\negmedspace\rightarrow\negmedspace\infty)\\
|
||||
\uline{Y}_{P,max}&=\infty & (\text{für } & f\negmedspace\rightarrow\negmedspace\infty) \qquad
|
||||
&\uline{Z}_{P,max}&=R_2 \quad &(\text{für } & f=0)\\
|
||||
\end{aligned}
|
||||
\end{align*}\\
|
||||
Zeichnen der Leitwertgeraden $\uline{Y}_P$; Konstruktion des Halbkreises für $\uline{Z}_P$
|
||||
\begin{align*}
|
||||
(G_2=20\,\milli\siemens\,\widehat{=}\,4\,\centi\metre : R_2 = 50\,\ohm\,\widehat{=}\, 5\,\centi\metre)
|
||||
\end{align*}
|
||||
Einzeichnen der $\uline{Z}_{P_o}$, $\uline{Z}_{P_u}$ Linien (Zeiger).\\
|
||||
\clearpage
|
||||
Berechnen der Gesamtschaltung $\uline{Y}$\\
|
||||
Graphisch wird der Widerstand $R_1$ addiert, durch verschieben der $\Im$ Achse um $1{,}5\,\centi\metre$ nach links.\\
|
||||
\begin{align*}
|
||||
\uline{Y}_{min}&=\frac{1}{R_1+R_2}=\frac{1}{65\,\ohm}=15{,}4\,\milli\siemens\,\widehat{=}\,3{,}08\,\centi\metre \qquad (f=0)\\
|
||||
\uline{Y}_{max}&=\frac{1}{R_1}=\frac{1}{15\,\ohm}=66{,}7\,\milli\siemens\,\widehat{=}\,13{,}33\,\centi\metre\qquad (f\rightarrow\infty)\\
|
||||
\end{align*}
|
||||
|
||||
Zeichnen des Leitwertkreises $\uline{Y}$ mit Radius $r$:
|
||||
\begin{align*}
|
||||
r&=\frac{\uline{Y}{max}-\uline{Y}_{min}}{2}\cdot \hspace{-1cm} \underbrace{\frac{1\,\centi\metre}{5\,\milli\siemens}}_{\text{Maßstabsumwandlung}}\hspace{-1cm}=5{,}12\,\centi\metre\\
|
||||
\end{align*}
|
||||
Der Mittelpunkt ergibt sich aus:
|
||||
\begin{align*}
|
||||
\uline{Y}_{min}+r=3{,}08\,\centi\metre+5{,}12\,\centi\metre=8{,}2\,\centi\metre
|
||||
\end{align*}
|
||||
zu messen ab neuer imaginärer Achse.\\
|
||||
|
||||
Einzeichnen der $\uline{Z}_o$, $\uline{Z}_u$ Linien (Zeiger).\\
|
||||
|
||||
Spiegeln der Zeiger $\uline{Z}_o$, $\uline{Z}_u$ an der reellen Achse liefert Schnittpunkt mit $\uline{Y}$. Diese sind entsprechend $\uline{Y}_o$, $\uline{Y}_u$.\\
|
||||
\clearpage
|
||||
$x=14\,\centi\metre; y=12\,\centi\metre$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](-2,-6)grid(12,6);
|
||||
\draw[thin](5,-6)--(5,6)(10,-6)--(10,6)(-2,5)--(12,5)(-2,5)--(12,5);
|
||||
\draw[thin,->](0,0)--(12.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,-6.5)--(0,6.5)node[above]{$\Im$};
|
||||
\draw[blue,->,ultra thick](0,0)--(5,0)node[below left]{$R_2$};
|
||||
\draw[blue,thin](2.5,0)circle(2.5cm);
|
||||
\fill[blue](2.5,0)circle(.075cm);
|
||||
\draw[red,->](0,0)--(4,0)node[below left]{$G_2$};
|
||||
\draw[red!50!blue,thin](6.67,0)circle(5.125cm);% Y-Kreis
|
||||
\fill[red!50!blue](6.67,0)circle(.075cm);
|
||||
\draw[red!50!blue](1.525,-.1)--(1.525,.1)(1.525,0)node[below]{$\uline{Y}_{min}$};
|
||||
\draw[red!50!blue](11.807,-.1)--(11.807,.1)(11.807,0)node[below]{$\uline{Y}_{max}$};
|
||||
\draw[red,thin](4,-6)--(4,6);
|
||||
\draw[red,|-|](4,1)--(4,5)node at (4.5,5){$f_o$};
|
||||
\draw[red]node at (4.5,1){$f_u$};
|
||||
\draw[red]node at (4.5,3){$\uline{Y}_P$};
|
||||
\draw[red]node at (4.5,-3){$\uline{Y}^*_P$};
|
||||
\draw[red]node at (6.5,4.5){$\uline{Y}$};
|
||||
\draw[red,|-|](4,-1)--(4,-5)node at (4.5,-5){$f_o$};
|
||||
\draw[red]node at (4.5,-0.75){$f_u$};
|
||||
\draw[blue,thin](0,0)--(8,-2)node at (3.0,-1.75){$\uline{Z}_P$; $\uline{Z}$};
|
||||
\draw[blue,thick,->](0,0)--(-14.036:4.851cm)node at(3,-.5){$\uline{Z}_{P_u}$};
|
||||
\draw[green!50!black,->](-1.5,0)--(-14.036:4.851cm)node at(3,-1.25){$\uline{Z}_u$};
|
||||
\draw[->,green!50!black](-1.5,0)--(1.9,-2.4)node at (-.5,-1){$\uline{Z}_o$};
|
||||
\draw[blue,thin](0,0)--(4,-5);
|
||||
\draw[->,blue](0,0)--(1.9,-2.4)node at(1.2,-1){$\uline{Z}_{P_o}$};
|
||||
\draw[blue,ultra thick](0:2.5cm)+(257:2.5cm)arc(257:332:2.5cm);%Mittelpunkt+Start arc Start:End:Radius
|
||||
\draw[blue]node at (5.5,-1.25){$f_u$};
|
||||
\draw[blue]node at (2,-2.75){$f_o$};
|
||||
\draw[blue!50!red,thin](-1.5,0)--(9.5,2)node at (2.5,1.5){$\uline{Y}$}; \draw[blue!50!red,thin](-1.5,0)--+(35.2:8.5cm);
|
||||
\draw[blue!50!red,ultra thick](0:6.67cm)+(148:5.125cm)arc(148:174.5:5cm);%Mittelpunkt+Start arc Start:End:Radius
|
||||
% \draw[blue!50!red]node at (1.25,.75){$f_u$};
|
||||
\draw[blue!50!red]node at (2,2.825){$f_o$};
|
||||
\draw[blue!20!red,ultra thick,->](-1.5,0)--+(35.2:4.72cm)node at (.5,2){$\uline{Y}(f_o)$};
|
||||
\draw[blue!50!red,ultra thick,->](-1.5,0)--+(10.3:3.15cm)node at (.5,.625){$\uline{Y}(f_u)$};
|
||||
\draw[blue!50!red]node at (5.5,1.25){$f_u$};
|
||||
% \draw[blue!50!red]node at (2.5,3.25){$f_o$};
|
||||
\draw[blue!50!red,thin,->](-1.5,-6.5)--(-1.5,6.5)node[above]{$\Im$ neu};
|
||||
\draw[blue!50!red,->](-1.5,0)--(0,0)node[below left]{$R_1$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
|
||||
|
||||
Ablesen der $\uline{Y}$ Werte und Maßstabsumrechnung $M=5\,\milli\siemens/\centi\metre$ ergibt Bereich zwischen $f_u$ und $f_o$.
|
||||
\begin{align*}
|
||||
\text{ablesen: } \uline{Y}(f_o)&=23{,}6\,\milli\siemens\cdot e^{(j35{,}2\,\degree)}\quad\text{aus Zeichnung $4{,}72\,\centi\metre$}\\
|
||||
\uline{I}&=\uline{Y}\cdot \uline{U}=\uuline{236\,\milli\ampere\cdot e^{(j35{,}2\,\degree)}}\\
|
||||
\uline{I}&=\uuline{(193+j136)\,\milli\ampere}\\
|
||||
\uline{Y}_P \text{ Gerade } \Rightarrow \uline{Z}_P \text{ Kreis } \\
|
||||
\uline{Z}_{P_u}, \uline{Z}_{P_o} \text{ Linie }\\
|
||||
R_1 \text{ addieren }\\
|
||||
\uline{Y}_{min} \text{ Kreis } \\
|
||||
\Rightarrow \uline{Y}(f_u); \uline{Y}(f_o)
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\subsubsection*{Kurzanleitung}
|
||||
(Siehe Merksätze zu Inversion im Script, hier graphisch veranschaulicht:)
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1.25]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw(0,-1)--(0,1)node[above]{$\Im$};
|
||||
\draw(0,0)--(2,0);
|
||||
\draw[red](1,-1)--(1,1);
|
||||
\draw[red,->](0,0)--(1,.75)node at(1.25,.75){$\uline{Y}_p$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]
|
||||
\draw(0,-1)--(0,1)node[above]{$\Im$};
|
||||
\draw(0,0)--(2,0);
|
||||
\draw[blue](.5,0)circle(.5cm);
|
||||
\draw[blue,->](0,0)--(.7,-.5cm)node at(1.25,-.5){$\uline{Z}_p$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=6cm,yshift=0cm]
|
||||
\draw(0,-1)--(0,1);
|
||||
\draw(-.5,0)--(2,0);
|
||||
\draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
|
||||
\draw[blue](.5,0)circle(.5cm);
|
||||
\draw[green!50!black,->](-.5,0)--(.7,-.5cm)node at(1,-.5){$\uline{Z}$}; \end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=9cm,yshift=0cm]
|
||||
\draw(0,-1)--(0,1);
|
||||
\draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
|
||||
\draw(-.5,0)--(2.5,0);
|
||||
\draw[blue](1.25,0)circle(1cm);
|
||||
\draw[blue!50!red,->](-.5,0)--(2.1,.5)node at(2.5,.5){$\uline{Y}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{enumerate}
|
||||
\item Leitwertortskuve der Parallelschaltung von $R_2$ und $C$ ergibt eine Gerade die im Abstand $G_2$ parallel zur imaginären Achse liegt.\\
|
||||
$G_2=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre$
|
||||
\item Grenzen $f_u: \uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre$, $f_o: \uline{Y}_P=25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre$ einzeichnen, auch für $\uline{Y}^*_P$
|
||||
%Grenzen eintragen für $\uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{ auch für $\uline{Y}^*_P$}$\\
|
||||
\item Inversion von $\uline{Y}_P$ liefert die Widerstandsortskurve $\uline{Z}_P$ ein Kreis durch den Ursprung.\\
|
||||
$\uline{Z}_{P_{min}}=0;\quad \uline{Z}_{P_{max}}=R_2=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\rightarrow\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}$
|
||||
\item Der Serienwiderstand $R_1$ wird durch verschieben der imaginären Achse addiert $R_1=15\,\ohm\,\widehat{=}\,1{,}5\,\centi\metre$ (Verschiebung nach links).
|
||||
\item Punktweise Inversion von $\uline{Z}$ liefert $\uline{Y}$ mit\\ $\uline{Y}_{min}=15{,}4\,\milli\siemens\,\widehat{=}\,3{,}08\,\centi\metre$ und $\uline{Y}_{max}=66{,}7\,\milli\siemens\,\widehat{=}\,13{,}3\,\centi\metre$,\\ wieder einen Kreis der jetzt nicht mehr durch den Ursprung geht.
|
||||
\item Ablesen von $\uline{Y}(f_o)=4{,}72\,\centi\metre$ und Winkel $35{,}2\,\degree \\ \Rightarrow \uline{Y}(f_o)=23{,}6\,\milli\siemens\cdot e^{(j35{,}2\,\degree)}$.
|
||||
\item Berechnen des Stroms \ldots \quad ;-)
|
||||
\end{enumerate}
|
||||
\clearpage
|
||||
}{}%
|
||||
94
ET2_L_B17_A2.tex
Normal file
94
ET2_L_B17_A2.tex
Normal file
@ -0,0 +1,94 @@
|
||||
\section{Leitwerts-, Widerstandsortskurve}
|
||||
Konstruieren Sie graphisch fur den dargestellten Zweipol
|
||||
die Leitwertsortskurve $\uline{Y}_1(p)$, die Widerstandsortskurve
|
||||
$\uline{Z}_1(p)$, und schlieslich die Widerstandsortskurve $\uline{Z}(p)$.\\
|
||||
Beziffern Sie jeweils die Punkte $p=0$; $p=1$; $p=3$ und den Grenzwert $p\rightarrow \infty$.\\[\baselineskip]
|
||||
Parameter $p$: $\omega=p\cdot \omega_0$ mit $\omega_0=1000\,\frac{1}{\second}$\\[\baselineskip]
|
||||
Maßstäbe: $2{,}5\,\milli\siemens\,\widehat{=}\,1\,\centi\metre$ : $10\,\ohm\,\widehat{=}\, 1\,\centi\metre$\\
|
||||
(Platzbedarf in x: $12\,\centi\metre$; in y: $14\,\centi\metre$)\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=.75cm]%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] {$100\,\ohm$};
|
||||
\draw [blue] node at(.5,-.125){\footnotesize$\underbrace{\phantom{\uline{Y}_1\text{; } \uline{Z}_1}}$};
|
||||
\draw [blue] node at(.5,-.2)[below]{\footnotesize$\uline{Y}_1\text{; } \uline{Z}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1.25cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$5\,\micro\farad$};
|
||||
\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] {$20\,\milli\henry$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
|
||||
\draw (-.5,0)--(0,0) (.2,1.25)--(0,1.25)--(0,.75)--(.2,.75) (.8,1.25)--(1,1.25)--(1,.75)--(.8,.75)(-.5,1)--(0,1) (-.5,0)--(2,0)--(2,1)--(1.8,1);
|
||||
\fill (-.5,0)circle(.025) (-.5,1)circle(.025);
|
||||
\draw [->,red] node at (-.5,.5)[left]{$\underline{Z}(p)\Rightarrow$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\uline{Y}_1(p)&=\frac{1}{R}+j\cdot p\cdot \omega_0\cdot C=(10+j\cdot p\cdot 5)\,\milli\siemens\\
|
||||
\uline{Z}_L(p)&=j\cdot p\cdot \omega_0\cdot L=+j\cdot p\cdot 20\,\ohm\\
|
||||
\uline{Z}(p)&=\uline{Z}_L(p)+\frac{1}{\uline{Y}_1(p)}
|
||||
\intertext{Nicht gefragt: Kontrollrechnung: (konjugiert komplex erweitern)}
|
||||
\uline{Z}(p)&=\uline{Z}_L(p)+\frac{1}{\uline{Y}_1(p)}=j\cdot p\cdot 20\,\ohm +\frac{1}{(10+j\cdot p\cdot 5)\,\milli\siemens}\\
|
||||
&=j\cdot p\cdot 20\,\ohm +\frac{1}{(10+j\cdot p\cdot 5)\,\milli\siemens}\cdot \frac{(10-j\cdot p\cdot 5)\cancel{\,\milli\siemens}}{(10-j\cdot p\cdot 5)\cancel{\,\milli\siemens}}\\
|
||||
&=\frac{10\cdot \power{10}{3}}{100+25\cdot p^2}\,\ohm+j\Big(20\cdot p-\frac{5\cdot \power{10}{3}\cdot p}{100+25\cdot p^2}\Big)\,\ohm
|
||||
\end{align*}
|
||||
\enlargethispage{1cm}
|
||||
%\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
%\centering
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,-8)grid(12,8);
|
||||
\draw[thin](5,-8)--(5,8)(10,-8)--(10,6)(0,5)--(12,5)(0,-5)--(12,-5);
|
||||
\draw[thin,->](0,0)--(12.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,-8)--(0,8.5)node[above]{$\Im$};
|
||||
\draw[red,->](0,0)--(4,0)node[above right]{$10\,\milli\siemens$};
|
||||
\draw[red](4,-8)--(4,8)node at (4,7.5)[right]{$\uline{Y}_1$};
|
||||
\draw[red]node at (4,-7.5)[right]{$\uline{Y}^*_1$};
|
||||
\draw [->](4.2,6.75)--(4.2,7.25)node at (4,6.5)[right]{$p$};
|
||||
\draw [->](4.2,-6.75)--(4.2,-7.25)node at (4,-6.5)[right]{$p$};
|
||||
\foreach \p in {-3,-2,...,3}
|
||||
\filldraw (4,2*\p)circle(.05cm)node at (4,2*\p)[below left]{$\p$};
|
||||
\draw[blue,thin](0,0)--(4,-6)(0,0)--(5,-5)(0,0)--(8,-4);
|
||||
\filldraw[blue](10,0)circle(0.05cm)node[above right]{$100\,\ohm$};
|
||||
\filldraw[blue](5,0)circle(0.05cm)node[below]{$M$};
|
||||
\draw[blue,thin](0:5cm)+(180:5cm)arc(180:360:5cm);%Mittelpunkt+Start arc Start:End:Radius
|
||||
\draw[blue,->](8,-4)--(8,-2)node[below left]{$\uline{Z}_L(1)$};
|
||||
\draw[blue,->](5,-5)--(5,-1)node[below left]{$\uline{Z}_L(2)$};
|
||||
\draw[blue,->](3.05,-4.6)--(3.05,1.4)node[below left]{$\uline{Z}_L(3)$};
|
||||
\filldraw node at (10,0)[below right]{$p=0$};
|
||||
\filldraw (8,-4)circle(.05cm)node at (8,-4)[below right]{$p=1$};
|
||||
\filldraw (5,-5)circle(.05cm)node at (5,-5)[below right]{$p=2$};
|
||||
\filldraw (3.05,-4.6)circle(.05cm)node at (3,-4.5)[below left]{$p=3$};
|
||||
\draw[color=blue!50!red, very thick,domain=0:5] plot[parametric,samples=100,id=ortskurve17-2] function{1000/(100+25*t*t),2*t-500*t/(100+25*t*t)};% Ortskurve Faktor 1/10 Ohm in cm;
|
||||
\draw[color=blue!50!red, very thick] node at (2.5, 4.25){$\uline{Z}(p)$};
|
||||
\draw[very thick] node at (13, 3){induktiv};
|
||||
\draw[very thick] node at (13, -3){kapazitiv};
|
||||
\draw[very thick] node at (10, 1){$f = 0$ rein ohmisch};
|
||||
\draw[very thick] node at (2, 6.75){$f\rightarrow\infty$ rein induktiv};
|
||||
\end{scope}
|
||||
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
% \draw[scale=0.5,domain=-3.141:3.141,smooth]
|
||||
%plot[parametric,id=parametric-example] function{t*sin(t),t*cos(t)};
|
||||
% \end{scope}
|
||||
\end{tikzpicture}
|
||||
%\end{align*}
|
||||
\vspace{-.5\baselineskip}
|
||||
\begin{enumerate}
|
||||
\item $\uline{Y}_1$ zeichnen mit $p$-Werten
|
||||
\item $\uline{Z}_{min}$, $\uline{Z}_{max}$ berechnen \\
|
||||
$\rightarrow \uline{Z}$-Halbkreis: $r=5\,\centi\metre$
|
||||
\item $p$-Werte auf $\uline{Z}$ einzeichnen
|
||||
\item $\uline{Z}_L(p)$ punktweise addieren
|
||||
\item $\uline{Z}(p)$ Kurve zeichnen
|
||||
\end{enumerate}
|
||||
\clearpage
|
||||
}{}%
|
||||
130
ET2_L_B17_A3.tex
Normal file
130
ET2_L_B17_A3.tex
Normal file
@ -0,0 +1,130 @@
|
||||
\section{Ortskurve}
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\setlength{\itemsep}{-0.2\baselineskip}
|
||||
\item Zeichnen Sie maßstäblich die Ortskurve für das Spannungsverhältnis $U_2/U_1$ in Abhängigkeit von der Frequenz $f$.
|
||||
\item Geben Sie die Grenzfrequenz der Schaltung an.
|
||||
\item Wie groß ist die Dämpfung des Vierpols für die Frequenz $f=1{,}2\,\kilo\hertz$ (falls in Vorlesung behandelt: in $\deci\bel$)
|
||||
\end{enumerate}
|
||||
$\qquad \,\, R=16\,\kilo\ohm$; $C=12\,\nano\farad$; $200\,\hertz\leq f \leq 1{,}2\,\kilo\hertz$
|
||||
\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] {$R$};
|
||||
\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=0cm]%Knotenpunkte
|
||||
\draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.5,0)--(1.5,0) (.8,1)--(1.5,1);
|
||||
\draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}_1$}; \fill (-.5,0)circle(.025) (-.5,1)circle(.025);
|
||||
\draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}_1$};
|
||||
\draw [->,blue] (1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$\underline{U}_2$}; \fill (1.5,0)circle(.025) (1.5,1)circle(.025);
|
||||
\draw [->,red] (1.4,1.1)--(1.1,1.1) node at (1.25,1.1)[above]{$\underline{I}_2=0$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung: (Platzbedarf in x: $5\,\centi\metre$; in y: $15\,\centi\metre$)\\[0.5\baselineskip]
|
||||
a) Spannungsverhältnis $U_2/U_1$ in Abhängigkeit von der Frequenz $f$
|
||||
\begin{align*}
|
||||
\uline{I}_1&=\uline{U}_1 \cdot \uline{Y} = \uline{U}_1 \cdot \frac{j\omega\cdot C}{1+j\omega\cdot C\cdot R}=\frac{\uline{U}_1}{R+\frac{1}{j\omega\cdot C}} \\
|
||||
\uline{U}_2&= \uline{I}_1\cdot \frac{1}{j\omega\cdot C}= \frac{\uline{U}_1}{(R+\frac{1}{j\omega\cdot C})\cdot j\omega\cdot C}=
|
||||
\frac{\uline{U}_1}{1+j\omega\cdot R\cdot C}\\
|
||||
\frac{\uline{U}_2}{\uline{U}_1}&=\frac{1}{1+j\omega\cdot R \cdot C}
|
||||
=\frac{1-j\omega RC}{1+(\omega RC)^2}\\[0.5\baselineskip]
|
||||
&\begin{tabular}{|l|l|l|}
|
||||
\hline
|
||||
% after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
|
||||
$f/Hz$&$2\pi\cdot f\cdot R\cdot C$&$\uline{U}_2/\uline{U}_1$\\
|
||||
\hline
|
||||
200 & 0{,}241&0{,}945-j0{,}228\\
|
||||
828{,}9&1&0{,}5-j0{,}5\\
|
||||
1200&1{,}448&0{,}323-j0{,}468\\
|
||||
\hline
|
||||
\end{tabular}
|
||||
\intertext{b) Grenzfrequenz}
|
||||
\omega_g\cdot R\cdot C&=1\\
|
||||
f_g&=\frac{1}{2\pi\cdot R\cdot C}=\uuline{828{,}9\,\hertz}
|
||||
\intertext{c) Dämpfung}
|
||||
f&=1{,}2\,\kilo\hertz\qquad a=20\cdot \lg\frac{\uline{U}_2}{\uline{U}_1}\\
|
||||
\frac{\uline{U}_2}{\uline{U}_1}&=0{,}323-j0{,}468=0{,}569\cdot e^{-j50{,}56\,\degree}\\
|
||||
a&=20\cdot \lg(0{,}569)=\uuline{-4{,}9\,\deci\bel}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,-7.5)grid [step=.5cm](5,7.5);
|
||||
\draw[thin,->](0,0)--(5.5,0)node[right]{$\Re \, \left\{\frac{\uline{U}_2}{\uline{U}_1}\right\}$};
|
||||
\draw[thin,->](0,-8)--(0,8)node[above]{$\Im \, \left\{\frac{\uline{U}_2}{\uline{U}_1}\right\}$};
|
||||
\foreach \y in {1.5,1,...,-1.5}
|
||||
\draw(0,5*\y)--(-.1,5*\y)node[left]{$\y$};
|
||||
\draw[red,thin](5,-7.5)--(5,7.5);
|
||||
\draw[red](4.9,5)--(5.1,5)node [right]{$f_g=828{,}9\,\hertz$};
|
||||
\draw[red](4.9,-5)--(5.1,-5)node [right]{$f_g=828{,}9\,\hertz$};
|
||||
\draw[red](4.9,1.205)--(5.1,1.205)node [right]{$f=200\,\hertz$};
|
||||
\draw[red](4.9,-1.205)--(5.1,-1.205)node [right] {$f=200\,\hertz$};
|
||||
\draw[red](4.9,7.24)--(5.1,7.24)node [right] {$f=1200\,\hertz$};
|
||||
\draw[red](4.9,-7.24)--(5.1,-7.24)node [right] {$f=1200\,\hertz$};
|
||||
\draw node at(6.125,6.2){$(1+j\omega RC)$};
|
||||
\foreach \x in {.5,1}
|
||||
\draw(5*\x,.1)--(5*\x,-.1)node at (5*\x+.1,-.25){$\x$};
|
||||
\fill[blue](2.5,0)circle(.05cm);
|
||||
\draw[blue](0:2.5cm)+(180:2.5cm)arc(180:360:2.5cm);%Mittelpunkt+Start arc Start:End:Radius
|
||||
\draw[blue!50!red,thin](0,0)--(5,-1.205)(0,0)--(5,-5)(0,0)--(5,-7.24);
|
||||
\filldraw[blue!50!red](4.73,-1.14)circle(.05)node [below left]{$200\,\hertz$};
|
||||
\filldraw[blue!50!red](2.5,-2.5)circle(.05)node [below]{$f_g$};
|
||||
\filldraw[blue!50!red](1.62,-2.35)circle(.05)node [below left]{$1{,}2\,\kilo\hertz$};
|
||||
\draw [blue] node at(4,-2.5){$\frac{\underline{U}_2}{\underline{U}_1}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
% (Siehe Merksätze zu Inversion im Script, hier graphisch veranschaulicht:)
|
||||
%\begin{align*}
|
||||
% \begin{tikzpicture}[very thick,scale=1.25]
|
||||
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
% \draw(0,-1)--(0,1)node[above]{$\Im$};
|
||||
% \draw(0,0)--(2,0);
|
||||
% \draw[red](1,-1)--(1,1);
|
||||
% \draw[red,->](0,0)--(1,.75)node at(1.25,.75){$\uline{Y}_p$};
|
||||
% \end{scope}
|
||||
% \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]
|
||||
% \draw(0,-1)--(0,1)node[above]{$\Im$};
|
||||
% \draw(0,0)--(2,0);
|
||||
% \draw[blue](.5,0)circle(.5cm);
|
||||
% \draw[blue,->](0,0)--(.7,-.5cm)node at(1.25,-.5){$\uline{Z}_p$};
|
||||
% \end{scope}
|
||||
% \begin{scope}[>=latex,very thick,xshift=6cm,yshift=0cm]
|
||||
% \draw(0,-1)--(0,1);
|
||||
% \draw(-.5,0)--(2,0);
|
||||
% \draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
|
||||
% \draw[blue](.5,0)circle(.5cm);
|
||||
% \draw[green!50!black,->](-.5,0)--(.7,-.5cm)node at(1,-.5){$\uline{Z}$}; \end{scope}
|
||||
% \begin{scope}[>=latex,very thick,xshift=9cm,yshift=0cm]
|
||||
% \draw(0,-1)--(0,1);
|
||||
% \draw[blue!50!red](-.5,-1)--(-.5,1)node[above]{$\Im$ neu};
|
||||
% \draw(-.5,0)--(2.5,0);
|
||||
% \draw[blue](1.25,0)circle(1cm);
|
||||
% \draw[blue!50!red,->](-.5,0)--(2.1,.5)node at(2.5,.5){$\uline{Y}$};
|
||||
% \end{scope}
|
||||
% \end{tikzpicture}
|
||||
%\end{align*}
|
||||
%Leitwertsortskuve der Parallelschaltung von $R_2$ und $C$ ergibt eine Gerade die im Abstand $G_2$ parallel zur imaginären Achse liegt.\\
|
||||
%$G_2=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre$\\
|
||||
%Grenzen eintragen für $\uline{Y}_P=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{ auch für $\uline{Y}^*_P$}$\\
|
||||
%Inversion von $\uline{Y}_P$ liefert die Widerstandsortskurve $\uline{Z}_P$ ein Kreis durch den Ursprung.\\
|
||||
%$R_2=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}$\\
|
||||
%Der Serienwiderstand $R_1$ wird durch verschieben der Koordinaten hinzugefügt.\\
|
||||
%Punktweise Inversion von $\uline{Z}$ ein Kreis der jetzt nicht mehr durch den Ursprung geht, liefert $\uline{Y}$, wieder einen Kreis.\\
|
||||
%$\uline{Y}\quad\text{Kreis um P(6.67,0) $r=5\,\centi\metre$, da $\uline{Y}_{min}$ und $\uline{Y}_{max}$ bekannt sind}$\\
|
||||
%\begin{align*}
|
||||
%R_2&=50\,\ohm\,\widehat{=}\, 5\,\centi\metre \quad\text{Kreis um P(2.5,0) $r=2{,}5\,\centi\metre$}\\
|
||||
%G_2&=20\,\milli\siemens\,\widehat{=}\, 4\,\centi\metre\\
|
||||
%\uline{Y}_P&=5\,\milli\siemens\,\widehat{=}\, 1\,\centi\metre (f_u)\quad 25\,\milli\siemens\,\widehat{=}\, 5\,\centi\metre (f_o)\quad\text{zeichnen, auch $\uline{Y}^*_P$}\\
|
||||
%\uline{Y}&\quad\text{Kreis um P(6.67,0) $r=5\,\centi\metre$, da $\uline{Y}_{min}$ und $\uline{Y}_{max}$ bekannt sind}\\
|
||||
%\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
104
ET2_L_B17_A4.tex
Normal file
104
ET2_L_B17_A4.tex
Normal file
@ -0,0 +1,104 @@
|
||||
\section{Stromortskurve}
|
||||
Konstruieren Sie die Stromortskurve $\uline{I}=f(p)$ zu der abgebildeten Schaltung\\
|
||||
für $0\leq p\leq 1$ !\\
|
||||
Es ist $Z_{RL}=p(R_0+jX_{L_0})$. Die Parameterwerte $p=0$; $0{,}25$; $0{,}5$; $0{,}75$ und $1$ sind zu markieren.\\
|
||||
Für welches $p$ wird $I=I_{max}$? Geben Sie diesen Stromwert an.\\
|
||||
Gegeben sind: $\uline{U}=U=10\,\volt$; $X_C=-3\,\kilo\ohm$; $R_0=6\,\kilo\ohm$; $X_{L_0}=8\,\kilo\ohm$.\\
|
||||
Maßstäbe: $1\,\kilo\ohm\,\widehat{=}\,1\,\centi\metre $; $50\,\micro\second\,\widehat{=}\, 1\,\centi\metre$\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.10) [above] {$C$};
|
||||
\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] {$R_{\phantom{L}}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.45,.0667) [above] {$X_{L}$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
|
||||
\draw (0,0)--(3,0)--(3,1)--(2.9,1);
|
||||
\draw [->,blue] (0,.9)--(0,.1)node at(0,.5)[right]{$\underline{U}$}; \fill (0,0)circle(.025) (0,1)circle(.025);
|
||||
\draw [->,red] (0,1.1)--(.4,1.1) node at (.25,1.1)[above]{$\underline{I}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick]%Variablen Pfeile
|
||||
\draw[->] (1.3,.75)--(1.7,1.25);
|
||||
\draw[->] (2.3,.75)--(2.7,1.25);
|
||||
\draw[dashed] (1.3,.75)--(2.3,.75);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung: (Platzbedarf in x: $11\,\centi\metre$; in y: $12\,\centi\metre$)\\[0.5\baselineskip]
|
||||
\begin{align*}
|
||||
\uline{I}(p)&=\uline{Y}(p)\cdot \uline{U}\\
|
||||
\intertext{$\uline{Z}(p)$ durch Vektoraddition der Widerstände zeichnen,}
|
||||
Z(p)&=\sqrt{R_O^2+X^2_{LO}}=\sqrt{6^2+8^2}\,\kilo\ohm=10\,\kilo\ohm\qquad \text{für }p=1
|
||||
\intertext{$\uline{Z}^*(p)$ durch Spiegelung an der reellen Achse zeichnen und Parameter $p$ einzeichnen.}
|
||||
X_C&=-3\,\kilo\ohm\,\widehat{=}\,-3\,\centi\metre\\
|
||||
R_0&=6\,\kilo\ohm\,\widehat{=}\,6\,\centi\metre\\
|
||||
X_{L_0}&=8\,\kilo\ohm\,\widehat{=}\,8\,\centi\metre\\
|
||||
\intertext{Senkrechte zu $\uline{Z}^*(p)$ durch den Ursprung zeichnen}
|
||||
\overline{0N}&=1{,}8\,\centi\metre\,\widehat{=}\,1{,}8\,\kilo\ohm\\
|
||||
\intertext{Invertieren ergibt Durchmesser des Kreises} \overline{0D}&=\frac{1}{\overline{0N}}=\frac{1}{1{,}8\,\kilo\ohm}=555{,}5\,\micro\siemens
|
||||
\,\widehat{=}\,11{,}1\,\centi\metre\\
|
||||
\intertext{Mittelpunkt bestimmen} \overline{0M}&=\frac{1}{2}\,\,\overline{0D}\,\widehat{=}\,5{,}55\,\centi\metre\\
|
||||
\text{$\uline{Y}(p)$ Kreis zeichnen. Max. Strom bei größtem Leitwert im Punkt D\newline (Durchmesser des Kreises = max. Abstand vom Ursprung)}
|
||||
\intertext{Ablesen von $p=0{,}24$ (Abstand zwischen $N(OD\,\cap\,\uline{Z}^*(p))$ und $\uline{Z}^*(p)|_{p=0}$)}
|
||||
%\uline{Z}^*(p)\text{ gibt }\Delta p=0{,}1\,\kilo\ohm\,\widehat{=}1\,\centi\metre,\text{ auf }
|
||||
%I_{max}\text{ für }p&=0{,}24\\
|
||||
I_{max}&=\uline{Y}(p)\cdot \uline{U}=555{,}5\,\micro\siemens\cdot 10\,\volt=\uuline{5{,}55\,\milli\ampere}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,-5)grid(10,10);
|
||||
\draw[thin,->](0,0)--(10.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,-5.5)--(0,10.5)node[above]{$\Im$};
|
||||
\foreach \y in {10,9,...,-5}
|
||||
\draw(0,\y)--(-.1,\y)node[left]{$\y$};
|
||||
\draw[red,->](0,0)--(0,-3)node at(.5,-1.5){$-jX_C$};
|
||||
\draw[red,->](0,-3)--(6,-3)node at(3,-2.75){$R$};
|
||||
\draw[red,->](6,-3)--(6,5)node at(6.5,1){$jX_{LO}$};
|
||||
\draw[red,->](0,-3)--(6,5)node at(4,1.25){$\uline{Z}(p)$};
|
||||
\draw[red,->](0,3)--(6,-5)node at(6,-4){$\uline{Z}^*(p)$};
|
||||
\draw[black!35!,thin](0,1.5)circle(1.5cm);
|
||||
\draw[blue,thin](0,0)--(36.87:11.1cm)node at(9.1,6.75){D};
|
||||
\draw[blue]node at(9,7.5){$\uline{Y}(p)$};
|
||||
\filldraw[blue](36.87:5.55)circle(0.05cm)node[left]{M};
|
||||
\filldraw[blue](0,0)--(8,6)node at(1,1.1){N};
|
||||
\draw[blue](36.87:5.55)circle(5.55cm);
|
||||
\filldraw[red!50!blue](0,3)circle(0.05cm)node [right]{\footnotesize{$p=0$}};
|
||||
\filldraw[red!50!blue](0,3)++(-53.13:2.5cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}25$}};
|
||||
\filldraw[red!50!blue](0,3)++(-53.13:5cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}5$}};
|
||||
\filldraw[red!50!blue](0,3)++(-53.13:7.5cm)circle(0.05cm)node [above right]{\footnotesize{$p=0{,}75$}};
|
||||
\filldraw[red!50!blue](0,3)++(-53.13:10cm)circle(0.05cm)node [right]{\footnotesize{$p=1$}};
|
||||
\filldraw[green!50!black](0,-3)circle(0.05cm)node [below right]{\footnotesize{$p=0$}};
|
||||
\filldraw[green!50!black](0,-3)++(53.13:2.5cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}25$}};
|
||||
\filldraw[green!50!black](0,-3)++(53.13:5cm)circle(0.05cm)node [below right]{\footnotesize{$p=0{,}5$}};
|
||||
\filldraw[green!50!black](0,-3)++(53.13:7.5cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}75$}};
|
||||
\filldraw[green!50!black](0,-3)++(53.13:10cm)circle(0.05cm)node [left]{\footnotesize{$p=1$}};
|
||||
\draw[red!50!blue,very thin](0,0)--(6,-5);
|
||||
\draw[red!50!blue,very thin](0,0)--(4.5,-3);
|
||||
\draw[red!50!blue,very thin](0,0)--(6,-2);
|
||||
\draw[red!50!blue,very thin](0,0)--(12,8);
|
||||
\draw[red!50!blue,very thin](0,0)--(0,6.7);
|
||||
\filldraw[red!50!blue,very thin](-40.5:2.45cm)circle(0.05cm)node [right]{\footnotesize{$p=1$}};
|
||||
\filldraw[red!50!blue,very thin](-34.4:3.6cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}75$}};
|
||||
\filldraw[red!50!blue,very thin](6,-2)circle(0.05cm)node [right]{\footnotesize{$p=0{,}5$}};
|
||||
\filldraw[red!50!blue,very thin](33.69:11.1cm)circle(0.05cm)node [right]{\footnotesize{$p=0{,}25$}};
|
||||
\filldraw[red!50!blue,very thin](90:6.666cm)circle(0.05cm)node [right]{\footnotesize{$p=0$}};
|
||||
\foreach \x in {0,1,...,10}
|
||||
\filldraw(\x,.1)--(\x,-.1)node at (\x,-.33){$\x$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Reihenfolge: $j\underline{X}_C; R; j\underline{X}_L; \underline{Z}(p)$; p-Werte; $\bot \underline{Z^*}(p)
|
||||
\Rightarrow \overline{ON}; \\
|
||||
\overline{OD}$ Durchmesser; Kreis um $\overline{OM} \Rightarrow \underline{Y(p)}$
|
||||
\clearpage
|
||||
}{}%
|
||||
123
ET2_L_B17_A5.tex
Normal file
123
ET2_L_B17_A5.tex
Normal file
@ -0,0 +1,123 @@
|
||||
\section{Widerstandstransformation}
|
||||
Ein Verbraucher mit $\uline{Z}_v=(6+j4)\,\kilo\ohm$ soll mit Hilfe von zwei Blindwiderständen so an eine Spannungsquelle mit dem Innenwiderstand $\uline{Z}_i=(3+j1{,}5)\,\kilo\ohm$ angepasst werden, dass er die größtmögliche Wirkleistung aufnimmt.\\
|
||||
Bestimmen Sie zeichnerisch die hinzuzuschaltenden Blindwiderstände (Art und Größe) einer möglichen Schaltung und skizzieren Sie ihre Zusammenschaltung mit $\uline{Z}_v$.\\
|
||||
Maßstab: $1\,\kilo\ohm \,\widehat{=}\,1\,\centi\metre$\\[\baselineskip]
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung: (Platzbedarf in x: $10\,\centi\metre$; in y: $20\,\centi\metre$) \\
|
||||
($14\,\centi\metre$ in y: reichen auch, wenn eine Linie durch die Rechnung geht)\\[0.5\baselineskip]
|
||||
Anpassung: Der transformierte Widerstand muss gleich dem konjugiert komplexen Innenwiderstand sein.\\[.5\baselineskip]
|
||||
$\uline{Z}^*_i=(3-j1{,}5)\,\kilo\ohm$\\[\baselineskip]
|
||||
\begin{minipage}[t]{0.45\textwidth}
|
||||
1. Möglichkeit
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_s$};
|
||||
\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] {$L_p$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Z}_v$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
|
||||
\draw (0,0)--(2,0)--(2,.1)(.9,1)--(2,1)--(2,.9);
|
||||
\draw node at(-.5,.5)[right]{$\underline{Z}^*_i\Rightarrow$};
|
||||
\fill (0,0)circle(.025) (0,1)circle(.025) (1,0)circle(.025) (1,1)circle(.025);
|
||||
\end{scope}
|
||||
\draw node at(1.5,-.25){$\underbrace{\phantom{xxxxxxxxxx}}_{\uline{Z}_p'}$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Imaginärteil anpassen:
|
||||
\begin{align*}
|
||||
\frac{1}{\underline{Z}'}&=\frac{1}{\underline{Z}_{v}}+\frac{1}{X_L{_p}}\\
|
||||
\intertext{Jetzt Zeichnung anfertigen um $X_{vp}$ und $X'_p$ zu bestimmen.}
|
||||
\frac{1}{X_L{_p}}&=\left(\overbrace{\frac{1}{6{,}3\,\kilo\ohm}}^{1/X'_p}-\overbrace{\frac{1}{13\,\kilo\ohm}}^{1/X_{vp}}\right)\\
|
||||
X_L{_p}&=\uuline{12{,}2\,\kilo\ohm}\\
|
||||
X_{C_s}&=\uuline{-5{,}6\,\kilo\ohm}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}[t]{0.45\textwidth}
|
||||
2. Möglichkeit
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_s$};
|
||||
\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] {$C_p$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$\uline{Z}_v$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
|
||||
\draw (0,0)--(2,0)--(2,.1)(.9,1)--(2,1)--(2,.9);
|
||||
\draw node at(-.5,.5)[right]{$\underline{Z}^*_i\Rightarrow$};
|
||||
\fill (0,0)circle(.025) (0,1)circle(.025) (1,0)circle(.025) (1,1)circle(.025);
|
||||
\end{scope}
|
||||
\draw node at(1.5,-.25){$\underbrace{\phantom{xxxxxxxxxx}}_{\uline{Z}_p''}$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
\frac{1}{\underline{Z}''_p}&=\frac{1}{\underline{Z}_{v}}+\frac{1}{X_{C_p}}\\
|
||||
\intertext{Im weiteren Schritt Zeichnung vervollständigen um $X''_p$ zu bestimmen.}
|
||||
\frac{1}{X_{C_p}}&=\left(\overbrace{\frac{1}{-6{,}3\,\kilo\ohm}}^{1/X''_p}-\overbrace{\frac{1}{13\,\kilo\ohm}}^{1/X_{vp}}\right)\\
|
||||
{X_{L_s}}&=\uuline{+2{,}6\,\kilo\ohm}\\
|
||||
X_{C_p}&=\uuline{-4{,}2\,\kilo\ohm}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
%\vfill
|
||||
\newpage
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,-7)grid(10,13);
|
||||
\draw[thin,->](0,0)--(10.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,-5.5)--(0,13.5)node[above]{$\Im$};
|
||||
\foreach \y in {13,12,...,-7}
|
||||
\draw(0,\y)--(-.1,\y)node[left]{$\y$};
|
||||
\foreach \x in {0,1,...,10}
|
||||
\draw(\x,0)--(\x,0)node[below]{$\x$};
|
||||
\draw[red,->](0,0)--(6,4)node at(3.5,2){$\uline{Z}_v$};
|
||||
\draw[black!50!,thin](6,4)+(214:.5cm)arc(214:304:.5cm);
|
||||
\fill[black!50!,thin](6,4)+(259:.25cm)circle(.05cm);
|
||||
\draw[red,->](0,0)--(3,-1.5)node at(1.75,-1.25){$\uline{Z}^*_i$};
|
||||
\filldraw[blue](4.3,0)circle(0.05cm)node[above]{M $\footnotesize(4.33)$};
|
||||
\draw(4.33,0)circle(4.33cm);
|
||||
\draw[blue](8.66,0)--(0,13)node [below right]{$X_{vp}\approx{13\,\kilo\ohm}$};
|
||||
\draw[blue,thick](8.66,0)--(0,6.3)node [right]{$X'_p=6{,}3\,\kilo\ohm$};
|
||||
\draw[blue,thick](8.66,0)--(0,-6.3)node [right]{$X''_p=-6{,}3\,\kilo\ohm$};
|
||||
\draw[red!50!blue,->,ultra thick](4.33,0)+(67.3:4.33cm)arc(67.3:108:4.33cm)node at(4.5,4.75){$X_{L_p}$};
|
||||
\draw[green!50!blue,->,ultra thick](4.33,0)+(67.3:4.33cm)arc(67.3:-108:4.33cm)node at(4.5,-4.75){$X_{C_p}$};
|
||||
\draw[red!50!blue,->,ultra thick](4.33,0)+(108:4.33cm)--(3,-1.5)node at(2.5,2.5){$X_{C_s}$};
|
||||
\draw[green!50!blue,->,ultra thick](4.33,0)+(-108:4.33cm)--(3,-1.5)node at(2.5,-2.5){$X_{L_s}$};
|
||||
\draw[magenta,->,ultra thick](0,0)--(3,4.14)node at(1,2){$\uline{Z}'$};
|
||||
\draw[green!75!red,->,ultra thick](0,0)--(3,-4.14)node at(1,-2){$\uline{Z}''$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=-.5cm]
|
||||
\fill[white](5,7)rectangle(10,13.5);
|
||||
\draw node at (5,13.25)[right]{1. Möglichkeit:};
|
||||
\draw[red] node at (5,12.75)[right]{1) $Z_v=(6+j4)\,\kilo\ohm$ und $\uline{Z}^*_i=(3-j1{,}5)\,\kilo\ohm$};
|
||||
\draw[blue] node at (5,12.25)[right]{2) $X_{vp}\bot Z_v\rightarrow X_{vp}\cap \Im\rightarrow X_{vp}\approx 13\,\kilo\ohm$};
|
||||
\draw node at (5,11.75)[right]{3) Kreis mit {\O}: $[0; X_{vp}\cap\Re]\qquad M(4.33)$};
|
||||
\draw[red!50!blue] node at (5,11.25)[right]{4) $L ||$ geschaltet $\rightarrow$ Kreissegment bis $\Re(\uline{Z}^*_i)$};
|
||||
\draw[red!50!blue] node at (5,10.75)[right]{5) $X_{C_s}$};
|
||||
\draw[blue] node at (5,10.25)[right]{6) $X'_p=6{,}3\,\kilo\ohm$};
|
||||
\draw[magenta] node at (5,9.75)[right]{7) $Z'$};
|
||||
\draw node at (5,9.25)[right]{2. Möglichkeit: (1--3 wie 1. Möglichkeit)};
|
||||
\draw[green!50!blue] node at (5,8.75)[right]{8) $X_{C_p}$ Kreissegment bis $\Re(\uline{Z}^*_i)$};
|
||||
\draw[green!50!blue] node at (5,8.25)[right]{9) $X_{L_s}=2{,}6\,\kilo\ohm$ abgelesen};
|
||||
\draw[blue] node at (5,7.75)[right]{10) $X''_p=-6{,}3\,\kilo\ohm$};
|
||||
\draw[green!75!red] node at (5,7.25)[right]{11) $Z''$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
97
ET2_L_B17_A6.tex
Normal file
97
ET2_L_B17_A6.tex
Normal file
@ -0,0 +1,97 @@
|
||||
\section{Brückenschaltung}
|
||||
Gegeben:\\[\baselineskip]
|
||||
$R_1=R_2=R_3=|X_C|=1\,\kilo\ohm$; $U=200\,\volt\cdot e^{j0\degree}$\\[\baselineskip]
|
||||
Gesucht:
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Spannung $U_{ab}$ zwischen den Klemmen $a-b$ nach Betrag und Phasenwinkel.
|
||||
\item Qualitatives Zeigerdiagramm aller Ströme und Spannungen.\\
|
||||
(Qualitativ, d.h. alle Bauteilwerte verschieden)
|
||||
\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] {$R_1$};
|
||||
\draw [->,red] (.95,-.1)--(.75,-.1) node at (.85,-.1)[right]{\footnotesize$I_{R}$};
|
||||
\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] {$R_3$};
|
||||
\draw [->,red] (.95,-.1)--(.75,-.1) node at (.85,-.1)[right]{\footnotesize$I_{C}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=.5cm,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] {$R_2$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{R_2}$};
|
||||
\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] {$C$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{C}$};
|
||||
\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]{$\underline{U}_{ab}$};
|
||||
\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]{$\underline{U}$};
|
||||
\draw node at (-.75,1){$+$};
|
||||
\draw node at (-.75,-1){$-$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung: (Platzbedarf in x: $11\,\centi\metre$; in y: $6\,\centi\metre$)\\
|
||||
a) Spannung $U_{ab}$
|
||||
\begin{align*}
|
||||
%R_i&=|X_c|=1\,\kilo\ohm\\
|
||||
%\uline{U}&=200\,\volt\\
|
||||
\uline{U}_{ab}&+\uline{U}_C-\uline{U}_{R_2}=0 \qquad \text{Masche}\\
|
||||
\uline{U}_{ab}&=\uline{U}_{R_2}-\uline{U}_C
|
||||
\intertext{Spannungsteiler}
|
||||
\uline{U}_{R_2}&=\uline{U}\cdot \frac{R}{2\cdot R}=\frac{\uline{U}}{2}=100\,\volt\\
|
||||
\uline{U}_C&=\uline{U}\cdot \frac{jX_C}{R+jX_C}=200\,\volt\cdot \frac{-j\cdot 1}{1-j\cdot 1}=200\,\volt(0{,}5-j0{,}5)=(100-j100)\,\volt\\
|
||||
\Rightarrow\uline{U}_{ab}&=100\,\volt-(100-j100)\,\volt=+j100\,\volt=\uuline{100\,\volt\cdot e^{+j90\degree}}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{b) Qualitatives Zeigerdiagramm (zu Schaltbild, beliebiges $X_C$)}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,0)grid(10,5);
|
||||
\draw[thin,->](0,0)--(10.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,0)--(0,5.5)node[above]{$\Im$};
|
||||
\foreach \x in {10,9,...,0}
|
||||
\draw(\x,0)--(\x,-.1)node[below]{$\x$};
|
||||
\foreach \y in {5,4,...,0}
|
||||
\draw(0,\y)--(-.1,\y)node[left]{$\y$};
|
||||
\draw[blue,->](0,0)--(10,0)node at(11.5,.375){$\uline{U}=\uline{U}_{R_1}+\uline{U}_{R_2}$};
|
||||
\draw[blue,->](0,0)--(5,0)node at(3.75,.25){$\uline{U}_{R_1}$};
|
||||
\draw[blue,->]node at(7.5,.25){$\uline{U}_{R_2}$};
|
||||
\draw[black!50!,thin](5,0)+(0:5cm)arc(0:180:5cm);
|
||||
\draw[black!50!,thin](2,4)+(-116.565:.5cm)arc(-116.565:-26.565:.5cm);
|
||||
\fill[black!50!,thin](2,4)+(-71.565:.25cm)circle(.05cm);
|
||||
\draw[blue!50!red,->](0,0)--(2,4)node at (1,3.5){$\uline{U}_{R_3}$};
|
||||
\draw[blue!50!red,->](2,4)--(10,0)node at (7,2){$\uline{U}_C$};
|
||||
\draw[blue!50!red,->](5,0)--(2,4)node at (4.5,1.5){$\uline{U}_{ab}$};
|
||||
\draw[red,->](0,0)--(1.5,3)node at (.45,1.5){$\uline{I}_{C}$};
|
||||
\draw[red,->](1.5,3)--(4,3)node at (2.5,2.75){$\uline{I}_{R}$};
|
||||
\draw[red,->](0,0)--(2.5,0)node at (1.5,.25){$\uline{I}_{R}$};
|
||||
\draw[magenta,->](0,0)--(4,3)node at (2.5,1.5){$\uline{I}$};
|
||||
\draw node at (-1.25,0){\textbf{+}};
|
||||
\draw node at (11.25,0){\textbf{--}};
|
||||
\draw node at (5.25,.375){\textbf{a}};
|
||||
\draw node at (2,4.25){\textbf{b}};
|
||||
\draw [blue]node at (10,5)[right]{\footnotesize{$\uline{U}=200\,\volt\,\widehat{=}\,10\,\centi\metre$}};
|
||||
\draw [blue]node at (10,4.5)[right]{\footnotesize{$\uline{U}_{R_1}=100\,\volt, \uline{U}_{R_2}=100\,\volt$}};
|
||||
\draw [red]node at (10,4)[right]{\footnotesize{$\uline{I}_{R}$}};
|
||||
\draw [red!50!blue]node at (10,3.5)[right]{\footnotesize{$\uline{U}_{C}\bot\,\uline{U}_{R_3}$ addiert sich zu $\uline{U}$}};
|
||||
\draw [red]node at (10,3)[right]{\footnotesize{$\uline{I}_{C}$ in Phase zu $\uline{U}_{R_3}$}};
|
||||
\draw [magenta]node at (10,2.5)[right]{\footnotesize{$\uline{I}=\uline{I}_{C}+\uline{I}_{R}$}};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Anmerkung:\\
|
||||
Wenn wie angegeben $|X_C|=R\Rightarrow \uline{U}_{R3}\bot\uline{U}_{C}\Rightarrow$ gleichschenkliges, rechtwinkliges Dreieck $\Rightarrow\uline{U}_{ab}\bot \Re$-Achse.
|
||||
\clearpage
|
||||
}{}%
|
||||
93
ET2_L_B17_A7.tex
Normal file
93
ET2_L_B17_A7.tex
Normal file
@ -0,0 +1,93 @@
|
||||
\section{Wechselstrombrücke}
|
||||
Gegeben ist die dargestellte Wechselstrombrücke, die zum Messen der Größe von $R_2$
|
||||
und $L_2$ dient. Dabei ist $R_1=1\,\kilo\ohm$, $R_3=R_4=2\,\kilo\ohm$ und $L_1=1\,\milli\henry$.\\
|
||||
Die Brücke ist bei einer Kreisfrequenz von $\omega=\power{10}{6}\,\power{\second}{-1}$ und $C_1=2\,\nano\farad$ abgeglichen.\\[\baselineskip]
|
||||
Berechnen Sie $R_2$ und $L_2$!
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\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] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C_1$};
|
||||
\draw[->](.3,-.2)--(.7,.2);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_3$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=.25cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=-.25cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_4$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%V-meter
|
||||
\draw (0,0)--(.367,0) (.633,0)--(1,0);
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw[->](.4,.1)--(.6,-.1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm]%Knotenpunkte
|
||||
\draw (.1,1)--(0,1)--(0,-.5)(0,0)--(1,0)
|
||||
(1.01,-.25)--(1,-.25)--(1,.25)--(1.01,.25)
|
||||
(1.99,-.25)--(2,-.25)--(2,.25)--(1.99,.25)
|
||||
(2,0)--(3,0)
|
||||
(3.9,1)--(4,1)--(4,-.5);
|
||||
\draw [->,blue] (.2,-.5)--(3.8,-.5)node at(3,-.5)[above]{$\underline{U}=20\,\volt$};
|
||||
\fill (0,-.5)circle(.025) (4,-.5)circle(.025);
|
||||
\end{scope}
|
||||
\draw node at (1.5,1.25)[above]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxxxxxxx}}$};
|
||||
\draw node at (3.5,1.25)[above]{$\overbrace{\phantom{xxxxx}}$};
|
||||
\draw node at (1.5,1.5)[above]{$\uline{Z}_1$};
|
||||
\draw node at (3.5,1.5)[above]{$\uline{Z}_3$};
|
||||
\draw node at (1.5,-.5)[below]{$\underbrace{\phantom{xxxxxxxxxxx}}$};
|
||||
\draw node at (3.5,-.5)[below]{$\underbrace{\phantom{xxxxx}}$};
|
||||
\draw node at (1.5,-.75)[below]{$\uline{Z}_2$};
|
||||
\draw node at (3.5,-.75)[below]{$\uline{Z}_4$};
|
||||
% \begin{scope}[>=latex,very thick]%Variablen Pfeile
|
||||
% \draw[->] (1.3,.75)--(1.7,1.25);
|
||||
% \draw[->] (2.3,.75)--(2.7,1.25);
|
||||
% \draw[dashed] (1.3,.75)--(2.3,.75);
|
||||
% \end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
R_P&=R_S+\frac{X^2_S}{R_S}\label{eq:rp177}\\
|
||||
X_P&=X_S+\frac{R^2_S}{X_S}\label{eq:xp177}
|
||||
\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\uline{Z}_1&=R_1+jX_{L_1}-jX_{C_1} &\uline{Z}_3&=R_3\\
|
||||
\quad\uline{Z}_2&=R_2 ||X_{L_2} &\uline{Z}_4&=R_4
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Brücke abgeglichen, wenn:}
|
||||
\frac{\uline{Z}_1}{\uline{Z}_2}&=\frac{\uline{Z}_3}{\uline{Z}_4}=\frac{R_3}{R_4}=1\\
|
||||
\uline{Z}_2&=\uline{Z}_1=R_1+j\omega L_1 -j\frac{1}{\omega C_1}\\
|
||||
&=1\,\kilo\ohm+j\Big(\underbrace{\power{10}{6} \,\cancel{\second^{-1}} \cdot \power{10}{-3}\,\ohm\cancel{\second}}_{1\,\kilo\ohm} -\underbrace{\frac{1}{\power{10}{6} \,\cancel{\second^{-1}}\cdot 2 \cdot \power{10}{-9}\,\frac{\cancel{\second}}{\ohm}}}_{\frac{1}{2}\,\kilo\ohm}\Big)=\uuline{(1+j0{,}5)\,\kilo\ohm}\\
|
||||
\end{align*}
|
||||
Um $R_2$ und $L_2$ zu bestimmen, die Reihenschaltung $\uline{Z}_2=R_{2S}+jX_{2S}=(1+j0{,}5)\,\kilo\ohm$ in Parallelwiderstände umrechnen.
|
||||
\begin{align*}
|
||||
\uline{Y_2}&=\frac{1}{\uline{Z_2}}=\frac{1}{(1+j0{,}5)\,\kilo\ohm}
|
||||
=(\underbrace{0{,}8}_{G_2}-\underbrace{j0{,}4}_{B_{L_2}})\,\milli\siemens\\
|
||||
R_{2}&=\frac{1}{G_2}=\frac{1}{0{,}8\,\milli\siemens}=\uuline{1{,}25\,\kilo\ohm}\\
|
||||
X_{L_2}&=\frac{-1}{B_{L_2}}=\frac{-1}{-0{,}4\,\milli\siemens}=\uline{+2{,}5\,\kilo\ohm}\\
|
||||
\text{aus }X_{L_2}&=\omega\cdot L \Rightarrow \\
|
||||
L_2&=\frac{X_{L_2}}{\omega}=\frac{2{,}5\,\kilo\ohm}{\power{10}{6}\cdot \power{\,\second}{-1}}=\uuline{2{,}5\,\milli\henry}\\
|
||||
\intertext{Alternativ mit Formeln \ref{eq:rp177} und \ref{eq:xp177}}
|
||||
R_{2P}&=R_{2S}+\frac{X^2_{2S}}{R_{2S}}\\
|
||||
X_{2P}&=X_{2S}+\frac{R^2_{2S}}{X_{2S}}\\
|
||||
R_2&=R_{2S}+\frac{X^2_{2S}}{R_{2S}}=(1+\frac{0{,}5^2}{1})\,\kilo\ohm=1{,}25\,\kilo\ohm\\
|
||||
X_{L_2}&=X_{2S}+\frac{R^2_{2S}}{X_{2S}}=(0{,}5+\frac{1^2}{0{,}5})\,\kilo\ohm=2{,}5\,\kilo\ohm
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
103
ET2_L_B17_A8.tex
Normal file
103
ET2_L_B17_A8.tex
Normal file
@ -0,0 +1,103 @@
|
||||
\section{Wechselstrombrücke}
|
||||
a) Zustand abgeglichenen Brücke $\mathbf{(U_{ab}=0})$\\
|
||||
Welches Bauteil muss fur $X$ eingesetzt werden, um diese Voraussetzung zu erfüllen?\\
|
||||
Berechnen Sie den Wert des Bauteils als Funktion von $R$ und $X_L$.\\[\baselineskip]
|
||||
b) Zustand nicht abgeglichene Brücke: $\mathbf{(U_{ab}\neq 0)}$\\[\baselineskip]
|
||||
Zeichnen Sie ein qualitatives Zeigerdiagramm aller eingezeichneten Spannungen\\
|
||||
(Bezug $\uline{U}=U\cdot e^{j0\,\degree}$), so dass $\uline{U}_{ab}$ auf $\uline{U}$ senkrecht steht (rechte Winkel müssen gekennzeichnet werden).\\
|
||||
Welches Bauteil muss fur $X$ eingesetzt werden, um diese Voraussetzung zu erfüllen?\\ Berechnen Sie den Wert dieses Bauteils als Funktion von $R$ und $X_L$ anhand der Beziehung zwischen den Zeigerlängen.
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2.5]
|
||||
\begin{scope}[>=latex,very thick,xshift=0.5cm,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,.35)--(.7,.35)node at(.5,.35)[right]{\footnotesize$\uline{U}_{R}$};
|
||||
\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] {$R$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{b}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=.5cm,yshift=-0cm,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);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{a}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,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] {$X$};
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{X}$};
|
||||
\draw node at (.5,0){?};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=-.5cm,yshift=-.5cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0)node at(.5,-.133)[right]{$\uline{U}$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$\uline{U}$};
|
||||
\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)(-.4,1)--(-.5,1)--(-.5,.5)(-.4,-1)--(-.5,-1)--(-.5,-.5);
|
||||
% \fill (-.5,1)circle(.025) (-.5,-1)circle(.025);
|
||||
\draw [->,blue] (.6,0)--(1.4,0)node at(1,0)[below]{$\underline{U}_{ab}$};
|
||||
\fill (.5,0)circle(.025) node at (.5,0)[left]{a};
|
||||
\fill (1.5,0)circle(.025)node at (1.5,0)[right]{b};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung: (Platzbedarf in x: $9\,\centi\metre$; in y: $\pm 5\,\centi\metre$)\\
|
||||
\begin{align*}
|
||||
\intertext{a) Zum Brückenabgleich muß $X$ eine Kapazität sein.}\\
|
||||
\intertext{Abgleichbedingung:}\\
|
||||
\frac{jX_L}{R}&=\frac{R}{jX_C}\\
|
||||
X&=\uuline{-\frac{R^2}{X_L}} \qquad\text{ neg. VZ $\Rightarrow$ Kapazität $X=X_C$} \\
|
||||
-\frac{1}{j\omega C}&=-\frac{R^2}{j\omega L}\Rightarrow \quad\text{oder}\quad C=\frac{L}{R^2}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
\intertext{b) Bedingung: $\uline{U}_{ab}\,\bot\,\uline{U}$; gleiche Zeigerlängen.}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[ultra thin,black!50!](0,-5)grid(8,5);
|
||||
\draw[thin,->](0,0)--(8.5,0)node[right]{$\Re$};
|
||||
\draw[thin,->](0,-5.5)--(0,5.5)node[above]{$\Im$};
|
||||
\foreach \y in {5,4,...,-5}
|
||||
\draw(0,\y)--(-.1,\y)node[left]{$\y$};
|
||||
\draw[black!50!,dashed](0,0)+(20:4.2cm)arc(20:-20:4.2cm);
|
||||
\draw[black!50!,dashed](8,0)+(160:4.2cm)arc(160:200:4.2cm);
|
||||
\draw[black!50!,dashed](4,1.5)--(4,-1.5);
|
||||
\fill(4,0)circle(0.075cm);
|
||||
\draw[blue,->](0,0)--(8,0)node at(7.5,.25){$\uline{U}$};
|
||||
\draw[black!50!](4,0)circle(4cm);
|
||||
\draw[blue,->](0,0)--(30:6.93)node at(2.5,1.75){$\uline{U}_a$};
|
||||
\draw[blue,->](0,0)--(-30:6.93)node at(2.5,-1.75){$\uline{U}_{b}$}; \draw[blue,->](30:6.93)--(-30:6.93)node at(5.5,-1.25){$\uline{U}_{ab}$};
|
||||
\draw[blue,->](30:6.93)--(8,0)node at(6.75,1.5){$\uline{U}_{R}$}; \draw[blue,->](-30:6.93)--(8,0)node at(6.75,-1.5){$\uline{U}_{X}$};
|
||||
\draw node at (6,3.75){a};
|
||||
\draw node at (6,-3.75){b};
|
||||
\draw[red](30:6.93)+(210:.5cm)arc(210:300:.5cm);
|
||||
\fill[red](30:6.93)+(255:.25cm)circle(.05cm);
|
||||
\draw[red](-30:6.93)+(150:.5cm)arc(150:60:.5cm);
|
||||
\fill[red](-30:6.93)+(105:.25cm)circle(.05cm);
|
||||
\draw[red](0:6)+(0:.5cm)arc(0:90:.5cm);
|
||||
\fill[red](0:6)+(45:.25cm)circle(.05cm);
|
||||
\draw node at(8.5,4.5)[right]{$\diamond$\ \ \footnotesize{$\uline{U}$}};
|
||||
\draw node at(8.5,4)[right]{$\phantom{\diamond x}$\footnotesize{Thaleskreis, da $\uline{U}_q \bot \uline{U}_R$ und $\uline{U}_b \bot \uline{U}_X$}};
|
||||
\draw node at(8.5,3.5)[right]{$\diamond$\ \footnotesize{$\uline{U}_a+\uline{U}_R=\uline{U}$}};
|
||||
\draw node at(8.5,3)[right]{$\phantom{\diamond x}$\footnotesize{$\uline{U}_a$ und $\uline{U}_R$ zeichnen}};
|
||||
\draw node at(8.5,2.5)[right]{$\diamond$\ \footnotesize{$\uline{U}_a+\uline{U}_{ab}-\uline{U}_b=0$ und $\uline{U}_{ab} \bot\uline{U}$}};
|
||||
\draw node at(8.5,2)[right]{$\phantom{\diamond x}$\footnotesize{$\uline{U}_{ab}$ und $\uline{U}_b$ zeichnen}};
|
||||
\draw node at(8.5,1.5)[right]{$\diamond$\ \footnotesize{$\uline{U}_b+\uline{U}_X=\uline{U}$}};
|
||||
\draw node at(8.5,1)[right]{$\phantom{\diamond x}$\footnotesize{$\uline{U}_X$ zeichnen}};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Bei der nicht abgeglichenen Brücke mit der Bedingung $\uline{U}_{ab}\,\bot\,\uline{U}$, muss $X$ eine Induktivität sein.\\
|
||||
|
||||
Rechnerisch:
|
||||
\begin{align*}
|
||||
|U_a|&=|U_b|\Rightarrow |X_L\cdot I_a|=|R\cdot I_b|\tag{1}\\
|
||||
|U_R|&=|U_X|\Rightarrow |R\cdot I_a|=|X\cdot I_b| \tag{2}\\
|
||||
\frac{X_L}{R}&=\frac{R}{X}\tag{1:2}\\
|
||||
X&=\uuline{+\frac{R^2}{X_L}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
88
ET2_L_B18_A1.tex
Normal file
88
ET2_L_B18_A1.tex
Normal file
@ -0,0 +1,88 @@
|
||||
\section {Übertrager im Leerlauf}
|
||||
$U_1=230\,\volt$, $R_1=5\,\ohm$, $I_1=10\,\ampere$, $U_2=100\,\volt$\\
|
||||
ausgangsseitiger Leerlauf\\
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie den Eigangswiderstand $\uline{Z}_1=\frac{\uline{U}_1}{\uline{I}_1}$ nach Betrag und Phase sowie die aufgenommene Wirk- und Blindleistung.
|
||||
\item Berechnen sie $\omega L_1$, $\omega L_2$ und $\omega M$ unter Annahme einer idealen Kopplung.
|
||||
\end{enumerate}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\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] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,-.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw [->,red] (1,.1)--(.75,.1)node at(.875,.1)[above right]{\footnotesize$\uline{I}_2=0$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(1,0)--(1,.2) (2.5,0)--(1.5,0)--(1.5,.2) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9);
|
||||
\fill (2.5,0)circle(0.025cm)(2.5,1)circle(0.025cm);
|
||||
\draw [->,blue] (2.5,.8)--(2.5,.2) node at (2.5,.5)[right]{$\uline{U}_2$};
|
||||
\draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$M$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{enumerate}
|
||||
\item Komplexer Eingangswiderstand\\
|
||||
Begriffe:\\
|
||||
$\uline{Z}$ Impedanz oder komplexer Widerstand\\
|
||||
$Z=|\uline{Z}|$ Scheinwiderstand \\
|
||||
\begin{minipage}[c]{0.25\textwidth}
|
||||
\begin{tikzpicture}[scale=.2]
|
||||
\draw[->](0,0)--(5,0)node[right]{$R_1$};
|
||||
\draw[->](5,0)--(5,22.45)node at(5,10)[right]{$\omega L_1$};
|
||||
\draw[->](0,0)--(77.44:23cm)node at(0,10)[left]{$\uline{Z}_1$};
|
||||
\draw[->](2,0)arc(0:77.4:2cm)node at(1,2)[right]{$\varphi_1$};
|
||||
\end{tikzpicture}
|
||||
\end{minipage}
|
||||
\begin{minipage}[c]{0.75\textwidth}
|
||||
\begin{align*}
|
||||
Z_1&=\frac{U_1}{I_1}=\frac{230\,\volt}{10\,\ampere}=23\,\ohm\\
|
||||
\text{mit }R_1&=Z_1\cdot \cos\varphi_1\Rightarrow\\
|
||||
\varphi_1&=\arccos\frac{R_1}{Z_1}=\arccos\frac{5\,\ohm}{23\,\ohm}=77{,}44\,\degree\\
|
||||
\uline{Z}_1&=Z_1\cdot e^{j\varphi_1}=\uuline{23\,\ohm\cdot e^{j77{,}44\,\degree}}\\
|
||||
P&=U_1\cdot I_1\cdot \cos\varphi_1=230\,\volt\cdot 10\,\ampere\cdot \cos(77{,}44\,\degree)=\uuline{500\,\watt}\\
|
||||
Q&=U_1\cdot I_1\cdot \sin\varphi_1=230\,\volt\cdot 10\,\ampere\cdot \sin(77{,}44\,\degree)=\uuline{2245\,var}\\
|
||||
\text{oder }\uline{S}&=P+jQ=\frac{\uline{U}_1^2}{\uline{Z_1}}=\frac{(230\,\volt)^2}{23\,\ohm\cdot e^{j77{,}44\,\degree}}=(\underbrace{500}_{P}-j\underbrace{2245}_{Q})\,\volt\ampere
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\clearpage
|
||||
\item Blindwiderstände und Kopplungswiderstand\\
|
||||
Begriffe:\\
|
||||
$L$ Selbstinduktivität\\
|
||||
$M$ Gegeninduktivität\\
|
||||
$X_L=\omega L$ Reaktanz oder Blindwiderstand\\
|
||||
$X_M=\omega M$ Kopplungswiderstand\\
|
||||
\begin{align*}
|
||||
\omega L_1&=Z_1\cdot \sin\varphi_1=23\,\ohm\cdot \sin(77{,}44\,\degree)=\uuline{22{,}45\,\ohm}\\
|
||||
U_2&=\omega M\cdot I_1\Rightarrow\\
|
||||
\omega M&=\frac{U_2}{I_1}=\frac{100\,\volt}{10\,\ampere}=\uuline{10\,\ohm}\\
|
||||
M&=\sqrt{L_1\cdot L_2}\\
|
||||
\omega M&=\sqrt{\omega L_1\cdot \omega L_2}\\
|
||||
\Rightarrow\omega L_2&=\frac{(\omega M)^2}{\omega L_1}=\frac{(10\,\ohm)^2}{22{,}45\,\ohm}=\uuline{4{,}45\,\ohm}
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
\clearpage
|
||||
}{}%
|
||||
204
ET2_L_B18_A2.tex
Normal file
204
ET2_L_B18_A2.tex
Normal file
@ -0,0 +1,204 @@
|
||||
\section {Übertrager mit kapazitiver Last}
|
||||
Von nebenstehender Schaltung ist gegeben:\\
|
||||
$\uline{U}_1=1\,\volt\cdot e^{j0\,\degree}$, $R_1=10\,\ohm$, $X_1=100\,\ohm$, $R_2=40\,\ohm$, $X_2=400\,\ohm$, $X_C=-200\,\ohm$, $X_M=40\,\ohm$\\
|
||||
|
||||
Gesucht\\
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Spannung $\uline{U}_2$ nach Betrag und Phase bei offenem Schalter S.
|
||||
\item Spannung $\uline{U}_2$ nach Betrag und Phase bei geschlossenem Schalter S.
|
||||
\item Größe und Richtung der über das Magnetfeld übertragenen Wirkleistung für den Fall b).
|
||||
\end{enumerate}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\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) [left] {$X_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.175,-.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_2$};
|
||||
\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,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(1,0)--(1,.2) (3,.2)--(3,0)--(1.5,0)--(1.5,.2) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9)(3,.8)--(3,1)--(2.8,1);
|
||||
\fill (2.5,0)circle(0.025cm)(2.5,1)circle(0.025cm)(2.8,1)circle(0.025cm);
|
||||
\draw [->,blue] (2.5,.8)--(2.5,.2) node at (2.5,.5)[left]{$\uline{U}_2$};
|
||||
\draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$X_M$};
|
||||
\draw[ultra thick](2.8,1)--+(150:.3cm)node at(2.7,1.2){$S$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{enumerate}
|
||||
\item Offener Schalter\\
|
||||
\begin{align*}
|
||||
\uline{I}_2&=0\quad\\
|
||||
&\text{Wegen Wicklungssinn $\uline{U}_2$ negativ (bzw. $X_M$ negativ)}\\
|
||||
\uline{U}_2&=jX_M\cdot \uline{I}_1=jX_M\cdot \frac{\uline{U}_1}{R+jX_1}=\frac{(-j40\cancel{\,\ohm})\cdot 1\,\volt}{(10+j100)\cancel{\,\ohm}}=(0{,}396-j0{,}04)\,\volt\\
|
||||
&=\uuline{0{,}398\,\volt\cdot e^{-j174{,}3\,\degree}}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\item Ersatzschaltbild: Schalter geschlossen\\
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw node at(.5,-.2){\footnotesize$10$};
|
||||
\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] {$X_1-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j140$};
|
||||
\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) [left] {$X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$\uline{U}_M$};
|
||||
\draw node at(.25,.1)[left]{\footnotesize$-j40$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_2-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j440$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw node at(.5,-.2){\footnotesize$40$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,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] {$X_C$};
|
||||
\draw node at(.25,-.05)[right]{\footnotesize$-j200$};
|
||||
\draw[->,red](.9,-.1)--(.6,-.1)node at(.75,-.1)[right]{$I_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0)--(4,.2) (3.8,1)--(4,1)--(4,.8);
|
||||
\fill (0,0)circle(0.025cm)(0,1)circle(0.025cm);
|
||||
\draw [->,blue] (3.8,.7)--(3.8,.3) node at (3.75,.5)[left]{$\uline{U}_2$};
|
||||
\draw [->,blue!50!red] (3.5,.7)--(2.5,.7) node at (3,.7)[above]{$\uline{U}'_2$};
|
||||
\draw [->,blue!50!red,thick] (3,.075)arc(270:0:.25cm) node at (3,.325){\scriptsize{Masche}};
|
||||
\draw [->,blue!50!red] node at (5,.7)[right]{$\uline{U}'_2$ \footnotesize{Kontrollrechnung}};
|
||||
\draw node at (0,.5)[left]{$\uline{Z}_{ges}\Rightarrow$};
|
||||
\draw node at (1,1.5)[above]{$\uline{Z}_1$};
|
||||
\draw node at (3,1.5)[above]{$\uline{Z}_2$};
|
||||
\draw node at (1,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.5)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.25)[below]{$\uline{Z}'=(40+j240)\,\ohm=243{,}3\,\ohm\cdot e^{j80{,}5\,\degree}$};
|
||||
\draw node at (3,-1)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.75)[below]{$\uline{Z}_{||}=\frac{(-j40)\cdot (40+j240)}{40+j200}\,\ohm=(1{,}54-j47{,}7)\,\ohm=47{,}72\,\ohm\cdot e^{-j88{,}2\,\degree}$};
|
||||
\end{scope}
|
||||
% \draw[dashed,blue!50!red](3,.9)--(3,.1)node at (3,.5)[right]{$P_2$};
|
||||
\end{tikzpicture}
|
||||
\begin{align*}
|
||||
\uline{Z}_{ges}&=\uline{Z}_1+\uline{Z}_{||}=(11{,}54+j92{,}3)\,\ohm=93{,}03\,\ohm\cdot e^{j82{,}87\,\degree}
|
||||
\intertext{Spannungsteiler}
|
||||
\uline{U}_M&=\uline{U}_1\cdot \frac{\uline{Z}_{||}}{\uline{Z}_{ges}}=1\,\volt\cdot e^{j0\,\degree}\cdot \frac{47{,}72\,\ohm\cdot e^{-j88{,}2\,\degree}}{93{,}03\,\ohm\cdot e^{j82{,}87\,\degree}}=(-0{,}51-j0{,}08)\,\volt=0{,}51\,\volt\cdot e^{-j171{,}1\,\degree}\\
|
||||
\uline{U}_2&=\uline{U}_M\cdot \frac{jX_C}{\uline{Z}'}=0{,}51\,\volt\cdot e^{-j171{,}1\,\degree}\cdot \frac{200\,\ohm\cdot e^{-j90\,\degree}}{243{,}3\,\ohm\cdot e^{j80{,}5\,\degree}}\\
|
||||
&=(0{,}4+j0{,}13)\,\volt=\uuline{0{,}42\,\volt\cdot e^{+j18{,}4\,\degree}}
|
||||
\end{align*}
|
||||
\item Wirkleistung
|
||||
\begin{align*}
|
||||
P_2&=I^2_2\cdot R_2\\
|
||||
\uline{I}_2&=\frac{\uline{U}_2}{jX_C}=\frac{0{,}42\,\volt\cdot e^{+j18{,}4\,\degree}}{-j200\,\ohm}=2{,}108\,\milli\ampere\cdot e^{+j108{,}4\,\degree}=(-0{,}666+j2)\,\milli\ampere\\
|
||||
P_2&=(2{,}108\,\milli\ampere)^2\cdot 40\,\ohm=1{,}78\cdot \power{10}{-4}\,\watt=\uuline{178\,\micro\watt}\\
|
||||
&\text{Von Primär- nach Sekundärseite}\\
|
||||
\intertext{Alternativ, in 2 Schritten}
|
||||
U_{R_2}&=I_2\cdot R_2=2{,}108\,\milli\ampere\cdot 40\,\ohm=84{,}32\,\milli\volt\\
|
||||
P_2&=U_{R_2}\cdot I_2=84{,}32\,\milli\volt\cdot 2{,}108\,\milli\ampere=178\micro\watt\\
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
Kontrolle: (über das Magnetfeld übertragenen Wirkleistung)
|
||||
\begin{align*}
|
||||
\uline{U}'_2&=\uline{U}_2 - \uline{U}_M=(0{,}4+j0{,}13)-(-0{,}51-j0{,}08))\,\volt=(0{,}91+j0{,}21)\,\volt=0{,}932\,\volt\cdot e^{j13{,}22\,\degree}\\
|
||||
&\text{oder auch aus den Impedanzen berechnet}\\
|
||||
\uline{U}'_2&=-\uline{I}_2\cdot [R_2+(\uline{X}_2-\uline{X}_M)]=-(-0{,}666+j2)\,\milli\ampere\cdot (40+j440)\,\ohm\\
|
||||
&=(0{,}907+j0{,}213)\,\volt=0{,}932\,\volt\cdot e^{j13{,}22\,\degree}\\[\baselineskip]
|
||||
&\text{mit }S_2=\uline{U}'_2\cdot \uline{I}^*_2\\[\baselineskip]
|
||||
P_2&=\Re\{\uline{U}'_2\cdot \uline{I}^*_2\}\\
|
||||
&=\Re\{(0{,}91+j0{,}21)\,\milli\volt\cdot (-0{,}666-j2)\cdot \power{10}{-3}\,\ampere\} \quad\text{Rechtwinklige Koordinaten oder}\\
|
||||
&=\Re\{0{,}932\,\volt\cdot e^{j13{,}22\,\degree}\cdot 2{,}108\,\milli\ampere\cdot e^{-j108{,}4\,\degree}\}\quad\text{Polar Koordinaten}\\
|
||||
&=\Re\{1{,}965\cdot e^{-j95{,}18\,\degree}\}\,\micro\volt\ampere
|
||||
=\Re\{\underbrace{-177{,}4}_{\text{EPS$\rightarrow$Verbraucher}}-j1{,}957\}\,\micro\volt\ampere\\
|
||||
&\text{\footnotesize{Erzeuger-Pfeil-System (EPS)}}
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=10]
|
||||
\draw[thin,black!50!,step=.1cm](-.1,-.3001)grid(1,.6);
|
||||
\draw[black!25!]+(18.42:.934cm)arc(18.42:198.42:.467cm); % Thales
|
||||
% \draw[black!25!](18.42:.467cm)circle(.467cm); % Thales
|
||||
\draw[<-,blue](0,0)++(18.4:.42cm)--+(8.9:.51cm)node at(.42,.175)[right]{$\uline{U}_M$};
|
||||
\draw[->,magenta](0,0)--(12.99:.934cm)node at(.7,.15)[below left]{$\uline{U}'_2$};
|
||||
\draw[->,red](0,0)--(108.46:.21cm)node [above]{$\uline{I}_2$};
|
||||
\draw[->,red](0,0)--(-108.4:.21cm)node [below]{$\uline{I}^*_2$};
|
||||
\draw[->,black!50!green](0,0)++(12.99:.934cm)--+(108.46:.084cm)node [below right]{$\uline{U}_{R2}=I_2\cdot R_2$};
|
||||
\draw[->,black!50!green](0,0)--(18.4:.935cm)node [above left]{$\uline{U}_{(X_2-X_M)}$};
|
||||
\draw[->,blue](0,0)--++(18.4:.42cm)node [above left]{$\uline{U}_2$};
|
||||
\draw node at(1.05,.55)[right]{$\uline{U}'_2=\uline{U}_2 - \uline{U}_M=0{,}932\cdot \,\volt\cdot e^{j13{,}22\,\degree}$};
|
||||
\draw node at(1.05,.45)[right]{$\uline{U}_{R_2}\quad||\quad\uline{I}_2$};
|
||||
\draw node at(1.05,.35)[right]{$\uline{U}_{(X_2-X_M)}\quad||\quad\uline{U}_2 \quad\bot\quad\uline{I}_2$};
|
||||
\draw node at(1.05,-.05)[right]{$S=\uline{U}'_2\cdot \uline{I}^*_2=1965\,\micro\volt\ampere$};
|
||||
\draw node at(1.05,-.15)[right]{$P=S\cdot \cos(-95{,}18\,\degree)=-177\micro\watt$};
|
||||
\draw node at(1.05,-.25)[right]{$Q=S\cdot \sin(-95{,}18\,\degree)=-1957\micro\volt\ampere r$};
|
||||
\draw[black!75!](0,0)++(108.4:.1cm)arc(108.4:12.99:.1cm)node at(90:.1cm) [above right]{$\varphi=-95{,}18\,\degree$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
%%%%delete
|
||||
%\uline{U}'_2&=-\uline{I}_2\cdot(\uline{X}_2-\uline{X}_M)=-(-0{,}666+j2)\,\milli\ampere\cdot (j440)\,\ohm=(0{,}88+j0{,}29)\,\volt\\
|
||||
%\uline{U}'_2+\uline{U}_M&=(0{,}88+j0{,}29)\,\volt+(-0{,}51-j0{,}08)\,\volt=(0{,}37+j0{,}21)\,\volt=0{,}425\,\volt\cdot e^{j29{,}58\,\degree}\\
|
||||
%P_2&=\Re\{(\uline{U}'_2+\uline{U}_M)\cdot \uline{I}^*_2\}\\
|
||||
%&=\Re\{(0{,}37+j0{,}21)\,\volt\cdot (-0{,}666-j2)\cdot \power{10}{-3}\,\ampere\}=\Re\{0{,}425\,\volt\cdot e^{j29{,}58\,\degree}\cdot 2{,}108\,\milli\ampere\cdot e^{-j108{,}4\,\degree}\}\\
|
||||
%&=\Re\{0{,}896\cdot e^{-j78{,}82\,\degree}\}\,\micro\volt\ampere=\Re\{\underbrace{+174}_{\text{EPS$\rightarrow$Quelle}}-j879\}\,\micro\volt\ampere\\[\baselineskip]
|
||||
%%
|
||||
%&\text{Langer Pfeil auch über XC}\\
|
||||
%\uline{U}'_2&=-\uline{I}_2\cdot [R_2+(\uline{X}_2-\uline{X}_M)+\uline{X}_C]=-(-0{,}666+j2)\,\milli\ampere\cdot (40+j240)\,\ohm\\
|
||||
%&=(0{,}507+j0{,}08)\,\volt=0{,}513\,\volt\cdot e^{j8{,}96\,\degree}\\
|
||||
%\uline{U}'_2+\uline{U}_M&=(0{,}507+j0{,}08)\,\volt+(-0{,}51-j0{,}08)\,\volt=0\\[\baselineskip]
|
||||
%&\text{Kurzer Pfeil + C2 ohne R2}\\
|
||||
%\uline{U}'_2&=-\uline{I}_2\cdot [R_2+(\uline{X}_2-\uline{X}_M)+\uline{X}_C]=-(-0{,}666+j2)\,\milli\ampere\cdot (+j240)\,\ohm\\
|
||||
%&=(0{,}480+j0{,}160)\,\volt=0{,}505\,\volt\cdot e^{j18{,}42\,\degree}\\
|
||||
%\uline{U}'_2+\uline{U}_M&=(0{,}507+j0{,}08)\,\volt+(-0{,}51-j0{,}08)\,\volt=(0{,}480+j0{,}160)\,\volt=0{,}505\,\volt\cdot e^{j18{,}44\,\degree}\\[\baselineskip]
|
||||
%&\text{Kurzer Pfeil}\\
|
||||
%\uline{U}'_2&=-\uline{I}_2\cdot (\uline{X}_2-\uline{X}_M)=-(-0{,}666+j2)\,\milli\ampere\cdot (+j440)\,\ohm\\
|
||||
%&=(0{,}880+j0{,}293)\,\volt=0{,}927\,\volt\cdot e^{-j18{,}42\,\degree}\\
|
||||
%\uline{U}'_2+\uline{U}_M&=(0{,}880+j0{,}293)\,\volt+(-0{,}51-j0{,}08)\,\volt
|
||||
%=(0{,}880+j0{,}293)\,\volt=0{,}927\,\volt\cdot e^{j18{,}42\,\degree}\\[\baselineskip]
|
||||
%P_2&=\Re\{(\uline{U}'_2+\uline{U}_M)\cdot \uline{I}^*_2\}\\
|
||||
%&=\Re\{(0{,}880+j0{,}293)\,\milli\volt\cdot (-0{,}666-j2)\cdot \power{10}{-3}\,\ampere\} \quad\text{Rechtwinklige Koordinaten oder}\\
|
||||
%&=\Re\{0{,}927\,\milli\volt\cdot e^{j18{,}42\,\degree}\cdot 2{,}108\,\milli\ampere\cdot e^{-j108{,}4\,\degree}\}\quad\text{Polar Koordinaten}\\
|
||||
%&=\Re\{1{,}954\cdot e^{-j89{,}98\,\degree}\}\,\micro\volt\ampere=\Re\{\underbrace{785}_{\text{EPS$\rightarrow$Quelle}}+j1954\}\,\micro\volt\ampere\\[\baselineskip]
|
||||
%&\text{seine alte Lösung }(0{,}4+j0{,}13)=0{,}4206\,\volt\cdot e^{-j18{,}0\,\degree}\\
|
||||
%\uline{U}''_x&=\uline{U}_2-\uline{U}_M=(0{,}4+j0{,}13)\,\volt-(-0{,}51-j0{,}08)\,\volt=(0{,}91+j0{,}21)\,\volt\\
|
||||
%P_2&=\Re\{(\uline{U}'_2+\uline{U}_M)\cdot \uline{I}^*_2\}=(-0{,}91-j0{,}21)\,\volt\cdot (-0{,}666-j2)\cdot \power{10}{-3}\,\ampere\\
|
||||
%&=\Re\{\underbrace{+178}_{\text{EPS$\rightarrow$Quelle}}+j1960\}\,\micro\volt\ampere\\[\baselineskip]
|
||||
% \draw[->,black!50!green](0,0)++(18.4:.42cm)--+(-171.1:.51cm)node[left]{$\uline{U}_M$};
|
||||
% \draw[->,black!75!green](0,0)++(18.4:.42cm)++(-171.1:.51cm)--(0:0cm)node[left]{$\uline{U}_2 - \uline{U}_M$};
|
||||
% \draw[->,black!85!green](0,0)--(8.4:.89cm)node[right]{$\uline{U}''_x$};
|
||||
% \draw[->](0,0)--(8.9:.51cm);%{$\uline{U}_M$};
|
||||
% \draw[->,black!50!green](0,0)--(-171.1:.51cm)node[left]{$\uline{U}_M$};
|
||||
% \draw[->,magenta](0,0)--(.91,.21)node at(.7,.15)[below left]{$\uline{U}'_2$};
|
||||
% \draw[->,black](0,0)++(12.99:.934cm)--+(108.46:.15cm)node [below right]{$\uline{U}_{R2}$};
|
||||
% \draw[->,black](0,0)--(22:.935cm)node [above left]{$\uline{U}_{(X_2-X_M)}$};
|
||||
186
ET2_L_B18_A3.tex
Normal file
186
ET2_L_B18_A3.tex
Normal file
@ -0,0 +1,186 @@
|
||||
\section {Übertrager mit Verbindung zum Eingang}
|
||||
\enlargethispage{3cm}
|
||||
Gegeben sind die Daten des abgebildeten Kreises:\\
|
||||
$R_1=28\,\ohm$, $X_1=96\,\ohm$, $R_2=14\,\ohm$, $X_M=60\,\ohm$, $X_2=400\,\ohm$, $\uline{I}_1=2\,\ampere\cdot e^{j0\,\degree}$\\
|
||||
Gesucht
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Spannung $\uline{U}_{ab}$ nach Betrag und Phase bei offenem Schalter S.
|
||||
\item Spannung $\uline{U}_S$ nach Betrag und Phase bei offenem Schalter S.
|
||||
\item Strom $\uline{I}_2$ und Spannung $\uline{U}_{ab}$ nach Betrag und Phase bei geschlossenem Schalter S.
|
||||
\item Größe und Richtung der über das Magnetfeld übertragenen Wirkleistung bei geschlossenem Schalter S.
|
||||
\end{enumerate}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
% \draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
|
||||
% \draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,0)--(.7,0) node at (.5,0)[left]{$\uline{U}_{ab}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw [->,red] (0.15,.1)--(.3,.1)node at(.225,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\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) [left] {$X_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule|
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.175,-.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw [->,red] (.85,.1)--(.7,.1)node at(.775,.1)[above]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(1,0)--(1,.2) (1.5,0)--(2.5,0)--(2.5,-.25)--(.1,-.25)--(.1,0) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9)
|
||||
(2.4,1)--(2.5,1)--(2.5,1.5)--(1.4,1.5)(1.1,1.5)--(.1,1.5)--(.1,1);
|
||||
\filldraw (0,1)circle(0.025cm)node[left]{a};
|
||||
\filldraw (0,0)circle(0.025cm)node[left]{b};
|
||||
\fill(.1,0)circle(0.025cm)(.1,1)circle(0.025cm) (1.1,1.5)circle(0.025cm)(1.4,1.5)circle(0.025cm);
|
||||
\draw [->,blue] (1,1.3)--(1.5,1.3) node at (1.25,1.3)[above]{$\uline{U}_S$};
|
||||
\draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.125)[below]{$X_M$};
|
||||
\draw[ultra thick](1.4,1.5)--+(150:.3cm)node at(1.4,1.7){$S$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
\vspace{-.5cm}
|
||||
Berechnung: Ersatzschaltbild Schalter offen
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw node at(.5,-.2){\footnotesize$28$};
|
||||
\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] {$X_1-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j156$};
|
||||
\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) [left] {$X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$U_M$};
|
||||
\draw node at(.25,.25){\footnotesize$-j60$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_2-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j460$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw node at(.5,-.2){\footnotesize$14$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0) (3.8,1)--(4,1)--(4,1.5)--(2.1,1.5)(1.9,1.5)--(.1,1.5)--(.1,1);
|
||||
\fill (4,0)circle(0.025cm)(.1,1)circle(0.025cm)
|
||||
(2.1,1.5)circle(.025cm)(1.9,1.5)circle(.025cm)
|
||||
(0,1)circle(0.025cm);
|
||||
\filldraw (0,1)circle(0.025cm)node[left]{a};
|
||||
\filldraw (0,0)circle(0.025cm)node[left]{b};
|
||||
\draw [->,blue] (0,.7)--(0,.3) node at (0,.5)[left]{$\uline{U}_{ab}$};
|
||||
\draw [->,blue] (.5,.7)--(1.5,.7) node at (1,.7)[below]{$\uline{U}_S$};
|
||||
\draw [->,blue] (1.5,1.85)--(2.5,1.85) node at (2,1.85)[above]{$\uline{U}_S$};
|
||||
\draw [->,blue] (3.5,.7)--(2.5,.7) node at (3,.7)[below]{$0\,\volt$};
|
||||
\draw[ultra thick](2.1,1.5)--+(150:.2cm)node at(2,1.7){$S$};
|
||||
\draw [->,red] (0.5,1.6)--(.75,1.6)node at(.625,1.6)[above]{\footnotesize$\uline{I}_S=0$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{enumerate}
|
||||
\item Spannung $\uline{U}_{ab}$ Schalter offen\\
|
||||
\vspace{-.5cm}
|
||||
\begin{align*}
|
||||
\uline{U}_{ab}&=[R_1+j(X_1-\cancel{X_M}+\cancel{X_M})]\cdot \uline{I}_1=(28+j96)\,\ohm\cdot 2\,\ampere=(56+j192)\,\volt\\
|
||||
&=\uuline{200\,\volt\cdot e^{j73{,}74\,\degree}}
|
||||
\end{align*}
|
||||
\item Spannung $\uline{U}_S$ Schalter offen\\
|
||||
\vspace{-.5cm}
|
||||
\begin{align*}
|
||||
&\text{Wegen gegensinnigem Wicklungssinn $X_M=-60\,\ohm$ }\\
|
||||
\uline{U}_S&=[R_1+j(X_1-X_M)]\cdot \uline{I}_1=(28+j156)\,\ohm\cdot 2\,\ampere=(56+j312)\,\volt\\
|
||||
&=\uuline{317\,\volt\cdot e^{j79{,}8\,\degree}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\enlargethispage{3cm}
|
||||
\item Strom $\uline{I}_2$ und Spannung $\uline{U}_{ab}$: Schalter geschlossen:
|
||||
\begin {align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw node at(.5,-.2){\footnotesize$28$};
|
||||
\draw [->,red] (0.15,.1)--(.3,.1)node at(.225,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\draw [->,red] (0.15,.6)--(.3,.6)node at(.225,.6)[above]{\footnotesize$\uline{I}_2$};
|
||||
\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] {$X_1-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j156$};
|
||||
\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) [left] {$X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$U_M$};
|
||||
\draw [->,red] (.25,-.1)--(.05,-.1)node at(.15,-.1)[right]{\footnotesize$\uline{I}=\uline{I}_1+\uline{I}_2$};
|
||||
\draw node at(.25,.25){\footnotesize$-j60$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_2-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j460$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw node at(.5,-.2){\footnotesize$14$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0) (3.8,1)--(4,1)--(4,1.5)--(.1,1.5)--(.1,1);
|
||||
\fill (4,0)circle(0.025cm)(.1,1)circle(0.025cm)
|
||||
(0,1)circle(0.025cm);
|
||||
\filldraw (0,1)circle(0.025cm)node[left]{a};
|
||||
\filldraw (0,0)circle(0.025cm)node[left]{b};
|
||||
\draw [->,blue] (0,.7)--(0,.3) node at (0,.5)[left]{$\uline{U}_{ab}$};
|
||||
\draw [->,blue] (.5,.7)--(1.5,.7) node at (1,.7)[below]{$\uline{U}_S$};
|
||||
\draw [->,blue] (3.5,.7)--(2.5,.7) node at (3,.7)[below]{$\uline{U}_S$};
|
||||
\draw [->,blue] (1,.4)--(1,.1) node at (1,.25)[right]{$\uline{U}_1$};
|
||||
\draw node at (1,2)[above]{$\uline{Z}_1$};
|
||||
\draw node at (3,2)[above]{$\uline{Z}_2$};
|
||||
\draw node at (1,2)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3,2)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxx}}_{}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end {align*}
|
||||
\vspace{-.5cm}
|
||||
Entweder Stromteiler oder, weil Strom $\uline{I}_1$ wie in b)\\
|
||||
\begin{align*}
|
||||
\uline{U}_S&=\uline{Z}_1\cdot \uline{I}_1=\uline{Z}_2\cdot \uline{I}_2\\
|
||||
\uline{I}_2&=\uline{I}_1\cdot \frac{\uline{Z}_1}{\uline{Z}_2}=2\,\ampere\cdot e^{j0\,\degree}\cdot \frac{(28+j156)\,\ohm}{(14+j460)\,\ohm}\\
|
||||
&=(0{,}681-j0{,}101)\,\ampere=\uuline{0{,}689\,\ampere\cdot e^{-j8{,}43\,\degree}}\\
|
||||
\uline{U}_{ab}&=\uline{U}_S+\uline{U}_M\\
|
||||
&=\uline{Z}_1\cdot \uline{I}_1+jX_M\cdot (\uline{I}_1+\uline{I}_2)\\
|
||||
&=(56+j312)\,\volt+\underbrace{\overbrace{(2+0{,}681-j0{,}101)\,\ampere}^{\uline{I}}\cdot (-j60)\,\ohm}_{\uline{U}_M=(-6{,}06-j160{,}81)\,\volt}\\
|
||||
&=(50+j151)\,\volt=\uuline{159\,\volt\cdot e^{+j71{,}7\,\degree}}\\
|
||||
\end{align*}
|
||||
\vspace{-1.5cm}
|
||||
\item Wirkleistung (über das Magnetfeld übertragen)
|
||||
\begin{align*}
|
||||
\uline{S}_1&=\uline{U}_1\cdot \uline{I}^*_1=[\uline{I}_1\cdot j(X_1-X_M)+\underbrace{\uline{I}\cdot jX_M}_{\uline{U}_M}]\cdot \uline{I}^*_1\\
|
||||
&=\underbrace{[\underbrace{2\,\ampere\cdot j156\,\ohm}_{j312\,\volt}+(-6{,}1-j161)\,\volt]}_{(-6{,}1+j151)\,\volt}\cdot 2\,\ampere=\uuline{(\overbrace{-12{,}2}^{\text{Quelle}}+j302)\,\volt\ampere}\\
|
||||
P&=\uuline{12{,}2\,\watt}\text{ Von Sekundär- nach Primärseite}
|
||||
\end{align*}
|
||||
\end{enumerate}
|
||||
Kontrolle:
|
||||
\begin{align*}
|
||||
\uline{S}_2&=(\uline{U}_M+\uline{I}_2\cdot j460\,\ohm)\cdot \uline{I}^*_2=(12{,}1+j108)\,\volt\ampere\\
|
||||
(P_2&=12{,}1\,\watt\text{ Verbraucher})
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
150
ET2_L_B18_A4.tex
Normal file
150
ET2_L_B18_A4.tex
Normal file
@ -0,0 +1,150 @@
|
||||
\section {Impedanzmatrix}
|
||||
%\enlargethispage{3cm}
|
||||
Geben Sie die Impedanzmatrix $(Z)$ des Vierpols an.\\
|
||||
$R_1=10\,\ohm$, $X_1=100\,\ohm$, $R_2=20\,\ohm$, $X_2=200\,\ohm$, $X_C=-200\,\ohm$, $X_M=120\,\ohm$\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0) node at (.5,-.133) [right] {$U_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\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) [left] {$X_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\fill(.825,-.075)circle(.033);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw [->,red] (1.5,.1)--(1.25,.1)node at(1.375,.1)[above]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2.5cm,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) [left] {$X_C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(1,0)--(1,.2) (3,0)--(1.5,0)--(1.5,.2) (.9,1)--(1,1)--(1,.9)(1.6,1)--(1.5,1)--(1.5,.9)(3,1)--(2.5,1);
|
||||
\fill (2.5,0)circle(0.025cm)(2.5,1)circle(0.025cm)(3,1)circle(0.025cm)(3,0)circle(0.025cm);
|
||||
\draw [->,blue] (3,.8)--(3,.2) node at (3,.5)[right]{$\uline{U}_2$};
|
||||
\draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$X_M$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
Umwandlung in T-Ersatzschaltbild\\
|
||||
%\begin{enumerate}
|
||||
%\item Offener Schalter\\
|
||||
%\begin{align*}
|
||||
%\uline{I}_2&=0\quad\text{Wegen Wicklungssinn $\uline{U}_2$ negativ (bzw. $X_M$ negativ)}\\
|
||||
%\uline{U}_2&=jX_M\cdot \uline{I}_1=jX_M\cdot \frac{\uline{U}_1}{(R+jX_1)}=\frac{(-j40)\,\ohm\cdot 1\,\volt}{(10+j100)\,\ohm}=(0{,}396-j0{,}04)\,\volt\\
|
||||
%&=\uuline{0{,}398\,\volt\cdot e^{-j174{,}3\,\degree}}\\
|
||||
%\end{align*}
|
||||
%\clearpage
|
||||
%\item Ersatzschaltbild\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw node at(.5,-.2){\footnotesize$10$};
|
||||
\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] {$X_1-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$-j20$};
|
||||
\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) [left] {$X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$U_M$};
|
||||
\draw node at(.25,.25){\footnotesize$j120$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_2-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j80$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw node at(.5,-.2){\footnotesize$20$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,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] {$X_C$};
|
||||
\draw node at(.25,-.2){\footnotesize$-j200$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0)--(4,.2) (3.8,1)--(4,1)--(4,.8)(4,0)--(4.5,0)(4,1)--(4.5,1);
|
||||
\fill (0,0)circle(0.025cm)(0,1)circle(0.025cm)(4.5,0)circle(0.025cm)(4.5,1)circle(0.025cm);
|
||||
\draw [->,blue] (0,.7)--(0,.3) node at (0,.5)[left]{$\uline{U}_1$};
|
||||
\draw [->,blue] (4.5,.7)--(4.5,.3) node at (4.5,.5)[right]{$\uline{U}_2$};
|
||||
% \draw [->,blue] (3.5,.7)--(2.5,.7) node at (3,.7)[below]{$\uline{U}'_2$};
|
||||
% \draw node at (0,.5)[left]{$\uline{Z}_{ges}\Rightarrow$};
|
||||
\draw node at (1,1.5)[above]{$\uline{Z}_1$};
|
||||
\draw node at (3,1.5)[above]{$\uline{Z}_2$};
|
||||
\draw node at (1,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3,1.5)[below]{$\overbrace{\phantom{xxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.5)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.25)[below]{$\uline{Z}'=(20-j120)\,\ohm$};
|
||||
\draw node at (3,-1)[above]{$\underbrace{\phantom{xxxxxxxxxxxxxxxxxxxxx}}_{}$};
|
||||
\draw node at (3.25,-.75)[below]{$\uline{Z}_{||}=X_M||\uline{Z}'=(720+j120)\,\ohm$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{Z}_{1}&=(10-j20)\,\ohm &\uline{Z}_{2}&=(20+j80)\,\ohm \\
|
||||
\uline{Z}'&=(20-j120)\,\ohm &\uline{Z}_{||}&=X_M||\uline{Z}'=\frac{j120\cdot (20-j120)}{\cancel{j120}+20-\cancel{j120}}\,\ohm=(720+j120)\,\ohm \\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
Vierpol:
|
||||
\begin{align*}
|
||||
\begin{bmatrix}
|
||||
\uline{U}_{1} \\
|
||||
\uline{U}_{2}
|
||||
\end{bmatrix}=
|
||||
\begin{bmatrix}
|
||||
\uline{Z}_{11} & \uline{Z}_{12} \\
|
||||
\uline{Z}_{21} & \uline{Z}_{22}
|
||||
\end{bmatrix}\cdot
|
||||
\begin{bmatrix}
|
||||
\uline{I}_{1} \\
|
||||
\uline{I}_{2}
|
||||
\end{bmatrix}
|
||||
\end{align*}
|
||||
|
||||
\begin{align*}
|
||||
\uline{U}_1&=\uline{Z}_{11}\cdot \uline{I}_1+\uline{Z}_{12}\cdot \uline{I}_2\\
|
||||
\uline{U}_2&=\uline{Z}_{21}\cdot \uline{I}_1+\uline{Z}_{22}\cdot \uline{I}_2\\
|
||||
\uline{Z}_{11}&=\frac{\uline{U}_1}{\uline{I}_1}\Big|_{\uline{I}_2=0}\\
|
||||
&=\uline{Z}_1+\uline{Z}_{||}=\uuline{(730+j100)\,\ohm}\\
|
||||
\uline{Z}_{12}&=\frac{\uline{U}_1}{\uline{I}_2}\Big|_{\uline{I}_1=0}\\[\baselineskip]
|
||||
\uline{I}_1&=0 \Rightarrow \text{ Leerlauf an Primärseite; Spannung an }\uline{U}_M=\uline{U}_1\\[\baselineskip]
|
||||
&\underline{I}_{X_M}=\uline{I}_2\cdot \frac{jX_C}{R_2+j(X_2-\cancel{X_M})+\cancel{jX_M}+jX_C}\quad\text{( Stromteiler)}\\
|
||||
\uline{U}_1&=jX_M\cdot \uline{I}_{X_M}=\uline{I}_2\cdot \underbrace{jX_M\cdot \frac{jX_C}{R_2+j(X_2+X_C)}}_{\uline{Z}_{12}}\\
|
||||
\uline{Z}_{12}&=\frac{-X_M\cdot X_C}{R_2+j(X_2+X_C)}=\frac{-120\cdot (-200)}{20-j0}\,\ohm=\uuline{+1200\,\ohm}\\
|
||||
\uline{Z}_{21}&=\uuline{\uline{Z}_{12}}\\
|
||||
\uline{Z}_{22}&=\frac{\uline{U}_2}{\uline{I}_2}\Big|_{\uline{I}_1=0}\\
|
||||
&=jX_C||(\uline{Z}_2)+jX_M)=\frac{-j200\cdot (20+j(80+120))}{20+\underbrace{j(80+120-200)}_{=0}}\,\ohm=\uuline{(2000-j200)\,\ohm}
|
||||
\end{align*}
|
||||
Impedanzmatrix:
|
||||
\begin{align*}
|
||||
\uuline{Z}&=
|
||||
\left[
|
||||
\begin{array}{cc}
|
||||
730+j100 & 1200 \\
|
||||
1200 & 2000-j200 \\
|
||||
\end{array}
|
||||
\right]\,\ohm
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
152
ET2_L_B18_A5.tex
Normal file
152
ET2_L_B18_A5.tex
Normal file
@ -0,0 +1,152 @@
|
||||
\section {Netztransformator}
|
||||
Von einem Netztransformator sind folgende Daten gegeben:\\[.5\baselineskip]
|
||||
Primärspannung $U_1=230\,\volt$; Frequenz $f=50\,\hertz$;\\
|
||||
Primärwindungszahl $N_1=784$ Windungen.\\[.5\baselineskip]
|
||||
Induktivität der Primärwicklung $L_1=5{,}66\,\henry$; \\
|
||||
Induktivität der Sekundärwicklung $L_2=1{,}42\,\henry$; \\[.5\baselineskip]
|
||||
Widerstand der Primärwicklung $R_1=800\,\ohm$;\\
|
||||
Widerstand der Sekundärwicklung $R_2=150\,\ohm$.\\[.5\baselineskip]
|
||||
Eisenquerschnitt $A_{Fe}=11\,\centi\square\metre$. Das Feld ist über dem Querschnitt $A_{Fe}$ homogen, die Streuung ist Null!
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Welche Spannung $U_2$ tritt an der Sekundärwicklung auf, wenn sie unbelastet ist, d.h. $I_2=0$ ist?
|
||||
\item Welchen Strom nimmt der Transformator bei sekundärseitigem Leerlauf auf?
|
||||
\item Welche Flussdichte $\widehat{B}_{Fe}$ tritt bei sekundärseitigem Leerlauf im Eisen auf?
|
||||
\end{enumerate}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\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] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0);% node at (.5,-.133) [right] {$U_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$j\omega M\cdot \uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=0cm,rotate=90]%Spannungsquelle
|
||||
\draw (0,0)--(1,0);% node at (.5,-.133) [right] {$U_1$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$j\omega M\cdot \uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_2$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3.5cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw [->,red] (1,.1)--(.75,.1)node at(.875,.1)[above]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(2,0)--(2,.2) (4.5,0)--(2.5,0)--(2.5,.2) (2,.8)--(2,1)--(1.8,1)(2.5,.8)--(2.5,1)--(2.6,1);
|
||||
\fill (0,0)circle(0.025cm)(0,1)circle(0.025cm)(4.5,0)circle(0.025cm)(4.5,1)circle(0.025cm);
|
||||
\draw [->,blue] (0,.8)--(0,.2) node at (0,.5)[left]{$\uline{U}_1$};
|
||||
\draw [->,blue] (4.5,.8)--(4.5,.2) node at (4.5,.5)[right]{$\uline{U}_2$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Ersatzschaltbild
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$};
|
||||
\draw node at(.5,-.2){\footnotesize$800$};
|
||||
\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] {$X_{L_1}-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$+j887$};
|
||||
\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) [left] {$X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[right]{\footnotesize$\uline{U}_M$};
|
||||
\draw node at(.25,.25){\footnotesize$j891$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_{L_2}-X_M$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw node at(.5,-.2){\footnotesize$-j445$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=3cm,yshift=1cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\draw node at(.5,-.2){\footnotesize$150$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0);
|
||||
\fill (0,0)circle(0.025cm)(0,1)circle(0.025cm)(4,0)circle(0.025cm)(4,1)circle(0.025cm);
|
||||
\draw [->,blue] (0,.7)--(0,.3) node at (0,.5)[left]{$\uline{U}_1$};
|
||||
\draw [->,blue] (4,.7)--(4,.3) node at (4,.5)[right]{$\uline{U}_2$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{enumerate}
|
||||
\item Spannung $U_2$
|
||||
\begin{align*}
|
||||
\uline{U}_1&=\uline{I}_1\cdot (R_1+j\omega\cdot L_1) +j\omega\cdot M\cdot \uline{I}_2\\
|
||||
\uline{U}_2&=\uline{I}_2\cdot (R_2+j\omega\cdot L_2)+j\omega\cdot M\cdot \uline{I}_1
|
||||
\end{align*}
|
||||
\footnotesize{Hinweis: Es kann auch $X_{L_1}=\omega\cdot L_1$ bzw. $X_M=\omega\cdot M$ verwendet werden.}\\
|
||||
\normalsize
|
||||
\clearpage
|
||||
\enlargethispage{2cm}
|
||||
\begin{align*}
|
||||
\text{mit }\uline{I}_2&=0\\
|
||||
\uline{U}_1&=\uline{I}_1\cdot (R_1+j\omega\cdot L_1)\\
|
||||
\uline{U}_2&=j\omega\cdot M\cdot \uline{I}_1 \\
|
||||
\Rightarrow\frac{\uline{U}_2}{\uline{U}_1}&=\frac{j\omega\cdot M}{R_1+j\omega\cdot L_1}\\
|
||||
U_2&=U_1\cdot \frac{\sqrt{(\omega\cdot M)^2}}{\sqrt{R^2_1+(\omega\cdot L_1)^2}}\qquad\text{(Betrag: }U_2=|\uline{U}_2|)\qquad \text{(1)}\\[.5\baselineskip]
|
||||
&\text{Streuung ist Null } \Rightarrow \text{Kopplungsfaktor }K=1=\frac{|M|}{\sqrt{L_1\cdot L_2}}\\
|
||||
\Rightarrow M&=\sqrt{L_1\cdot L_2}=\sqrt{5{,}66\cdot 1{,}42}\,\henry=2{,}835\,\henry\\
|
||||
X_{L_1}&=\omega\cdot L_1=2\cdot \pi\cdot 50\,\cancel{\power{\second}{-1}}\cdot 5{,}66\,\volt\cancel{\second}\per\ampere=1{,}778\,\kilo\ohm\quad\text{\footnotesize{(Zur Vollständigkeit $X_{L_2}=446\,\ohm$)}}\\
|
||||
X_{M}&=\omega\cdot M=2\cdot \pi\cdot 50\,\cancel{\power{\second}{-1}}\cdot 2{,}835\,\volt\cancel{\second}\per\ampere=891\,\ohm\\
|
||||
&\text{aus (1):} \Rightarrow U_2=\uline{U}_1\cdot \frac{X_M}{\sqrt{R^2_{1}+X^2_{L_1}}}=230\,\volt\cdot \frac{891}{\sqrt{800^2+1778^2}}=\uuline{105{,}1\,\volt}
|
||||
\end{align*}
|
||||
\item Stromaufnahme bei Leerlauf
|
||||
\vspace{-.25cm}
|
||||
\begin{align*}
|
||||
\uline{I}_1&=\frac{\uline{U}_1}{R_1+j\omega L_1}\\
|
||||
I_1&=\frac{U_1}{\sqrt{R^2_{1}+X^2_{L_1}}}=\frac{230\,\volt}{\sqrt{800^2+1778^2}}\,\frac{1}{\ohm}=\uuline{118\,\milli\ampere}\quad\text{(Betrag)}\\
|
||||
\end{align*}
|
||||
\begin{minipage}[c]{.62\textwidth}
|
||||
\item Flussdichte, Sekundärspule spielt keine Rolle bei $I_2=0$
|
||||
\vspace{-.25cm}
|
||||
\begin{align*}
|
||||
\psi&=L_1\cdot i=N_1\cdot \phi=N_1\cdot B\cdot A\\
|
||||
B&=\frac{L_1\cdot i}{N_1\cdot A}\\
|
||||
\widehat{B}&=\frac{L_1\cdot \widehat{I}}{N_1\cdot A_{Fe}}=\frac{5{,}66\,\volt\second\per\cancel{\ampere}\cdot \sqrt{2}\cdot 118\cdot \power{10}{-3}\,\cancel{\ampere}}{784\cdot 11\cdot \underbrace{\power{10}{-4}}_{(\power{10}{-2})^2}\,\metre^2}=\uuline{1{,}095\,\tesla}
|
||||
\end{align*}
|
||||
\end{minipage}%
|
||||
\begin{minipage}[c]{.33\textwidth}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1.0]
|
||||
\draw(0,0)rectangle(1.5,1.5);
|
||||
\draw(.25,.25)rectangle(1.25,1.25);
|
||||
\draw[->,red](.25,1.375)--(.5,1.375)node at(.375,1.5)[above]{$\phi$};
|
||||
\draw[black!75!](1,1.25)--(1,1.5)node [above]{$A_{Fe}$};
|
||||
\draw[red!70!blue](0,.5)--(-.5,.5)(0,1)--(-.5,1)node [left]{$\uline{I}_1$};
|
||||
\fill[red!70!blue](0,.5)rectangle(.25,1)node at(.25,.75)[right]{$N_1$};
|
||||
\draw[red!70!blue](1.5,.5)--(2,.5)(1.5,1)--(2,1)node [right]{$\uline{I}_2=0$};
|
||||
\fill[red!70!blue](1.25,.5)rectangle(1.5,1)node at(1.25,.75)[left]{$N_2$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
Bemerkung:
|
||||
\begin{align*}
|
||||
U_{L_1}&=\frac{2\pi}{\sqrt{2}}\cdot f\cdot N_1\cdot \widehat{B}\cdot A_{Fe}=I_1\cdot \omega\cdot L_1=209{,}7\,\volt\\
|
||||
\widehat{B}&=\frac{2\pi}{\sqrt{2}}\cdot f\cdot \frac{U_1}{N_1\cdot A_{Fe}}=1{,}2\,\tesla \text{ ist falsch, gilt nur für idealen Transformator! }
|
||||
\end{align*}
|
||||
\footnotesize{Warum? Hier ist der Widerstand $R_1$ nicht berücksichtigt!\\
|
||||
Rechnung oben nur mit Beträgen, nicht im Komplexen.\\ $\uline{U}_1=\uline{U}_{R_1}+\uline{U}_{L_1}=(94{,}4+j209{,}7)\,\volt=230\,\volt\cdot e^{-j114{,}3}$}
|
||||
\end{enumerate}
|
||||
\clearpage
|
||||
}{}%
|
||||
75
ET2_L_B18_A6.tex
Normal file
75
ET2_L_B18_A6.tex
Normal file
@ -0,0 +1,75 @@
|
||||
\section {3-Phasen Spannungssystem}
|
||||
Ein symmetrisches 3-Phasen Spannungssystem mit der Phasenlage 1-2-3 speist einen unsymmetrischen
|
||||
Verbraucher mit den Impedanzen\\
|
||||
$\uline{Z}_1=R_1$; $\quad\uline{Z}_2=R_2+jX_{L2}$; $\quad\uline{Z}_3=jX_{L3}$;\\
|
||||
$U_{12}=400\,\volt\cdot e^{j30\degree}$; $\quad R_2=50\,\ohm$; $\quad X_{L2}=30\,\ohm$; $\quad X_{L3}=25\,\ohm$\\
|
||||
Der Leistungsmesser zeigt $1323\,\watt$ an. Berechnen Sie den Strom $\uline{I}_N$.
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,xshift=0cm,yshift=1.5cm]%Wattmeter
|
||||
\draw (0,0)--(.367,0) (.633,0)--(1,0) node at (.5,0) {W};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw(.25,0)--(.25,.25)--(.5,.25)--(.5,.133)(.5,-.133)--(.5,-1.5);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=1cm,yshift=1.5cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_1$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=1cm,yshift=1.cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_2$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=1cm,yshift=.5cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$\uline{Z}_3$};
|
||||
\draw [->,red] (0,.1)--(.25,.1)node at(.125,.1)[above]{\footnotesize$\uline{I}_3$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(2,0)--(2,1.5)--(1.8,1.5);% (4.5,0)--(2.5,0)--(2.5,.2) (2,.8)--(2,1)--(1.8,1)(2.5,.8)--(2.5,1)--(2.6,1);
|
||||
\filldraw(0,0)circle(0.025cm)node [left]{$N$};
|
||||
\filldraw(0,.5)circle(0.025cm)(1,.5)--(0,.5)node [left]{$L3$};
|
||||
\filldraw(0,1)circle(0.025cm)(1,1)--(0,1)node [left]{$L2$};
|
||||
\filldraw(0,1.5)circle(0.025cm)node at(0,1.5) [left]{$L1$};
|
||||
\fill(.5,0)circle(0.025cm)(2,.5)circle(0.025cm)(2,1)circle(0.025cm);
|
||||
\draw [->,red] (1.25,.1)--(1,.1) node at (1.125,.1)[above]{$\uline{I}_N$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
\draw[black!25!,very thin,step=.5cm](-1,-1)grid(1,1);
|
||||
\draw[->]++(-180:1cm)--++(30:1.732cm)node [above right]{$2$};
|
||||
\draw[->]++(-180:1cm) ++(30:1.732cm)-- ++(270:1.732cm)node [below]{$3$};
|
||||
\draw[->]++(-180:1cm) ++(330:1.732cm)--(-180:1cm)node [left]{$1$};
|
||||
\draw[->](-180:1cm)--(0,0)node [right]{$N$};
|
||||
\draw[->](60:1cm)--(0,0);
|
||||
\draw[->](300:1cm)--(0,0);
|
||||
\draw[blue] node at(-.35,.6){$\uline{U}_{12}$};
|
||||
\draw[blue] node at(-.35,-.6){$\uline{U}_{31}$};
|
||||
\draw[blue] node at(.75,0){$\uline{U}_{23}$};
|
||||
\draw[blue] node at(-.35,.15){$\uline{U}_{1}$};
|
||||
\draw[blue] node at(.35,.25){$\uline{U}_{2}$};
|
||||
\draw[blue] node at(.35,-.25){$\uline{U}_{3}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
%\clearpage
|
||||
\enlargethispage{2\baselineskip}
|
||||
\begin{align*}
|
||||
\uline{I}_N&=\uline{I}_1+\uline{I}_2+\uline{I}_3\qquad \text{(Um Ströme zu berechnen, Spannungen ermitteln)}\\
|
||||
\uline{U}_1&=\frac{\uline{U}_{12}}{\sqrt{3}}=\frac{400\,\volt}{\sqrt{3}}=230{,}9\,\volt\cdot e^{j0\,\degree}\\
|
||||
\uline{U}_2&=230{,}9\,\volt\cdot e^{-j120\,\degree}\\
|
||||
\uline{U}_3&=230{,}9\,\volt\cdot e^{+j120\,\degree}\\
|
||||
P_{Anzeige}&=I_1\cdot U_1\cdot \cos(\varphi_1)\qquad\text{ mit }\varphi_1=0\text{, da }\uline{Z}_1=R_1\\
|
||||
I_1&=\frac{P_{Anzeige}}{U_1}=\frac{1323\,\watt}{230{,}9\,\volt}=5{,}73\,\ampere \qquad\Rightarrow\uline{I}_1=(5{,}73+j0)\,\ampere\\[.5\baselineskip]
|
||||
\uline{I}_2&=\frac{\uline{U}_2}{\uline{Z}_2}=\frac{230{,}9\,\volt\cdot e^{-j120\,\degree}}{(50+j30)\,\ohm}=\frac{230{,}9\,\volt\cdot e^{-j120\,\degree}}{58{,}31\,\ohm\cdot e^{j30{,}96\,\degree}}=3{,}96\,\ampere\cdot e^{-j150{,}96\,\degree}=(-3{,}463-j1{,}923)\,\ampere\\
|
||||
\uline{I}_3&=\frac{\uline{U}_3}{\uline{Z}_3}=\frac{230{,}9\,\volt\cdot e^{j120\,\degree}}{25\,\ohm\cdot e^{j90\,\degree}}=9{,}24\,\ampere\cdot e^{j30\,\degree}=(8+j4{,}62)\,\ampere\\[.5\baselineskip]
|
||||
\uline{I}_N&=\uline{I}_1+\uline{I}_2+\uline{I}_3=(10{,}3+j2{,}7)\,\ampere=\uuline{10{,}6\,\ampere\cdot e^{j14{,}7\,\degree}}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
116
ET2_L_B18_A7.tex
Normal file
116
ET2_L_B18_A7.tex
Normal file
@ -0,0 +1,116 @@
|
||||
\section {3-Phasen System mit unsymmetrischem Verbraucher}
|
||||
Ein symmetrisches 3-Phasen System mit der Phasenlage 1-2-3 speist einen unsymmetrischen
|
||||
Verbraucher. Gegeben sind:\\
|
||||
$\uline{U}_{31}=380\,\volt\cdot e^{j0\,\degree}$; $R_1=60\,\ohm$; $\uline{X}_{L_1}=20\,\ohm$; $R_2=100\,\ohm$; $\uline{X}_{C_2}=-80\,\ohm$; $X_{L_3}=50\,\ohm$;\\[\baselineskip]
|
||||
Es ist die Anzeige des Leistungsmessinstrumentes zu berechnen!
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=4]
|
||||
\begin{scope}[>=latex,xshift=0cm,yshift=1.5cm]%Wattmeter
|
||||
\draw (0,0)--(.367,0) (.633,0)--(1,0) node at (.5,0) {$W$};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw(.25,0)--(.25,.25)--(.5,.25)--(.5,.133)(.5,-.133)--(.5,-1);
|
||||
\end{scope}
|
||||
\begin{scope}[xshift=1cm,yshift=.5cm,rotate=60,scale=.577]
|
||||
\begin{scope}[>=latex,xshift=0cm,yshift=0cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.1330667) [left] {$R_1$};
|
||||
\draw [<-,red] (.8,.1)--(1.2,.1)node at(1,.14)[left]{$\uline{I}_{12}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Spule -
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.1330667) [ left] {$L_1$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\end{scope}
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2.01cm,yshift=.75cm,rotate=120,scale=.577]%Spule
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,-.1330667) [ right] {$L_3$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (1.2,-.1)--(.8,-.1)node at(1,-.14)[right]{$\uline{I}_{31}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=1.289cm,yshift=.5cm,scale=.577]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1.289cm,yshift=.25cm,scale=.577]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.1133) [above] {$C_2$};
|
||||
% \draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_{C2}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(2.154,0)--(2.154,.5)(.9,.5)--(1.289,.5) (1.289,.5)--(1.289,.25)(1.867,.5)--(1.867,.25);
|
||||
\filldraw(0,0)circle(0.025cm)node [left]{$L3$};
|
||||
\filldraw(0,.5)circle(0.025cm)(1,.5)--(0,.5)node [left]{$L2$};
|
||||
\filldraw(0,1.5)circle(0.025cm)node at(0,1.5) [left]{$L1$};
|
||||
\fill(.5,.5)circle(0.025cm)(1,.5)circle(0.025cm)(1.289,.5)circle(0.025cm)(1.867,.5)circle(0.025cm)(2.154,.5)circle(0.025cm)(1.577,1.5)circle(0.025cm);
|
||||
\draw [->,blue] (.8,1.4)--(.8,0.6) node at (.8,1)[right]{$\uline{U}_{12}$};
|
||||
\draw [->,red] (.9,1.6)--(1.1,1.6) node at (1,1.6)[above]{$\uline{I}_1$};
|
||||
\draw +(1,1.5)--(1.577,1.5)--(2.154,.5cm)--(1.75,.5);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
%\begin{align*}
|
||||
% \begin{tikzpicture}[scale=2,rotate=-150]
|
||||
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]
|
||||
% \draw[black!25!,very thin,step=.5cm](-1,-1)grid(1,1);
|
||||
% \draw[->]++(-180:1cm)--++(30:1.732cm)node [above right]{$2$};
|
||||
% \draw[->]++(-180:1cm) ++(30:1.732cm)-- ++(270:1.732cm)node [below]{$3$};
|
||||
% \draw[->]++(-180:1cm) ++(330:1.732cm)--(-180:1cm)node [left]{$1$};
|
||||
% \draw[->](-180:1cm)--(0,0)node [right]{$N$};
|
||||
% \draw[->](60:1cm)--(0,0);
|
||||
% \draw[->](300:1cm)--(0,0);
|
||||
% \draw[blue] node at(-.35,.6){$\uline{U}_{12}$};
|
||||
% \draw[blue] node at(-.35,-.6){$\uline{U}_{31}$};
|
||||
% \draw[blue] node at(.75,0){$\uline{U}_{23}$};
|
||||
% \draw[blue] node at(-.35,.15){$\uline{U}_{1}$};
|
||||
% \draw[blue] node at(.35,.25){$\uline{U}_{2}$};
|
||||
% \draw[blue] node at(.35,-.25){$\uline{U}_{3}$};
|
||||
% \end{scope}
|
||||
% \end{tikzpicture}
|
||||
%\end{align*}
|
||||
%%\clearpage
|
||||
\begin{minipage}[c]{0.25\textwidth}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1]
|
||||
\draw[->,blue](0:0)--(0:1cm)node [right]{$\uline{U}_{31}$};
|
||||
\draw[->,blue](0:0)--(120:1cm)node [above]{$\uline{U}_{23}$};
|
||||
\draw[->,blue](0:0)--(-120:1cm)node [below]{$\uline{U}_{12}$};
|
||||
\draw[->,red](0:0)--(141.2:.577cm)node [above left]{$\uline{I}_1$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
%\hfill
|
||||
\begin{minipage}[l]{0.75\textwidth}
|
||||
\begin{align*}
|
||||
\uline{U}_{31}&=380\,\volt\cdot e^{j0\,\degree}\\
|
||||
\uline{U}_{12}&=380\,\volt\cdot e^{-j120\,\degree}\\
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\begin{align*}
|
||||
P&=\Re\{\uline{U}_{12}\cdot I^*_1\}\\
|
||||
\uline{I}_1&=\uline{I}_{12}-\uline{I}_{31}\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{I}_{12}&=\frac{\uline{U}_{12}}{R_1+jX_{L_1}}
|
||||
=\frac{\overbrace{380\,\volt\cdot e^{-j120\,\degree}}^{(-190-j329{,}09)\,\volt}}{\underbrace{63,246\,\ohm\cdot ^{j18{,}435\,\degree}}_{(60+j20)\,\ohm}}=6{,}01\,\ampere\cdot e^{-j138{,}4\,\degree}=(-4{,}495-j3{,}986)\,\ampere\\
|
||||
\uline{I}_{31}&=\frac{\uline{U}_{31}}{jX_{L_3}}=\frac{380\,\volt\cdot e^{j0\,\degree}}{j50\,\ohm}=-j\frac{380\,\volt}{50\,\ohm}=-j7{,}6\,\ampere \qquad\tag{$\frac{1}{j}=-j$}\\
|
||||
\uline{I}_1&=[(-4{,}495-j3{,}986)-(-j7{,}6)]\,\ampere\\ &=(-4{,}495+j3{,}614)\,\ampere=5{,}77\,\ampere\cdot e^{j141{,}2\,\degree}\\[\baselineskip]
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2.5]
|
||||
\draw[->,blue](0:0)--(0:1cm)node [right]{$\uline{U}_{31}$};
|
||||
\draw[->,blue](0:0)--(120:1cm)node [above]{$\uline{U}_{23}$};
|
||||
\draw[->,blue](0:0)--(-120:1cm)node [below]{$\uline{U}_{12}$};
|
||||
\draw[->,red!50!blue](0:0)--(-138.4:.601cm)node [below]{$\uline{I}_{12}$};
|
||||
\draw[->,red!50!blue](0:0)--(-90:.76cm)node [below]{$\uline{I}_{31}$};
|
||||
\draw[<-,red!75!blue](-138.4:.601cm)-- ++(90:.76cm)node [below left]{$\uline{I}_{31}$};
|
||||
\draw[->,red](0:0)--(141.2:.577cm)node [above left]{$\uline{I}_1$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\uline{S}&=\uline{U}_{12}\cdot \uline{I}^*_1=\underbrace{380\,\volt\cdot e^{-j120\,\degree}}_{(-190-j329{,}09)\,\volt}\cdot \underbrace{5{,}77\,\ampere\cdot e^{-j141{,}2\,\degree}}_{(-4{,}495-j3{,}614)\,\ampere}\\
|
||||
&=2192{,}6\,\volt\ampere\cdot e^{j98{,}8\,\degree}=(\underbrace{-335{,}28}_{P}+\underbrace{j2165{,}92}_{jQ})\,\volt\ampere\\
|
||||
P&=S\cdot \cos(\varphi)=2192{,}6\,\volt\ampere\cdot \underbrace{\cos(98{,}8\,\degree)}_{-0{,}153}=\uuline{-335{,}44\,\watt}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
108
ET2_L_B18_A8.tex
Normal file
108
ET2_L_B18_A8.tex
Normal file
@ -0,0 +1,108 @@
|
||||
\section {Strangströme 3-Phasen System mit unsymmetrischem Verbraucher}
|
||||
Ein symmetrisches 3-Phasen System mit der Phasenlage 1-2-3 speist einen unsymmetrischen
|
||||
Verbraucher. Gegeben sind:\\[\baselineskip]
|
||||
$\uline{U}_{12}=400\,\volt\cdot e^{j30\,\degree}$; $\quad \uline{X}_C=-100\,\ohm$; $\quad \uline{X}_{L_1}=125\,\ohm$; $\quad X_{L_2}=60\,\ohm$;$\quad R=80\,\ohm$\\[\baselineskip]
|
||||
Berechnen Sie die $3$ Strangströme, den Leiterstrom $\uline{I}_1$ und die Anzeige des Leistungsmessers!
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\draw[black!15!,very thin](0,0)grid(4,3);
|
||||
\begin{scope}[>=latex,very thick,xshift=2.268cm,yshift=1.995cm,scale=1,rotate=30]%Kondensator
|
||||
\draw (-1,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(2,0)node at (.5,.1133) [left] {$X_C$};
|
||||
\draw [<-,red] (1.15,.1)--(1.55,.1)node at(1.25,.1)[left]{$\uline{I}_C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=2.268cm,yshift=1.005cm,rotate=-30,scale=1]%Spule
|
||||
\draw (-1,0)--(.3,0) (.7,0)--(2,0)node at (.5,-.1330667) [ left] {$X_{L_1}$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (-.3,-.1)--(-.7,-.1)node at(-.5,-.14)[left]{$\uline{I}_L$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=4cm,yshift=2cm,rotate=90]%Wattmeter
|
||||
\draw (0,0)--(.367,0) (.633,0)--(1,0) node at (.5,0) {W};
|
||||
\draw (.5,0)circle(.133);
|
||||
\draw(.45,-.125)--(.45,-1)--(0,-1)--(-2,-1)--(-2,0)(.55,-.125)--(.55,-1)--(1,-1)--(1,0);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_{L_2}$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [->,red] (.8,-.1)--(1.2,-.1)node at(1,-.1)[right]{\footnotesize$\uline{I}_{RL}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=4cm,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] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(4,0)(0,1.5)--(1.433,1.5)(0,3)--(4,3);
|
||||
\filldraw(0,0)circle(0.033cm)node [left]{L3};
|
||||
\filldraw(0,1.5)circle(0.033cm)node [left]{L2};
|
||||
\filldraw(0,3)circle(0.033cm)node [left]{L1};
|
||||
\draw [->,blue] (0,2.8)--(0,1.7)node at(0,2.25)[right]{\footnotesize$\uline{U}_{12}$};
|
||||
\draw [->,blue] (0,1.3)--(0,.2)node at(0,.75)[right]{\footnotesize$\uline{U}_{23}$};
|
||||
\fill(1.433,1.5)circle(0.033cm)(4,3)circle(0.033cm)(4,2)circle(0.033cm)(4,1)circle(0.033cm)(4,0)circle(0.033cm);
|
||||
\draw [->,red] (.8,3.1)--(1.2,3.1)node at(1,3.1)[above]{\footnotesize$\uline{I}_1$};
|
||||
\draw [red!50!blue,very thick]node at (4,3)[above]{u};
|
||||
\draw [red!50!blue,very thick]node at (4,0)[below]{w};
|
||||
\draw [red!50!blue,very thick]node at (1.433,1.5)[above left]{v};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{minipage}[c]{0.35\textwidth}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1.5]
|
||||
\draw[thin](0,0)--(1,0);
|
||||
\draw[->,blue](0:0)--(30:1cm)node [right]{$\uline{U}_{12}$};
|
||||
\draw[->,blue](0:0)--(-90:1cm)node [below]{$\uline{U}_{23}$};
|
||||
\draw[->,blue](0:0)--(150:1cm)node [left]{$\uline{U}_{31}$};
|
||||
\draw[->,red](0:0)--(-153.2:.48cm)node [left]{$\uline{I}_1$};
|
||||
\draw[thin](0:.7cm)arc(0:30:.7cm)node at(10:1cm){\footnotesize$30\,\degree$};
|
||||
\draw[thin](30:.7cm)arc(30:150:.7cm)node at(90:1cm){\footnotesize$120\,\degree$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\begin{minipage}[c]{0.35\textwidth}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=1.5]
|
||||
\draw[thin](0,0)--(1,0);
|
||||
\draw[thin](0:0cm)--+(0:1cm);%node [right]{$\uline{U}_{12}$};
|
||||
\draw[thin](0:1cm)--+(-60:1cm);%node [below]{$\uline{U}_{23}$};
|
||||
\draw[thin](0:1cm)--+(60:1cm);%node [left]{$\uline{U}_{31}$};
|
||||
\draw[->,blue](0:0cm)--+(30:1.732cm)node [right]{$\uline{U}_{12}$};
|
||||
\draw[->,blue](30:1.732cm)--+(-90:1.732cm)node [below]{$\uline{U}_{23}$};
|
||||
\draw[->,blue](-30:1.732cm)--(0:0cm)node [left]{$\uline{U}_{31}$};
|
||||
\draw[thin](0:.7cm)arc(0:30:.7cm)node at(15:1cm){\footnotesize$30\,\degree$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
%%\hfill
|
||||
\begin{minipage}[l]{0.3\textwidth}
|
||||
\begin{align*}
|
||||
\uline{U}_{12}&=400\,\volt\cdot e^{j30\,\degree}\\
|
||||
\uline{U}_{31}&=400\,\volt\cdot e^{j150\,\degree}\\
|
||||
\uline{U}_{31}&=400\,\volt\cdot e^{j270\,\degree}=400\,\volt\cdot e^{-j90\,\degree}\\
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
Strangströme:
|
||||
\begin{align*}
|
||||
\uline{I}_C&=\frac{\uline{U}_{12}}{jX_C}=\frac{400\,\volt\cdot e^{j30\,\degree}}{100\,\ohm\cdot e^{-j90\,\degree}}=\uuline{4\,\ampere\cdot e^{j120\,\degree}}=\uuline{(-2+j3{,}46)\,\ampere}\\
|
||||
\uline{I}_L&=\frac{\uline{U}_{23}}{jX_{L_1}}=\frac{400\,\volt\cdot e^{-j90\,\degree}}{125\,\ohm\cdot e^{j90\,\degree}}=\uuline{3{,}2\,\ampere\cdot e^{-j180\,\degree}}=-3{,}2\,\ampere\\
|
||||
\uline{I}_{RL}&=\frac{\uline{U}_{31}}{R+jX_{L_2}}=\frac{400\,\volt\cdot e^{j150\,\degree}}{(80+j60)\,\ohm}=\frac{400\,\volt\cdot e^{j150\,\degree}}{100\,\ohm\cdot e^{j36{,}9\,\degree}}\\
|
||||
&=\uuline{4\,\ampere\cdot e^{j113{,}1\,\degree}}=\uuline{(-1{,}57+j3{,}68)\,\ampere}\\
|
||||
\intertext{Leiterströme:}
|
||||
\uline{I}_1&=\uline{I}_C-\uline{I}_{RL}=(-0{,}439-j0{,}214)\,\ampere=\uuline{0{,}48\,\ampere\cdot e^{-j153\,\degree}}\\
|
||||
\text{Zur Vollständigkeit}\\
|
||||
\uline{I}_2&=\uline{I}_L-\uline{I}_C=(-1{,}2-j3{,}46)\,\ampere=\uuline{3{,}66\,\ampere\cdot e^{j109{,}1\,\degree}}\\
|
||||
\uline{I}_3&=\uline{I}_{RL}-\uline{I}_L=(1{,}63+j3{,}68))\,\ampere=\uuline{4{,}03\,\ampere\cdot e^{j66{,}1\,\degree}}
|
||||
\intertext{Anzeige der Wirkleistung:}
|
||||
P&=\Re\{\uline{U}_{31}\cdot I^*_{RL}\}\\
|
||||
&=\Re\{400\,\volt\cdot e^{j150\,\degree}\cdot 4\,\ampere\cdot e^{-j113{,}1\,\degree}\}\\
|
||||
&=\Re\{1600\,\volt\ampere\cdot e^{j37\,\degree}\}\\
|
||||
&=\Re\{(1280-j960)\,\volt\ampere\}=\uuline{1280\,\watt}\\
|
||||
\intertext{oder}
|
||||
P&=U_{31}\cdot I_{31}\cdot \cos(\varphi_{_{31}})=400\,\volt\cdot 4\,\ampere\cdot \cos(36{,}9\,\degree)=\uuline{1280\,\watt}\\
|
||||
\intertext{oder}
|
||||
P&=I^2_{RL}\cdot R=(4\,\ampere)^2\cdot 80\,\ohm=\uuline{1280\,\watt}\qquad \text{(Betrag von $\uline{I}_{RL}$!)}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
51
ET2_L_B19_A1.tex
Normal file
51
ET2_L_B19_A1.tex
Normal file
@ -0,0 +1,51 @@
|
||||
\section {Resonanzfrequenz Zweipol}
|
||||
Berechnen Sie die Resonanzfrequenz des abgebildeten Zweipols
|
||||
$L=12\,\milli\henry$, $C_1=2\,\micro\farad$, $R=160\,\ohm$
|
||||
\vspace{-.5cm}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\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]
|
||||
\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=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw (0,0)--(2,0)--(2,.2) (1,1)--(2,1)--(2,.9);
|
||||
\fill (0,0)circle(0.05cm)(0,1)circle(0.05cm);
|
||||
% \draw [->,blue] (2.5,.8)--(2.5,.2) node at (2.5,.5)[right]{$\uline{U}_2$};
|
||||
% \draw[<->](1.5,1.125)arc(60:120:.5cm)node at(1.25,1.25)[above]{$M$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{align*}
|
||||
\text{Falsch ist: } f_{res}&=\frac{1}{2\pi\sqrt{L\cdot C}}=\frac{1}{2\pi\sqrt{12\cdot \power{10}{-3}\,\ohm\second\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\ohm}}}=\uline{1027\,\frac{1}{\second}}\\
|
||||
&\text{Gilt nur für } R\rightarrow\infty \text{, reine Reihen- oder Parallelschaltung.}\\[.5\baselineskip]
|
||||
\text{Bei Resonanz: }\Im\, (\uline{Z})&=0 \\
|
||||
\uline{Z}=jX_L+(R||jX_C)&=j\omega L+\frac{R\cdot \frac{-j}{\omega C}}{R-j\cdot \frac{1}{\omega C}}\underbrace{\cdot \frac{R+j\frac{1}{\omega C}}{R+j\frac{1}{\omega C}}}_{\text{konj. komplex erweitern}}\hspace{-.75cm}=j\omega L+\frac{R^2\cdot \frac{-j}{\omega C}+R\frac{1}{(\omega C)^2}}{R^2+\frac{1}{(\omega C)^2}}\\
|
||||
&=\underbrace{\frac{\frac{R}{(\omega C)^2}}{R^2+\frac{1}{(\omega C)^2}}}_{\Re\,= \text{ Widerstand bei Resonanz, }\omega_{res}}+j\underbrace{\Big(\omega L-\frac{R^2}{\omega C\big(R^2+(\frac{1}{\omega C})^2\big)}\Big)}_{\Im =0\text{ bei }\omega_{res}}\\
|
||||
&\omega L-\frac{R^2}{\omega C\left(R^2+\frac{1}{(\omega C)^2}\right)}=0\\
|
||||
\Im\quad\Rightarrow\, &\, \omega_{res}\cdot \left[L(\omega_{res}\cdot C\cdot \left(R^2+\frac{1}{(\omega_{res}\cdot C)^2}\right)\right]=R^2\\
|
||||
&\omega^2_{res}\cdot L\cdot C\cdot R^2+\frac{L}{C}=R^2\\
|
||||
\Rightarrow\, &\omega^2_{res}=\frac{R^2-\frac{L}{C}}{L\cdot C\cdot R^2}=\frac{1}{L\cdot C}-\frac{1}{(R\cdot C)^2}=\frac{1}{12\,\milli\henry\cdot 2\,\micro\farad}-\frac{1}{(160\,\ohm\cdot 2\,\micro\farad)^2}\\
|
||||
&=\frac{1}{12\cdot \power{10}{-3}\,\cancel{\ohm}\second\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\cancel{\ohm}}}-\frac{1}{160\,\cancel{\ohm}\cdot 2\cdot \power{10}{-6}\,\frac{\second}{\cancel{\ohm}}}\\
|
||||
&=4{,}17\cdot \power{10}{7}\power{\second}{-2}-9{,}77\cdot \power{10}{6} \power{\second}{-2}
|
||||
=3{,}19\cdot \power{10}{7}\cdot \power{\second}{-2}\\
|
||||
\omega_{res}&=\sqrt{\omega^2_{res}}=5648\cdot \power{\second}{-1}\\
|
||||
f_{res}&=\frac{\omega_{res}}{2\pi}=\uuline{899\,\hertz}\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\intertext{Nicht gefragt Resonanzwiderstand: }\\
|
||||
Z&=\frac{R}{1+(\omega \cdot C\cdot R)^2}=\frac{160\,\ohm}{(5648\cdot \cancel{\power{\second}{-1}}\cdot 2\cdot \power{10}{-6}\,\cancel{\frac{\second}{\ohm}}\cdot 160\,\cancel{\ohm})^2+1}=\uuline{37{,}5\,\ohm}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
105
ET2_L_B19_A2.tex
Normal file
105
ET2_L_B19_A2.tex
Normal file
@ -0,0 +1,105 @@
|
||||
\section {RLC-Reihenschwingkreis}
|
||||
Von einem RLC-Reihenschwingkreis ist die Abhängigkeit $I(f)$ gegeben, siehe Kennlinie. Der Schwingkreis wird von einer konstanten sinusförmigen Spannung gespeist mit \\
|
||||
$U=100\,\volt$.\\
|
||||
Bestimmen Sie die Bauelemente $R$, $L$, und $C$!\\
|
||||
Sie dürfen auch mit der Näherung Güte $\gg 1$ rechnen.
|
||||
\begin{align*}
|
||||
\tikzstyle{every pin}=[fill=white,draw=black,font=\footnotesize]
|
||||
\tikzstyle{every axis legend}+=[at={(0.935,0.1)},anchor=south east,inner sep=0pt]%
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\begin{axis}[xlabel=$f (Hz)$,ylabel=$I (A)$,title={Kennlinie },grid=major,xminorgrids=false,yminorgrids=false,xmin=300,xmax=600,ymin=0,ymax=10]
|
||||
\addplot[color=red,smooth,thick]
|
||||
plot coordinates {
|
||||
(300,1.15)
|
||||
(350,1.85)
|
||||
(400,3.63)
|
||||
(432.2,7.07)
|
||||
%(445,9.26)
|
||||
%(450,9.84)
|
||||
(452,9.96)
|
||||
(454,10)
|
||||
(456,9.96)
|
||||
(477.15,7.07)
|
||||
(500,4.57)
|
||||
(550,2.49)
|
||||
(600,1.73)
|
||||
};
|
||||
\addplot[color=blue]
|
||||
plot coordinates {
|
||||
(432.2,7.07)
|
||||
(477.15,7.07)
|
||||
};
|
||||
\axispath\node[coordinate,pin=below right:f res]
|
||||
at (axis cs:454,10) {};
|
||||
\axispath\node[coordinate,pin=left:7.07A]
|
||||
at (axis cs:430,7.07) {};
|
||||
\axispath\node[coordinate,pin=below:B]
|
||||
at (axis cs:454,7.07) {};
|
||||
\end{axis}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand
|
||||
\draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%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=2cm,yshift=0cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\fill (0,0)circle(0.025cm)(3,0)circle(0.025cm);
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
\text{Ablesen: } I_{max}&=10\,\ampere\qquad f_{res}=454\,\hertz\\
|
||||
R&=\frac{U}{I_{res}}=\frac{100\,\volt}{10\,\ampere}=\uuline{10\,\ohm}\qquad\text{Bei Resonanz: Nur Spannung über R, da $\Im=0$}\\
|
||||
I&=\frac{I_{max}}{\sqrt{2}}=\frac{10\,\ampere}{\sqrt{2}}=7{,}07\,\ampere\Rightarrow\Delta f=B\approx 45\,\hertz\qquad\text{Bandbreite-Grenzfrequenzen}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
&C \text{ und } L \text{ bestimmen}\\
|
||||
f_{res}&=\frac{1}{2\pi\sqrt{L\cdot C}}=454\,\hertz\tag{1}\\
|
||||
\text{Näherung }B&\approx \frac{f_{res}}{Q_S}\Rightarrow Q_S\approx \frac{f_{res}}{B}=\frac{454\,\hertz}{45\,\hertz}=10{,}1\\
|
||||
Q_S&=\frac{1}{R}\cdot \sqrt{\frac{L}{C}}=10{,}1\qquad\text{Gleichung für L und C}\tag{2}\\
|
||||
\sqrt{L}&=\underbrace{\frac{1}{2\pi \cdot f_{res}\cdot \sqrt{C}}}_{\text{aus (1)}}
|
||||
=\underbrace{Q_S\cdot \sqrt{C}\cdot R}_{\text{aus (2)}}\\
|
||||
C&=\frac{1}{2\pi \cdot f_{res}\cdot Q_S\cdot R}=\frac{1}{2\pi \cdot 454\,\hertz\cdot 10{,}1\cdot 10\,\ohm}=3{,}47\cdot \power{10}{-6}\,\frac{\ampere\second}{\volt}=\uuline{3{,}47\,\micro\farad}\\
|
||||
L&=(\sqrt{L})^2=Q^2_S\cdot C\cdot R^2=10{,}1^2\cdot 3{,}47\,\micro\frac{\ampere\second}{\volt}\cdot (10\,\ohm)^2=\uuline{35{,}4\,\milli\henry}
|
||||
\intertext{Alternative: 2. Punkt auf der Kurve z.B. $4\,\ampere$ bei $405\,\hertz$ ergibt 2 Gleichungen.}
|
||||
\omega&=2\pi\cdot f\qquad\Rightarrow\omega=\omega_{405}=2545;\qquad\omega_{res}=2853\\
|
||||
Z_{405}&=\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}=\frac{U}{I_{405}}=\frac{100\,\volt}{4\,\ampere}=25\,\ohm\\
|
||||
&\Rightarrow \,(\omega L-\frac{1}{\omega C})^2=Z^2-R^2\\
|
||||
&\omega L-\frac{1}{\omega C}=\pm \sqrt{Z^2-R^2}\\
|
||||
&\Rightarrow L=\frac{1}{\omega}\cdot \Big(\pm \sqrt{Z^2-R^2}+\frac{1}{\omega C}\Big)\\
|
||||
L&=\frac{1}{\omega^2_{res}\cdot C}=\Big(\pm \frac{\sqrt{Z^2-R^2}}{\omega}+\frac{1}{\omega^2 C}\Big)\quad\text{(Formelsammlung: $\omega_{res}=\frac{1}{\sqrt{LC}}$)}\\
|
||||
&\Big(\frac{1}{\omega^2_{res}}-\frac{1}{\omega^2}\Big)\cdot \frac{1}{C}
|
||||
=\pm\frac{\sqrt{Z^2-R^2}}{\omega}\quad\text{(nach C auflösen)}\\
|
||||
\end{align*}
|
||||
\begin{align*}
|
||||
C&=\frac{\Big(\frac{1}{\omega^2_{res}}
|
||||
-\frac{1}{\omega^2}\Big)\cdot \omega}{\pm \sqrt{Z^2-R^2}}
|
||||
=\frac
|
||||
{\bigg(
|
||||
\overbrace{
|
||||
\frac{1}{\Large(2835\cdot \frac{1}{\second}\Large)^2}
|
||||
-\frac{1}{\Large(2545\cdot \frac{1}{\second}\Large)^2}
|
||||
}^{31{,}54\cdot \power{10}{-9}}
|
||||
\bigg)
|
||||
\cdot 2545\cdot \frac{1}{\second}}
|
||||
{\pm\sqrt{(25\,\ohm)^2-(10\,\ohm)^2}}
|
||||
=\frac{-80{,}3\cdot \power{10}{6}}{\underbrace{\pm}_{- gilt} \sqrt{525}}=\uuline{3{,}5\,\micro\farad}\\
|
||||
L&=\frac{1}{\omega^2_{res}\cdot C}=\frac{1}{(2835\,\frac{1}{\second})^2\cdot 3{,}5\,\micro\farad}=\frac{1}{28{,}07\,\frac{\ampere}{\volt\second}}=\uuline{35{,}6\milli\henry}
|
||||
\end{align*}
|
||||
Rundungsfehler durch Ablesung und Näherung
|
||||
\clearpage
|
||||
}{}%
|
||||
157
ET2_L_B19_A3.tex
Normal file
157
ET2_L_B19_A3.tex
Normal file
@ -0,0 +1,157 @@
|
||||
\section {Effektivwert und Klirrfaktor}
|
||||
\begin{minipage}[c]{.7\textwidth}
|
||||
Bild 1 zeigt einen Teil aus dem Ersatzschaltbild eines Transformators, aus dem hervorgeht,
|
||||
dass sich der Leerlaufstrom $i_0(t)$ zusammensetzt aus dem (verzerrten) Magnetisierungsstrom
|
||||
$i_\mu(t)$ und dem Strom $i_{Fe}(t)$, der die Eisenverluste repräsentiert.
|
||||
\end{minipage}
|
||||
\begin{minipage}[c]{.3\textwidth}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1.5]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spule |
|
||||
\draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_{1h}$};
|
||||
\fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
\draw [<-,red] (.75,.1)--(.95,.1)node at(.85,.1)[left]{\footnotesize$i_{\mu}(t)$};
|
||||
\end{scope}
|
||||
\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] {$R_{Fe}$};
|
||||
\draw [<-,red] (.75,-.1)--(.95,-.1)node at(.85,-.1)[right]{\footnotesize$i_{Fe}(t)$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
\draw(0,.1)--(0,0)--(1,0)--(1,.1)(0,.9)--(0,1)--(1,1)--(.9,1)(.5,1)--(.5,1.5)(.5,0)--(.5,-.5);
|
||||
\fill (0.5,0)circle(0.025cm)(.5,1)circle(0.025cm);
|
||||
\draw [<-,red] (.6,-.35)--(.6,-.15)node at(.6,-.25)[right]{\footnotesize$i_{0}(t)$};
|
||||
\end{scope}
|
||||
\draw node at (.5,-1){Bild 1};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
Bild 2 zeigt die zeitlichen Verläufe von $i_\mu(t)$ und $i_{Fe}(t)$, welche durch folgende Fourier-Reihen approximiert werden können:\\
|
||||
$i_\mu(t)=10\,\milli\ampere\cdot \cos(\omega t) + 2,88\,\milli\ampere\cdot \cos(3\omega t)$;\\ $i_{Fe}(t)=-4\,\milli\ampere\cdot \sin(\omega t)$\\[\baselineskip]
|
||||
Der resultierende, in Bild 3 dargestellte Leerlaufstrom ist die Summe:\\
|
||||
$i_0(t)= i_\mu(t)+i_{Fe}(t)$
|
||||
\begin{align*}
|
||||
% \begin{tikzpicture}[scale=1.5]
|
||||
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spule |
|
||||
% \draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$L_{1h}$};
|
||||
% \fill (.3,-0.0667)rectangle(.7,0.0667);
|
||||
% \draw [<-,red] (.75,.1)--(.95,.1)node at(.85,.1)[left]{\footnotesize$i_{\mu}(t)$};
|
||||
% \end{scope}
|
||||
% \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] {$R_{Fe}$};
|
||||
% \draw [<-,red] (.75,-.1)--(.95,-.1)node at(.85,-.1)[right]{\footnotesize$i_{Fe}(t)$};
|
||||
% \end{scope}
|
||||
% \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen.
|
||||
% \draw(0,.1)--(0,0)--(1,0)--(1,.1)(0,.9)--(0,1)--(1,1)--(.9,1)(.5,1)--(.5,1.5)(.5,0)--(.5,-.5);
|
||||
% \fill (0.5,0)circle(0.025cm)(.5,1)circle(0.025cm);
|
||||
% \draw [<-,red] (.6,-.35)--(.6,-.15)node at(.6,-.25)[right]{\footnotesize$i_{0}(t)$};
|
||||
% \end{scope}
|
||||
%\draw node at (.5,-1){Bild 1};
|
||||
% \end{tikzpicture}
|
||||
\begin{tikzpicture}[scale=1.25]
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=0]
|
||||
\foreach \ii in {5} { % Enter Number of Decades in x
|
||||
\foreach \jj in {2.5} { % Enter Number of Decades in y
|
||||
\foreach \i in {1,2,...,\ii} {
|
||||
\foreach \j in {0,1,2,...,\jj} {
|
||||
\draw[black!50!, step=0.5] (0,-.5) grid (\ii,\jj); % Draw Sub Linear grid
|
||||
}}% End Log Grid
|
||||
\draw[black!80!] (0,-.5) grid (\ii,\jj); % Draw Linear grid
|
||||
\draw [->,thick] (0,-.5)--(0,\jj+.25) node (yaxis) [above] {$i\,[\milli\ampere]$};
|
||||
\draw [->,thick] (0,1)--(\ii+.25,1) node (xaxis) [right] {$\omega t$}; % Draw axes
|
||||
\draw node at (0,-.5)[below]{$0$};
|
||||
\draw node at (2,-.5)[below]{$\pi$};
|
||||
\draw node at (4,-.5)[below]{$2\pi$};
|
||||
\foreach \y in {-10,0,10}% y Axis Label:
|
||||
\node [anchor=east] at(0,\y/10+1){$\y$};
|
||||
}}
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=1cm]
|
||||
\draw[color=red,thick,domain=0:5,smooth,samples=100] plot[id=iomega] function{1*cos(.5*3.14*x)+.288*cos(1.5*3.14*x)};
|
||||
\draw[color=blue,thick,domain=0:5,smooth,samples=100] plot[id=iFe] function{-.4*sin(.5*3.14*x)};
|
||||
\draw[red] node at (.825,1.25) {{\footnotesize $i_{\mu}(t)$}};
|
||||
\draw[blue] node at (2.5,.75) {{\footnotesize $i_{Fe}(t)$}};
|
||||
\end{scope}
|
||||
\draw node at (2.25,-1)[below]{Bild 2};
|
||||
\end{tikzpicture}
|
||||
\begin{tikzpicture}[scale=1.25]
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=0]
|
||||
\foreach \ii in {5} { % Enter Number of Decades in x
|
||||
\foreach \jj in {2.5} { % Enter Number of Decades in y
|
||||
\foreach \i in {1,2,...,\ii} {
|
||||
\foreach \j in {0,1,2,...,\jj} {
|
||||
\draw[black!50!, step=0.5] (0,-.5) grid (\ii,\jj); % Draw Sub Linear grid
|
||||
}}% End Log Grid
|
||||
\draw[black!80!] (0,-.5) grid (\ii,\jj); % Draw Linear grid
|
||||
\draw [->,thick] (0,-.5)--(0,\jj+.25) node (yaxis) [above] {$i\,[\milli\ampere]$};
|
||||
\draw [->,thick] (0,1)--(\ii+.25,1) node (xaxis) [right] {$\omega t$}; % Draw axes
|
||||
\draw node at (0,-.5)[below]{$0$};
|
||||
\draw node at (2,-.5)[below]{$\pi$};
|
||||
\draw node at (4,-.5)[below]{$2\pi$};
|
||||
\foreach \y in {-10,0,10}% y Axis Label:
|
||||
\node [anchor=east] at(0,\y/10+1){$\y$};
|
||||
}}
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=1cm]
|
||||
\draw[color=red!50!blue,thick,domain=0:5,smooth,samples=100] plot[id=iomega] function{1*cos(.5*3.14*x)+.288*cos(1.5*3.14*x)-.4*sin(.5*3.14*x)};
|
||||
\draw[red!50!blue] node at (.825,1.25) {{\footnotesize $i_0(t)$}};
|
||||
\end{scope}
|
||||
\draw node at (2.25,-1)[below]{Bild 3};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie Effektivwert und Klirrfaktor von $i_\mu(t)$
|
||||
\item Berechnen Sie Effektivwert und Klirrfaktor von $i_0(t)$
|
||||
\end{enumerate}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
k&=\frac{\sqrt{\sum \limits_{n=2}^{\infty}A^2_n}}{\sqrt{\sum \limits_{n=1}^{\infty}A^2_n}}=\frac{\text{Effektivwert der Oberschwingungen}}{\text{Effektivwert des Gesamtsignals}}
|
||||
\end{align}
|
||||
\clearpage
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\intertext{a) Effektivwert $I_{\mu}$ und Klirrfaktor $k_{\mu}$}
|
||||
I_{\mu}&=\sqrt{I_{\mu,\omega}^{\phantom{\mu}2}+I_{\mu,3\omega}^{\phantom{\mu}2}}
|
||||
=\sqrt{\left(\frac{\widehat{i}_{\mu,\omega}}{\sqrt{2}}\right)^2
|
||||
+\left(\frac{\widehat{i}_{\mu,3\omega}}{\sqrt{2}}\right)^2}
|
||||
=\frac{1}{\sqrt{2}}\cdot \sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}36\,\milli\ampere}\\
|
||||
%I_{\mu}&=\sqrt{I_{\mu,\omega}^{\phantom{\mu}2}+I_{\mu,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}+\widehat{i}_{\mu,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}36\,\milli\ampere}\\
|
||||
k_{\mu}&=\frac{I_{\mu,3\omega}}{I_{\mu}}
|
||||
=\frac{\widehat{i}_{\mu,3\omega}\cancel{/\sqrt{2}}}{\sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}
|
||||
+\widehat{i}_{\mu,3\omega}^{\phantom{\mu}2}}\cancel{/\sqrt{2}}}
|
||||
=\frac{2{,}88\,\milli\ampere}{\sqrt{(10\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}}=\uuline{0{,}277}=\uuline{27{,}7\%}\\
|
||||
\end{align*}
|
||||
\begin{minipage}[c]{.8\textwidth}
|
||||
\begin{align*}
|
||||
\intertext{b) Effektivwert $I_0$ und Klirrfaktor $k_0$}
|
||||
\widehat{i}_{0,3\omega}&=\widehat{i}_{\mu,3\omega}=2{,}88\,\milli\ampere\\
|
||||
\intertext{Nulldurchgang von $\widehat{i}_{Fe}$ bei den Spitzenwerten}\\
|
||||
\widehat{i}_{0,\omega}&=\sqrt{\widehat{i}_{\mu,\omega}^{\phantom{\mu}2}+\widehat{i}_{Fe,\omega}^{\phantom{Fe}2}}\\
|
||||
&=\sqrt{(10\,\milli\ampere)^2+(4\,\milli\ampere)^2}=10{,}77\,\milli\ampere\\[\baselineskip]
|
||||
I_0&=\frac{1}{\sqrt{2}}\cdot \sqrt{\widehat{i}_{0,\omega}^{\phantom{\mu}2}+\widehat{i}_{0,3\omega}^{\phantom{\mu}2}}=\frac{1}{\sqrt{2}}\cdot \sqrt{(10{,}77\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}=\uuline{7{,}88\,\milli\ampere}\\
|
||||
k_0&=\frac{\widehat{i}_{0,3\omega}\cancel{/\sqrt{2}}}{\sqrt{\widehat{i}_{0,\omega}^{\phantom{\mu}2}+\widehat{i}_{0,3\omega}^{\phantom{\mu}2}}\cancel{/\sqrt{2}}}
|
||||
=\frac{2{,}88\,\milli\ampere}{\sqrt{(10{,}77\,\milli\ampere)^2+(2{,}88\,\milli\ampere)^2}}=\uuline{0{,}258}=\uuline{25{,}8\%}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\begin{minipage}[c]{.2\textwidth}
|
||||
\begin{tikzpicture}[scale=1.2]
|
||||
\begin{scope}[>=latex, xshift=2cm, yshift=5]
|
||||
\draw[dashed](0,-2)rectangle(1,0);
|
||||
\draw[->,blue](0,0)--(1,0)node[right]{$\widehat{i}_{Fe,\omega}$};
|
||||
\draw[->,red](0,0)--(0,-2)node[below]{$\widehat{i}_{\mu,\omega}$};
|
||||
\draw[->,red!50!blue](0,0)--(1,-2)node[below right]{$\widehat{i}_{0,\omega}$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{minipage}\\[\baselineskip]
|
||||
|
||||
\uline{Nicht gefragt}\\[\baselineskip]
|
||||
Typische Klirrfaktoren:\\
|
||||
Rechteckschwingung $33\%$\\
|
||||
Sprache noch verständlich $10\%$\\
|
||||
Max. HiFi Verstärker $1\%$\\
|
||||
Guter HiFi Verstärker $0{,}1\%$\\
|
||||
|
||||
Weiteres unter \url{http://de.wikipedia.org/wiki/Klirrfaktor}
|
||||
\clearpage
|
||||
}{}%
|
||||
60
ET2_L_B19_A4.tex
Normal file
60
ET2_L_B19_A4.tex
Normal file
@ -0,0 +1,60 @@
|
||||
\section {Klirrfaktor}
|
||||
Am Eingang liegt die Spannung\\
|
||||
$u_e(t)=2\,\volt+3\,\volt\cdot \sin(2\pi\cdot 50\,\power{\second}{-1}\cdot t) + 4\,\volt\cdot \sin(2\pi\cdot 100\,\power{\second}{-1}\cdot t)$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[very thick,scale=2]
|
||||
\begin{scope}[>=latex,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);
|
||||
% \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{L}$};
|
||||
\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] {$C$};
|
||||
% \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\footnotesize$U_{C1}$};
|
||||
\end{scope}
|
||||
\draw(0,0)--(1.5,0)(1,1)--(1.5,1);
|
||||
\fill(0,0)circle(.05cm)(0,1)circle(.05cm)(1.5,0)circle(.05cm)(1.5,1)circle(.05cm);
|
||||
\draw[->,blue](0,.9)--(0,.1)node at(0,.5)[left]{$u_e$};
|
||||
\draw[->,blue](1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$u_a$};
|
||||
\draw[->,red](1.5,1.1)--(1.3,1.1)node at(1.2,1.1)[above]{$i_a=0$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie den Klirrfaktor $k_a$ der Ausgangsspannung $u_a$
|
||||
\item Berechnen Sie den Effektivwert $U_a$
|
||||
\end{enumerate}
|
||||
$L=100\,\milli\henry$, $C=250\,\micro\farad$
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
k&=\frac{\sqrt{\sum \limits_{n=2}^{\infty}A^2_n}}{\sqrt{\sum \limits_{n=1}^{\infty}A^2_n}}=\frac{\text{Effektivwert der Oberschwingungen}}{\text{Effektivwert des Gesamtsignals}}\\
|
||||
k&=\sqrt{\frac{A^2_2+A^2_3+\cdots+A^2_n}{A^2_1+A^2_2+\cdots+A^2_n}}\\
|
||||
A_0&\quad\text{Gleichanteil}\notag\\
|
||||
A_1&\quad\text{Grundschwingung}\notag\\
|
||||
A_2\cdots A_n&\quad\text{Oberwellen}\notag
|
||||
\end{align}
|
||||
\clearpage
|
||||
Berechnung:
|
||||
\begin{align*}
|
||||
\intertext{a) Klirrfaktor am Ausgang\ $k_a$}
|
||||
U_{a0}&=U_{e0}=2\,\volt\qquad\text{Gleichanteil ist ohne Einfluß auf den Klirrfaktor}\\[\baselineskip]
|
||||
\frac{\uline{U}_a}{\uline{U}_e}&=\frac{jX_C}{jX_L+jX_C}=\frac{X_C}{X_L+X_C}\qquad\text{Wechselspannung}\\[\baselineskip]
|
||||
\text{Für }&50\,\hertz\text{ Grundwelle}\\
|
||||
X_{C1}&=-\frac{1}{2\pi f\cdot C}=\frac{-1}{2\pi\cdot 50\,\cancel{\frac{1}{\second}}\cdot 250\cdot \power{10}{-6}\,\frac{\ampere\cancel{\second}}{\volt}}=-12{,}73\,\ohm\\
|
||||
X_{L1}&=2\pi f\cdot L=2\pi\cdot 50\,\cancel{\frac{1}{\second}}\cdot 0{,}1\cdot \,\frac{\volt\cancel{\second}}{\ampere}=31{,}42\,\ohm\\
|
||||
U_{a1}&=\frac{-12{,}73\,\ohm}{31{,}42\,\ohm-12{,}73\,\ohm}\cdot U_{e1}=-0{,}6811\cdot U_{e1}\\
|
||||
U_{a1}&=|-0{,}6811\cdot U_{e1}|=0{,}6811\cdot \frac{3\,\volt}{\sqrt{2}}=1{,}445\,\volt\quad \text{Effektivwert}\\
|
||||
U^2_{a1}&=2{,}088\,\volt^2\\[\baselineskip]
|
||||
\text{Für }&100\,\hertz\text{ 1. Oberwelle}\\
|
||||
X_{C2}&=\frac{1}{2}\cdot X_{C1}=-6{,}365\,\ohm\\
|
||||
X_{L2}&=2X_{L1}=62{,}84\,\ohm\\
|
||||
U_{a2}&=\left|\frac{-6{,}365\,\ohm}{62{,}84\,\ohm-6{,}365\,\ohm}\right|\cdot \frac{4\,\volt}{\sqrt{2}}=0{,}1127\cdot \frac{4\,\volt}{\sqrt{2}}=0{,}3188\,\volt\quad \text{Effektivwert}\\
|
||||
U^2_{a2}&=0{,}1016\,\volt^2\\[\baselineskip]
|
||||
k_a&=\sqrt{\frac{U^2_{a2}}{U^2_{a1}+U^2_{a2}}}=\sqrt{\frac{0{,}1016\,\volt^2}{2{,}088\,\volt^2+0{,}1016\,\volt^2}}
|
||||
=\uuline{0{,}215}=\uuline{21{,}5\%}
|
||||
\intertext{b) Effektivwert $U_a$}
|
||||
U_a&=\sqrt{U^2_{a0}+U^2_{a1}+U^2_{a2}}=\sqrt{(2\,\volt)^2+2{,}088\,\volt^2+0{,}1016\,\volt^2}=\uuline{2{,}49\,\volt}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
93
ET2_L_B19_A5.tex
Normal file
93
ET2_L_B19_A5.tex
Normal file
@ -0,0 +1,93 @@
|
||||
\section {Momentanspannung}
|
||||
An die Schaltung wird die Spannung
|
||||
$u_E(t)=U_0+\widehat{u}_1\cdot \cos(\omega t+\varphi_1) + \widehat{u}_2\cdot \cos(2\omega t+\varphi_2)$
|
||||
angelegt.\\[.5\baselineskip]
|
||||
Berechnen Sie dies Spannung $u_C$ zur Zeit $t=T$\\
|
||||
\begin{minipage}[c]{.5\textwidth}
|
||||
\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] {$R_1$};
|
||||
% \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$};
|
||||
\end{scope}
|
||||
\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] {$R_2$};
|
||||
\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] {$C$};
|
||||
\end{scope}
|
||||
\draw(0,0)--(2.5,0)(1,1)--(2.5,1);
|
||||
\fill(0,0)circle(.025cm)(0,1)circle(.025cm)(2.5,0)circle(.025cm)(2.5,1)circle(.025cm);
|
||||
\draw[->,blue](0,.9)--(0,.1)node at(0,.5)[left]{$u_E$};
|
||||
\draw[->,blue](2.5,.9)--(2.5,.1)node at(2.5,.5)[right]{$u_C$};
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\begin{minipage}[c]{.5\textwidth}
|
||||
\begin{align*}
|
||||
R_1&=12\,\ohm\quad R_2=20\,\ohm\\
|
||||
C&=25\,\nano\farad\quad \omega=961\cdot \power{10}{3}\,\frac{1}{\second}\\
|
||||
U_0&=6\,\volt\quad \widehat{u}_1=7\,\volt\quad \widehat{u}_2=3\,\volt\\
|
||||
\varphi_1&=60\,\degree\quad \varphi_2=135\,\degree\quad T=2\,\micro\second
|
||||
\end{align*}
|
||||
\end{minipage}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\[.5\baselineskip]
|
||||
Jede Frequenz für sich betrachten; Überlagerung der Momentanwerte\\[.5\baselineskip]
|
||||
$u_E(t)=6\,\volt+7\,\volt\cdot \cos(961\cdot 10^3\,\cdot 1\per\second\cdot t+60\degree)+3\,\volt\cdot \cos(2\cdot 961\cdot 10^3\,\cdot 1\per\second\cdot t+135\degree)$\\
|
||||
a) Gleichspannung - Spannungsteiler
|
||||
\begin{align*}
|
||||
%\intertext{a) Gleichspannung - Spannungsteiler}
|
||||
u_{C0}&=U_0\cdot \frac{R_2}{R_1+R_2}=6\,\volt\cdot \frac{20\,\ohm}{32\,\ohm}=3{,}75\,\volt
|
||||
\intertext{b) allgemein:}
|
||||
%\vspace{-1.5cm}
|
||||
\uline{U_C}&=\uline{U}_E\cdot \frac{\uline{Z}_{||}}{\uline{Z}_{||}+R_1}\\
|
||||
\uline{Z}_{||}&=\frac{R_2\cdot jX_C}{R_2+jX_C}
|
||||
\intertext{c) $1\cdot \omega$: Transformation in komplexe Ebene: $u(t)\rightarrow \uline{U}$}
|
||||
\uline{U}_{E1}&=\frac{7\,\volt}{\sqrt{2}}\cdot e^{j60\,\degree}=4{,}95\,\volt\cdot e^{j60\,\degree}=(2{,}47+j4{,}29)\,\volt\\
|
||||
X_{C1}&=\frac{-1}{\omega C}=-41{,}6\,\ohm\\
|
||||
\uline{Z}_{||1}&=\frac{20\,\ohm\cdot (-j41{,}6\,\ohm)}{20\,\ohm-j41{,}6\,\ohm}=\frac{-j832\,\ohm^2}{20\,\ohm-j41{,}6\,\ohm}\\
|
||||
&=\frac{828\,\ohm\cancel{^2}\cdot e^{-j90}}{45{,}98\cancel{\,\ohm}\cdot e^{-j64{,}22}}=18{,}05\,\ohm\cdot e^{-j25{,}67}=(16{,}27-j7{,}82)\,\ohm\\
|
||||
\uline{Z}_{||1}+R1&=(16{,}27-j7{,}82+12)\,\ohm=(28{,}27-j7{,}82)\,\ohm=29{,}33\,\ohm\cdot e^{-j15{,}52}\\
|
||||
\uline{U}_{C1}&=\uline{U}_{E1}\cdot \frac{\uline{Z}_{||1}}{\uline{Z}_{||1}+R_1}
|
||||
%&= 4{,}95\,\volt\cdot e^{j60\,\degree}\cdot \frac{(16{,}2-j7{,}834)\cancel{\,\ohm}}{(28{,}2-j7{,}834)\cancel{\,\ohm}}
|
||||
=4{,}95\,\volt\cdot e^{j60\,\degree}\cdot \frac{18{,}05\cancel{\,\ohm}\cdot e^{-j25{,}67}}{29{,}33\cancel{\,\ohm}\cdot e^{-j15{,}46\,\degree}}\\
|
||||
&=3{,}04\,\volt\cdot e^{j49{,}79\,\degree}=(1{,}97+j2{,}32)\,\volt\\
|
||||
U_{C1}&=3{,}04\,\volt\quad\text{Effektivwert}\\
|
||||
\varphi_{C1}&=49{,}79\,\degree=0{,}868\,\radian\quad\text{Umwandlung wegen } (\omega t+\varphi_{C1})
|
||||
\end{align*}
|
||||
\enlargethispage{2cm}
|
||||
\clearpage
|
||||
\begin{align*}
|
||||
\intertext{d) $1\cdot \omega$ Rücktransformation: mit $t=T$}
|
||||
\omega t+\varphi_{C1}&=\omega \cdot T+\varphi_{C1}=961\cdot \power{10}{3}\,\cancel{\frac{1}{\second}}\cdot 2\cdot \power{10}{-6}\,\cancel{\second}+0{,}868\,\radian=2{,}791\,\radian =159{,}9\,\degree\\
|
||||
u_{C1}(t=T)&=\sqrt{2}\cdot U_{C1}\cdot \cos(\omega T+\varphi_{C1})=\sqrt{2}\cdot 3{,}04\,\volt\cdot \cos(159{,}9\,\degree)\\
|
||||
&=4{,}3\,\volt\cdot (-0{,}939)=-4{,}037\,\volt
|
||||
\intertext{e) $2\cdot \omega$: Transformation in komplexe Ebene: $u(t)\rightarrow \uline{U}$}
|
||||
\uline{U}_{E2}&=\frac{3\,\volt}{\sqrt{2}}\cdot e^{j135\,\degree}=2{,}121\,\volt\cdot e^{j135\,\degree}=(-1{,}5+j1{,}5)\,\volt\\
|
||||
X_{C2}&=\frac{1}{2}\cdot X_{C1}=-20{,}8\,\ohm\\
|
||||
\uline{Z}_{||2}&=\frac{20\,\ohm\cdot (-j20{,}8\,\ohm)}{20\,\ohm-j20{,}8\,\ohm}=\frac{-j414\,\ohm^2}{20\,\ohm-j20{,}8\,\ohm}\\
|
||||
&=\frac{414\,\ohm\cancel{^2}\cdot e^{-j90}}{28{,}78\cancel{\,\ohm}\cdot e^{-j45{,}99}}=14{,}41\,\ohm\cdot e^{-j43{,}87}=(10{,}39-j9{,}992)\,\ohm\\
|
||||
\uline{Z}_{||2}+R1&=(10{,}39-j9{,}992+12)\,\ohm=(22{,}34-j9{,}992)\,\ohm=24{,}51\,\ohm\cdot e^{-j24\,\degree}\\
|
||||
\uline{U}_{C2}&=\uline{U}_{E2}\cdot \frac{\uline{Z}_{||2}}{\uline{Z}_{||2}+R_1}\\
|
||||
%&= 2{,}121\,\volt\cdot e^{j135\,\degree}\cdot
|
||||
&=\uline{U}_{E2}\cdot
|
||||
\frac{(10{,}39-j9{,}992)\cancel{\,\ohm}}{(22{,}39-j9{,}992)\cancel{\,\ohm}}
|
||||
=2{,}121\,\volt\cdot e^{j135\,\degree}\cdot \frac{14{,}38\cdot e^{-j44{,}01}}{24{,}48\cdot e^{-j24{,}1\,\degree}}\\
|
||||
&=1{,}247\,\volt\cdot e^{j115{,}1\,\degree}=(-0{,}530+j1{,}129)\,\volt\\
|
||||
U_{C2}&=1{,}247\,\volt\quad\text{Effektivwert}\\
|
||||
\varphi_{C2}&=115{,}1\,\degree=2{,}009\,\radian\quad\text{Umwandlung wegen } (2\cdot \omega t+\varphi_{C2})
|
||||
\intertext{f) $2\cdot \omega$ Rücktransformation: mit $t=T$}
|
||||
2\cdot \omega t+\varphi_{C2}&=2\omega\cdot T+\varphi_{C2}=2\cdot 961\cdot \power{10}{3}\,\cancel{\frac{1}{\second}}\cdot 2\cdot \power{10}{-6}\,\cancel{\second}+2{,}009\,\radian\\
|
||||
&=5{,}854\,\radian =335{,}4\,\degree\\
|
||||
u_{C2}(t=T)&=\sqrt{2}\cdot U_{C2}\cdot \cos(2\cdot \omega T+\varphi_{C2})=\sqrt{2}\cdot 1{,}247\,\volt\cdot \cos(335{,}4\,\degree)\\
|
||||
&=1{,}762\,\volt\cdot 0{,}909=1{,}602\,\volt
|
||||
\intertext{g) Überlagerung:}
|
||||
u_C(t=T=2\,\micro\second)&=u_{C0}(T)+u_{C1}(T)+u_{C2}(T)=(3{,}75\,\volt-4{,}037\,\volt+1{,}602\,\volt)\\
|
||||
&=\uuline{1{,}315\,\volt}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
29
ET2_L_B19_A6.tex
Normal file
29
ET2_L_B19_A6.tex
Normal file
@ -0,0 +1,29 @@
|
||||
\section {Nichtlinears Bauelement}
|
||||
Für ein nichtlinears Bauelement gilt: $i=a\cdot u^2$ \\
|
||||
Berechnen Sie den Klirrfaktor des Stromes, wenn die Spannung $u(t)=U_0+\widehat{u}\cdot \sin(\omega t)$ angelegt wird.\\[\baselineskip]
|
||||
$a=20\,\milli\ampere\per\volt^2;\quad U_0=1{,}5\,\volt;\quad \widehat{u}=1{,}2\,\volt; \quad \omega=1500\,\power{\second}{-1}$\\[\baselineskip]
|
||||
Es gilt $\sin^2\alpha=0{,}5\cdot (1-\cos(2\alpha))$\\[\baselineskip]
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\
|
||||
\begin{align*}
|
||||
k_i&=\frac{I_2}{\sqrt{I^2_1+I^2_2}}\qquad\text{$\omega$ ohne Bedeutung}\\
|
||||
(\alpha&=\omega t)\\
|
||||
i(t)&=a\cdot (U_0+\widehat{u}\cdot \sin\alpha)^2=a\cdot (U^2_0+2\cdot U_0\cdot \widehat{u}\cdot \sin\alpha+\widehat{u}^2 \overbrace{\cdot \sin^2\alpha)}^{0,5-0,5\cdot cos(2\alpha)}\\
|
||||
&=\overbrace{a\cdot U^2_0}^{DC}+\underbrace{2\cdot a\cdot U_0\cdot \widehat{u}}_{\widehat{i}_1}\cdot \sin\alpha+\overbrace{0{,}5\cdot a\cdot \widehat{u}^2}^{DC}-\underbrace{0{,}5\cdot a\cdot \widehat{u}^2}_{\widehat{i}_2}\cdot \cos(2\alpha)\qquad\text{DC=Gleichanteil, ohne Einfluß}\\[\baselineskip]
|
||||
k_i&=\frac{\frac{1}{\sqrt{2}}\cdot 0{,}5\cdot a\cdot \widehat{u}^2}{\sqrt{\frac{1}{2}\cdot 4\cdot a^2\cdot U^2_0\cdot \widehat{u}^2+\frac{1}{2}\cdot \frac{1}{4}\cdot a^2\cdot \widehat{u}^4}}
|
||||
=\frac{\cancel{\frac{1}{\sqrt{2}}}\cdot 0{,}5\cdot \cancel{a}\cdot \widehat{u}\cancel{^2}}
|
||||
{\cancel{\frac{1}{\sqrt{2}}\cdot a\cdot \widehat{u}}\cdot \sqrt{4\cdot U^2_0\cdot +\frac{1}{4}\cdot \widehat{u}^2}}\\
|
||||
&=\frac{0{,}5\cdot \widehat{u}}{\sqrt{4\cdot U^2_0+\frac{1}{4}\cdot \widehat{u}^2}}=\frac{0{,}5\cdot 1{,}2\,\volt}{\sqrt{4\cdot 1{,}5^2\,\volt^2+\frac{1}{4}\cdot 1{,}2^2\,\volt^2}}=\uuline{0{,}1961}=\uuline{19{,}61\%}
|
||||
\end{align*}
|
||||
alternativ:
|
||||
\begin{align*}
|
||||
I_1&=\frac{\widehat{i}_1}{\sqrt{2}}=50{,}91\,\milli\ampere\\
|
||||
I_2&=\frac{\widehat{i}_2}{\sqrt{2}}=10{,}18\,\milli\ampere\\
|
||||
k_i&=\frac{I_2}{\sqrt{I^2_1+I^2_2}}=\frac{10{,}18\,\milli\ampere}{\sqrt{(50{,}91\,\milli\ampere)^2
|
||||
+(10{,}18\,\milli\ampere)^2}}=\uuline{0{,}1961}=\uuline{19{,}61\%}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
68
ET2_L_B19_A7.tex
Normal file
68
ET2_L_B19_A7.tex
Normal file
@ -0,0 +1,68 @@
|
||||
\section {Wirkleistung Zweipol}
|
||||
An einem Zweipol liegt die dargestellte Spannung $u$. Es fließt der dargestellte Rechteckstrom $i$. Spannung und Strom sind periodisch und haben die gleiche Periodendauer.
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\begin{scope}[>=latex, xshift=-4cm, yshift=0]
|
||||
\draw(0,-1)rectangle(2,1);
|
||||
\draw(-.5,.75)--(0,.75);
|
||||
\draw(-.5,-.75)--(0,-.75);
|
||||
\draw[->,blue,thick](-.5,.5)--(-.5,-.5)node at(-.5,0)[left]{$u$};
|
||||
\draw[->,red,thick](-.5,.9)--(-.1,.9)node at(-.3,.9)[above]{$i$};
|
||||
\draw node at(1,0){Zweipol};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex, xshift=0cm, yshift=0]
|
||||
\draw[thin,black!50!](0,-1)grid(8,1);
|
||||
\draw[color=blue,thick,domain=0:8,samples=100] plot[id=sin19_7] function{sin(1.047*x)} node at (2,1) [above] {$u$};
|
||||
\draw[red, very thick] (0,-.5)--(1,-.5)--(1,.5)--(4,.5)--(4,-.5)--(7,-.5)--(7,.5)--(7,.5)--(8,.5)node at(4,.5)[above right]{i};
|
||||
\draw node at(0,1)[left]{$u_{max}$};
|
||||
\draw node at(0,.5)[left]{$iI_{max}$};
|
||||
\draw node at(0,-.5)[left]{$-i_{max}$};
|
||||
\draw node at(0,-1)[left]{$-u_{max}$};
|
||||
\draw node at(1,-1)[below]{$\frac{\pi}{3}$};
|
||||
\draw node at(4,-1)[below]{$\frac{4}{3}\pi$};
|
||||
\draw node at(7,-1)[below]{$\frac{7}{3}\pi$};
|
||||
\draw[->,thick,black](0,-1)--(0,1.25);
|
||||
\draw[->,thick,black](0,0)--(8.25,0)node [right]{$\omega t$};
|
||||
\draw node at(4,-1.75)[below]{$u_{max}=300\,\volt\qquad i_{max}=6\,\ampere$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
Berechnen Sie die Wirkleistung, die der Zweipol aufnimmt. \\[\baselineskip]
|
||||
Hinweis: Unterschiedliche Lösungsverfahren sind möglich.
|
||||
falls benötigt: Der Strom $i$ kann durch folgende Fourier Reihe dargestellt werden:
|
||||
\begin{align*}
|
||||
i&=\frac{24\,\ampere}{\pi}\left[\sin\left(\omega t-\frac{\pi}{3}\right)+\frac{1}{3}\sin\left( 3\left(\omega t-\frac{\pi}{3}\right)\right)+\frac{1}{5}\sin\left(5\left(\omega t-\frac{\pi}{3}\right)\right)+\cdots\right]
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
%\begin{align}
|
||||
%\intertext{Formeln:}
|
||||
%\end{align}
|
||||
Berechnung:\\[\baselineskip]
|
||||
Elegante Lösung mit Fourier-Reihe:\\
|
||||
Für die Leistungsaufnahme des Zweipols ist nur der Stromanteil entscheidend, der die gleiche Frequenz wie die Spannung besitzt, d.h. nur
|
||||
\begin{align*}
|
||||
P&=U\cdot I_\omega\cdot \cos\varphi\\
|
||||
i_\omega(t)&=\frac{24\,\ampere}{\pi}\sin(\omega t\underbrace{-\frac{\pi}{3}}_{\varphi})\\
|
||||
I_\omega&=\frac{24\,\ampere}{\pi\cdot \sqrt{2}}\\
|
||||
\Rightarrow\quad P&=U\cdot I_\omega\cdot \cos\varphi=\frac{300\,\volt}{\sqrt{2}}\cdot \frac{24\,\ampere}{\pi\sqrt{2}}\cdot \underbrace{\cos\left(-\frac{\pi}{3}\right)}_{0{,5}}=\uuline{573\,\watt}\\
|
||||
\end{align*}
|
||||
oder Standardlösung
|
||||
\begin{align*}
|
||||
p(t)&=u(t)\cdot i(t)\\
|
||||
P&=\frac{1}{T}\int_0^Tp(t)\cdot dt=\frac{1}{T}\int_0^Tu(t)\cdot i(t)\cdot dt=\frac{1}{2\pi}\int_0^{2\pi}u(\omega t)\cdot i(\omega t)\cdot d(\omega t)\\
|
||||
&\text{(Verschiebung Start- und Endwert für Integration)}\\
|
||||
&=\frac{1}{2\pi}\left[\int_{\frac{\pi}{3}}^{\frac{4}{3}\pi}6\,\ampere\cdot 300\,\volt\cdot \sin(\omega t)d(\omega t)+\int_{\frac{4}{3}\pi}^{\frac{7}{3}\pi}-6\,\ampere\cdot 300\,\volt\cdot \sin(\omega t)d(\omega t)\right]\\
|
||||
&=\frac{1}{2\pi}\Big[1800\,\volt\ampere\cdot \Big(\underbrace{-\cos\Big(\frac{4}{3}\pi\Big)}_{0{,}5}+\underbrace{\cos\Big(\frac{\pi}{3}\Big)}_{0{,}5}\Big)
|
||||
-1800\,\volt\ampere\cdot \Big(\underbrace{-\cos\Big(\frac{7}{3}\pi\Big)}_{-0{,}5}+\underbrace{\cos\Big(\frac{4}{3}\pi\Big)}_{-0{,}5}\Big)\Big]\\
|
||||
&=\frac{1800\,\volt\ampere}{\pi}=\uuline{573\,\watt}\\[\baselineskip]
|
||||
\end{align*}
|
||||
oder sinusförmiger Strom unter Berücksichtigung des Formfaktors
|
||||
\begin{align*}
|
||||
&\text{Für sinusförmigen Strom:}\\
|
||||
P_{\sin}&=U\cdot I\cdot \cos\varphi=\frac{300\,\volt}{\sqrt{2}}\cdot \frac{6\,\ampere}{\sqrt{2}}\cdot \cos(60\,\degree)=450\,\watt (\sin) =636\,\watt\\
|
||||
&\text{Für den rechteckförmigen Strom muß der Formfaktor berücksichtigt werden:}\\
|
||||
F&=\frac{\pi}{2\sqrt{2}}=1{,}11\\
|
||||
P&=\frac{P_{\sin}}{F}=\frac{636\,\watt}{1{,}11}=\uuline{573\,\watt}\\
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
92
ET2_L_B20_A1.tex
Normal file
92
ET2_L_B20_A1.tex
Normal file
@ -0,0 +1,92 @@
|
||||
\section {Ringspule}
|
||||
Durch das Zentrum einer Ringspule mit rechteckigem Querschnitt (Länge
|
||||
$l=12\,\milli\metre$, $r_i=10\,\milli\metre$, $r_a=20\,\milli\metre$;
|
||||
Windungszahl $N= 1800$) wird ein Leiter geführt, in dem ein Wechselstrom fließt.
|
||||
($I_{eff}=40\,\ampere$; $f=50\,\hertz$).\\
|
||||
Berechnen Sie den Effektivwert $U$ der Spannung $u$.\\
|
||||
$\mu=\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2.5, z=0.2pt]
|
||||
\begin{scope}[>=latex, xshift=0, yshift=0]
|
||||
\foreach \z in {0,0.2,...,1}
|
||||
\fill [black!25!](0,0,\z cm) circle (0.7cm);% Zylinder außen
|
||||
\draw [black!75!](0,0,1 cm) circle (0.7cm);%Hintere äußere Ringkante
|
||||
%\foreach \z in {0,0.2,...,0.5}
|
||||
%\fill [black!50!](0,0,\z cm) circle (0.5cm);% Zylinder innen
|
||||
\fill [black!25!](0,0) circle (0.7cm);% Stirnseitenfüllung
|
||||
\fill [black!35](0,0) circle (0.5cm); %Füllung innen
|
||||
\begin{scope}%Clip Füllung
|
||||
\clip (0,0,1cm) circle (0.5cm);%Clip mit hinterer inneren Ringkante
|
||||
\fill [black!0!](0,0) circle (0.5cm);%Innere Ringkante
|
||||
\end{scope}
|
||||
\begin{scope}%Clip hintere innere Kante
|
||||
\clip (0,0) circle (0.5cm);%Clip mit hinterer inneren Ringkante
|
||||
\draw [black!75!](0,0,1cm) circle (0.5cm);%Hintere innere Ringkante
|
||||
\end{scope}
|
||||
\draw [black!75!](0,0) circle (0.5cm);% Innere Ringkante
|
||||
\draw [black!75!](0,0) circle (0.7cm);% Äußere Ringkante
|
||||
\foreach \a in {-45,-30,-15,15,30,45,60,75,90,105,120,135} %Rechte Drähte
|
||||
\draw [red!60!black, ultra thick] (\a:0.5)--(\a:0.7)--+(45:.3);
|
||||
\foreach \a in {150,165,...,300} %Linke Drähte
|
||||
\draw [red!60!black, ultra thick] (\a:0.7)--(\a:0.5)--+(45:.3);
|
||||
\draw [red!60!black, ultra thick] (0.5,0,0)--(1.5,0,0)circle(0.01);%Anschußdrähte
|
||||
\draw [red!60!black, ultra thick] (0.7,0,1cm)--(1.5,0,1cm)circle(0.01);%Anschußdrähte
|
||||
\draw [->,blue, thick] (1.6,0,1cm) -- (1.6,0,0) node at (1.6,0,.5cm)[right] {$u$};
|
||||
\draw [blue, ultra thick] (0,0,3.5cm) -- (0,0,7cm);
|
||||
\draw [->,red, thick] (0,0,7.5cm) -- (0,0,8cm)node [right] {$i$};
|
||||
\draw [blue, ultra thick] (0,0,1.75cm) -- (0,0,-5cm)node [above left]{Leiter};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\begin{tikzpicture}[scale=1.4]%Querschnitt
|
||||
\begin{scope}[>=latex,xshift=0,yshift=0]
|
||||
\fill [black!25!] (0,0)rectangle(1,1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=0,yshift=3cm]
|
||||
\fill [black!25!] (0,0)rectangle(1,1);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=0,yshift=0cm]
|
||||
\draw [very thick] (0,0)rectangle(1,4);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=-1cm,yshift=1.9cm]
|
||||
\fill [blue!50!] (0,0)rectangle(3,.2);
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,xshift=0cm,yshift=4cm]%Hilfslinien
|
||||
\draw [very thin] (-1,0)--(0,0)--(0,.5) (1,0)--(1,.5)(-.5,-1)--(0,-1);
|
||||
\draw [very thin, dashed] (-1.5,-2)--(2.5,-2);
|
||||
\draw [->,very thin] (-.9,-2)--(-.9,0)node at(-.9,-1)[right]{$r_a$};
|
||||
\draw [->,very thin] (-.4,-2)--(-.4,-1)node at(-.4,-1.5)[right]{$r_i$};
|
||||
\draw [<->,very thin] (0,0.4)--(1,0.4)node at(.5,0.4)[above]{$l$};
|
||||
\draw node at(.5,-.5){$\mu=\mu_0$};
|
||||
\draw [->,red,thin] (1.25,-1.75)--(1.5,-1.75) node [right]{$i$};
|
||||
\end{scope}
|
||||
\end{tikzpicture}
|
||||
\end{align*}
|
||||
\ifthenelse{\equal{\toPrint}{Lösung}}{%
|
||||
\begin{align}
|
||||
\intertext{Formeln:}
|
||||
u&=N\cdot \frac{d\phi(t)}{dt}
|
||||
\end{align}
|
||||
Berechnung:
|
||||
Strom im Leiter bewirkt Fluss durch die Spule:
|
||||
\begin{align*}
|
||||
\phi&=\widehat{\phi}_0\cdot \sin(\omega t)\\
|
||||
\widehat{\phi}_0&=\int_{r_i}^{r_a}l\cdot \widehat{B}(r)\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \int_{r_i}^{r_a}\frac{1}{r}\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \ln\frac{r_a}{r_i}\\
|
||||
\text{mit }\widehat{i}&=\sqrt{2}\cdot I\\
|
||||
\Rightarrow \widehat{\phi}_0&=\frac{1{,}26\cdot \power{10}{-6}\, \frac{\volt\second}{\ampere\metre}\cdot 12\cdot \power{10}{-3}\,\metre\cdot \sqrt{2}\cdot 40\,\ampere}{2\pi}\cdot \ln(2)=9{,}44\cdot \power{10}{-8}\,\volt\second\\
|
||||
\phi_{eff}&=\frac{\widehat{\phi}_0}{\sqrt{2}}=6{,}672\cdot \power{10}{-8}\,\volt\second\\
|
||||
\intertext{Veränderlicher Fluss induziert Spannung:}
|
||||
u&=N\cdot \frac{d\phi(t)}{dt}=N\cdot \widehat{\phi}_0\cdot \omega\cdot \cos(\omega t)=\widehat{u}\cdot \cos(\omega t)\\
|
||||
\widehat{u}&=N\cdot \phi_0\cdot \omega=1800\cdot 9{,}44\cdot \power{10}{-8}\,\volt\second\cdot 2\pi\cdot 50\,\power{\second}{-1}=53{,}36\,\milli\volt\\
|
||||
U&=\frac{\widehat{u}}{\sqrt{2}}=\frac{53{,}36\,\milli\volt}{\sqrt{2}}=\uuline{37{,}73\,\milli\volt}
|
||||
\intertext{Anmerkung:}
|
||||
u_i&=2\cdot a\cdot r\cdot N\cdot B\cdot \omega\cdot \sin(\omega t)\\
|
||||
&=2\cdot a\cdot r\cdot N\cdot \frac{\Phi}{A}\cdot \omega\cdot \sin(\omega t)\\
|
||||
&\text{mit $2\cdot a\cdot r=A$}\\
|
||||
&=N\cdot \Phi \cdot \omega\cdot \sin(\omega t)\\
|
||||
&=\widehat{u}_i\cdot \sin(\omega t)\\
|
||||
\intertext{oder}
|
||||
U&=\mu_0\cdot l\cdot \ln\frac{r_a}{r_i}\cdot N\cdot f\cdot I\\
|
||||
&=1{,}26\cdot \power{10}{-6}\,\frac{\volt\cancel{\second}}{\cancel{\ampere}\cancel{\metre}}\cdot 12\cdot \power{10}{-3}\,\cancel{\metre}\cdot \ln\frac{20\,\cancel{\milli\metre}}{10\,\cancel{\milli\metre}}\cdot 1800\cdot 50\,\cancel{\frac{1}{\second}}\cdot 40\,\cancel{\ampere}=\uuline{37{,}73\,\milli\volt}
|
||||
\end{align*}
|
||||
\clearpage
|
||||
}{}%
|
||||
192
ET2_L_B20_A2.tex
Normal file
192
ET2_L_B20_A2.tex
Normal file
@ -0,0 +1,192 @@
|
||||
\section {Netzwerk Wirk- und Blindanteil}
|
||||
$\uline{U}=4{,}2\,\volt\cdot e^{j30\,\degree};\quad f=120\,\kilo\hertz ;\quad \uline{I}_0= 2{,}7\milli\ampere \cdot e^{j110\,\degree}$\\
|
||||
$R=1{,}3\,\kilo\ohm; \quad L=3{,}6\,\milli\henry; \quad C=1{,}5\,\nano\farad$.
|
||||
\renewcommand{\labelenumi}{\alph{enumi})}
|
||||
\begin{enumerate}
|
||||
\item Berechnen Sie den Strom \uline{I} und seinen Wirk- und Blindanteil $\uline{I}_{\textrm{w}}$ bzw. $\uline{I}_{\textrm{b}}$.
|
||||
\item Berechnen Sie die Wirk und Blindleistung an den Anschlussklemmen und an der Stromquelle $\uline{I}_0$.
|
||||
\end{enumerate}
|
||||
\vspace{-.5cm}
|
||||
\begin{align*}
|
||||
\begin{tikzpicture}[scale=2]
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]
|
||||
\draw [->,blue](.8,0)--(.2,0)node at(.5,0)[left]{\footnotesize$\uline{U}$};
|
||||
\end{scope}
|
||||
\begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Kondensator -
|
||||
\draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$};
|
||||
\draw [<-,red] (.25,0)--(.05,0)node at(.15,0)[above]{$\footnotesize\uline{I}$};
|
||||
\end{scope}
|
||||
\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] {$R$};
|
||||
\draw [<-,red] (.75,0)--(.95,0)node at(.85,0)[right]{\footnotesize$\uline{I}_R$};
|
||||