Niebler
2 years ago
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ET2_Deckblatt.tex
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| \pagenumbering{roman} | |||
| %\ifthenelse{\equal{\toPrint}{Lösung}}{\input{ET2_Deckblatt}}{\input{ET2_Deckblatt_A}} | |||
| \begin{titlepage} | |||
| \vspace*{3cm} | |||
| \begin{center} | |||
| {\bfseries \large {Technische Hochschule Nürnberg \\ Georg Simon Ohm}\\[\baselineskip] | |||
| Übung Grundlagen der Elektrotechnik 2}\\[\baselineskip] | |||
| % Prüfen ob Musterlösung oder nicht, wenn ja, dann mit Dozentenkommentar, ansonsten nur Aufgabenstellung für Studierende | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}% | |||
| {% | |||
| {\normalsize Prof.\,Dr.\,C.\,Niebler}\\[2\baselineskip] | |||
| \textcolor{red}{Nur für Dozentengebrauch, nicht zur Weitergabe an Studenten.}\\[\baselineskip] | |||
| {\large \textbf{Dozenten Exemplar mit Lösungen für}}\\[2\baselineskip] | |||
| \textbf{\ProfName}\\[2\baselineskip] | |||
| %\copyright\ Dr. Christine Niebler\\ | |||
| }% Dozentenausgabe | |||
| {}% Studentenausgabbe | |||
| \end{center} | |||
| \end{titlepage} | |||
| \thispagestyle{empty} | |||
| \cleardoublepage | |||
| \clearpage | |||
| \pagenumbering{arabic} | |||
| \enlargethispage{1cm} | |||
| %\chead{Inhaltsverzeichnis} | |||
| \footnotesize{\tableofcontents} | |||
| \clearpage | |||
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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 | |||
| entering extended mode | |||
| **C:/Users/Niebler/Documents/Vorlesungen/ETechnik/TeX/UeET2_20130923/ET2_Deckbl | |||
| att_A.tex | |||
| (C:/Users/Niebler/Documents/Vorlesungen/ETechnik/TeX/UeET2_20130923/ET2_Deckbla | |||
| tt_A.tex | |||
| LaTeX2e <2011/06/27> | |||
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ET2_Deckblatt_A.tex
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| \pagenumbering{roman} | |||
| \begin{titlepage} | |||
| \vspace*{3cm} | |||
| \begin{center} | |||
| {\bfseries \Large {Technische Hochschule Nürnberg\\ | |||
| Georg Simon Ohm\\[\baselineskip] | |||
| {Übung Grundlagen der Elektrotechnik 2}}\\[\baselineskip] | |||
| \normalsize{Prof.\,Dr.\,C.\,Niebler}\\[\baselineskip] | |||
| \Large\textbf{Aufgabenstellung}}\\[\baselineskip] | |||
| % SS 2013}\\[\baselineskip] | |||
| \end{center} | |||
| \end{titlepage} | |||
| \thispagestyle{empty} | |||
| \cleardoublepage | |||
| \clearpage | |||
| \pagenumbering{arabic} | |||
| \enlargethispage{1cm} | |||
| %\chead{Inhaltsverzeichnis} | |||
| %\footnotesize{ | |||
| \tableofcontents | |||
| %} | |||
| \clearpage | |||
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ET2_LS_B12_A1.tex
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| \section{Blitzableiter} | |||
| Eine $l=4\,\metre$ lange Verbindungsleitung, deren Leiter voneinander einen Abstand $d = 2\,\centi\metre$ haben, ist im Abstand $D=3\,\metre$ parallel zu einem Blitzableiter verlegt. (Siehe Skizze).\\ | |||
| $\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$ | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Welche Spannung $u$ (Betrag !) wird induziert, wenn der | |||
| Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt? | |||
| \item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !) | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom | |||
| \draw [->] (0,4) -- (0,0) node [below] {$i$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung | |||
| \draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$}; | |||
| \draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D | |||
| \draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$}; | |||
| \draw [dashed] (3.1,.5)--(3.1,-.5); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d | |||
| \draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$}; | |||
| \draw [<-] (0.2,0) -- (0.7,0); | |||
| \end{scope} | |||
| \begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l | |||
| \draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$}; | |||
| \draw (0,0) -- (-0.5,0); | |||
| \draw (0,3) -- (-0.5,3); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| Formeln: | |||
| \begin{align} | |||
| % \intertext{Formeln:} | |||
| u&=-N\cdot \frac{d\Phi}{dt}\\ | |||
| \Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\ | |||
| H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r} | |||
| \end{align} | |||
| Magnetischer Fluss $\Phi$; Flussdichte $B$; Feldstärke $H$;\\ | |||
| Magnetische Permeabilität $\mu=\mu_0\cdot \mu_r$; In Luft $\mu_r=1$\\ | |||
| \clearpage | |||
| Berechnung: | |||
| \begin{align*} | |||
| u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\qquad u=-\frac{d\Phi}{dt}\\ | |||
| \Phi&=\int{\vec{B}\cdot \vec{dA}} \qquad \text{mit $B\bot A$}\qquad \Phi=B\cdot A=\underbrace{\mu\cdot H}_{B}\cdot \underbrace{l\cdot d}_{A}\\ | |||
| H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\ | |||
| \phantom{blablabla}bla\\ | |||
| \vphantom{u=-\frac{y}{x}}..\\ | |||
| u&=-\frac{d\Phi}{dt}\hspace{-.5cm}\underbrace{=}_{linearer Anstieg}\hspace{-.5cm}-\frac{\Phi}{t}=-\mu\cdot l\cdot d\cdot \frac{i}{2\cdot \pi\cdot D\cdot t}=-\frac{\mu\cdot i\cdot l\cdot d}{2\cdot \pi\cdot D\cdot t}\\ | |||
| % } %end pahntom | |||
| &=-\frac{1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)\cdot 15000\,\ampere\cdot 4\,\metre\cdot 0{,}02\,\metre}{2\cdot \pi \cdot 3\,\metre\cdot 0{,}6\cdot \power{10}{-6}\,\second}=\uuline{-134\,\volt}\\[\baselineskip] | |||
| &\text{Leitung symmetrisch zur Mittellinie bei } 2{,}99\,\metre \text{ und } 3{,}01\,\metre\\ | |||
| \text{mit }&\Phi=\frac{\mu\cdot i\cdot l}{2\cdot \pi}\cdot\ln\frac{D+d/2}{D-d/2}\\ | |||
| u&=-\frac{\mu_0\cdot 15\cdot \power{10}{3}\,\ampere\cdot 4\,\metre}{0{,}6\cdot \power{10}{-6}\,\second\cdot 2\cdot \pi}\cdot \underbrace{\ln\frac{3{,}01\,\metre}{2{,}99\,\metre}}_{6{,}66\cdot \power{10}{-3}}=-133{,}7\,\volt\\ | |||
| \text{\uline{Lentzsche Regel:}} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2.5] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$}; | |||
| \draw [<-,blue] (.2,.35)--(.8,.35)node at(.5,.35)[above]{\footnotesize$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=.25cm,yshift=-.25cm]% Leiterschleife | |||
| \draw (0,0)--(0,-2)--(.5,-2)--(.5,0); | |||
| \draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.05) node [above right] {\footnotesize$+$}; | |||
| \fill (1,0)circle(.05) node [above left] {\footnotesize$-$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=-1cm]% O. | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$\frac{d\Phi}{dt}$}; | |||
| \fill (.5,0)circle(.05); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,green!50!black,xshift=0cm,yshift=-1.33cm]% Ox | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$}; | |||
| \draw [black]node at (1.5,-1)[below]{Wirkung $\frac{d\Phi_{ind}}{dt}$ entgegen Ursache $\frac{d\Phi}{dt}$}; | |||
| \draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=3cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.05) node [above right] {\footnotesize$+$}; | |||
| \fill (1,0)circle(.05) node [above left] {\footnotesize$-$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=-0.25cm]% Leiterschleife | |||
| \draw (0,0)--(0,-2)--(.5,-2)--(.5,0); | |||
| \draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=-2.25cm]%Stromquelle | |||
| \draw [green!50!black](.25,0)--(.75,0); | |||
| \draw node at (.5,.133) [above] {$i_{}$}; | |||
| \draw [green!50!black](.5,0)circle(.133); | |||
| \draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$i_{ind}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,green!50!black,xshift=3cm,yshift=-1.33cm]% Ox | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$}; | |||
| \draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1.25cm]% | |||
| \draw [->,red] (-2.5,.5)--(-2.5,-.5) node at(-2.5,0)[right]{$\frac{di}{dt}$}; | |||
| \draw [<-,green!50!black] (.5,.5)--(.5,-.5) node at(.5,0)[left]{$i_{ind}$}; | |||
| \draw [<-,red!75!black] (-1.75,.5)--(-1.75,-.5) node at(-1.75,0)[left]{$i_{ind}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| $u$ hat negatives Vorzeichen, da Polung von $i_{ind}$ entgegen Bezugspfeilen aus Skizze\\ | |||
| ($u_R = -u$) | |||
| \clearpage | |||
| }{}% | |||
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ET2_L_B12_A1.tex
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| \section{Blitzableiter} | |||
| Eine $l=4\,\metre$ lange Verbindungsleitung, deren Leiter voneinander einen Abstand $d = 2\,\centi\metre$ haben, ist im Abstand $D=3\,\metre$ parallel zu einem Blitzableiter verlegt. (Siehe Skizze).\\ | |||
| $\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$ | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Welche Spannung $u$ (Betrag !) wird induziert, wenn der | |||
| Blitzstrom $i$ linear in $0{,}6\,\micro\second$ auf $15\,\kilo\ampere$ ansteigt? | |||
| \item Ist die Spannung während dieser Zeit positiv oder negativ? (Begründung !) | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] % Blitzstrom | |||
| \draw [->] (0,4) -- (0,0) node [below] {$i$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick, xshift=3cm, yshift=0.5cm] % Leitung | |||
| \draw (0,3) circle (0.025) -- (0,0) -- (0.2,0) -- (0.2,3) circle (0.025); % node at (0.1,3) [above ] {$u$}; | |||
| \draw [<-] (0,3.1) -- (.2,3.1) node at (0.1,3.1) [above] {$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,xshift=0cm,yshift=2cm] % Abstand D | |||
| \draw [<->] (0,0) -- (3.1,0) node at (1.55,0) [above] {$D$}; | |||
| \draw [dashed] (3.1,.5)--(3.1,-.5); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,xshift=3cm,yshift=1cm] % Pfeile d | |||
| \draw [->] (-.5,0) -- (0,0) node at (0.5,0) [above] {$d$}; | |||
| \draw [<-] (0.2,0) -- (0.7,0); | |||
| \end{scope} | |||
| \begin{scope}[black!75!,>=latex,thick,xshift=4.125cm,yshift=0.5cm] % Länge l | |||
| \draw [<->] (-.125,3) -- (-.125,0) node at (0,1.5) [right] {$l$}; | |||
| \draw (0,0) -- (-0.5,0); | |||
| \draw (0,3) -- (-0.5,3); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| Formeln: | |||
| \begin{align} | |||
| % \intertext{Formeln:} | |||
| u&=-N\cdot \frac{d\Phi}{dt}\\ | |||
| \Phi&=B\cdot A=\mu\cdot H\cdot A=\int{\vec{B}\cdot \vec{dA}}\\ | |||
| H&=\frac{i\cdot N}{l} \qquad \text{für Zylinder: }\qquad H=\frac{i\cdot N}{2\cdot \pi\cdot r} | |||
| \end{align} | |||
| Magnetischer Fluss $\Phi$; Flussdichte $B$; Feldstärke $H$;\\ | |||
| Magnetische Permeabilität $\mu=\mu_0\cdot \mu_r$; In Luft $\mu_r=1$\\ | |||
| \clearpage | |||
| Berechnung: | |||
| \begin{align*} | |||
| u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\qquad u=-\frac{d\Phi}{dt}\\ | |||
| \Phi&=\int{\vec{B}\cdot \vec{dA}} \qquad \text{mit $B\bot A$}\qquad \Phi=B\cdot A=\underbrace{\mu\cdot H}_{B}\cdot \underbrace{l\cdot d}_{A}\\ | |||
| H&=\frac{i}{2\cdot \pi\cdot D} \qquad \text{Abhängigkeit vom Abstand $d$ vernachlässigbar $d\ll D$. }\\ | |||
| u&=-\frac{d\Phi}{dt}\hspace{-.5cm}\underbrace{=}_{linearer Anstieg}\hspace{-.5cm}-\frac{\Phi}{t}=-\mu\cdot l\cdot d\cdot \frac{i}{2\cdot \pi\cdot D\cdot t}=-\frac{\mu\cdot i\cdot l\cdot d}{2\cdot \pi\cdot D\cdot t}\\ | |||
| &=-\frac{1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)\cdot 15000\,\ampere\cdot 4\,\metre\cdot 0{,}02\,\metre}{2\cdot \pi \cdot 3\,\metre\cdot 0{,}6\cdot \power{10}{-6}\,\second}=\uuline{-134\,\volt}\\[\baselineskip] | |||
| &\text{Leitung symmetrisch zur Mittellinie bei } 2{,}99\,\metre \text{ und } 3{,}01\,\metre\\ | |||
| \text{mit }&\Phi=\frac{\mu\cdot i\cdot l}{2\cdot \pi}\cdot\ln\frac{D+d/2}{D-d/2}\\ | |||
| u&=-\frac{\mu_0\cdot 15\cdot \power{10}{3}\,\ampere\cdot 4\,\metre}{0{,}6\cdot \power{10}{-6}\,\second\cdot 2\cdot \pi}\cdot \underbrace{\ln\frac{3{,}01\,\metre}{2{,}99\,\metre}}_{6{,}66\cdot \power{10}{-3}}=-133{,}7\,\volt\\ | |||
| \text{\uline{Lentzsche Regel:}} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2.5] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$}; | |||
| \draw [<-,blue] (.2,.35)--(.8,.35)node at(.5,.35)[above]{\footnotesize$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=.25cm,yshift=-.25cm]% Leiterschleife | |||
| \draw (0,0)--(0,-2)--(.5,-2)--(.5,0); | |||
| \draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.05) node [above right] {\footnotesize$+$}; | |||
| \fill (1,0)circle(.05) node [above left] {\footnotesize$-$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=0cm,yshift=-1cm]% O. | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$\frac{d\Phi}{dt}$}; | |||
| \fill (.5,0)circle(.05); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,green!50!black,xshift=0cm,yshift=-1.33cm]% Ox | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$}; | |||
| \draw [black]node at (1.5,-1)[below]{Wirkung $\frac{d\Phi_{ind}}{dt}$ entgegen Ursache $\frac{d\Phi}{dt}$}; | |||
| \draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,red,xshift=3cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.05) node [above right] {\footnotesize$+$}; | |||
| \fill (1,0)circle(.05) node [above left] {\footnotesize$-$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3.25cm,yshift=-0.25cm]% Leiterschleife | |||
| \draw (0,0)--(0,-2)--(.5,-2)--(.5,0); | |||
| \draw [dashed] (0,0)--(-.25,.25)(.5,0)--(.75,.25); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=-2.25cm]%Stromquelle | |||
| \draw [green!50!black](.25,0)--(.75,0); | |||
| \draw node at (.5,.133) [above] {$i_{}$}; | |||
| \draw [green!50!black](.5,0)circle(.133); | |||
| \draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$i_{ind}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,green!50!black,xshift=3cm,yshift=-1.33cm]% Ox | |||
| \draw (.5,0)circle(.133)node at (1,0)[right]{$-\frac{d\Phi_{ind}}{dt}$}; | |||
| \draw [very thick](.5,0)--+(45:.133) (.5,0)--+(135:.133)(.5,0)--+(225:.133)(.5,0)--+(315:.133); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=-1.25cm]% | |||
| \draw [->,red] (-2.5,.5)--(-2.5,-.5) node at(-2.5,0)[right]{$\frac{di}{dt}$}; | |||
| \draw [<-,green!50!black] (.5,.5)--(.5,-.5) node at(.5,0)[left]{$i_{ind}$}; | |||
| \draw [<-,red!75!black] (-1.75,.5)--(-1.75,-.5) node at(-1.75,0)[left]{$i_{ind}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| $u$ hat negatives Vorzeichen, da Polung von $i_{ind}$ entgegen Bezugspfeilen aus Skizze\\ | |||
| ($u_R = -u$) | |||
| \clearpage | |||
| }{}% | |||
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| \section{Drahtschleife} | |||
| Eine Drahtschleife $N=1$ wird von einem zeitlich veränderlichen Fluss durchsetzt.\\ | |||
| $\Phi(t)=\Phi_0\cdot (1-e^{-t/T})$ mit $\Phi_0=60\cdot \power{10}{-6 | |||
| }\,\volt\second$ und $T=1\,\milli\second$. | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Berechnen Sie die Spannung $u$ für $t=0{,}5\cdot T$! | |||
| \item Geben Sie die Polarität der Anschlussklemmen der Drahtschleife a-b für diesen Zeitpunkt an und begründen Sie ihre Angabe! | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \begin{scope}[>=latex,red,ultra thick,xshift=0,yshift=0] | |||
| \draw (0,0) -- (0,1); | |||
| \draw [dashed] (0,1) -- (0,2); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick,xshift=0,yshift=2cm] | |||
| \draw (0,0) ellipse (2cm and 1cm); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick,xshift=1.9cm,yshift=1.75cm] | |||
| \filldraw [black!0!] (0,0.02) rectangle (0.2,0.48); | |||
| \draw [fill] (1pt,0) -- (1,0) circle (2pt) node [right] {b}; | |||
| \draw [fill] (1pt,0.5cm) -- (1,0.5) circle (2pt) node [right] {a}; | |||
| \draw node at (1.5,0.25)[right] {$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] | |||
| \draw [->] (0,2) -- (0,4) node [above] {$\Phi(t)$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| u&=-N\cdot \frac{d\Phi}{dt} | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \intertext{a) Spannung $u$ für $t=0{,}5\cdot T$} | |||
| u&=-N\cdot \frac{d\Phi}{dt} \qquad \text{mit $N=1$ }\\ | |||
| &=-\frac{d\left(\Phi_0\cdot \left(1-e^{-\frac{t}{T}}\right)\right)}{dt}\\ | |||
| &=-\Phi_0\cdot \frac{d}{dt} \left(1-e^{-\frac{t}{T}}\right)\\ | |||
| &=-\Phi_0\cdot \Bigg[-e^{-\frac{t}{T}}\cdot \hspace{-.7cm}\underbrace{ \left(-\frac{1}{T}\right)}_{\mathrm{Nachdifferenzieren}}\hspace{-.5cm}\Bigg]\\ | |||
| &\text{mit $t=0{,}5\cdot T\qquad T=1\,\milli\second$}\\ | |||
| u&=-60\cdot \power{10}{-6}\,\volt\second\cdot \left[-e^{-0{,}5}\cdot \left(-\frac{1}{1\,\milli\second}\right)\right]=\uuline{-36{,}39\,\milli\volt} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{b) Polarität der Spannung $u$ für $t=0{,}5\cdot T$} | |||
| \frac{d\Phi}{dt}&>0 \Rightarrow \text{Linke-Hand für Stromrichtung}\\ | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \foreach \xs in {0} { % Enter Start value of x label | |||
| \foreach \ys in {0} { % Enter Start value of y label | |||
| \foreach \ii in {10} { % Enter Number of Decades in x | |||
| \foreach \jj in {6} { % Enter Number of Decades in y | |||
| \foreach \xe in {5} { % Enter End value of x label | |||
| \foreach \ye in {60} { % Enter End value of y label | |||
| \foreach \i in {1,2,...,\ii} { | |||
| \foreach \j in {1,2,...,\jj} { | |||
| }}% End Log Grid | |||
| \draw[black!80!] (0,0) grid (\ii,\jj); % Draw Linear grid | |||
| \draw [<->,thick] (0,\jj+.2) node (yaxis) [above] {$\Phi\,[\micro\volt\second]$} |- (\ii+.2,0) node (xaxis) [right] {$t\,[\milli\second]$}; % Draw axes | |||
| \draw [thick] (-1.25,\jj+.2) node (yaxis) [above] {$u\,[\milli\volt]$} ; % Draw axes | |||
| \foreach \x in {\xs,.5,...,\xe}% x Axis Label: | |||
| \node [black,anchor=north] at(\x*2-\xs,0){$\x$}; | |||
| \foreach \y in {\ys,10,...,\ye}% y Axis Label: | |||
| \node [red,anchor=east] at(0,\y/10-\ys){$\y$}; | |||
| \foreach \y in {\ys,10,...,\ye}% y Axis Label: | |||
| \node [blue,anchor=east] at(-1,\y/10-\ys){$-\y$}; | |||
| }}}}}} | |||
| \draw [black,thick,dashed] (0,0) -- (2,6) -- (2,0.3); | |||
| \draw [blue,thick,dashed] (1,3.64) -- (-.75,3.64) node [left]{$-36{,}39\,\milli\volt$}; | |||
| \draw[color=red, very thick,domain=0:10] plot[id=tau] function{6*(1-exp(-x/2))}; | |||
| \draw[color=blue, very thick,domain=0:10] plot[id=tau] function{6*(exp(-x/2))}; | |||
| \draw [black] node at (2,0)[above] {$T$}; | |||
| \draw [red] node at (2,3.5) [right]{$\Phi=\Phi_0\cdot \left(1-e^{-t/T}\right)$}; | |||
| \draw [blue] node at (2,2.5) [right]{$u=-\Phi_0\cdot \left(\frac{e^{-t/T}}{T}\right)$}; | |||
| \draw node at (0,-1) [right]{Induzierte Spannung bei zunehmendem Fluss}; | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{\uline{Lenzsche Regel:}} | |||
| \text{Klemme a $\Rightarrow$ $+$}\\ | |||
| \text{Klemme b $\Rightarrow$ $-$}\\ | |||
| %F_{\sqcup\sqcap}\\ %-------------------Versuch mit Symbolen-------------- | |||
| %\ding{43}\\ | |||
| %\oplus\ominus\odot\\ | |||
| %\circlearrowright\\ | |||
| %\lightning\\ | |||
| %\curvearrowright\\ | |||
| %\leftthumbsup\\ | |||
| %\SmallSquare\\ | |||
| %\SmallTriangleUp\\ %-------------------Versuch mit Symbolen-------------- | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \begin{scope}[>=latex,red,ultra thick,xshift=0,yshift=0] | |||
| \draw (0,0) -- (0,1); | |||
| \draw [dashed] (0,1) -- (0,2); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick,xshift=0,yshift=2cm] | |||
| \draw (0,0) ellipse (2cm and 1cm); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,blue,very thick,xshift=1.9cm,yshift=1.75cm] | |||
| \filldraw [black!0!] (0,0.02) rectangle (0.2,0.48); | |||
| \draw [fill] (1pt,0) -- (1,0) circle (2pt) node [below] {b $-$}; | |||
| \draw [fill] (1pt,0.5cm) -- (1,0.5) circle (2pt) node [above] {a $+$}; | |||
| \draw node at (1.5,0.25)[right] {$u$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,red, ultra thick,xshift=0,yshift=0] | |||
| \draw [->] (0,2) -- (0,4) node [above] {$\Phi(t)$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=5cm,yshift=1.5cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$}; | |||
| \draw [<-,blue] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$U_{R}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=2cm] | |||
| \draw [dashed] (0,.25)--(2,.5) (0,-.25)--(2,-.5); | |||
| \draw [->,red] (0,.25) +(7.2:.75) -- +(7.2:1.25) node [above left]{$i$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \section{Spannungsverlauf} | |||
| Gegeben ist die dargestellte Spannung: | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \begin{scope}[>=latex,thick] | |||
| \draw [->](0,0) -- (6,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,-1.25) -- (0,1.25) node [above] {$u\,[\volt]$}; | |||
| \draw [red,very thick](0,0)--(1,1)--(1,0)--(1.5,0)--(1.5,-1) | |||
| --(2.5,-1)--(2.5,0)--(3.5,1)--(3.5,0)--(4,0)--(4,-1)--(5,-1)--(5,0)--(6,1); | |||
| \foreach \x in {10,20,...,50} | |||
| \draw (\x/10,0) -- (\x/10,-0.2) node[anchor=north] {$\x$}; | |||
| \foreach \y in {-10,0,10} | |||
| \draw (0,\y/10) -- (-0.2,\y/10) node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Ermitteln Sie die Frequenz der Grundschwingung! | |||
| \item Berechnen Sie den Gleichrichtwert der Spannung! | |||
| \item Berechnen Sie den Effektivwert der Spannung! | |||
| \item Berechnen Sie den Formfaktor der Spannung! | |||
| \item Nun wird die dargestellte Spannung an einen Ohmschen Widerstand von $100\,\ohm$ angelegt. Welche Verlustleistung tritt im Widerstand auf?\\ | |||
| \end{enumerate} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| \overline{|u|}&=\frac{1}{T}\cdot \int_{t=0}^{T}{|u(t)|\cdot dt}&\text{Gleichrichtwert}\\ | |||
| U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&\\ | |||
| F&=\frac{U}{\overline{|u|}}=\frac{\text{Effektivwert}}{\text{Gleichrichtwert}} | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \intertext{a) Grundschwingung mit $T=25\,\milli\second$:} | |||
| f&=\frac{1}{T}=40\,\hertz\\ | |||
| \intertext{b) Gleichrichtwert der Spannung:} | |||
| \overline{|u|}&=\frac{1}{T}\cdot (F_{\triangle} +F_{\sqcup\hspace{-.2cm}\sqcap})=\frac{1}{25\,\milli\second}\cdot (\frac{1}{2}\cdot 10\,\volt\cdot 10\,\milli\second+10\,\volt\cdot 10\,\milli\second)=\frac{150\,\volt\cdot \milli\second}{25\milli\second}=\uuline{6\,\volt}\\ | |||
| \intertext{c) Effektivwert der Spannung:} | |||
| U&=\sqrt{\frac{1}{T}\int_{0}^{T}{u^2}\cdot dt}\\ | |||
| U^2&=\frac{1}{T}\left(\int_{0}^{10\,\milli\second}{\left(\frac{10\,\volt}{\power{10}{-2}\,\second}\cdot t\right)^2\cdot dt}+\int_{15\,\milli\second}^{25\,\milli\second}{(-10\,\volt)^2\cdot dt}\right)\\ | |||
| &=\frac{1}{25\,\milli\second}\left(\frac{100\,\square\volt}{\power{10}{-4}\,\square\second}\cdot \left[\frac{t^3}{3}\right]_{0}^{10\,\milli\second}+100\,\square\volt\cdot \big[t\big]_{15\,\milli\second}^{25\,\milli\second}\right)\\ | |||
| &=\frac{100\,\square\volt}{25\,\milli\second}\left(\power{10}{4}\frac{1}{\,\square\second}\cdot \frac{1}{3}\cdot \power{10}{-6}\,\cubic\second+10\,\milli\second\right)=53{,}33\,\square\volt\\ | |||
| U&=\sqrt{53{,}33}\,\volt=\uuline{7{,}30\,\volt} | |||
| \intertext{d) Formfaktor der Spannung:} | |||
| F&=\frac{U}{\overline{|u|}}=\frac{7{,}30\,\volt}{6\,\volt}=\uuline{1{,}22} | |||
| \intertext{e) Verlustleistung im Widerstand:} | |||
| P&=\frac{U^2}{R}=\frac{53{,}33\,\square\volt}{100\,\ohm}=\uuline{0{,}533\,\watt} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Phasenanschnitt} | |||
| Berechnen Sie den Effektivwert dieser sinusförmigen Spannung mit Phasenanschnitt.\\ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \begin{scope}[>=latex,thick] | |||
| \draw [ultra thin] (0,-2)grid(6,2); | |||
| \draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$}; | |||
| \draw[color=red,very thick,domain=0.6:2,smooth,samples=100](0,0)--(0.6,0)--(0.6,1.618) | |||
| plot[id=sina]function{2*sin(.5*3.14*x)}; | |||
| \draw[color=red,very thick,domain=0.6:2,smooth,samples=100]plot[id=sinb]function{2*sin(0.5*3.14*x)}; | |||
| \draw[color=red,very thick,domain=2.6:4,smooth,samples=100] (2,0)--(2.6,0)-- (2.6,-1.618) plot[id=sinc] function{2*(sin(0.5*3.14*x))}; | |||
| \draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618)plot[id=sind]function{2*sin(0.5*3.14*x)}; | |||
| \foreach \x in {10,20,30} | |||
| \draw(\x/5,0)--(\x/5,-0.2) node[anchor=north]{$\x$}; | |||
| \draw (3/5,0)--(3/5,-0.2)node[anchor=north]{$3$}; | |||
| \foreach \y in {-400,-200,...,400} | |||
| \draw (0,\y/200)--(-0.2,\y/200)node[anchor=east]{$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}&\\ | |||
| \sin^2\alpha&=\frac{1}{2}(1-\cos 2\alpha) | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1] | |||
| \begin{scope}[>=latex,thick] | |||
| \draw [ultra thin] (0,-2)grid(6,2); | |||
| \draw [->](0,0) -- (6.5,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,-2.25) -- (0,2.25) node [above] {$u\,[\volt]$}; | |||
| \draw[color=red,very thick,domain=0.6:2,smooth,samples=100] (0,0)--(0.6,0)--(0.6,1.618) plot[id=sin12a53] function{2*sin(0.5*3.14*x)}; | |||
| \draw[color=red,very thick,domain=2.6:4,smooth,samples=100] (2,0)--(2.6,0)--(2.6,-1.618) plot[id=sin12a54] function{2*(sin(0.5*3.14*x))}; | |||
| \draw[color=red,very thick,domain=4.6:6,smooth,samples=100] (4,0)--(4.6,0)--(4.6,1.618) plot[id=sin12a55] function{2*sin(0.5*3.14*x)}; | |||
| \draw[color=blue,very thick,dashed, domain=0:4,smooth,samples=100] plot[id=sin12a56] function{2*sin(0.5*3.14*x)}; | |||
| \foreach \x in {10,20,30} | |||
| \draw (\x/5,0) -- (\x/5,-0.2) node[anchor=north] {$\x$}; | |||
| \draw (3/5,0) -- (3/5,-0.2)node[anchor=north] {$3$}; | |||
| \foreach \y in {-400,-200,0,200,400} | |||
| \draw (0,\y/200) -- (-0.2,\y/200) node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{Periodendauer $T=20\,\milli\second$, da Symmetrie in einer Periode. Betrachtung nur einer Sinus-Halbwelle mit $\frac{1}{2}\cdot T=10\,\milli\second$} | |||
| \omega&=2\pi f=\frac{2\pi}{T}=\frac{2\pi}{20\,\milli\second}=314\,\frac{1}{\second}\\ | |||
| u(t)&= | |||
| \begin{cases} | |||
| 0&\text{ für }t=0\ldots 3\,\milli\second\\ | |||
| 400\,\volt\cdot \sin(\omega t)=400\,\volt\cdot \sin(314\,\frac{1}{\second}\cdot t) &\text{ für }t=3\,\milli\second\ldots 10\,\milli\second\\ | |||
| \end{cases}\\[\baselineskip] | |||
| U^2&=\frac{1}{T/2}\int_{3\,\milli\second}^{10\,\milli\second}{(400\,\volt\cdot \sin(\omega t))^2\cdot dt}\\ | |||
| &=\frac{1}{T/2}\cdot (400\,\volt)^2\int_{3\,\milli\second}^{10\,\milli\second}{\sin^2(\omega t)\cdot dt}\\[\baselineskip] | |||
| &\text{mit }\sin^2\alpha=\frac{1}{2}(1-\cos 2\alpha)\\[\baselineskip] | |||
| U^2&=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot\frac{1}{2}\cdot \left(\int_{3\,\milli\second}^{10\,\milli\second}{1\, dt} -\int_{3\,\milli\second}^{10\,\milli\second}{\cos(2 \omega t)\cdot dt}\right)\\ | |||
| &=\frac{(400\,\volt)^2}{10\,\milli\second}\cdot \frac{1}{2}\cdot \left(\Big[t\Big]_{3\,\milli\second}^{10\,\milli\second}-\left[\sin(2 \omega t)\cdot \frac{1}{2\omega} \right]_{3\,\milli\second}^{10\,\milli\second}\right)\\ | |||
| &=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{1}{2\cdot \omega}\cdot \underbrace{\left(sin(\cancel{2}\cdot \frac{2\pi}{\cancel{20\,\milli\second}}\cdot \cancel{10\,\milli\second}\right)}_{sin(2\pi)=0}-sin\left(2\cdot \frac{\pi}{20\,\milli\second}\cdot 3\,\milli\second\right)\right)\\ | |||
| &=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot \left(7\,\milli\second-\frac{20\,\milli\second}{4\pi}\cdot \underbrace{(-sin(0{,}6\pi))}_{-0{,}951}\right)\\ | |||
| &=\frac{(400\,\volt)^2}{20\,\milli\second}\cdot 8{,}51\,\milli\second=68109\,\square\volt\\[\baselineskip] | |||
| U&=400\,\volt\cdot \sqrt{\frac{8{,}51\,\milli\second}{20\,\milli\second}}=\uuline{261\,\volt}\\ | |||
| \intertext{Zum Vergleich: Sinus ohne Phasenanschnitt hätte einen Effektivwert von} | |||
| U_{(sin)}&=\frac{400\,\volt}{\sqrt{2}}=283\,\volt | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Rechteckspannung} | |||
| Gegeben ist eine periodische Rechteckspannung mit der Periodendauer von $10\,\milli\second$.\\ | |||
| Berechnen Sie den Effektivwert, wenn der arithmetische Mittelwert gleich Null ist.\\ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=0.5] | |||
| \begin{scope}[>=latex,thick] | |||
| \draw [->](0,-2) -- (10.75,-2) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$}; | |||
| \draw [<->,blue, very thick] (-.5,0)--(-.5,5) node at (0,2.5)[right]{$5\,\volt$}; | |||
| \draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0) | |||
| --(10,5)--(10.5,5); | |||
| \foreach \x in {0,1,...,10} | |||
| \draw (\x,-2) -- (\x,-2.2) node[anchor=north] {$\x$}; | |||
| \foreach \y in {0,1,...,5} | |||
| \draw (0,\y) -- (-0.2,\y);% node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| \overline{u}&=\frac{1}{T}\cdot \int_{t=0}^{T}{u(t)\cdot dt}&\text{Arithmetischer Mittelwert}\\ | |||
| U&=U_{\textrm{eff}}=\sqrt{\frac{1}{T}\cdot \int_{t=0}^{T}{u^2(t)\cdot dt}}&\text{Effektivwert}& | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \intertext{a) Arithmetischer Mittelwert $\overline{u}=0$} | |||
| &\text{$\Rightarrow$ Fläche ober- und unterhalb der Nulllinie muß gleich sein!}\\ | |||
| &\text{$\Rightarrow$ Wo ist die Nulllinie?} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=0.5] | |||
| \begin{scope}[>=latex,thick] | |||
| \draw [->](0,0) -- (10.75,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,0) -- (0,5.5) node [above] {$u\,[\volt]$}; | |||
| \fill [black!15!] (0,3.5)rectangle(7,5) (7,3.5)rectangle(10,0); | |||
| \draw [red,very thick](-0.5,0)--(0,0)--(0,5)--(7,5)--(7,0)--(10,0) | |||
| --(10,5)--(10.5,5); | |||
| \draw [blue,very thick] (0,3.5)--(10,3.5) node [right]{NULL}; | |||
| \draw [->,blue,very thick] (3.5,3.5)--(3.5,5) node at(3.5,4.25)[right]{$u_1$}; | |||
| \draw [<-,blue,very thick] (8.5,0)--(8.5,3.5) node at(8.5,1.75)[right]{$u_2$}; | |||
| \foreach \x in {0,1,...,10} | |||
| \draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$}; | |||
| \foreach \y in {-3.5,-2.5,...,1.5} | |||
| \draw (0,\y+3.5) -- (-0.2,\y+3.5) node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \enlargethispage{1cm} | |||
| \begin{align*} | |||
| &u_1+(-u_2)=5\,\volt\rightarrow u_2=-(5\,\volt-u_1)\\ | |||
| &\text{Fläche:}\\ | |||
| &\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\ | |||
| &\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\ | |||
| &u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\ | |||
| &u_1\cdot 7\,\milli\second-(5\,\volt-U_1)\cdot 3\,\milli\second=0\\ | |||
| &(7\,\milli\second+3\,\milli\second)\cdot u_1=15\,\volt\cdot \milli\second\\ | |||
| &u_1=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uuline{1{,}5\,\volt}\\ | |||
| &u_2=-(5\,\volt-1{,}5\,\volt)=\uuline{-3{,}5\,\volt} | |||
| \intertext{Alternativ mit Beträgen} | |||
| &|u_1|+|u_2|=5\,\volt\rightarrow |u_2|=5\,\volt-|u_1|\\ | |||
| &\text{Fläche:}\\ | |||
| &\frac{1}{T}\int_{0}^{T}{u(t)\cdot dt}\stackrel{!}{=}0\\ | |||
| &\frac{1}{T}\cdot \left(\int_{0}^{7\,\milli\second}{u_1\cdot dt}-\int_{7\,\milli\second}^{10\,\milli\second}{u_2\cdot dt}\right)\stackrel{!}{=}0\\ | |||
| &u_1\cdot 7\,\milli\second-u_2\cdot 3\,\milli\second=0\\ | |||
| &u_1\cdot 7\,\milli\second-(5\,\volt-|u_1|)\cdot 3\,\milli\second=0\\ | |||
| &(7\,\milli\second+3\,\milli\second)\cdot |u_1|=15\,\volt\cdot \milli\second\\ | |||
| &|u_1|=\frac{15\,\volt\cdot \milli\second}{10\,\milli\second}=\uline{1{,}5\,\volt}\qquad | |||
| |u_2|=5\,\volt-1{,}5\,\volt)=\uline{3{,}5\,\volt}\\ | |||
| &\text{Da $u_1$ positives Vorzeichen in der Skizze hat, muss $u_2$ ein negatives Vorzeichen erhalten.}\\ | |||
| \Rightarrow& u_1=\uuline{1{,}5\,\volt} \qquad u_2=\uuline{-3{,}5\,\volt} | |||
| \intertext{b) Effektivwert} | |||
| U^2&=U^2_{\textrm{eff}}=\frac{1}{T}\int{\left(u(t)\right)^2\cdot dt}\\ | |||
| &=\frac{1}{10\,\milli\second}\left(\int_{0}^{7\,\milli\second}{(1{,}5\,\volt)^2\cdot dt}+\int_{7\,\milli\second}^{10\,\milli\second}{(-3{,}5\,\volt)^2\cdot dt}\right)\\ | |||
| &=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot \big[t\big]_{0}^{7\,\milli\second}+12{,}25\,\volt^2\cdot \big[t\big]_{7\,\milli\second}^{10\,\milli\second}\right)\\ | |||
| &=\frac{1}{10\,\milli\second}\cdot \left(2{,}25\,\volt^2\cdot 7\,\milli\second+12{,}25\,\volt^2\cdot (10\,\milli\second-7\,\milli\second)\right)=5{,}25\,\volt^2\\ | |||
| U&=\sqrt{5{,}25\,\volt^2}=\uuline{2{,}29\,\volt} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Scheinersatzwiderstände} | |||
| Der Eingangswiderstand eines linearen Zweipols beträgt bei der Frequenz $f=800\,\hertz$\\ | |||
| $Z=600\,\ohm$, sein Phasenwinkel ist $\varphi=30\,\degree$ induktiv. | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Berechnen Sie die Schaltungselemente $R_r$ und $L_r$ der gleichwertigen Reihenersatzschaltung! | |||
| \item Berechnen Sie die Schaltungselemente $R_p$ und $L_p$ der gleichwertigen Parallelersatzschaltung! | |||
| \item Wie ändern sich die Scheinersatzwiderstände (Betrag und Phase) beider Ersatzschaltungen, wenn die Frequenz $f'= 600\,\hertz$ beträgt? | |||
| \end{enumerate} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| \underline{Z}& &\text{Scheinwiderstand (Impedanz)}\\ | |||
| Z&=|\underline{Z}| &\text{Betrag des Scheinwiderstandes}\\ | |||
| X&=\omega\cdot L &\text{Blindwiderstand (Reaktdanz)}\\ | |||
| B&=-\frac{1}{\omega\cdot L} &\text{Blindleitwert (Suszepdanz)} | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \intertext{a) Widerstandsebene:} | |||
| R_r&=Z\cdot \cos(\varphi_r)=600\,\ohm\cdot \cos(30\degree)=\uuline{520\,\ohm}\\ | |||
| X_r&=Z\cdot \sin(\varphi_r)=600\,\ohm\cdot \sin(30\degree)=300\,\ohm\\ | |||
| L_r&=\frac{X_r}{\omega}=\frac{X_r}{2\pi f}=\frac{300\,\ohm}{2\pi\cdot 800\,\frac{1}{\second}}=\uuline{60\,\milli\henry} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_r$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{R_r}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Spule - | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L_r$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$u_{L_r}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.025) node [above left] {\footnotesize$1$}; | |||
| \fill (2,0)circle(.025) node [above right] {\footnotesize$2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick, xshift=3cm, yshift=-.25cm] | |||
| \draw [->,thin](0,0)--(1,0)node[right]{$R$}; | |||
| \draw [->](0,0)--(.52,0)node at (.26,0)[below]{$R_r$}; | |||
| \draw [->,thin](0,0)--(0,.5)node[above]{$X$}; | |||
| \draw [->](.52,0)--(.52,.3)node at (.52,.15)[right]{$X_r$}; | |||
| \draw [->](0:0)--(30:.6)node at (30:.3)[above left]{$Z$}; | |||
| \draw [->,red,thin] (0:.26) arc (0:30:.26cm) node at (15:.26) [right] {$\varphi_r$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{b) Leitwertebene:} | |||
| Y&=\frac{1}{Z}=\frac{1}{600\,\ohm}=1{,}667\,\milli\siemens \text{; }\qquad\varphi_p=-\varphi_r=-30\degree\\ | |||
| G_p&=Y\cdot \cos(\varphi_p)=1{,}667\,\milli\siemens\cdot \cos(-30\degree) | |||
| =1{,}443\,\milli\siemens\Rightarrow R_p=\frac{1}{G_p}=\uuline{693\,\ohm}\\ | |||
| B_p&=Y\cdot \sin(\varphi_p)=1{,}667\,\milli\siemens\cdot \sin(-30\degree) | |||
| =-0{,}833\,\milli\siemens\Rightarrow X_p=-\frac{1}{B_p}=1200\,\ohm\\ | |||
| %&\text{mit }B_p=-\frac{1}{\omega\cdot L_p}\Rightarrow \\ | |||
| L_p&=-\frac{1}{\omega\cdot B_p}=-\frac{1}{2\pi f\cdot B_p}=\frac{-1}{2\pi\cdot 800\,\frac{1}{\second}\cdot (-0{,}833\cdot \power{10}{-3}\,\frac{1}{\ohm})}=\uuline{239\,\milli\henry} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$G_p$}; | |||
| \draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$i_{G_p}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Spule | | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$B_p$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \draw [<-,red] (.3,.2)--(.7,.2)node at(.5,.2)[left]{\footnotesize$i_{B_p}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=-1cm,yshift=0cm]%End Knoten | |||
| \draw (2,.5)--(2,0)--(0,0); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=-1cm,yshift=1cm]%End Knoten | |||
| \draw (2,-.5)--(2,0)--(0,0); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=-1cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,1)circle(.025) node [above left] {\footnotesize$1$}; | |||
| \fill (0,0)circle(.025) node [above left] {\footnotesize$2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick, xshift=2cm, yshift=.5cm] | |||
| \draw [->,thin](0,0)--(1,0)node[right]{$G$}; | |||
| \draw [->](0,0)--(.52,0)node at (.26,0)[above]{$G_p$}; | |||
| \draw [->,thin](0,-.5)--(0,.5)node[above]{$jB$}; | |||
| \draw [->](.52,0)--(.52,-.3)node at (.52,-.15)[right]{$B_p$}; | |||
| \draw [->](0:0)--(-30:.6)node at (-30:.3)[below left]{$Y$}; | |||
| \draw [->,red,thin] (0:.26) arc (0:-30:.26cm) node at (-15:.26) [right] {$\varphi_p$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{c) Frequenz $f'$ \newline Reihenschaltung:} | |||
| R'_r&\stackrel{!}{=} R_r=520\,\ohm\\ | |||
| X'_r&=\omega'\cdot L_r=2\pi \cdot 600\,\frac{1}{\second}\cdot 0{,}06\,\ohm\second=226\,\ohm\\ | |||
| Z'_r&=\sqrt{R'^2_r+X'^2_r}=\sqrt{520^2+226^2}\,\ohm=\uuline{567\,\ohm}\\ | |||
| %\varphi'&=\arctan\frac{\Im\mathfrak m}{\Re\mathfrak e}=\arctan\frac{X'_r}{R'_r}=\arctan\frac{226\,\ohm}{520\,\ohm}=\uuline{23{,}5\,\degree} | |||
| \varphi'_r&=\arctan\frac{\Im}{\Re}=\arctan\frac{X'_r}{R'_r}=\arctan\frac{226\,\ohm}{520\,\ohm}=\uuline{23{,}5\,\degree} | |||
| \intertext{Parallelschaltung:} | |||
| G'_p&\stackrel{!}{=} G_p=1{,}443\,\milli\siemens\\ | |||
| B'_p&=\frac{-1}{\omega'\cdot L_p}=\frac{-1}{2\pi \cdot 600\,\frac{1}{\second}\cdot 0{,}239\,\ohm\second}=\frac{-1}{901\,\ohm}=-1{,}11\,\milli\siemens\\ | |||
| Y'_p&=\sqrt{G'^2_r+B'^2_r}=\sqrt{1{,}443^2+(-1{,}11)^2}\,\milli\siemens=1{,}82\,\milli\siemens\\ | |||
| Z'_p&=\frac{1}{Y'_p}=\uuline{549\,\ohm}\\ | |||
| \varphi'_p&=\arctan\frac{-1{,}11\,\milli\siemens}{1{,}443\,\milli\siemens}=\uuline{-37{,}6\,\degree} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \section{Blindleistungskompensation} | |||
| Die Daten der beiden Verbraucher am Einphasen-Wechselstromnetz sind:\\ | |||
| Heizwiderstand $R_H$: | |||
| Aufgenommene Leistung $P_H=1,5\,\kilo\watt$\\ | |||
| Motor $M$: | |||
| Aufgenommene Leistung $P_{auf}=2,5\,\kilo\watt$\\ | |||
| Leistungsfaktor $\cos\varphi=0,7$\\ | |||
| $U_N=230\,\volt$; $f=50\,\hertz$ | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Wie groß ist der dem Netz entnommene Strom $I_N$? | |||
| \item Welche Phasenverschiebung hat der Strom $\uline{I}_N$ zu der Spannung $\uline{U}_N$? | |||
| \item Welche Kapazität muss ein Kondensator, parallel zu den Verbrauchern geschaltet, haben, damit der Blindstrom voll kompensiert wird? | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=.75] | |||
| \begin{scope}[>=latex,very thick] | |||
| \draw (0,0) circle (1) node [above]{$M$}; | |||
| \draw node at (0,0) [below]{$1 \approx$}; % F E H L E R | |||
| \draw (180:1)--(-2,0)--(-2,5)circle(.025) (0:1)--(2,0)--(2,5)circle(.025) ; | |||
| \draw (-1,2.5)--(-2,2.5)circle(.025) (1,2.5)--(2,2.5)circle(.025) (-1,2.2 )rectangle (1,2.8); | |||
| \draw (0,2.5)node{$R_H$}; | |||
| \draw [->,blue](-1.8,5)--(1.8,5) node at (0,5) [above]{$\underline{U}_N$;$f$}; | |||
| \draw [->,red](-1.8,4.8)--(-1.8,2.7) node at (-1.8,3.75) [right]{$\underline{I}_N$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1.5] | |||
| \begin{scope}[>=latex,very thick] | |||
| \draw [->,red!50!blue](0:0)--(45.6:3.571)node [above left]{$\underline{S}_M$}; | |||
| \draw [->,red!50!blue](0:0)--(0:2.5)node at (1.25,0)[above]{$P_M$}; | |||
| \draw [->](0:0)--(32.52:4.744)node [right]{$\underline{S}_{gesamt}$}; | |||
| \draw [->](45.6:3.571)--+(0:1.5)node at (3.25,2.551)[above]{$P_H$}; | |||
| \draw [->,red!50!blue](2.5,0)--+(90:2.551)node at (2.5,1.25)[right]{$Q_M$}; | |||
| \draw [red!50!blue]node at (1.5,.75)[below]{Motor}; | |||
| \draw [->](2.5,0)--(4,0)node at (3.25,0)[above]{$P_H$}; | |||
| \draw [->,red!50!blue](4,0)--+(90:2.551)node at (4,1.25)[right]{$Q_M$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex, xshift=0, yshift=0] | |||
| \draw [black!50!] (0,0)grid(4,3); | |||
| \foreach \x in {0,1,...,4}% x Axis Label: | |||
| \node [anchor=north] at(\x,0){$\x$}; | |||
| \foreach \y in {0,1,...,3}% y Axis Label: | |||
| \node [anchor=east] at(0,\y){$\y$}; | |||
| \draw [->,thick](0,0)--(4.25,0) node [right]{$P\,[\kilo\watt]$}; | |||
| \draw [->,thick](0,0)--(0,3.25) node [above]{$Q\,[\kilo\var]$}; | |||
| \end{scope} | |||
| \end{tikzpicture}\\ | |||
| \centering \underline{S}_{gesamt}=\sqrt{(P_M+P_H)^2+Q_M^2} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{a) Nennstrom} | |||
| \varphi&=\arccos(0{,}7)=45{,}6\degree\\ | |||
| S_M&=\frac{P_{auf}}{0{,}7}=\uline{3571\,\volt\ampere} &\text{Motor Scheinleistung}\\ | |||
| Q_M&=S_M\cdot \sin\varphi=3571\,\volt\ampere\cdot 0{,}7141=\uline{2551\,\var} &\text{Motor Blindleistung}\\ | |||
| P&=P_{auf}+P_H=\uline{4\,\kilo\watt} &\text{Gesamte Wirkleistung}\\ | |||
| S&=\sqrt{P^2+Q^2_M}=\uline{4744\,\volt\ampere} &\text{Gesamte Scheinleistung}\\ | |||
| I_N&=\frac{S}{U_N}=\frac{4744\,\volt\ampere}{230\,\volt}=\uuline{20{,}63\,\ampere} | |||
| \intertext{b) Phasenverschiebung} | |||
| \varphi_N&=\arccos\frac{P}{S}=\arccos\frac{4000\,\watt}{4744\,\volt\ampere}=\uuline{32,52\degree} | |||
| \intertext{c) Kompensation} | |||
| |Q_C|&=Q_M=2551\,\var=U^2_N\cdot \omega\cdot C\\ | |||
| C&=\frac{2551\,\var}{2\pi\cdot 50\,\frac{1}{\second}\cdot (230\,\volt)^2}=\uuline{153{,}6\,\micro\farad} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Energieübertragung} | |||
| Die Skizze zeigt ein System zur elektrischen Energieübertragung bestehend aus Quelle, | |||
| Leitung und Verbraucher. Das System soll mit einem parallel geschalteten Kondensator $X_C$ so | |||
| optimiert werden, dass die Leitungsverluste $P_{VRL}$ minimal werden.\\ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Spannungsquelle | | |||
| \draw (0,0)--(1,0)node at(.5,-.133)[right]{\footnotesize{$\underline{U}_q=100\,\volt\cdot e^{j0}$}}; | |||
| \draw (.5,0)circle(.133); | |||
| \draw [<-,blue] (.3,.2)--(.7,.2)node at (.5,.2)[left]{\footnotesize$\underline{U}_q$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {\footnotesize{$\underline{Z}_i=(1+2j)\,\ohm$}}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {\footnotesize{$R_L=1\,\ohm$}}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2.5cm,yshift=0cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [above right] {\footnotesize{$\underline{Z}_V=(10+j5)\,\ohm$}}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3.5cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {\footnotesize{$jX_C$}}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw [dashed](2.5,0)--(3.5,0)--(3.5,.2) (2.5,1)--(3.5,1)--(3.5,.8); | |||
| \draw [dashed](0,0)--(2.5,0) (2.5,1)--(3.5,1)--(3.5,.8); | |||
| \draw (0,0)--(2.5,0) (2,1)--(2.5,1); | |||
| \draw (0,.2)--(0,0)--(0,.2) (0,.8)--(0,1)--(.2,1); | |||
| \draw node at(.5,0) [below] {\footnotesize Quelle}; | |||
| \draw node at(1.5,0) [below] {\footnotesize Leitung}; | |||
| \draw node at(2.5,0) [below] {\footnotesize Verbraucher}; | |||
| \draw [very thin, dashed] (1,-.2)--(1,1.2)(2,-.2)--(2,1.2); | |||
| \draw [->,red](2.125,1.125)--(2.375,1.125) node [right] {\footnotesize{$\underline{I}$}}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte | |||
| \draw (0,0)circle(.035); | |||
| \fill [white](0,0)circle(.025); | |||
| \draw (1,0)circle(.035); | |||
| \fill [white](1,0)circle(.025); | |||
| \fill (1.5,0)circle(.025); | |||
| \fill (2.5,0)circle(.025); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte | |||
| \draw (0,0)circle(.035); | |||
| \fill [white](0,0)circle(.025); | |||
| \draw (1,0)circle(.035); | |||
| \fill [white](1,0)circle(.025); | |||
| \fill (1.5,0)circle(.025); | |||
| \fill (2.5,0)circle(.025); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Bestimmen Sie $X_C$ so, dass der Blindleistungsbedarf des Verbrauchers verschwindet. | |||
| \item Berechnen Sie die Verlustleistung $P_{VRL}$ der Leitung und die Wirkleistung $P_W$ im Verbraucher. | |||
| \end{enumerate} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| Berechnung:\\[\baselineskip] | |||
| a) Verbraucher $\uline{Z}_V\,||\,X_C$, daher Ersatzschaltbild für $\uline{Z}_V$ (ESB) in Parallelform erforderlich | |||
| \begin{align*} | |||
| \underline{Z}_V&=R_V+jX_V=(10+j5)\,\ohm\,\,\quad\text{ Scheinwiderstand, entspricht einer Reihenschaltung}\\[\baselineskip] | |||
| Z^2_V&=R_V\cdot R_p=R^2_V+X^2_V\,\qquad\qquad\text{Umwandlung in Parallel-ESB}\\ | |||
| R_p&=R_V+\frac{X^2_V}{R_V}=(10+\frac{25}{10})\,\ohm=12{,}5\,\ohm\\[\baselineskip] | |||
| Z^2_V&=X_V\cdot X_{L_p}=R^2_V+X^2_V\,\,\quad\qquad\text{Umwandlung in Parallel-ESB}\\ | |||
| X_{L_p}&=X_V+\frac{R^2_V}{X_V}=(5+\frac{100}{5})\,\ohm=25\,\ohm | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_p$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spule | | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$jX_{L_p}$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$jX_C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw (0,0)--(3,0)--(3,.2) (0,1)--(3,1)--(3,.8); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Knotenpunkte | |||
| \fill (0,0)circle(.025); | |||
| \fill (1,0)circle(.025); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Knotenpunkte | |||
| \fill (0,0)circle(.025); | |||
| \fill (1,0)circle(.025); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Knotenpunkte | |||
| \draw (0,0)circle(.05); | |||
| \fill (0,0)circle(.025); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Knotenpunkte | |||
| \draw (0,0)circle(.05); | |||
| \fill (0,0)circle(.025); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick] | |||
| \draw node at (0,0.5) [left] {$\underline{Z}'\Rightarrow$}; | |||
| \end{scope} | |||
| \draw node at(1.5,0)[below]{Verbraucher $Z_V$}; | |||
| % \end{tikzpicture} | |||
| % \begin{tikzpicture}[scale=.5,xshift=15cm,yshift=-2cm] | |||
| \begin{scope}[>=latex,very thick,scale=.25,xshift=18cm,yshift=2cm] | |||
| \draw [->](0,0)--(1,0)node [right]{$R_p$}; | |||
| \draw [->](1,0)--(1,2)node [above]{$jX_{Lp}$}; | |||
| \draw [->](1,0)--(1,-2)node[below]{$jX_C$}; | |||
| \draw [->,red!50!blue](0,0)--(1,2)node at (.5,1)[left]{$Z_V$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \underline{Z}'&=\underline{Z}_V || jX_C\\ | |||
| \frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\frac{1}{jX_{Lp}}+\frac{1}{jX_C}\\ | |||
| \intertext{Leitungsverluste sind minimal, wenn die Blindleistung $=0$ wird (Kompensation)} | |||
| \frac{1}{\underline{Z}'}&=\frac{1}{R_p}+\cancel{\frac{1}{jX_{Lp}}}+\cancel{\frac{1}{jX_C}}\qquad\Rightarrow \underline{Z}'=R_p\\ | |||
| \Im(\underline{Z}')&=0 \quad\text{oder}\quad |X_C| \stackrel{!}{=} |X_{L_p}|\text{ also}\\ | |||
| %\Re(\underline{Z}')&=R_p\\ | |||
| X_C&=-X_{L_p}=\uuline{-25\,\ohm}\\ | |||
| \underline{Z}_{ges}&=\underline{Z}_i+R_L+\underline{Z}'\\ | |||
| \underline{Z}_{ges}&=\underline{Z}_i+R_L+R_p=(1+j2+1+12{,}5)\,\ohm=(14{,}5+j2)\,\ohm\\ | |||
| |\underline{Z}_{ges}|&=\sqrt{14{,}5^2+2^2}\,\ohm=14{,}64\,\ohm\\ | |||
| I&=\frac{U}{|Z_{ges}|}=\frac{100\,\volt}{14{,}64\,\ohm}=6{,}83\,\ampere\\ | |||
| \intertext{b) Verlust- und Wirkleistung} | |||
| P_{VR_L}&=I^2\cdot R_L=(6{,}83\,\ampere)^2\cdot 1\,\ohm=\uuline{46{,}7\,\watt}\\ | |||
| P_W&=I^2\cdot R_p=(6{,}83\,\ampere)^2\cdot 12{,}5\,\ohm=\uuline{583\,\watt} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Wechselstrommotor} | |||
| Ein Einphasen- Wechselstrommotor liegt an einer Spannung von $230\,\volt - 50\,\hertz$ und gibt eine Leistung von $2\,\kilo\watt$ ab, wobei sein Wirkungsgrad $\eta= 80\%$ und sein Leistungsfaktor $\cos\varphi=0,7$ beträgt. | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Wie groß ist die Stromaufnahme des Motors? | |||
| \item Welche Kapazität muss parallelgeschaltet werden, um eine Blindstromkompensation auf | |||
| $cos\varphi=0{,}9$ zu erreichen? | |||
| \item Wie groß ist der dem Netz bei $cos\varphi=0{,}9$ entnommene Strom? | |||
| \end{enumerate} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| S&=U\cdot I\qquad Scheinleistung\\ | |||
| P_{el}&=S\cdot\cos(\varphi)\qquad Wirkleistung\\ | |||
| P_{ab}&=\eta\cdot P_{el}\\ | |||
| \cos\varphi&=\frac{P}{S} | |||
| \end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \intertext{a) Stromaufnahme:} | |||
| P_{el}&=\frac{P_{ab}}{\eta}=\frac{2\,\kilo\watt}{0{,}8}=2{,}5\,\kilo\watt\\ | |||
| S&=\frac{P_{el}}{\cos(\varphi)}=\frac{2{,}5\,\kilo\watt}{0{,}7}=3{,}571\,\kilo\volt\ampere\\ | |||
| I&=\frac{S}{U}=\frac{3{,}571\,\kilo\volt\ampere}{230\,\volt}=\uuline{15{,}5\,\ampere} | |||
| \end{align*} | |||
| \clearpage | |||
| b) Kapazität: | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex, xshift=0, yshift=0] | |||
| \draw [black!50!,very thin](0,0)grid(2,2); | |||
| \draw [->,red,very thick] (2.525,0)--(2.525,2.55)node [right]{$Q$}; | |||
| \draw [->,thick](0:0)--(0:2.5)node [right]{$P$}; | |||
| \draw [->,thick] (0:0)--(90:2.55)node [above]{$Q$}; | |||
| \draw [->,very thick](0:0)--(25.8:2.778) node [above left] {$\underline{S}'$}; | |||
| \draw [->,very thick] (0,0)--(45.6:3.571) node [above left] {$\underline{S}$}; | |||
| \draw [->,blue,ultra thick] (45.6:3.571)--(25.8:2.778)node [above right] {$Q_C$}; | |||
| \draw [->,black!50!green, very thick] (2.5,0)--(2.5,1.209)node [below right]{$Q'$}; | |||
| \draw (0:1)arc(0:45.6:1)node [right]{$45{,}6\degree$}; | |||
| \draw (0:1.75)arc(0:25.8:1.75)node [below right]{$25{,}8\degree$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \varphi&=\arccos(0{,}7)=45{,}6\degree\\ | |||
| {\color{red}Q}&=S\cdot \sin(45{,}6\degree)=3{,}571\,\kilo\volt\ampere \cdot 0{,}714=2{,}55\,\kilo \var\\ | |||
| \varphi'&=\arccos(0{,}9)=25{,}86\degree\\ | |||
| S'&=\frac{P}{\cos \varphi'}=\frac{2{,}5\,\kilo\watt}{0{,}9}=2{,}778\,\kilo\volt\ampere\\ | |||
| {\color{black!50!green}Q'}&=S'\cdot \sin \varphi '=2{,}778\,\kilo\volt\ampere \cdot 0{,}435=1{,}209\,\kilo \var\\ | |||
| \intertext{für Kompensation muß gelten:} | |||
| Q+Q_C-Q'&=0 \\ | |||
| \Rightarrow Q_C=Q'-Q&=1{,}209\,\kilo \var -2{,}55\,\kilo \var=-1{,}341\,\kilo \var\\ | |||
| {\color{blue}|Q_C|}&=\frac{U^2}{|X_C|}\quad\Rightarrow |X_C|=\frac{U^2}{|Q_C|}=\frac{(230\,\volt)^2}{1341\,\var}=39{,}4\,\ohm=\frac{1}{\omega C}\Rightarrow\\ | |||
| C&=\frac{1}{\omega |X_C|}=\frac{1}{2\pi\cdot 50\frac{1}{\,\second}\cdot 39{,}4\,\frac{\,\volt}{\ampere}}=8{,}06\cdot\power{10}{-5}\,\frac{\ampere\second}{\volt}=\uuline{80{,}6\,\micro\farad}\\ | |||
| \intertext{c) Stromaufnahme bei $\cos\varphi=0{,}9$:} | |||
| S'&=U\cdot I'\Rightarrow I'=\frac{S'}{U}=\frac{2778\,\volt\ampere}{230\,\volt}=\uuline{12{,}1\,\ampere} | |||
| \end{align*} | |||
| Nicht auf $cos(\varphi)=1$ kompensieren, da dann Schwingkreis ! | |||
| \clearpage | |||
| }{}% | |||
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| \section{Parallelschaltung von L und C} | |||
| An der Parallelschaltung von $L$ und $C$ liegt die Spannung $u(t)$ (siehe Diagramm).\\ | |||
| Bei $t=0$ ist $i_L=0$.\\ | |||
| Berechnen Sie den Strom $i$ bei $t=t_2$! | |||
| \begin{align*} | |||
| U_0&=5\,\volt\\ | |||
| t_1&=3\milli\second\\ | |||
| t_2&=5\milli\second\\ | |||
| L&=6\,\milli\henry\\ | |||
| C&=100\,\micro\farad | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule | |||
| \draw (0,0)--(.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$L$}; | |||
| \fill (.2,-0.1)rectangle(.8,0.1); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Kondensator | |||
| \draw (0,0)--(.45,0) (.45,-.2)--(.45,.2) (.55,-.2)--(.55,.2) (.55,0)--(1,0)node at(.75,.05)[above]{$C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw (0,2)--(0,0)--(.1,0) (1,2)--(1,0)--(.9,0); | |||
| \fill (0,2)circle(.05) (1,2)circle(.05); | |||
| \draw [->,blue] (.3,2)--(.7,2) node at (.5,2)[below]{$u(t)$}; | |||
| \draw [->,red] (.2,1.75)--(.2,1.25) node at (.2,1.5)[right]{$i(t)$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,scale=.5,xshift=4cm,yshift=.5cm] | |||
| \draw [very thin,step=0.5cm](0,0)grid(5,2.5); | |||
| \draw [->](0,0) -- (5.5,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,0) -- (0,3) node [above] {$u\,[\volt]$}; | |||
| \draw [blue,very thick](0,0)--(3,2.5)--(5,2.5)node at (1.5,1.25)[right]{$u(t)$}; | |||
| \foreach \x in {1,2,...,5} | |||
| \draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$}; | |||
| \foreach \y in {0,5} | |||
| \draw (0,\y/2) -- (-0.2,\y/2) node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| i_C&=C\cdot \frac{du}{dt}\\ | |||
| u_L&=L\cdot \frac{di}{dt}\\ | |||
| \text{KNP: }\sum i&=0 | |||
| \end{align} | |||
| Berechnung:\\ | |||
| \begin{align*} | |||
| i(t)&=i_C(t)+i_L(t)\\ | |||
| i_C(t)&=C\cdot \frac{du}{dt}\tag{1}\label{eq:paralleschaltung-ic}\\ | |||
| u_L(t)&=L\cdot \frac{di}{dt}\tag{2}\label{eq:paralleschaltung-ul}\\ | |||
| \end{align*} | |||
| \begin{align*} | |||
| \text{aus \ref{eq:paralleschaltung-ic}}\quad\, i_C(t_2)&=0 \quad \text{zum Zeitpunkt }t_2=5\,\milli\second \quad \frac{du}{dt}=0\\ | |||
| &\Rightarrow i(t_2)=i_L(t_2)\\ | |||
| \text{aus \ref{eq:paralleschaltung-ul}}\qquad\, di_L&=\frac{1}{L}\cdot u_L\cdot dt \qquad \Big|\int \\ | |||
| \big[i_L(t)\big]_{t_a}^{t_b}&=i_L(t_b)-i_L(t_a)=\frac{1}{L}\int_{t_a}^{t_b}{u_L\cdot dt}\\[\baselineskip] | |||
| \text{für }0&\leq t \leq t_1\\ | |||
| i_L(t_1)-\underbrace{i_L(t=0)}_{0}&=\frac{1}{L}\int_{0}^{t_1}{\frac{U_0}{t_1}\cdot t\cdot dt}=\frac{U_0}{L\cdot t_1}\left[\frac{t^2}{2}\right]_{0}^{t_1}=\frac{U_0\cdot t_1}{2\cdot L}\\ | |||
| i_L(t_1)&=\frac{5\,\volt\cdot 3\cdot \,\milli\second}{2\cdot 6\cdot \frac{\,\milli\volt\second}{\ampere}}=1{,}25\,\ampere\\[\baselineskip] | |||
| \text{für }t_1&\leq t \leq t_2\\ | |||
| i_L(t_2)-i_L(t_1)&=\frac{1}{L}\int_{t_1}^{t_2}{U_0\cdot dt}=\frac{U_0}{L}\cdot (t_2-t_1)\\ | |||
| i_L(t_2)&=1{,}25\,\ampere+\frac{5\,\volt\cdot 2\cdot \power{10}{-3}\,\second}{6\cdot \power{10}{-3}\frac{\,\volt\second}{\ampere}}=1{,}25\,\ampere+1{,}67\,\ampere=\uuline{2{,}92\,\ampere} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{Alternativ Graphisch: $i_L \textrm{ ist proportional zur \textit{Fläche}}\cdot \frac{1}{L}+\textrm{const.}$} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,thick,scale=.5,xshift=4cm,yshift=.5cm] | |||
| \fill [black!10!](0,0)--(3,2.5)--(3,0)--(0,0); | |||
| \fill [black!25!](3,2.5)--(5,2.5)--(5,0)--(3,0); | |||
| \draw [very thin,step=0.5cm](0,0)grid(5,2.5); | |||
| \draw [->](0,0) -- (5.5,0) node [right] {$t\,[\milli\second]$}; | |||
| \draw [->](0,0) -- (0,3) node [above] {$u\,[\volt]$}; | |||
| \draw [blue,very thick](0,0)--(3,2.5)--(5,2.5)node at (1.5,1.25)[right]{$u(t)$}; | |||
| \foreach \x in {1,2,...,5} | |||
| \draw (\x,0) -- (\x,-0.2) node[anchor=north] {$\x$}; | |||
| \foreach \y in {0,5} | |||
| \draw (0,\y/2) -- (-0.2,\y/2) node[anchor=east] {$\y$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| i(t_2)&=\frac{1}{L}\cdot \underbrace{\left(\frac{U_0\cdot t_1}{2}+U_0(t_2-t_1)\right)}_{\textrm{Fläche}} =\frac{1}{6\,\milli\henry}\cdot \left(\frac{5\,\volt\cdot 3\,\milli\second}{2}+ 5\,\volt\cdot (5-3)\,\milli\second\right)\\ | |||
| &=\frac{(7{,}5+10)\cdot \power{10}{-3}\,\volt\second}{6\cdot \power{10}{-3}\,\frac{\volt\second}{\ampere}}= \uuline{2{,}92\,\ampere} | |||
| \end{align*} | |||
| \footnotesize{Warum ist $i_L(t_2)-i_L(t_1)\not= 0$? Strom ändert sich noch, nur Spannung ist konstant.\\ | |||
| Wenn $u=konstant \rightarrow$ Strom steigt unendlich an. $u=L\frac{di}{dt} \Rightarrow i(t)=1/L\int u(t)dt$} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Werte $R_L$ und $L$ einer Spule} | |||
| Aus den drei gemessenen sinusförmigen Spannungen $U$, $U_N$, und $U_{SP}$ lassen sich die Werte $R_L$ und $L$ einer Spule bestimmen. | |||
| \begin{align*} | |||
| U=100\,\volt\\ | |||
| U_N=60\,\volt\\ | |||
| U_{SP}=70\,\volt\\ | |||
| R_N=60\,\ohm\\ | |||
| f = 50\,\hertz | |||
| \end{align*} | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Zeichnen Sie ein qualitatives Zeigerdiagramm der Spannungen! | |||
| \item Bestimmen Sie $R_L$ und $L$! | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Widerstand | |||
| \draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_N$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$U_N$}; | |||
| \draw (0,1)--(0,0)--(.1,0) (3,1)--(3,0)--(2.9,0);%anschuß und Füllt die Ecken der Verbindung! | |||
| \fill (0,1)circle (0.025) (3,1)circle (0.025); | |||
| \draw [->,blue](.2,1)--(2.8,1)node at (1.5,1)[below]{\footnotesize$U$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm]%Widerstand | |||
| \draw (0,0)--(0.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$R_L$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm]%Spule | |||
| \draw (0,0)--(.2,0) (.2,-0.1)rectangle(.8,0.1) (.8,0)--(1,0)node at (.5,.1) [above] {$L$}; | |||
| \fill (.2,-0.1)rectangle(.8,0.1); | |||
| \draw [->,blue] (-.7,-.2)--(.7,-.2) node at (0,-.2)[below]{\footnotesize$U_{SP}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| \begin{align*} | |||
| \intertext{Berechnung:} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=.5] | |||
| \begin{scope}[>=latex,very thick, xshift=0, yshift=0] | |||
| \draw [black!25!,very thin](0,0)grid(8,7); | |||
| \draw [->,red, thick](0,0)--(8,0)node [right] {$\underline{I}$}; | |||
| \draw [->](0,0)--(6,0)node at(3,0)[below] {$\underline{U}_N$}; | |||
| \draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{U}_{SP}$}; | |||
| \draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{U}$}; | |||
| \draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$\underline{U}_{R_L}$}; | |||
| \draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$\underline{U}_{L}$}; | |||
| \draw [black!50!](33:10)arc(33:53:10)node [left]{$\underline{U}=100\,\volt\widehat{=}10\,\centi\metre$}; | |||
| \draw [black!50!](6,0)+(90:7)arc(90:70:7)node [right]{$\uline{U}_{SP}=70\,\volt\widehat{=}7\,\centi\metre$}; | |||
| \draw [black!50!](3,-1)node[below]{$\underline{U}_N=60\,\volt\widehat{=}6\,\centi\metre$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick, xshift=12cm, yshift=2cm] | |||
| \draw node at(0,2)[right]{$I$ zeichnen}; | |||
| \draw node at(0,1)[right]{$u_N || I$}; | |||
| \draw node at(0,0)[right]{Mit Zirkel $U$ und $U_{SP}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| I&=\frac{U_N}{R_N}=\frac{60\,\volt}{60\,\ohm}=\uuline{1\,\ampere}\\ | |||
| \underline{U}_{SP}&=70\,\volt=\sqrt{U^2_{RL}+U^2_L}\\ | |||
| \end{align*} | |||
| \clearpage | |||
| Widerstandsoperatoren:\\ | |||
| \footnotesize{Impedanzdreieck wie Spannungsdreieck} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=.5] | |||
| \begin{scope}[>=latex,very thick, xshift=0, yshift=0] | |||
| \draw [black!25!,very thin](0,0)grid(8,7); | |||
| \draw [->](0,0)--(6,0)node at(3,0)[below] {$R_N$}; | |||
| \draw [->](6,0)--(7.25,6.887)node at(6.5,3.5)[left] {$\underline{Z}_{SP}$}; | |||
| \draw [->](0,0)--(7.25,6.887)node at(3.5,3.5)[left] {$\underline{Z}$}; | |||
| \draw [->,blue](6,0)--(7.25,0)node at(6.5,0)[below] {$R_L$}; | |||
| \draw [->,blue](7.25,0)--(7.25,6.887)node at(7,3.5)[right] {$X_L$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| Z_{SP}&=\frac{U_{SP}}{I}=\frac{70\,\volt}{1\,\ampere}=70\,\ohm \quad \text{\footnotesize{(Nur Effektivwerte - ohne Winkel)}}\\%=\sqrt{R^2_L+X^2_L}\\ | |||
| Z^2_{SP}&=R^2_L+X^2_L=(70\,\ohm)^2\\ | |||
| X^2_L&=(70\,\ohm)^2-R^2_L \tag{1}\\[\baselineskip] | |||
| Z&=\frac{U}{I}=\frac{100\,\volt}{1\,\ampere}=100\,\ohm\\ | |||
| Z^2&=(R_N+R_L)^2+X^2_L=(100\,\ohm)^2\\ | |||
| X^2_L&=(100\,\ohm)^2-(R_N+R_L)^2\\ | |||
| &=(100\,\ohm)^2-(R^2_N+2\cdot R_N\cdot R_L+R^2_L) \tag{2}\\[\baselineskip] | |||
| (70\,\ohm)^2-\cancel{R^2_L}&=(100\,\ohm)^2-R^2_N-2\cdot R_N\cdot R_L-\cancel{R^2_L}\tag{$1$ in $2$}\\ | |||
| 2\cdot R_N\cdot R_L&=(100\,\ohm)^2-R^2_N-(70\,\ohm)^2\\ | |||
| R_L&=\frac{(100\,\ohm)^2-R^2_N-(70\,\ohm)^2}{2\cdot R_N}=\frac{(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2}{2\cdot 60\,\ohm}\\ | |||
| &=\frac{1500\,\ohm^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm}\\ | |||
| %(100\,\ohm)^2&=(60\,\ohm)^2+2\cdot 60\,\ohm\cdot R_L+\cancel{R^2_L}+(70\,\ohm)^2 -\cancel{R^2_L}\\ | |||
| %2\cdot 60\,\ohm\cdot R_L&=(100\,\ohm)^2-(60\,\ohm)^2-(70\,\ohm)^2=1500(\,\ohm)^2\\ | |||
| %R_L&=\frac{1500(\,\ohm)^2}{2\cdot 60\,\ohm}=\uuline{12{,}5\,\ohm} \tag{in $1$}\\ | |||
| \text{in (1) }\qquad X_L&=\sqrt{(70\,\ohm)^2-(12{,}5\,\ohm)^2}=68{,}87\,\ohm\\ | |||
| L&=\frac{X_L}{\omega}=\frac{68{,}87\,\ohm}{2\pi\cdot 50\,\frac{1}{\second}}=\uuline{0{,}219\,\henry} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Zeigerdiagramm} | |||
| Gegeben ist die Ausgangsspannung $U_a = 5\,\volt \cdot e^{j0\degree}$ und $R_1=R_2=\frac{1}{\omega C_1}=\frac{1}{\omega C_2}=1\,\kilo\ohm$\\ | |||
| Zeichnen Sie ein maßstäbliches Zeigerdiagramm aller Spannungen und aller Ströme!\\ | |||
| Maßstäbe: $1\,\centi\metre\,\widehat{=}\, 1\,\volt \text{;}\quad 1\,\centi\metre\,\widehat{=}\, 1\,\milli\ampere$ | |||
| Entnehmen Sie dem Zeigerdiagramm Betrag und Phasenwinkel der Spannung $U_e$ ! | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C_1$}; | |||
| \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C_2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick]%Knotenpunkte | |||
| \draw (0,0)--(2.5,0) (2,1)--(2.5,1); | |||
| \fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025); | |||
| \draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$}; | |||
| \draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| e^{j\varphi}&=\cos\varphi+j\sin\varphi \quad\text{Eulersche Formel} | |||
| \end{align} | |||
| \footnotesize{R-C Ketten sind u.a. ein Ersatzbild für Leitungen (Kapazität pro Längeneinheit)}\\ | |||
| Berechnung: | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_1$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_1}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C_1$}; | |||
| \draw [<-,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{\tiny$\underline{U}_{1}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_2$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\tiny$\underline{U}_{R_2}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C_2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick]%Knotenpunkte | |||
| \draw (0,0)--(2.5,0) (2,1)--(2.5,1); | |||
| \fill (0,0)circle(.025) (2.5,0)circle(.025) (0,1)circle(.025) (2.5,1)circle(.025); | |||
| \draw [->,blue] (0,.9)--(0,.1) node at (0,.5)[right]{$\underline{U}_{e}$}; | |||
| \draw [->,blue] (2.5,.9)--(2.5,.1) node at (2.5,.5)[right]{$\underline{U}_{a}$}; | |||
| \draw [->,red] (.05,1.1)--(.25,1.1) node at (.1,1.1)[above]{$\underline{I}_{e}$}; | |||
| \draw [->,red] (2.2,1.1)--(2.4,1.1) node at (2.3,1.1)[above]{$\underline{I}_{a}=0$}; | |||
| \draw [->,red] (1.1,.9)--(1.1,.7) node at (1.1,.8)[right]{$\underline{I}_{1}$}; | |||
| \draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{2}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,thick,xshift=.4cm,yshift=.5cm]%Masche | |||
| \draw [->,red!50!blue] (270:.15)arc(270:-60:.15) node at (0,0){$M_1$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \underline{U}_a&=5\,\volt\cdot e^{j0\degree}\\ | |||
| \underline{I}_a&=0\text{, da kein Lastwiderstand angeschlossen ist!}\\ | |||
| \underline{I}_2&=\frac{\underline{U}_a}{\underline{jX}_2} | |||
| =\frac{5\,\volt\cdot e^{j0\degree}}{1\kilo\ohm\cdot e^{-j90\degree}} | |||
| =5\,\milli\ampere\cdot e^{j90\degree}=j5\,\milli\ampere | |||
| \quad\text{(Strom eilt vor)}\quad 5\,\milli\ampere\;\angle +90\degree\\ | |||
| \underline{U}_{R_2}&=R_2\cdot \underline{I}_2=1\,\kilo\ohm\cdot e^{-j90\degree}=1\,\kilo\ohm\cdot 5\,\milli\ampere\;\angle +90\degree=5\,\volt\;\angle +90\,\degree\\ | |||
| \underline{U}_1&=\underline{U}_a+\underline{U}_{R_2}=5\,\volt+j5\,\volt\quad\text{(Vektoren addieren)}\quad =\sqrt{2}\cdot 5\,\volt\;\angle +45\degree \sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}\\\underline{I}_1&=\frac{\underline{U}_1}{\underline{jX}_1}=\frac{\sqrt{2}\cdot 5\,\volt\cdot e^{j45\degree}}{ 1\kilo\ohm\cdot e^{-j90\degree}}=\sqrt{2}\cdot 5\,\milli\ampere\cdot e^{j135\degree}=5\cdot (-1+j)\,\milli\ampere\quad\text{(Strom eilt $90\degree$ vor) } 5\,\milli\ampere\;\angle +135\degree\\ | |||
| \underline{I}_e&=\underline{I}_1+\underline{I}_2=\sqrt{2}\cdot 5\,\milli\ampere\;\angle +135\degree+5\,\milli\ampere\;\angle +90\degree\quad\text{(Vektoren addieren)}\\[\baselineskip] | |||
| \end{align*} | |||
| \begin{align*} | |||
| \intertext{Jetzt zeichnen oder rechnerisch: (jedoch aufwendiger)} | |||
| \underline{I}_e&=I_1\cdot (\cos\varphi+j\sin\varphi)+I_2\cdot (\cos\varphi+ j\sin\varphi)\\ | |||
| &=I_1\cdot (\cos135\degree+j\sin 135\degree)+I_2\cdot (\cos 90\degree+ j\sin 90\degree)\\ | |||
| &=|\underline{I}_1|\cdot (-\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}})+|\underline{I}_2|\cdot (0+j)\\ | |||
| &=[\cancel{\sqrt{2}}\cdot 5\cdot (-\frac{1}{\cancel{\sqrt{2}}}+j\frac{1}{\cancel{\sqrt{2}}})+j\cdot 5]\,\milli\ampere=(-5+j10\,\milli\ampere)\\ | |||
| &\quad |\underline{I}_e|= \sqrt{10^2+5^2}\,\milli\ampere=11{,}18\,\milli\ampere \\ | |||
| &\quad\tan\varphi=\frac{\Im}{\Re}=\tan\frac{10}{-5}=\tan-2\Rightarrow\varphi=\arctan\frac{-2}=-1{,}107\,\rad\,\widehat{=}\,-63{,}435\degree\\ | |||
| &\quad\text{(Definitionsbereich $\tan\varphi [-\pi/2\cdots\pi/2]$ beachten!)}\\ | |||
| &=11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere\\[\baselineskip] | |||
| \underline{U}_{R_1}&=R_1\cdot \underline{I}_e\\ | |||
| &\quad\text{(Nur zur Vollständigkeit) }\underline{U}_{R_1}=1\,\kilo\ohm\cdot 11{,}18\cdot e^{j116{,}565\degree}\,\milli\ampere=11{,}18\cdot e^{j116{,}565\degree}\,\volt\\ | |||
| \underline{U}_e&=\underline{U}_{R_1}+\underline{U}_1=\uuline{15\,\volt\cdot e^{+j90\degree}}\quad\text{(Vektoren addieren)} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=.5] | |||
| \begin{scope}[>=latex] | |||
| \draw [very thin,black!50!](-5,0)grid(5,15); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick] | |||
| \draw [->,blue] (0:0)--(90:15) node at (90:7.5)[right]{$\underline{U}_e$}; | |||
| \draw [->,blue] (0:0)--(0:5) node at (0:2.5)[below]{$\underline{U}_a$}; | |||
| \draw [->,blue] (0:0)--(45:7.07) node at (45:3.54)[right]{$\underline{U}_1$}; | |||
| \draw [->,blue,ultra thick] (0:0)--(-5,10) node at (-2.5,5)[right]{$\underline{U}_{R_1}$}; | |||
| \draw [->,blue] (-5,10)--(0,15) node at (-2.5,12.5)[left]{$\underline{U}_{1}$}; | |||
| \draw [->,blue,ultra thick] (0:0)--(90:5) node at (90:3.5)[right]{$\underline{U}_{R_2}$}; | |||
| \draw [->] (5,0)--(5,5) node at (5,3.5)[right]{$\underline{U}_{R_2}$}; | |||
| \draw [->,red] (0:0)--(135:7.07) node at (135:3.54)[left]{$\underline{I}_1$}; | |||
| \draw [->,red,thick] (0:0)--(90:5) node at (90:3.5)[left]{$\underline{I}_2$}; | |||
| \draw [->] (135:7.07)--+(90:5) node at (-5,7.5)[right]{$\underline{I}_2$}; | |||
| \draw [->,red,thick] (0:0)--(-5,10) node at (-2.5,5)[left]{$\underline{I}_{e}$}; | |||
| \end{scope} | |||
| \draw node at (8,10)[right]{Reihenfolge$U_a\, I_2, U_{R_2},U_1, I_1, I_e, U_{R_1})$}; | |||
| \draw node at (8,9)[right]{$U_a\widehat{=}\,5\,\centi\metre\angle 0\,\degree\quad(5+j0)$}; | |||
| \draw node at (8,8)[right]{$U_{R_2}\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$}; | |||
| \draw node at (8,7)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(5+j5)$}; | |||
| \draw node at (8,6)[right]{$I_1\widehat{=}\,7{,}07\,\centi\metre\angle 135\,\degree\quad(-5+j5)$}; | |||
| \draw node at (8,5)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(0+j5)$}; | |||
| \draw node at (8,4)[right]{$I_2\widehat{=}\,5\,\centi\metre\angle 90\,\degree\quad(-5+j5)\rightarrow(0+j5)$ addiert zu $I_1$}; | |||
| \draw node at (8,3)[right]{$I_e\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree(-5+j10)$}; | |||
| \draw node at (8,2)[right]{$U_{R_1}\widehat{=}\,11{,}18\,\centi\metre\angle 116{,}5\,\degree\quad(-5+j10)$}; | |||
| \draw node at (8,1)[right]{$U_1\widehat{=}\,7{,}07\,\centi\metre\angle 45\,\degree\quad(-5+j10)\rightarrow(5+j5)$ addiert zu $U_{R_1}$}; | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \sum M_1=0=U_{R_1}+U_1-U_e\\ | |||
| U_e=15\,\volt \cdot e^{j90\degree} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| @@ -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 | |||
| }{}% | |||
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| \section{Brückenschaltung} | |||
| %\enlargethispage{0cm} | |||
| Zeichnen Sie zu der abgebildeten Brückenschaltung ein \underline{maßstäbliches} Zeigerdiagramm aller Ströme und Spannungen.\\ | |||
| Entnehmen Sie dem Zeigerdiagramm die Spannung $U_a$ und geben Sie von dieser Spannung Betrag und Phasenwinkel an.\\[\baselineskip] | |||
| \uline{Maßstäbe: }$ 10\,\volt\,\widehat{=}\,1\centi\metre\qquad 0,2\,\ampere\,\widehat{=}\,1\,\centi\metre $ (Platzbedarf in x und y $15\,\centi\metre$)\\ | |||
| $R_1=100\,\ohm \qquad R_2=80\,\ohm \qquad X_L=200\,\ohm \qquad X_C=-120\,\ohm\qquad\uline{U}=150\,\volt\cdot e^{j0\degree}$ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=2] | |||
| \begin{scope}[>=latex,very thick,xshift=0.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$}; | |||
| \draw [<-,blue] (.3,.5)--(.7,.5)node at(.5,.5)[right]{\footnotesize$\uline{U}_{R_1}$}; | |||
| \draw [<-,red] (.05,.0)--(.25,.0)node at(.15,.0)[right]{\footnotesize$\uline{I}_{1}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$}; | |||
| \draw [<-,blue] (.3,.5)--(.7,.5)node at(.5,.5)[right]{\footnotesize$\uline{U}_{R_2}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=.5cm,yshift=-1cm,rotate=90]%Spule | | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at(.5,-.0667) [right] {$X_L$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1.5cm,yshift=-1cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$X_C$}; | |||
| \draw [->,red] (.35,0)--(.15,0) node at (.25,0)[right]{\footnotesize$\uline{I}_2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick]%Knotenpunkte | |||
| \draw (1.5,.8)--(1.5,1)--(-.5,1) (1.5,-.8)--(1.5,-1)--(-.5,-1); | |||
| \fill (-.5,1)circle(.025) (-.5,-1)circle(.025); | |||
| \draw [->,blue] (.6,0)--(1.4,0)node at(1,0)[below]{$\underline{U}_{a}$}; | |||
| % \draw node at(-1,-.5)[left]{$\underline{Z}\Rightarrow$}; | |||
| \fill (.5,0)circle(.025) node at (.5,0)[left]{A}; | |||
| \fill (1.5,0)circle(.025)node at (1.5,0)[right]{B}; | |||
| \draw [->,blue] (-.5,.8)--(-.5,-.8) node at (-.5,0)[left]{$\underline{U}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %%begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| Berechnung: | |||
| \begin{align*} | |||
| Z_1&=\sqrt{R^2_1+X^2_L}=\sqrt{100^2+200^2}\,\ohm=223{,}6\,\ohm\\ | |||
| Z_2&=\sqrt{R^2_2+X^2_C}=\sqrt{80^2+(-120)^2}\,\ohm=144{,}2\,\ohm\\ | |||
| \varphi_1&=\arctan\frac{\Im}{\Re}=\arctan\frac{200\,\ohm}{100\,\ohm}=\arctan 2=63{,}4\,\degree\\ | |||
| \varphi_2&=\arctan\frac{-120\,\ohm}{80\,\ohm}=\arctan (-1{,}5)=-56{,}3\,\degree\\ | |||
| \uline{Z}_1&=Z_1\cdot e^{j\varphi_1}=223{,}6\,\ohm\cdot e^{j63{,}4\,\degree}\\ | |||
| \uline{Z}_2&=Z_2\cdot e^{j\varphi_2}=144{,}2\,\ohm\cdot e^{-j56{,}3\,\degree}\\ | |||
| \uline{I}_1&=\frac{U}{Z_1}=\frac{150\,\volt\cdot e^{j0\degree}}{223{,}6\,\ohm\cdot e^{j63{,}4\,\degree}}=0{,}67\,\ampere\cdot e^{-j63{,}4\,\degree} \approx 3{,}4\,\centi\metre\\ | |||
| \uline{I}_2&=\frac{U}{Z_2}=\frac{150\,\volt\cdot e^{j0\degree}}{144{,}2\,\ohm\cdot e^{-j56{,}3\,\degree}}=1{,}04\,\ampere\cdot e^{j56{,}3\,\degree} \approx 5{,}2\,\centi\metre\\ | |||
| \uline{I}&=\uline{I_1}+\uline{I_2}\\ | |||
| \uline{U}_{R_1}&=I_1\cdot R_1= 0{,}671\,\ampere\cdot e^{j63{,}4\,\degree}\cdot 100\,\ohm=67{,}1\,\volt\cdot e^{-j63{,}4\,\degree}\\ | |||
| \uline{U}_{R_2}&=I_2\cdot R_2=1{,}04\,\ampere\cdot 80\,\ohm=83{,}2\,\volt\cdot e^{j56{,}3\,\degree}\\ | |||
| \uline{U}_a&=\uline{U}_{R_2}-\uline{U}_{R_1}\\ | |||
| \uline{U}_L&=X_L\cdot \uline{I}_1=200\,\ohm\cdot e^{j90\degree}0{,}67\,\ampere\cdot e^{-j63{,}4\,\degree}=134\,\volt\cdot e^{j26{,}6\,\degree}\\ | |||
| \uline{U}_C&=X_C\cdot \uline{I}_2=120\,\ohm\cdot e^{-j90\degree}\cdot 1{,}04\,\ampere\cdot e^{j56{,}3\,\degree}=124{,}8\,\volt\cdot e^{-j33{,}7\,\degree} | |||
| \end{align*} | |||
| \clearpage | |||
| Zeigerdiagramm Teil 1: (Maßstäblich) | |||
| \enlargethispage{2\baselineskip} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=.6] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw [black!25!,very thin](0,-8)grid(15,8); | |||
| \draw [black!50!,thin,dashed](16,0)--(18,0); | |||
| \draw [black!50!,thin,dashed](3,-6)--(5,-6); | |||
| \draw [black!50!,thin](7.5,0)circle(7.5); | |||
| \draw [->,blue](0,0)--(15,0)node [right]{$\uline{U}$}; | |||
| \draw [->,blue](0,0)--(-63.4:6.71)node [left]{$\uline{U}_{R_1}$}; | |||
| \draw [->,blue](0,0)--(56.3:8.32)node [left]{$\uline{U}_{R_2}$}; | |||
| \draw [->,blue!50!red](0,0)+(-63.4:6.71)--(15,0)node at(15,-1.5)[below]{$\uline{U}_L=134\,\volt\cdot e^{j26{,}63{,}4\,\degree}$}; | |||
| \draw [->,blue!50!red](0,0)+(56.3:8.32)--(15,0)node at(15,2)[above]{$\uline{U}_C=124{,}8\,\volt\cdot e^{-j33{,}73\,\degree}$}; | |||
| \draw [->,blue](0,0)+(-63.4:6.71)--(56.3:8.32)node at (4,2)[right]{$\uline{U}_{AB}=\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$}; | |||
| \draw node at (3,-6)[below]{$A$}; | |||
| \draw node at (4.8,7)[above]{$B$}; | |||
| \draw node at (16,0)[above right]{$\Re (A)$}; | |||
| \draw [black!50!](0,0)++(-63.4:6.71)+(26.6:1)arc(26.6:116.6:1);%Rechter Winkel bei UR1 | |||
| \fill [black!50!](0,0)++(-63.4:6.71)+(71.6:.5)circle(0.1); | |||
| \draw [black!50!](0,0)++(56.3:8.32)+(-33.7:1)arc(-33.7:-123.7:1);%Rechter Winkel bei UR2 | |||
| \fill [black!50!](0,0)++(56.3:8.32)+(-78.7:.5)circle(0.1); | |||
| \draw [<->,black!50!](0,0)+(-63.4:1)arc(-63.4:53.3:1);%Winkel UR1 UR2 | |||
| \draw node at (0.8,0)[below right]{$\varphi_1=-63{,}4\,\degree$}; | |||
| \draw node at (0.8,0)[above right]{$\varphi_2=53{,}3\,\degree$}; | |||
| \draw [->,black!75!](3,-6)+(0:2)arc(0:83:2)node at (5,-5.5)[right]{$\varphi_{\uline{U}} =83\,\degree$};%U - UAB | |||
| \draw node at (19,5) [right]{$y=15\,\centi\metre$}; | |||
| \draw node at (19,4) [right]{Kreis da}; | |||
| \draw node at (19,3) [right]{$\uline{U}_{R_2}+\uline{U}_{C}=\uline{U}$}; | |||
| \draw node at (19,2) [right]{$\uline{U}_{R_1}+\uline{U}_{L}=\uline{U}$}; | |||
| \draw node at (19,1) [right]{$\uline{U}_{R_2}\bot \uline{U}_{C}$}; | |||
| \draw node at (19,0) [right]{$\uline{U}_{R_1}\bot \uline{U}_{L}$}; | |||
| \draw node at (19,-1) [right]{A und B einzeichnen}; | |||
| \draw node at (19,-2) [right]{Messen $\uline{U}_{AB}=\uline{U}_a$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| Zeigerdiagramm vollständig: | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=.6] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw [black!25!,very thin](0,-8)grid(15,8); | |||
| \draw [black!50!,thin,dashed](16,0)--(18,0); | |||
| \draw [black!50!,thin,dashed](3,-6)--(5,-6); | |||
| \draw [black!50!, thin](7.5,0)circle(7.5); | |||
| \draw [->,blue](0,0)--(15,0)node [right]{$\uline{U}$}; | |||
| \draw [->,blue](0,0)--(-63.4:6.71)node [left]{$\uline{U}_{R_1}$}; | |||
| \draw [->,blue](0,0)--(56.3:8.32)node [left]{$\uline{U}_{R_2}$}; | |||
| \draw [->,blue!50!red](0,0)+(-63.4:6.71)--(15,0)node at(12,-1.5)[below]{$\uline{U}_L$}; | |||
| \draw [->,blue!50!red](0,0)+(56.3:8.32)--(15,0)node at(12,2)[above]{$\uline{U}_C$}; | |||
| \draw [->,blue](0,0)+(-63.4:6.71)--(56.3:8.32)node at (4,2)[right]{$\uline{U}_{AB}=\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$}; | |||
| \draw [black!50!](0,0)++(-63.4:6.71)+(26.6:1)arc(26.6:116.6:1);%Rechter Winkel bei UR1 | |||
| \fill [black!50!](0,0)++(-63.4:6.71)+(71.6:.5)circle(0.1); | |||
| \draw [black!50!](0,0)++(56.3:8.32)+(-33.7:1)arc(-33.7:-123.7:1);%Rechter Winkel bei UR2 | |||
| \fill [black!50!](0,0)++(56.3:8.32)+(-78.7:.5)circle(0.1); | |||
| \draw [->,red](0,0)--(-63.4:3.34)node [left]{$\uline{I}_1$}; | |||
| \draw [->,red](0,0)--(56.3:5.2)node [left]{$\uline{I}_2$}; | |||
| \draw [<-,red,thin](0:0)++(56.3:5.2)++(-63.4:3.34)--(-63.4:3.34)node at(3,-.5) [left]{$\uline{I}_2$}; | |||
| \draw [<-,red,ultra thick](0:0)++(56.3:5.2)++(-63.4:3.34)--(0:0)node at (6,.6)[above]{$\uline{I}=\uline{I}_1+\uline{I}_2$}; | |||
| \draw [<-,black](0,0)+(17.0:3)arc(17.0:0:3)node at (1.0,1.5)[right]{$\varphi_{\uline{I}}=17\,\degree$};%Winkel I | |||
| \draw node at (3,-6)[below]{$A$}; | |||
| \draw node at (4.8,7)[above]{$B$}; | |||
| \draw node at (16,0)[above right]{$\Re (A)$}; | |||
| \draw [<->,black!50!](0,0)+(-63.4:1)arc(-63.4:53.3:1);%Winkel UR1 UR2 | |||
| \draw node at (.7,-.35)[below right]{$\varphi_1$}; | |||
| \draw node at (.7,.35)[above right]{$\varphi_2$}; | |||
| \draw [->,black!75!](3,-6)+(0:2)arc(0:83:2)node at (5,-5.5)[right]{$\varphi_{\uline{U}} =83\,\degree$};%U - UAB | |||
| \draw node at (20.8,5) [right]{$\uline{I}_1 \approx 3{,}4\,\centi\metre$}; | |||
| \draw node at (20.8,4) [right]{$\uline{I}_2 \approx 5{,}2\,\centi\metre$}; | |||
| \draw node at (20.8,3) [right]{$\uline{I}=\uline{I_1}+\uline{I_2}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \section{Zeigerdiagramm Netzwerk} | |||
| Zeichnen Sie für das abgebildete Netzwerk ein maßstäbliches Zeigerdiagramm aller Spannungen und Ströme.\\ | |||
| Wie groß muß der Widerstand $R_1$ sein damit der Strom $I$ der Spannung $U$ um $30\,\degree$ nacheilt?\\[\baselineskip] | |||
| $\varphi_u - \varphi_i = 30\,\degree$\\ | |||
| $R_2 = |X_L| = |X_C| = 10\,\kilo\ohm$\\ | |||
| $\uline{I}_C = 1\,\milli\ampere\cdot e^{j90\,\degree}$\\[\baselineskip] | |||
| \uline{Maßstäbe:}\\ | |||
| $1\,\volt \,\widehat{=}\,0{,}8\centi\metre$\\ | |||
| $1\,\milli\ampere \,\widehat{=}\,5\centi\metre$ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=3] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Spule - | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$L$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick]%Knotenpunkte | |||
| \draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(2,0)--(2,.2) (.8,1)--(2,1)--(2,.8); | |||
| \draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$}; | |||
| \fill (-.5,0)circle(.025) (-.5,1)circle(.025); | |||
| \draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$}; | |||
| \draw [->,red] (1.1,.9)--(1.1,.6) node at (1.1,.75)[right]{$\underline{I}_C$}; | |||
| \draw [->,black!50!] (1.5,.9)--(1.5,.1)node at(1.5,.5)[right]{$\underline{U}_{R_2}$}; | |||
| \draw [->,black!50!] (1.4,1)--(1.6,1) node at (1.5,1)[above]{$\underline{I}_2$}; | |||
| \draw [->,black!50!] (.3,.9)--(.7,.9) node at (.5,.9)[below]{$\underline{U}_L$}; | |||
| \draw [->,black!50!] (.75,1)--(.95,1) node at (.85,1)[above]{$\underline{I}_L$}; | |||
| \draw [->,black!50!] (0,.95)--(0,.75) node at (0,.85)[left]{$\underline{I}_{R_1}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align*} | |||
| \intertext{Berechnung: (Platz in $x=\pm10\,\centi\metre$ und $x=10\,\centi\metre$} | |||
| R_2=|X_C|&=10\,\kilo\ohm\text{ (Stomteiler, mit gleichem Betrag des Stroms)}\\ | |||
| \text{mit }|\uline{I}_{R_2}| &=|\uline{I}_C|=1\,\milli\ampere\\ | |||
| \uline{I}_C &= 1\,\milli\ampere\cdot e^{j90\,\degree}\Rightarrow\uline{I}_{R_2}=1\,\milli\ampere\cdot e^{j0\,\degree}\text{ $(\uline{I}_C$ eilt vor)}\\ | |||
| \uline{U}_{R_2}&=R_2\cdot \uline{I}_{R_2}=10\,\cancel{\kilo}\ohm\cdot 1\,\cancel{\milli}\ampere\cdot e^{j0\,\degree}=10\,\volt\cdot e^{j0\,\degree}\,\widehat{=}\,8\centi\metre\\ | |||
| \uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C =1\,\milli\ampere\cdot e^{j0\,\degree}+1\,\milli\ampere\cdot e^{j90\,\degree}=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\\ | |||
| \uline{U}_L&=\uline{I}_L\cdot j\cdot X_L=1{,}41\,\milli\ampere\cdot e^{j45\,\degree}\cdot j10\,\kilo\ohm =14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\,11{,}3\centi\metre\text{ $(\uline{U}_L$ voreilend)}\\ | |||
| \uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}=14{,}1\,\volt\cdot e^{j135\,\degree}+10\,\volt\cdot e^{j0\,\degree} | |||
| =(-10+j10+10)\,\volt=10\,\volt\cdot e^{j90\,\degree}\\ | |||
| \text{Zeichnen: }\varphi_u-\varphi_i&=30\,\degree\text{ deshalb $30\,\degree$, Linie zeichnen, Schnittpunkt mit } \uline{I}_L+\uline{I}_{R_1}\Rightarrow\uline{I}=2\,\milli\ampere\cdot e^{j30\,\degree}\\ | |||
| \text{Ablesen: }\qquad\,\,\uline{I}_{R_1}&=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\ | |||
| R_1&=\frac{\uline{U}}{\uline{I}_{R_1}}=\frac{10\,\volt\cancel{\cdot e^{j90\,\degree}}}{0{,}72\,\milli\ampere\cancel{\cdot e^{j90\,\degree}}}=\uuline{13{,}89\,\kilo\ohm}\\ | |||
| \end{align*} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=.7] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm] | |||
| \draw [black!15!,very thin](-10,0)grid(10,10); | |||
| \draw [->,black!25!](-10,0)--(10.5,0); | |||
| \draw [->,black!25!](0,0)--(0,10.5); | |||
| \draw node at (10.5,0)[right]{$\Re$}; | |||
| \draw node at (0,10.5)[above]{$\Im$}; | |||
| \draw [black!75!,thick](0:0)--(60:11)node [right]{$30\,\degree$ Linie}; | |||
| \draw [->,blue] (0,0)--(90:8)node [below right]{$\underline{U}$}; | |||
| \draw [->,red] (0,0)--(90:5)node [below right]{$\underline{I}_C$}; | |||
| \draw [->,red] (0,0)--(0:5)node [above left]{$\underline{I}_R$}; | |||
| \draw [->,red] (0,0)--(45:7.07)node [below right]{$\underline{I}_L$}; | |||
| \draw [->,red!50!blue] (5,5)--(5,8.65)node [below right]{$\underline{I}_{R_1}=0{,}72\,\milli\ampere\,\widehat{=}\,3{,}6\centi\metre$}; | |||
| \draw [->,red!50!blue](0:0)--(60:10)node at (3,7.8) [above]{$\uline{I}=2\,\milli\ampere$}; | |||
| \draw [->,blue] (0,0)--(135:11.28)node [below left]{$\underline{U}_L$}; | |||
| \draw [->,blue] (-8,8)--(0,8)node [above left]{$\underline{U}_{R_2}$}; | |||
| \draw [->,black!75!](0:0)+(90:2)arc(90:60:2)node at(.6,2)[above]{$30\,\degree$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \uline{I}_{R_2}&=1\,\milli\ampere\cdot e^{j0\,\degree} \,\widehat{=}\ 5\,\centi\metre\\ | |||
| \uline{I}_{C}&=1\,\milli\ampere\cdot e^{j90\,\degree} \,\widehat{=}\ 5\,\centi\metre\\ | |||
| \uline{I}_L&=\uline{I}_{R_2}+\uline{I}_C \,\widehat{=}\ 7{,}05\,\centi\metre\\ | |||
| \uline{U}_{L}&=14{,}1\,\volt\cdot e^{j135\,\degree}\,\widehat{=}\ 11{,}3\centi\metre \quad \uline{U}_{L}\bot \uline{I}_{L}\\ | |||
| \uline{R}_{R_2}&=10\,\volt\cdot e^{j0\,\degree} \,\widehat{=}\ 8\centi\metre\\ | |||
| \uline{U}&=\uline{U}_{L}+\uline{U}_{R_2}\\ | |||
| &\text{Gerade für I, $30\degree $ nacheilend}\\ | |||
| \uline{I}_{L}&+\uline{I}_{R_1}=\uline{I};\qquad \uline{I}_{R_1}||\uline{U}\\ | |||
| \rightarrow\ &\text{ ablesen } 3{,}6\,\centi\metre\ \widehat{=}\ \uline{I}_{R_1}=0{,}72\,\milli\ampere\cdot e^{j90\,\degree}\\ | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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ET2_L_B16_A8.tex
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| @@ -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 | |||
| }{}% | |||
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| @@ -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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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]{$\uline{U}_{(X_2-X_M)}$}; | |||
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| @@ -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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \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 | |||
| }{}% | |||
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| \section {Ringspule} | |||
| Durch das Zentrum einer Ringspule mit rechteckigem Querschnitt (Länge | |||
| $l=12\,\milli\metre$, $r_i=10\,\milli\metre$, $r_a=20\,\milli\metre$; | |||
| Windungszahl $N= 1800$) wird ein Leiter geführt, in dem ein Wechselstrom fließt. | |||
| ($I_{eff}=40\,\ampere$; $f=50\,\hertz$).\\ | |||
| Berechnen Sie den Effektivwert $U$ der Spannung $u$.\\ | |||
| $\mu=\mu_0=1{,}26\cdot \power{10}{-6}\,\volt\second\per(\ampere\metre)$ | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=2.5, z=0.2pt] | |||
| \begin{scope}[>=latex, xshift=0, yshift=0] | |||
| \foreach \z in {0,0.2,...,1} | |||
| \fill [black!25!](0,0,\z cm) circle (0.7cm);% Zylinder außen | |||
| \draw [black!75!](0,0,1 cm) circle (0.7cm);%Hintere äußere Ringkante | |||
| %\foreach \z in {0,0.2,...,0.5} | |||
| %\fill [black!50!](0,0,\z cm) circle (0.5cm);% Zylinder innen | |||
| \fill [black!25!](0,0) circle (0.7cm);% Stirnseitenfüllung | |||
| \fill [black!35](0,0) circle (0.5cm); %Füllung innen | |||
| \begin{scope}%Clip Füllung | |||
| \clip (0,0,1cm) circle (0.5cm);%Clip mit hinterer inneren Ringkante | |||
| \fill [black!0!](0,0) circle (0.5cm);%Innere Ringkante | |||
| \end{scope} | |||
| \begin{scope}%Clip hintere innere Kante | |||
| \clip (0,0) circle (0.5cm);%Clip mit hinterer inneren Ringkante | |||
| \draw [black!75!](0,0,1cm) circle (0.5cm);%Hintere innere Ringkante | |||
| \end{scope} | |||
| \draw [black!75!](0,0) circle (0.5cm);% Innere Ringkante | |||
| \draw [black!75!](0,0) circle (0.7cm);% Äußere Ringkante | |||
| \foreach \a in {-45,-30,-15,15,30,45,60,75,90,105,120,135} %Rechte Drähte | |||
| \draw [red!60!black, ultra thick] (\a:0.5)--(\a:0.7)--+(45:.3); | |||
| \foreach \a in {150,165,...,300} %Linke Drähte | |||
| \draw [red!60!black, ultra thick] (\a:0.7)--(\a:0.5)--+(45:.3); | |||
| \draw [red!60!black, ultra thick] (0.5,0,0)--(1.5,0,0)circle(0.01);%Anschußdrähte | |||
| \draw [red!60!black, ultra thick] (0.7,0,1cm)--(1.5,0,1cm)circle(0.01);%Anschußdrähte | |||
| \draw [->,blue, thick] (1.6,0,1cm) -- (1.6,0,0) node at (1.6,0,.5cm)[right] {$u$}; | |||
| \draw [blue, ultra thick] (0,0,3.5cm) -- (0,0,7cm); | |||
| \draw [->,red, thick] (0,0,7.5cm) -- (0,0,8cm)node [right] {$i$}; | |||
| \draw [blue, ultra thick] (0,0,1.75cm) -- (0,0,-5cm)node [above left]{Leiter}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \begin{tikzpicture}[scale=1.4]%Querschnitt | |||
| \begin{scope}[>=latex,xshift=0,yshift=0] | |||
| \fill [black!25!] (0,0)rectangle(1,1); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,xshift=0,yshift=3cm] | |||
| \fill [black!25!] (0,0)rectangle(1,1); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,xshift=0,yshift=0cm] | |||
| \draw [very thick] (0,0)rectangle(1,4); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,xshift=-1cm,yshift=1.9cm] | |||
| \fill [blue!50!] (0,0)rectangle(3,.2); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,xshift=0cm,yshift=4cm]%Hilfslinien | |||
| \draw [very thin] (-1,0)--(0,0)--(0,.5) (1,0)--(1,.5)(-.5,-1)--(0,-1); | |||
| \draw [very thin, dashed] (-1.5,-2)--(2.5,-2); | |||
| \draw [->,very thin] (-.9,-2)--(-.9,0)node at(-.9,-1)[right]{$r_a$}; | |||
| \draw [->,very thin] (-.4,-2)--(-.4,-1)node at(-.4,-1.5)[right]{$r_i$}; | |||
| \draw [<->,very thin] (0,0.4)--(1,0.4)node at(.5,0.4)[above]{$l$}; | |||
| \draw node at(.5,-.5){$\mu=\mu_0$}; | |||
| \draw [->,red,thin] (1.25,-1.75)--(1.5,-1.75) node [right]{$i$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| u&=N\cdot \frac{d\phi(t)}{dt} | |||
| \end{align} | |||
| Berechnung: | |||
| Strom im Leiter bewirkt Fluss durch die Spule: | |||
| \begin{align*} | |||
| \phi&=\widehat{\phi}_0\cdot \sin(\omega t)\\ | |||
| \widehat{\phi}_0&=\int_{r_i}^{r_a}l\cdot \widehat{B}(r)\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \int_{r_i}^{r_a}\frac{1}{r}\cdot dr=\frac{\mu_0\cdot l\cdot \widehat{i}}{2\pi}\cdot \ln\frac{r_a}{r_i}\\ | |||
| \text{mit }\widehat{i}&=\sqrt{2}\cdot I\\ | |||
| \Rightarrow \widehat{\phi}_0&=\frac{1{,}26\cdot \power{10}{-6}\, \frac{\volt\second}{\ampere\metre}\cdot 12\cdot \power{10}{-3}\,\metre\cdot \sqrt{2}\cdot 40\,\ampere}{2\pi}\cdot \ln(2)=9{,}44\cdot \power{10}{-8}\,\volt\second\\ | |||
| \phi_{eff}&=\frac{\widehat{\phi}_0}{\sqrt{2}}=6{,}672\cdot \power{10}{-8}\,\volt\second\\ | |||
| \intertext{Veränderlicher Fluss induziert Spannung:} | |||
| u&=N\cdot \frac{d\phi(t)}{dt}=N\cdot \widehat{\phi}_0\cdot \omega\cdot \cos(\omega t)=\widehat{u}\cdot \cos(\omega t)\\ | |||
| \widehat{u}&=N\cdot \phi_0\cdot \omega=1800\cdot 9{,}44\cdot \power{10}{-8}\,\volt\second\cdot 2\pi\cdot 50\,\power{\second}{-1}=53{,}36\,\milli\volt\\ | |||
| U&=\frac{\widehat{u}}{\sqrt{2}}=\frac{53{,}36\,\milli\volt}{\sqrt{2}}=\uuline{37{,}73\,\milli\volt} | |||
| \intertext{Anmerkung:} | |||
| u_i&=2\cdot a\cdot r\cdot N\cdot B\cdot \omega\cdot \sin(\omega t)\\ | |||
| &=2\cdot a\cdot r\cdot N\cdot \frac{\Phi}{A}\cdot \omega\cdot \sin(\omega t)\\ | |||
| &\text{mit $2\cdot a\cdot r=A$}\\ | |||
| &=N\cdot \Phi \cdot \omega\cdot \sin(\omega t)\\ | |||
| &=\widehat{u}_i\cdot \sin(\omega t)\\ | |||
| \intertext{oder} | |||
| U&=\mu_0\cdot l\cdot \ln\frac{r_a}{r_i}\cdot N\cdot f\cdot I\\ | |||
| &=1{,}26\cdot \power{10}{-6}\,\frac{\volt\cancel{\second}}{\cancel{\ampere}\cancel{\metre}}\cdot 12\cdot \power{10}{-3}\,\cancel{\metre}\cdot \ln\frac{20\,\cancel{\milli\metre}}{10\,\cancel{\milli\metre}}\cdot 1800\cdot 50\,\cancel{\frac{1}{\second}}\cdot 40\,\cancel{\ampere}=\uuline{37{,}73\,\milli\volt} | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| \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$\uline{I}_R$}; | |||
| \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)[right]{\footnotesize$\uline{I}_L$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Stromquelle | |||
| \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0); | |||
| \draw [->,blue](.8,.2)--(.2,.2)node at(.5,.2)[left]{\footnotesize$\uline{U}_{RL}$}; | |||
| \draw (.5,0)circle(.133); | |||
| \draw [->,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{\footnotesize$\uline{I}_0$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. | |||
| \draw (.8,1)--(3,1)--(3,.9) (0,0)--(3,0)--(3,.2); | |||
| \fill(0,0)circle(0.05cm)(0,1)circle(0.05cm); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| Berechnung:\\ | |||
| \begin{align*} | |||
| \text{a)}\quad X_L&=\omega L=2\pi\cdot f\cdot 3{,}6\,\milli\henry=2714\,\ohm\\ | |||
| X_C&=\frac{-1}{\omega C}=\frac{-1}{2\pi\cdot f\cdot 1{,}5\,\nano\farad}=-884\,\ohm\\ | |||
| \uline{Z}_{RL}&=\frac{R\cdot jX_L}{R+jX_L}=\frac{1{,}3\,\kilo\ohm\cdot j2714\,\ohm}{1{,}3\,\kilo\ohm+j2714\,\ohm}=(1057+j506)\,\ohm=1172\,\ohm\cdot e^{j25{,}59\,\degree} | |||
| \end{align*} | |||
| Serienersatzschaltbild: | |||
| \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] {$X_C$}; | |||
| \draw [<-,red] (.25,0)--(.05,0)node at(.15,0)[above]{$\footnotesize\uline{I}$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[below]{\footnotesize$\uline{U}_C$}; | |||
| \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] {$\uline{Z}_{RL}$}; | |||
| \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{RL}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Spannungsquelle | |||
| \draw (0,0)--(1,0); | |||
| \draw (.5,0)circle(.133); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [<-,blue] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{\footnotesize$\uline{U}_0$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. | |||
| \draw (1.8,1)--(2,1)--(2,.9) (0,0)--(2,0)--(2,.2); | |||
| \fill(0,0)circle(0.05cm)(0,1)circle(0.05cm); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \begin{tikzpicture}[scale=.5] | |||
| \begin{scope}[>=latex,very thick,xshift=8cm,yshift=0cm] | |||
| \draw[black!25!,very thin](-3,0)grid(5,5); | |||
| \draw[thick](-3,0)--(5,0) (0,0)--(0,5); | |||
| \draw[->,blue](0,0)--(30:4.25cm)node[above right]{$\uline{U}$}; | |||
| \draw[->,red](0,0)--(110:2.7cm)node[above]{$\uline{I}_0$}; | |||
| \draw[->,blue](0,0)--(135.6:3.173cm)node[above left]{$\uline{U}_0$}; | |||
| \draw[->,red](0,0)--(18.5:5.25cm)node[below right]{$\uline{I}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \uline{U}_0&=\uline{I}_0\cdot \uline{Z}_{RL}=2{,}7\milli\ampere \cdot e^{j110\,\degree}\cdot 1172\,\ohm\cdot e^{j25{,}59\,\degree}\\ | |||
| &=3{,}164\,\volt\cdot e^{j135{,}6\,\degree}=(-2{,}26+j2{,}21)\,\volt\\ | |||
| \uline{Z}_{RLC}&=\uline{Z}_{RL}+jX_C=(1060+j507)\,\ohm-j884\,\ohm\\ | |||
| &=(1057-j378)\,\ohm=1122{,}6\,\ohm\cdot e^{-j19{,}67\,\degree}\\ | |||
| \intertext{Masche:} | |||
| \uline{I}&\cdot \uline{Z}_{RLC}+\uline{U}_0-\uline{U}=0\\ | |||
| \uline{I}&=\frac{\uline{U}-\uline{U}_0}{\uline{Z}_{RLC}}=\frac{\overbrace{(3{,}637+j2{,}1)\,\volt}^{4{,}2\,\volt\cdot e^{j30\,\degree}}-(-2{,}26+j2{,}21)\,\volt}{1122{,}6\,\ohm\cdot e^{-j19{,}67\,\degree}}=\frac{5{,}90\,\volt\cdot e^{-j1{,}06\,\degree}}{1122{,}6\,\ohm\cdot e^{-j19{,}67\,\degree}}\\ | |||
| &=\uuline{5{,}25\milli\ampere\cdot e^{j18{,}5\,\degree}}=\uuline{(4{,}98+j1{,}67)\milli\ampere} | |||
| \end{align*} | |||
| \clearpage | |||
| Parallelersatzschaltbild: (Wirkstrom in Phase, Blindstrom senkrecht zur Spannung) | |||
| \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 [<-,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$\uline{I}_W$}; | |||
| \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] {$X$}; | |||
| \draw [<-,red] (.75,0)--(.95,0)node at(.85,0)[right]{\footnotesize$\uline{I}_B$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. | |||
| \draw (0,1)--(2,1)--(2,.9) (0,0)--(2,0)--(2,.2); | |||
| \fill(0,0)circle(0.05cm)(0,1)circle(0.05cm); | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \begin{tikzpicture}[scale=.5] | |||
| \begin{scope}[>=latex,very thick,xshift=8cm,yshift=0cm] | |||
| \draw[black!25!,very thin](-3,0)grid(5,5); | |||
| \draw[thick](-3,0)--(5,0) (0,0)--(0,5); | |||
| \draw[->,blue!45!](0,0)--(30:4.25cm)node[above right]{$\uline{U}$}; | |||
| \draw[->,red!45!](0,0)--(110:2.7cm)node[above]{$\uline{I}_0$}; | |||
| \draw[->,blue!45!](0,0)--(135.6:3.173cm)node[above left]{$\uline{U}_0$}; | |||
| \draw[->,red!45!](0,0)--(18.5:5.25cm)node[below right]{$\uline{I}$}; | |||
| \draw[->,red](0,0)--(30:5.14cm)node at (2.5,2){$\uline{I}_W$}; | |||
| \draw[->,red](30:5.14cm)--+(-60:1.05cm)node at (5.5,2.25){$\uline{I}_B$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| |\uline{I}_{\textrm{w}}|&=I\cdot \cos(\underbrace{\varphi_u}_{30\,\degree}-\underbrace{\varphi_i}_{18{,}5\,\degree})=5{,}14\,\milli\ampere\\ | |||
| \Rightarrow & \uline{I}_{\textrm{w}}=\uuline{5{,}14\,\milli\ampere\cdot e^{j30\,\degree}}\\ | |||
| |\uline{I}_b|&=I\cdot \sin(\varphi_u-\varphi_i)=1{,}05\,\milli\ampere\\ | |||
| \Rightarrow & \uline{I}_b=\uuline{1{,}05\,\milli\ampere\cdot e^{-j60\,\degree}}\\ | |||
| \end{align*} | |||
| \begin{minipage}[b]{.5\textwidth} | |||
| b) | |||
| \begin{align*} | |||
| \uline{U}_C&=\uline{I}\cdot jX_C=4{,}64\,\volt\cdot e^{-j71{,}5\,\degree}\\ | |||
| \uline{U}_{RL}&=\uline{U}-\uline{U}_C=6{,}85\,\volt\cdot e^{j71{,}6\,\degree}\\ | |||
| \uline{I}_R&=\frac{\uline{U}_{RL}}{R}=5{,}27\,\milli\ampere\cdot e^{j71{,}6\,\degree}\\ | |||
| \uline{I}_L&=\frac{\uline{U}_{RL}}{jX_L}=2{,}52\,\milli\ampere\cdot e^{-j18{,}4\,\degree}\\ | |||
| \end{align*} | |||
| Probe: | |||
| \begin{align*} | |||
| I+I_0&=5{,}85\,\milli\ampere\cdot e^{j46\,\degree}\\ | |||
| I_R+I_L&=5{,}85\,\milli\ampere\cdot e^{j46\,\degree}\\ | |||
| &\text{stimmt überein.} | |||
| \end{align*} | |||
| \end{minipage} | |||
| \begin{minipage}[c]{.5\textwidth} | |||
| \begin{tikzpicture}[scale=.75] | |||
| \begin{scope}[>=latex,very thick,xshift=8cm,yshift=0cm] | |||
| \draw[black!25!,very thin](-3,-5)grid(5,7); | |||
| \draw[thick](-3,0)--(5,0) (0,-5)--(0,7); | |||
| \draw[->,red!66!](0,0)--(110:2.7cm)node[above]{$\uline{I}_0$}; | |||
| \draw[->,blue!25!](0,0)--(135.6:3.173cm)node[above left]{$\uline{U}_0$}; | |||
| \draw[->,red!66!](0,0)--(18.5:5.25cm)node[below right]{$\uline{I}$}; | |||
| \draw[->,red!25!](0,0)--(30:5.14cm)node at (2.5,2){$\uline{I}_W$}; | |||
| \draw[->,red!25!](30:5.14cm)--+(-60:1.05cm)node at (5.5,2.25){$\uline{I}_B$}; | |||
| \draw[->,blue!66!](0,0)--(30:4.25cm)node[above right]{$\uline{U}$}; | |||
| \draw[->,blue](0,0)--(-71.5:4.64cm)node[right]{$\uline{U}_C$}; | |||
| \draw[->,blue](30:4.25cm)--+(108.5:4.64cm)node[below right]{$-\uline{U}_C$}; | |||
| \draw[->,blue](0,0)--(71.6:6.85cm)node[below left]{$\uline{U}_{RL}$}; \draw[->,red](0,0)--(71.6:5.27cm)node[below left]{$\uline{I}_R$}; | |||
| \draw[->,red](71.6:5.27cm)--(46:5.85cm)node[above left]{$\uline{I}_L$}; | |||
| \draw[<-,red!50!blue]+(46:5.85cm)--(18.5:5.25cm)node at(5,3){$\uline{I}_0$}; \draw[->,red!50!blue](0,0)--(46:5.85cm)node at(2,3.25){\footnotesize$\uline{I}+\uline{I}_0$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{minipage}\\ | |||
| \clearpage | |||
| \begin{align*} | |||
| \intertext{Leistung an Klemme:} | |||
| \uline{S}_K&=\uline{U}\cdot \uline{I}^*=4{,}2\,\volt\cdot e^{j30\,\degree}\cdot 5{,}25\milli\ampere\cdot e^{-j18{,}5\,\degree}=(\underbrace{21{,}6}_{P_K}+\underbrace{j4{,}4}_{Q_K})\,\milli\volt\ampere\\ | |||
| P_K&=\uuline{21{,}6\,\milli\watt}\\ | |||
| Q_K&=\uuline{4{,}4\,\milli\var}\\ | |||
| \intertext{Leistung der Stromquelle (Pfeilsystem wie oben):} | |||
| S_I&=\uline{U}_{RL}\cdot I^*_0=6{,}85\,\volt\cdot e^{j71{,}6\,\degree}\cdot 2{,}7\milli\ampere \cdot e^{-j110\,\degree}=(\underbrace{14{,}5}_{P_I}\underbrace{-j11{,}5}_{Q_I})\,\milli\volt\ampere\\ | |||
| P_I&=\uuline{14{,}5\,\milli\watt}\\ | |||
| Q_I&=\uuline{-11{,}5\,\milli\var}\\ | |||
| \intertext{Gesamtleistung: Summe aus Quell- und Klemmleistung} S&=S_K+S_I=(21{,}6+j4{,}4+14{,}5-j11{,}5)\,\milli\volt\ampere=36{,}1\,\milli\watt-j7{,}1\,\milli\volt\ampere | |||
| \intertext{Probe: Wirkleistung: } | |||
| P&=\frac{U^2_{RL}}{R}=\frac{(6{,}85\,\volt)^2}{1300\,\ohm}=36{,}1\,\milli\watt \quad\text{(Übereinstimmung)} | |||
| \intertext{Probe: Blindleistung: } | |||
| C:\quad Q_C&=X_C\cdot I^2=-884\,\ohm\cdot(5{,}25\,\milli\ampere)^2=-24{,}4\,\milli\var\\ | |||
| L:\quad Q_L&=\frac{U^2_{RL}}{X_L}= \frac{(6{,}85\,\volt)^2}{2714\,\ohm}=+17{,}3\,\milli\var\\ | |||
| Q&=Q_C+Q_L=(-24{,}4+17{,}3)\,\milli\var=-7{,}1\,\milli\var \quad\text{(Übereinstimmung)}\\ | |||
| \end{align*} | |||
| \clearpage | |||
| }{}% | |||
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| @@ -0,0 +1,185 @@ | |||
| \section {Gleichungen in Matrizenschreibweise} | |||
| Die Spannungen am gegebenen Netzwerk | |||
| sollen mit Hilfe des Knotenpotenzialverfahrens | |||
| berechnet werden. Stellen Sie die Gleichungen | |||
| in Matrizenschreibweise auf.\\ | |||
| $R=2\,\kilo\ohm;\quad C=5\,\nano\farad \quad L=2{,}5\,\milli\henry; $\\ | |||
| $f=63{,}662\,\kilo\hertz;$\\ | |||
| $\uline{I}_1=7\,\milli\ampere;\quad \uline{U}_2=3\,\volt\cdot e^{j90\,\degree} \quad \uline{I}_3= 5\milli\ampere \cdot e^{-j90\,\degree}$ | |||
| \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] {$L$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,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] {$R$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=2cm,rotate=90]%Spannungsquelle | | |||
| \draw (0,0)--(1,0); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [->,blue] (.3,.2)--(.7,.2) node at (.5,.2)[left]{$\uline{U}_{2}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=3cm,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=2cm]%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=2cm]%Kondensator - | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,.133) [above] {$C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=2cm]%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=3cm,yshift=0.5cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=2.5cm,rotate=90] | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=4cm,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] {$C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=4cm,yshift=2.5cm,rotate=90]%Stromquelle | |||
| \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_3$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. | |||
| \draw (0,.2)--(0,0)--(4,0)--(4,.7) (0,3.8)--(0,4)--(4,4)--(4,3.4)(4,1.2)--(4,2.7) | |||
| (3,0)--(3,.7)(3,1.2)--(3,2.7)(3,2)--(4,2)(3,3.2)--(3,4)(1,0)--(1,-.2)(.9,-.2)--(1.1,-.2); | |||
| \fill(0,2)circle(0.05cm)(3,2)circle(0.05cm)(4,2)circle(0.05cm)(3,0)circle(0.05cm) | |||
| (4,0)circle(0.05cm)(3,4)circle(0.05cm)(4,4)circle(0.05cm)(0,4)circle(0.05cm)(1,0)circle(0.05cm); | |||
| \draw(0,4)--+(135:.5cm)(4,0)--+(225:.5cm); | |||
| \draw[<-,red](0,4)--+(135:.25cm)node[above right]{$\uline{I}_1$}; | |||
| \draw[->,red](4,0)--+(225:.25cm)node[below right]{$\uline{I}_0$}; | |||
| \end{scope} | |||
| \draw[magenta](0,2)node[left]{$1$}; | |||
| \draw[magenta](4,2)node[right]{$3$}; | |||
| \draw[magenta](3,4)node[above]{$2$}; | |||
| \draw[magenta](3,0)node[below]{$0$}; | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\begin{align} | |||
| %\intertext{Formeln:} | |||
| %\end{align} | |||
| Berechnung: | |||
| \begin{align*} | |||
| \begin{tikzpicture}[scale=1.5] | |||
| \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); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [->,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_1$}; | |||
| \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] {$\uline{Y}_{10}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm,rotate=90]%Stromquelle | |||
| \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_2$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,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] {$\uline{Y}_{12}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,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] {$\uline{Y}_{23}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=4cm,yshift=1cm,rotate=90]%Stromquelle | |||
| \draw (0,0)--(.367,0) (.5,-.133)--(.5,.133) (.633,0)--(1,0); | |||
| \draw (.5,0)circle(.133); | |||
| \draw [<-,red] (.3,-.2)--(.7,-.2) node at (.5,-.2)[right]{$\uline{I}_3$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=1cm]%Widerstand | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [below] {$\uline{Y}_{13}$}; | |||
| \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] {$\uline{Y}_{30}$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm]%Fehlstellen Eckverbindungen. | |||
| \draw (0,.7)--(0,0)--(4,0)--(4,.2) (0,1.3)--(0,2)--(4,2)--(4,1.8)(1,1)--(2,1) | |||
| (3,1)--(4,1); | |||
| \draw[magenta](1,1)node[left]{$1$}; | |||
| \draw[magenta](4,1)node[right]{$3$}; | |||
| \draw[magenta](2.5,2)node[above]{$2$}; | |||
| \draw[magenta](2.5,0)node[below]{$0$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \begin{align*} | |||
| \left[\begin{array}{c}\uline{Y}\\ \end{array} \right]\cdot | |||
| \left[\begin{array}{c}\uline{U}\\ \end{array}\right] | |||
| &=\left[\begin{array}{c} -\sum{\uline{I}_q}\\ \end{array}\right]\\ | |||
| \end{align*} | |||
| $\uline{I}_0\stackrel{!}{=}\uline{I}_1$ wegen Gleichgewicht. Abfließende Quellströme positiv. | |||
| \clearpage | |||
| \begin{align*} | |||
| G&=\frac{1}{R}=0{,}5\,\milli\siemens\\ | |||
| \omega&=2\pi\cdot f=0{,}4\cdot \power{10}{-6}\cdot \frac{1}{\second}\\ | |||
| X_C&=\frac{-1}{\omega C}=-0{,}5\,\kilo\ohm\\ | |||
| \Rightarrow\quad B_C&=\omega C=2{,}0\,\milli\siemens\\ | |||
| X_L&=\omega L=1{,}0\,\kilo\ohm\\ | |||
| \Rightarrow\quad B_L&=\frac{-1}{\omega L}=-1{,}0\,\milli\siemens\\[\baselineskip] | |||
| \uline{Y}_{10}&=\frac{1}{R+jX_L}=\frac{1}{(2+j1)\,\kilo\ohm}&=(0{,}4-j0{,}2)\,\milli\siemens\\ | |||
| \uline{Y}_{30}&=G+jB_C&=(0{,}5+j2{,}0)\,\milli\siemens\\ | |||
| \uline{Y}_{13}&=\frac{1}{R+j(X_L+X_C)}=\frac{1}{[2+j(1-0{,}5)]\,\kilo\ohm}&=(0{,}4706-j0{,}1176)\,\milli\siemens\\ | |||
| \uline{Y}_{12}&=jB_C=2\,\milli\siemens\cdot e^{j90\,\degree}&=(0+j2{,}0)\,\milli\siemens\\ | |||
| \uline{Y}_{23}&=G&=(0{,}5+j0)\,\milli\siemens\\[\baselineskip] | |||
| \uline{I}_1&=7\,\milli\ampere \cdot e^{j0\,\degree}&=(7-j0)\,\milli\ampere\\ | |||
| \uline{I}_2&=\uline{Y}_{12}\cdot \uline{U}_2=2\,\milli\siemens\cdot e^{j90\,\degree}\cdot 3\,\volt\cdot e^{j90\,\degree}=6\,\milli\ampere \cdot e^{j180\,\degree}&=(-6+j0)\,\milli\ampere\\ | |||
| \uline{I}_3&=5\,\milli\ampere \cdot e^{-j90\,\degree}&=(0-j5)\,\milli\ampere\\ | |||
| \end{align*} | |||
| %\scriptsize | |||
| \begin{align*} | |||
| &\left( | |||
| \begin{array}{ccc} | |||
| \uline{Y}_{10}+\uline{Y}_{12}+\uline{Y}_{13} & -\uline{Y}_{12} & -\uline{Y}_{13} \\ | |||
| -\uline{Y}_{12} & \uline{Y}_{12}+\uline{Y}_{23} & -\uline{Y}_{23} \\ | |||
| -\uline{Y}_{13} & -\uline{Y}_{23} & \uline{Y}_{30}+\uline{Y}_{13}+\uline{Y}_{23} \\ | |||
| \end{array} | |||
| \right) | |||
| \left( | |||
| \begin{array}{c} | |||
| \uline{U}_{10} \\ | |||
| \uline{U}_{20} \\ | |||
| \uline{U}_{30} \\ | |||
| \end{array} | |||
| \right)= | |||
| \left( | |||
| \begin{array}{c} | |||
| \uline{I}_2\\ | |||
| \uline{I}_1-\uline{I}_2-\uline{I}_3\\ | |||
| \uline{I}_3\\ | |||
| \end{array} | |||
| \right)\\ | |||
| \end{align*} | |||
| \scriptsize | |||
| \begin{align*} | |||
| &\left( | |||
| \begin{array}{ccc} | |||
| (0{,}8706+j1{,}6824) & (0-j2) & (-0{,}4706+j0{,}1176) \\ | |||
| (0-j2) & (0{,}5+j2{,}0) & (0{,}5+j0) \\ | |||
| (-0{,}4706-j0{,}1176) & (0{,}5+j0) & (1{,}4706+j1{,}8823) \\ | |||
| \end{array} | |||
| \right)\,\milli\siemens | |||
| \left( | |||
| \begin{array}{c} | |||
| \uline{U}_{10} \\ | |||
| \uline{U}_{20} \\ | |||
| \uline{U}_{30} \\ | |||
| \end{array} | |||
| \right)= | |||
| \left( | |||
| \begin{array}{c} | |||
| (6-j0)\\ | |||
| (13+j5)\\ | |||
| (0-j5)\\ | |||
| \end{array} | |||
| \right)\,\milli\ampere | |||
| \end{align*} | |||
| \normalsize | |||
| \clearpage | |||
| }{}% | |||
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ET2_L_Ergebnisse.tex
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| @@ -0,0 +1,254 @@ | |||
| % \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \newcounter{acounter} | |||
| \setcounter{acounter}{1} | |||
| \newcounter{bcounter} | |||
| \newcounter{bsubcounter} | |||
| \setcounter{bcounter}{12} | |||
| \setcounter{bsubcounter}{0} | |||
| \section{Ergebnisse} | |||
| %% Blatt 12 | |||
| \textbf{\arabic{acounter} {Blitzableiter}} %\textbf{\arabic{acounter} Blitzableiter} \textbf{\arabic{acounter} {Blitzableiter}} | |||
| \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $u=-133{,}7\,\volt$\\ | |||
| \textbf{\arabic{acounter} Drahtschleife} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $u=-36{,}39\,\milli\volt$\\ | |||
| \textbf{\arabic{acounter} Metallstab} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $|B_z|=0{,}398\,\frac{\volt\second}{\square\metre}$\\ | |||
| \textbf{\arabic{acounter} Spannungsverlauf} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $f=40\,\hertz$\\ | |||
| $\overline{|u|}=6\,\volt$\\ | |||
| $U=7{,}30\,\volt$\\ | |||
| $F=1{,}22$\\ | |||
| $P=0{,}533\,\watt$\\ | |||
| \textbf{\arabic{acounter} Phasenanschnitt} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $U=261\,\volt$\\ | |||
| \textbf{\arabic{acounter} Rechteckspannung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $u_1=1{,}5\,\volt \qquad u_2=-3{,}5\,\volt\\ | |||
| U=2{,}29\,\volt$\\ | |||
| %% Blatt 13 | |||
| \textbf{\arabic{acounter} Scheinersatzwiderstände} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter} \setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $R_r=520\,\ohm\\ | |||
| L_r=60\,\milli\henry | |||
| G_p=693\,\ohm\\ | |||
| L_p=239\,\milli\henry\\ | |||
| \varphi'_r=23{,}5\,\degree\\ | |||
| \varphi'_p=-37{,}6\,\degree$\\ | |||
| \textbf{\arabic{acounter} Verbraucherleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $p(t)=1079\,\watt+1318\,\volt\ampere\cdot \sin(2\omega t+0{,}96)\\ | |||
| S=1318\,\volt\ampere\\ | |||
| P=1079\,\watt\\ | |||
| Q=-756\,\mathrm{var}$\\ | |||
| \textbf{\arabic{acounter} Blindleistungskompensation} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $I_N=20{,}63\,\ampere\\ | |||
| \varphi_N=32,52\degree\\ | |||
| C=153{,}6\,\micro\farad$\\ | |||
| \textbf{\arabic{acounter} Energieübertragung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $X_C=-25\,\ohm\\ | |||
| P_{VR_L}=46{,}7\,\watt\\ | |||
| P_W=583\,\watt $\\ | |||
| \textbf{\arabic{acounter} Wechselstrommotor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $I=15{,}5\,\ampere\\ | |||
| C=80{,}6\,\micro\farad\\ | |||
| I'=12{,}1\,\ampere $\\ | |||
| \clearpage | |||
| \textbf{\arabic{acounter} Parallelschaltung von L und C} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $i_L(t_2)=2{,}92\,\ampere $\\ | |||
| \textbf{\arabic{acounter} Werte $R_L$ und $L$ einer Spule} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $I=1\,\ampere\\ | |||
| R_L=12{,}5\,\ohm\\ | |||
| L=0{,}219\,\henry $\\ | |||
| %% Blatt 14 | |||
| \textbf{\arabic{acounter} Zeigerdiagramm} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $\underline{U}_e=15\,\volt\cdot e^{+j90\degree}$\\ | |||
| \textbf{\arabic{acounter} Gesamtwiderstand} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| Induktiv\\ | |||
| \textbf{\arabic{acounter} Brückenschaltung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{U}_a=130\,\volt\cdot e^{j83\,\degree}$\\ | |||
| \textbf{\arabic{acounter} Zeigerdiagramm Netzwerk} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $R_1=13{,}89\,\kilo\ohm$\\ | |||
| \textbf{\arabic{acounter} Blind- Wirk- und Scheinleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{Z}=84{,}85\,\ohm\cdot e^{-j45\,\degree}=(60-j60)\,\ohm\\ | |||
| \uline{I}=2{,}71\,\ampere\cdot e^{j45\,\degree}=(1{,}916+j1{,}916)\,\ampere\\ | |||
| \uline{I}_{R_C}=1{,}21\,\ampere\cdot e^{-j18{,}4\,\degree}=(1{,}150-j0{,}383)\,\ampere\\ | |||
| \uline{I}_{C}=2{,}42\,\ampere\cdot e^{+j71{,}6\,\degree}=(0{,}766+j2{,}30)\,\ampere\\ | |||
| \uline{U}_{R_C}=242\,\volt\cdot e^{-j18{,}4\,\degree}\\ | |||
| \uline{U}_{R_L}=54{,}2\,\volt\cdot e^{j45\,\degree}\\ | |||
| \uline{U}_L=54.2\,\volt\cdot e^{j135\,\degree}\\ | |||
| S=623\,\volt\ampere\\ | |||
| P=447\,\watt\\ | |||
| Q=-447\,\volt\ampere r $\\ | |||
| %% Blatt 15 | |||
| \textbf{\arabic{acounter} Komplexe Wechselstromrechnung Netzwerk Strom} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I_2}=53{,}92\,\milli\ampere\cdot e^{j176{,}32\,\degree}$\\ | |||
| \textbf{\arabic{acounter} Übergang Zeitabhängige zu Komplexen Größen} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}_L=34{,}45\,\milli\ampere\cdot e^{j50{,}83\,\degree}\\ | |||
| i_L(T)=-25{,}79\,\milli\ampere$\\ | |||
| \textbf{\arabic{acounter} Leitwert} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $R=500\,\ohm\\ | |||
| B_C=2{,}175\,\milli\siemens$\\ | |||
| \textbf{\arabic{acounter} Strom L-R-C} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}=0{,}4058\,\ampere\cdot e^{j67{,}4\,\degree}=(0{,}156+j0{,}375)\,\ampere $\\ | |||
| \textbf{\arabic{acounter} Überlagerungsmethode} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}_C=(578{,}41+j279{,}99)\,\milli\ampere=642{,}6\,\milli\ampere\cdot e^{+j25{,}38\,\degree}$\\ | |||
| \textbf{\arabic{acounter} Momentan Leistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $u_R(T)=-114{,}1\,\milli\volt\\ | |||
| u_L(T)=-50{,}5\,\milli\volt\\ | |||
| u_C(T)=119{,}7\,\milli\volt\\ | |||
| p(T)=0{,}427\,\milli\watt$\\ | |||
| \clearpage | |||
| %% Blatt 16 | |||
| \textbf{\arabic{acounter} CLR Netzwerk} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $S=243{,}32\,\milli\volt\ampere\\ | |||
| P=141{,}50\,\milli\watt\\ | |||
| Q=-197{,}93\,\milli\,\var$\\ | |||
| \textbf{\arabic{acounter} Wirkleistung vs. Blindleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| --\\ | |||
| \textbf{\arabic{acounter} Wirkleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $R_v=17{,}24\,\ohm\\ | |||
| C=19{,}14\,\micro\farad\\ | |||
| P_{v\text{,}max}=14{,}5\,\milli\watt$ | |||
| \textbf{\arabic{acounter} Abgebbare Wirkleistung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $F_ \% =-78{,}8\, \%$\\ | |||
| \textbf{\arabic{acounter} Wirkleistung Spannungsquelle} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $P=-16{,}31\,\milli\watt$\\ | |||
| \textbf{\arabic{acounter} Dualitätskonstante} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $L_2=100\,\milli\henry\\ | |||
| C_2=5\,\micro\farad$ | |||
| \textbf{\arabic{acounter} Dualitätskonstante verlustbehaftete Bauelemente} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $R_G=1\,\milli\henry\\ | |||
| G_R=2\,\milli\siemens\\ | |||
| C_L=100\,\nano\farad$\\ | |||
| \textbf{\arabic{acounter} Vierpol Y-Parameter} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{Y}_L=\frac{1}{j\omega\cdot L}=-j\frac{1}{\omega\cdot L}\\ | |||
| \uline{Y}_{11}=\frac{1}{3}\cdot \uline{Y}_L\\ | |||
| \uline{Y}_{12}=-\frac{1}{6}\cdot \uline{Y}_L\\ | |||
| \uline{Y}_{22}=\uline{Y}_{11}=\frac{1}{3}\cdot \uline{Y}_L=\\ | |||
| \uline{Z}_{11}=\uline{Z}_{24}= 4\cdot j\omega\cdot L\\ | |||
| \uline{Z}_{12}=\uline{Z}_{21}= 2\cdot j\omega\cdot L$\\ | |||
| \textbf{\arabic{acounter} Spannung Vierpol} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{U}_0=130{,}4\,\milli\volt$\\ | |||
| %% Blatt 17 | |||
| \textbf{\arabic{acounter} Stromortskurve} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}=236\,\milli\ampere\cdot e^{(j35{,}2\,\degree)}=(193+j136)\,\milli\ampere$\\ | |||
| \textbf{\arabic{acounter} Leitwerts-, Widerstandsortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| -- | |||
| \textbf{\arabic{acounter} Ortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $f_g=828{,}9\,\hertz\\ | |||
| a=-4{,}9\,\deci\bel$\\ | |||
| \textbf{\arabic{acounter} Stromortskurve} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $I_{max}=5{,}55\,\milli\ampere $\\ | |||
| \textbf{\arabic{acounter} Widerstandstransformation} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $X_L{_p}=12{,}2\,\kilo\ohm\\ | |||
| X_{C_s}=-5{,}6\,\kilo\ohm\\ | |||
| {X_{L_s}}=+2{,}6\,\kilo\ohm\\ | |||
| X_{C_p}=-4{,}2\,\kilo\ohm $\\ | |||
| \clearpage | |||
| \textbf{\arabic{acounter} Brückenschaltung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{U}_{ab}=+j100\,\volt=100\,\volt\cdot e^{+j90\degree} $ | |||
| \textbf{\arabic{acounter} Wechselstrombrücke} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $ \uline{Z}_2=(1+j0{,}5)\,\kilo\ohm\\ | |||
| R_2=1{,}25\,\kilo\ohm\\ | |||
| L_2=2{,}5\,\milli\henry$\\ | |||
| \textbf{\arabic{acounter} Wechselstrombrücke} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $X=+\frac{R^2}{X_L}$\\ | |||
| %% Blatt 18 | |||
| \textbf{\arabic{acounter} Übertrager im Leerlauf} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): }\\ | |||
| $Z_1=23\,\ohm\cdot e^{j77{,}44\,\degree}\\ | |||
| P=500\,\watt\\ | |||
| Q=2245\,var\\ | |||
| \omega L_1=22{,}45\,\ohm\\ | |||
| \omega M=10\,\ohm\\ | |||
| \omega L_2=4{,}45\,\ohm$\\ | |||
| \textbf{\arabic{acounter} Übertrager mit kapazitiver Last} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{U}_2=0{,}398\,\volt\cdot e^{-j174{,}3\,\degree}\\ | |||
| \uline{U}_2=0{,}42\,\volt\cdot e^{+j18{,}4\,\degree}\\ | |||
| P_2=178\,\micro\watt $\\ | |||
| \textbf{\arabic{acounter} Übertrager mit Verbindung zum Eingang} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{U}_{ab}=200\,\volt\cdot e^{j73{,}74\,\degree}\\ | |||
| \uline{U}_S=317\,\volt\cdot e^{j79{,}8\,\degree}\\ | |||
| \uline{I}=0{,}689\,\ampere\cdot e^{-j8{,}43\,\degree}\\ | |||
| \uline{U}_{ab}=159\,\volt\cdot e^{+j71{,}7\,\degree}\\ | |||
| P=12{,}2\,\watt$\\ | |||
| \textbf{\arabic{acounter} Impedanzmatrix} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{Z}_{11}=(730+j100)\,\ohm\\ | |||
| \uline{Z}_{12}=\uline{Z}_{21}=1200\,\ohm\\ | |||
| \uline{Z}_{22}=(2000-j200)\,\ohm $\\ | |||
| \textbf{\arabic{acounter} Netztransformator} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $U_2=105{,}1\,\volt\\ | |||
| I_1=118\,\milli\ampere\\ | |||
| \widehat{B}=1{,}095\,\tesla$\\ | |||
| \textbf{\arabic{acounter} 3-Phasen Spannungssystem} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}_N=10{,}6\,\ampere\cdot e^{j14{,}7\,\degree} $\\ | |||
| \textbf{\arabic{acounter} 3-Phasen System mit unsymmetrischem Verbraucher} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $P=-335{,}44\,\watt $\\ | |||
| \textbf{\arabic{acounter} Strangströme 3-Phasen System mit unsymmetrischem | |||
| Verbraucher} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}_C=4\,\ampere\cdot e^{j120\,\degree}=(-2+j3{,}46)\,\ampere\\ | |||
| \uline{I}_{RL}=4\,\ampere\cdot e^{j113{,}1\,\degree}=(-1{,}57+j3{,}68)\,\ampere\\ | |||
| \uline{I}_1=0{,}48\,\ampere\cdot e^{-j153\,\degree}\\ | |||
| \uline{I}_2=3{,}66\,\ampere\cdot e^{j109{,}1\,\degree}\\ | |||
| \uline{I}_3=4{,}03\,\ampere\cdot e^{j66{,}1\,\degree}\\ | |||
| P=1280\,\watt $ | |||
| \clearpage | |||
| %% Blatt 19 | |||
| \textbf{\arabic{acounter} Resonanzfrequenz Zweipol} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): } | |||
| $f_{res}=899\,\hertz\\ | |||
| Z=37{,}5\,\ohm $\\ | |||
| \textbf{\arabic{acounter} RLC-Reihenschwingkreis} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $R=10\,\ohm\\ | |||
| C=3{,}47\,\micro\farad\\ | |||
| L=35{,}4\,\milli\henry$\\ | |||
| \textbf{\arabic{acounter} Effektivwert und Klirrfaktor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $I_{\mu}=7{,}36\,\milli\ampere\\ | |||
| k_{\mu}=27{,}7\%\\ | |||
| I_0=7{,}88\,\milli\ampere\\ | |||
| k_0=25{,}8\%$\\ | |||
| \textbf{\arabic{acounter} Klirrfaktor} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $k_a=21{,}5\%\\ | |||
| U_a=2{,}49\,\volt$\\ | |||
| \textbf{\arabic{acounter} Momentanspannung} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $u_C(t=T=2\,\micro\second)=1{,}315\,\volt $\\ | |||
| \textbf{\arabic{acounter} Nichtlinears Bauelement} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $k_i=19{,}61\%$\\ | |||
| \textbf{\arabic{acounter} Wirkleistung Zweipol} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $P=573\,\watt$\\ | |||
| %% Blatt 20 | |||
| \textbf{\arabic{acounter} Ringspule} \stepcounter{bcounter}\stepcounter{acounter}\textbf{(B\arabic{bcounter}\setcounter{bsubcounter}{1}A\arabic{bsubcounter}): } | |||
| $U=37{,}73\,\milli\volt$\\ | |||
| \textbf{\arabic{acounter} Netzwerk Wirk- und Blindanteil} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| $\uline{I}=5{,}25\milli\ampere\cdot e^{j18{,}5\,\degree}=(4{,}98+j1{,}67)\milli\ampere\\ | |||
| \uline{I}_{\textrm{w}}=5{,}14\,\milli\ampere\cdot e^{j30\,\degree}\\ | |||
| \uline{I}_b=1{,}05\,\milli\ampere\cdot e^{-j60\,\degree}\\ | |||
| P_K=21{,}6\,\milli\watt\\ | |||
| Q_K=4{,}4\,\milli\var\\ | |||
| P_I=14{,}5\,\milli\watt\\ | |||
| Q_I=-11{,}5\,\milli\var $\\ | |||
| \textbf{\arabic{acounter} Gleichungen in Matrizenschreibweise} \stepcounter{acounter}\textbf{(B\arabic{bcounter}\stepcounter{bsubcounter}A\arabic{bsubcounter}): }\\ | |||
| \scriptsize | |||
| \begin{align*} | |||
| &\left( | |||
| \begin{array}{ccc} | |||
| (0{,}8706+j1{,}6824) & (0-j2) & (-0{,}4706+j0{,}1176) \\ | |||
| (0-j2) & (0{,}5+j2{,}0) & (0{,}5+j0) \\ | |||
| (-0{,}4706-j0{,}1176) & (0{,}5+j0) & (1{,}4706+j1{,}8823) \\ | |||
| \end{array} | |||
| \right)\,\milli\siemens | |||
| \left( | |||
| \begin{array}{c} | |||
| \uline{U}_{10} \\ | |||
| \uline{U}_{20} \\ | |||
| \uline{U}_{30} \\ | |||
| \end{array} | |||
| \right)= | |||
| \left( | |||
| \begin{array}{c} | |||
| (6-j0)\\ | |||
| (13+j5)\\ | |||
| (0-j5)\\ | |||
| \end{array} | |||
| \right)\,\milli\ampere | |||
| \end{align*}\\ | |||
| \clearpage | |||
| % }{}% | |||
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| \documentclass[twoside,a4paper]{scrartcl} | |||
| \usepackage[T1]{fontenc} | |||
| \usepackage[ansinew]{inputenc} | |||
| \usepackage[ngerman]{babel} | |||
| %\usepackage{textcomp} % dazugeladen wegen Spezial Zeichen | |||
| \usepackage{pifont} % dazugeladen wegen Spezial Zeichen | |||
| %\usepackage{ifsym} % dazugeladen wegen Spezial Zeichen | |||
| %\usepackage{dingbat} % dazugeladen wegen Spezial Zeichen | |||
| %\usepackage{mathabx} % nicht vorhanden | |||
| \usepackage{amsfonts} | |||
| \usepackage{amssymb} | |||
| \usepackage{amsmath} | |||
| \usepackage[amssymb,thickspace, thickqspace]{SIunits}% see SIunits manual.. Load SIunits package after amssymb package. | |||
| \usepackage{graphicx} | |||
| \usepackage{xcolor} | |||
| \usepackage{xspace} | |||
| \usepackage{scrpage2} | |||
| \usepackage{tikz} | |||
| \usetikzlibrary{decorations.pathmorphing,decorations.pathreplacing,decorations.shapes} | |||
| \usepackage{xkeyval} %see tikzpgf manual page 54 | |||
| %\usetikzlibrary{patterns} | |||
| \usepackage{pgf} | |||
| %\usepackage{pgfbaseplot} | |||
| %\usepackage{pgfmath} | |||
| \usepackage{pgfplots}% neu Feuersänger | |||
| \usetikzlibrary {arrows} | |||
| \SIunits[thickspace] % , thickqspace (geht nicht) | |||
| \usepackage{sistyle} %% wegen amssymb gestrichen !? | |||
| \usepackage{slashed} | |||
| \usepackage{cancel} | |||
| \usepackage {ulem} | |||
| \numberwithin{equation}{section} | |||
| \usepackage{ifthen} | |||
| % \newcommand\toPrint{xLösung}% print Lösung (Lösung) ein (xLösung) druckt nur die Aufgaben | |||
| \newcommand{\var}{\mathrm{var}} | |||
| \usepackage{pdfpages} | |||
| \usepackage{hyperref} | |||
| \hypersetup{colorlinks=true, | |||
| pdfpagemode=none, | |||
| pdftitle={ET2 Übungen}, | |||
| pdfsubject={ET2 Übungen}, | |||
| pdfauthor={Prof.\,Dr.\,C.\,Niebler}, | |||
| pdfkeywords={Elektrotechnik Uebung}, | |||
| pdffitwindow=true, | |||
| pdfstartview=FitH | |||
| } | |||
| %%\setkomafont{pagehead}{% | |||
| %%\footnotesize\sffamily | |||
| %%} | |||
| %% | |||
| %%\pagestyle{scrheadings} | |||
| %%\lehead{\pagemark} | |||
| %%\cehead{Georg-Simon-Ohm Hochschule Nürnberg} | |||
| %%\rehead{} | |||
| %%\lohead{} | |||
| %%\cohead{Georg-Simon-Ohm Hochschule Nürnberg} | |||
| %%\rohead{\pagemark} | |||
| %%\lefoot{} | |||
| %%\cefoot{C. Niebler -- 2008} | |||
| %%\refoot{} | |||
| %%\lofoot{} | |||
| %%\cofoot{C. Niebler -- 2008} | |||
| %%\rofoot{} | |||
| %%%\cfoot{\pagemark} | |||
| %\usepackage{geometrie} | |||
| %{geometry:a4paper} | |||
| %{geometry:left=5cm} | |||
| %{geometry:right=3cm} | |||
| \usepackage{scrtime} | |||
| %\newcommand{\footlinetext}{\footnotesize \textsf{\color{gray} Niebler, C.: | |||
| % Grundlagen Elektrotechnik -- Übung 2, GSO-Hochschule, SS\,2008;\hfill\today\ \thistime \normalsize}} | |||
| % \newcommand{\footlinetext}{\footnotesize \textsf{\color{gray} | |||
| % Technische Hochschule Nürnberg Georg Simon Ohm\hfill Übung Grundlagen Elektrotechnik 2\hspace{3em} SS\,2013 \\ | |||
| % Dr. Christine Niebler \hfill\today \normalsize}} | |||
| \pagestyle{scrheadings} | |||
| \ohead{\input{ohmlogoTH}} %% mit TIKZ | |||
| \chead{} | |||
| \ihead{\footnotesize{Fachbereich Elektro-\\ | |||
| Feinwerk- und Informationstechnik}} | |||
| \cfoot{\tiny{Übung Grundlagen Elektrotechnik 2 (\myVersion)}} | |||
| \ifoot{\tiny{Prof.\,Dr.\,C.\,Niebler}} | |||
| \ofoot{\pagemark} | |||
| \setlength{\parindent}{0pt} | |||
| \setlength{\parsep}{\baselineskip} | |||
| \usepackage[titles]{tocloft} | |||
| \setlength{\cftbeforesecskip}{0.5ex} | |||
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| % \chead{} | |||
| % \ihead{\pagemark} | |||
| % \ifoot{\footlinetext} | |||
| % \cfoot{} | |||
| % \ofoot{} | |||
| % \setlength{\parindent}{0pt} | |||
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| \thispagestyle{empty} | |||
| {\Large Vorwort:} | |||
| \paragraph{1. Ausgabe --- SoSe 2008} | |||
| Ein wichtiges Element beim Studium ist es, sich die Grundlagen zu verinnerlichen, um diese dann in der Praxis fachgerecht anwenden zu können. | |||
| Diese Unterlage wurde aus einer handschriftlichen Loseblattsammlung entwickelt. Die Lösungen wurden in logische Schritte eingeteilt, damit es den Studenten das Verständnis erleichtert. Wichtig ist es auch um die Übungen effizient vorbereiten zu können. | |||
| % Die Aufgaben wurden mit SPICE simuliert und die Lösungen verifiziert.\\ | |||
| \paragraph{2. Ausgabe --- SoSe 2013} | |||
| Die Erfahrungen aus der Vorlesung/Übung wurde in der Ausgabe eingebracht. | |||
| \paragraph{3. Ausgabe --- SoSe 2014} | |||
| Anpassung der Überschriftennummern an Blatt und Aufgabennummer, zur leichteren Gruppierung in Themenfelder. | |||
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| \makeatletter | |||
| \AddToShipoutPicture{ | |||
| \setlength{\@tempdimb}{.75\paperwidth} | |||
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| {\fontsize{3cm}{3cm}\selectfont{\ProfName}}}} | |||
| } | |||
| } | |||
| \makeatother | |||
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| \providecommand\hyper@newdestlabel[2]{} | |||
| \catcode`"\active | |||
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| \@writefile{toc}{\contentsline {section}{\numberline {20/2{}}Netzwerk Wirk- und Blindanteil}{124}{section.58}} | |||
| \@writefile{toc}{\contentsline {section}{\numberline {20/3{}}Gleichungen in Matrizenschreibweise}{127}{section.59}} | |||
| \@writefile{toc}{\contentsline {section}{\numberline {}Ergebnisse}{129}{section.60}} | |||
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Main_ET2_V2.i.gnuplot
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| set table "Main_ET2_V2.i.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:5] .85*cos(.5*3.14*x-1.57) | |||
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Main_ET2_V2.iFe.gnuplot
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| set table "Main_ET2_V2.iFe.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:5] -.4*sin(.5*3.14*x) | |||
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Main_ET2_V2.iomega.gnuplot
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| set table "Main_ET2_V2.iomega.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:5] 1*cos(.5*3.14*x)+.288*cos(1.5*3.14*x)-.4*sin(.5*3.14*x) | |||
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Main_ET2_V2.log
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Main_ET2_V2.ortskurve17-2.gnuplot
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| set table "Main_ET2_V2.ortskurve17-2.table"; set format "%.5f" | |||
| set samples 100; set parametric; plot [t=0:5] 1000/(100+25*t*t),2*t-500*t/(100+25*t*t) | |||
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Main_ET2_V2.out
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| \BOOKMARK [1][-]{section.1}{Blitzableiter}{}% 1 | |||
| \BOOKMARK [1][-]{section.2}{Drahtschleife}{}% 2 | |||
| \BOOKMARK [1][-]{section.3}{Metallstab}{}% 3 | |||
| \BOOKMARK [1][-]{section.4}{Spannungsverlauf}{}% 4 | |||
| \BOOKMARK [1][-]{section.5}{Phasenanschnitt}{}% 5 | |||
| \BOOKMARK [1][-]{section.6}{Rechteckspannung}{}% 6 | |||
| \BOOKMARK [1][-]{section.7}{Scheinersatzwiderst\344nde}{}% 7 | |||
| \BOOKMARK [1][-]{section.8}{Verbraucherleistung}{}% 8 | |||
| \BOOKMARK [1][-]{section.9}{Blindleistungskompensation}{}% 9 | |||
| \BOOKMARK [1][-]{section.10}{Energie\374bertragung}{}% 10 | |||
| \BOOKMARK [1][-]{section.11}{Wechselstrommotor}{}% 11 | |||
| \BOOKMARK [1][-]{section.12}{Parallelschaltung von L und C}{}% 12 | |||
| \BOOKMARK [1][-]{section.13}{Werte RL und L einer Spule}{}% 13 | |||
| \BOOKMARK [1][-]{section.14}{Zeigerdiagramm}{}% 14 | |||
| \BOOKMARK [1][-]{section.15}{Gesamtwiderstand}{}% 15 | |||
| \BOOKMARK [1][-]{section.16}{Br\374ckenschaltung}{}% 16 | |||
| \BOOKMARK [1][-]{section.17}{Zeigerdiagramm Netzwerk}{}% 17 | |||
| \BOOKMARK [1][-]{section.18}{Blind- Wirk- und Scheinleistung}{}% 18 | |||
| \BOOKMARK [1][-]{section.19}{Komplexe Wechselstromrechnung Netzwerk Strom}{}% 19 | |||
| \BOOKMARK [1][-]{section.20}{\334bergang Zeitabh\344ngige zu Komplexen Gr\366\337en}{}% 20 | |||
| \BOOKMARK [1][-]{section.21}{Leitwert}{}% 21 | |||
| \BOOKMARK [1][-]{section.22}{Strom L-R-C}{}% 22 | |||
| \BOOKMARK [1][-]{section.23}{\334berlagerungsmethode}{}% 23 | |||
| \BOOKMARK [1][-]{section.24}{Momentan Leistung}{}% 24 | |||
| \BOOKMARK [1][-]{section.25}{CLR Netzwerk}{}% 25 | |||
| \BOOKMARK [1][-]{section.26}{Wirkleistung vs. Blindleistung}{}% 26 | |||
| \BOOKMARK [1][-]{section.27}{Wirkleistung}{}% 27 | |||
| \BOOKMARK [1][-]{section.28}{Abgebbare Wirkleistung}{}% 28 | |||
| \BOOKMARK [1][-]{section.29}{Wirkleistung Spannungsquelle}{}% 29 | |||
| \BOOKMARK [1][-]{section.30}{Dualit\344tskonstante}{}% 30 | |||
| \BOOKMARK [1][-]{section.31}{Dualit\344tskonstante verlustbehaftete Bauelemente}{}% 31 | |||
| \BOOKMARK [1][-]{section.32}{Vierpol Y-Parameter}{}% 32 | |||
| \BOOKMARK [1][-]{section.33}{Spannung Vierpol}{}% 33 | |||
| \BOOKMARK [1][-]{section.34}{Stromortskurve}{}% 34 | |||
| \BOOKMARK [1][-]{section.35}{Leitwerts-, Widerstandsortskurve}{}% 35 | |||
| \BOOKMARK [1][-]{section.36}{Ortskurve}{}% 36 | |||
| \BOOKMARK [1][-]{section.37}{Stromortskurve}{}% 37 | |||
| \BOOKMARK [1][-]{section.38}{Widerstandstransformation}{}% 38 | |||
| \BOOKMARK [1][-]{section.39}{Br\374ckenschaltung}{}% 39 | |||
| \BOOKMARK [1][-]{section.40}{Wechselstrombr\374cke}{}% 40 | |||
| \BOOKMARK [1][-]{section.41}{Wechselstrombr\374cke}{}% 41 | |||
| \BOOKMARK [1][-]{section.42}{\334bertrager im Leerlauf}{}% 42 | |||
| \BOOKMARK [1][-]{section.43}{\334bertrager mit kapazitiver Last}{}% 43 | |||
| \BOOKMARK [1][-]{section.44}{\334bertrager mit Verbindung zum Eingang}{}% 44 | |||
| \BOOKMARK [1][-]{section.45}{Impedanzmatrix}{}% 45 | |||
| \BOOKMARK [1][-]{section.46}{Netztransformator}{}% 46 | |||
| \BOOKMARK [1][-]{section.47}{3-Phasen Spannungssystem}{}% 47 | |||
| \BOOKMARK [1][-]{section.48}{3-Phasen System mit unsymmetrischem Verbraucher}{}% 48 | |||
| \BOOKMARK [1][-]{section.49}{Strangstr\366me 3-Phasen System mit unsymmetrischem Verbraucher}{}% 49 | |||
| \BOOKMARK [1][-]{section.50}{Resonanzfrequenz Zweipol}{}% 50 | |||
| \BOOKMARK [1][-]{section.51}{RLC-Reihenschwingkreis}{}% 51 | |||
| \BOOKMARK [1][-]{section.52}{Effektivwert und Klirrfaktor}{}% 52 | |||
| \BOOKMARK [1][-]{section.53}{Klirrfaktor}{}% 53 | |||
| \BOOKMARK [1][-]{section.54}{Momentanspannung}{}% 54 | |||
| \BOOKMARK [1][-]{section.55}{Nichtlinears Bauelement}{}% 55 | |||
| \BOOKMARK [1][-]{section.56}{Wirkleistung Zweipol}{}% 56 | |||
| \BOOKMARK [1][-]{section.57}{Ringspule}{}% 57 | |||
| \BOOKMARK [1][-]{section.58}{Netzwerk Wirk- und Blindanteil}{}% 58 | |||
| \BOOKMARK [1][-]{section.59}{Gleichungen in Matrizenschreibweise}{}% 59 | |||
| \BOOKMARK [1][-]{section.60}{Ergebnisse}{}% 60 | |||
BIN
Main_ET2_V2.pdf
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Main_ET2_V2.sin12a53.gnuplot
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| set table "Main_ET2_V2.sin12a53.table"; set format "%.5f" | |||
| set samples 100; plot [x=0.6:2] 2*sin(0.5*3.14*x) | |||
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Main_ET2_V2.sin12a54.gnuplot
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| set table "Main_ET2_V2.sin12a54.table"; set format "%.5f" | |||
| set samples 100; plot [x=2.6:4] 2*(sin(0.5*3.14*x)) | |||
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Main_ET2_V2.sin12a55.gnuplot
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| set table "Main_ET2_V2.sin12a55.table"; set format "%.5f" | |||
| set samples 100; plot [x=4.6:6] 2*sin(0.5*3.14*x) | |||
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Main_ET2_V2.sin12a56.gnuplot
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| set table "Main_ET2_V2.sin12a56.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:4] 2*sin(0.5*3.14*x) | |||
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Main_ET2_V2.sin19_7.gnuplot
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| set table "Main_ET2_V2.sin19_7.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:8] sin(1.047*x) | |||
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Main_ET2_V2.sina.gnuplot
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| set table "Main_ET2_V2.sina.table"; set format "%.5f" | |||
| set samples 100; plot [x=0.6:2] 2*sin(.5*3.14*x) | |||
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Main_ET2_V2.sinb.gnuplot
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| set table "Main_ET2_V2.sinb.table"; set format "%.5f" | |||
| set samples 100; plot [x=0.6:2] 2*sin(0.5*3.14*x) | |||
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Main_ET2_V2.sinc.gnuplot
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| set table "Main_ET2_V2.sinc.table"; set format "%.5f" | |||
| set samples 100; plot [x=2.6:4] 2*(sin(0.5*3.14*x)) | |||
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Main_ET2_V2.sind.gnuplot
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| set table "Main_ET2_V2.sind.table"; set format "%.5f" | |||
| set samples 100; plot [x=4.6:6] 2*sin(0.5*3.14*x) | |||
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Main_ET2_V2.synctex.gz
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Main_ET2_V2.tau.gnuplot
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| @@ -0,0 +1,2 @@ | |||
| set table "Main_ET2_V2.tau.table"; set format "%.5f" | |||
| set samples 25; plot [x=0:10] 6*(exp(-x/2)) | |||
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Main_ET2_V2.tex
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| \input{ET2_Style} | |||
| \newcommand{\myVersion}{V2.1} | |||
| % 2008- v1.0 initial version | |||
| % 2013-09-23 v2.0 verbesserte Ausgabe 2013 | |||
| % 2014-03-17 v2.1 added Blatt und Aufgabennummer in Überschriften, manuelle Kopfzeilenbeschriftung entfernt | |||
| % Abstand nach Nummerierung in TOC anpassen | |||
| \usepackage{tocstyle} | |||
| \settocfeature{spaceafternumber}{0.5cm} | |||
| \newcommand\toPrint{Lösung}% print Lösung (Lösung) ein (xLösung) druckt nur die Aufgaben | |||
| %\newcommand\toPrint{xLösung}% print Lösung (Lösung) ein (xLösung) druckt nur die Aufgaben | |||
| \newcommand{\ProfName}{} %% Watermark Text | |||
| \renewcommand{\ProfName}{Prof.\,Dr.\,C.\,Niebler} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,T.\,Giesler} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,G.\,Sztefka} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,R.\,Janker} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,M.\,Zwanger} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,W.\,Kühnel} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,S.\,May} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,M.\,Chowanetz} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,S.\,Walter} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,J.\,Wohlrab} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,A.\,Eizenhöfer} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,G.\,Siegmund} | |||
| % \renewcommand{\ProfName}{Dipl.-Ing.\,P.\,Wiegner} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,K.\,Schmidt} | |||
| % \renewcommand{\ProfName}{Prof.\,Dr.\,B.\,Klehn} | |||
| % \renewcommand{\ProfName}{Dr.\,H.\,Brünner} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{\input{ET2_Wasser}}{} | |||
| %Anlgegen der Zaehlervariablen | |||
| \newcounter{blatt} | |||
| \newcounter{aufgabe} | |||
| % Vorbelegung ab welchem Blatt die Nummerierung beginnt. | |||
| \setcounter{blatt}{11} | |||
| \setcounter{aufgabe}{0} | |||
| % Befehle zum Ausführen einer automatischen Zählung. wenn ein neues Blatt beginnt, muss die Aufgabenzählung wieder bei 1 beginnen. | |||
| \newcommand{\Aufgabennr}{\arabic{blatt}/\arabic{aufgabe}} | |||
| \newcommand{\incrementAufgabennr}{\stepcounter{aufgabe}} | |||
| \newcommand{\incrementBlatt}{\stepcounter{blatt}\setcounter{aufgabe}{0}} | |||
| \renewcommand{\thesection}{\Aufgabennr{}} % Fortlaufender Zähler \arabic{section} | |||
| % % % \usepackage{draftwatermark} | |||
| \begin{document} | |||
| \graphicspath{{Grafik/}{../}} | |||
| % % % % \SetWatermarkText{niebler} | |||
| % % % % \SetWatermarkText{chowanetz} | |||
| % % % % \SetWatermarkText{giesler} | |||
| % % % % \SetWatermarkText{sztefka} | |||
| % % % \SetWatermarkText{ } | |||
| % % % % \SetWatermarkText{ revision } | |||
| % % % % | |||
| % % % \SetWatermarkAngle{75} | |||
| % % % % \SetWatermarkColor[rgb]{0.95,0.95,0.95} | |||
| % % % \SetWatermarkLightness{.98} %% dark=0 light=1 %% alternativ bei Grau | |||
| % % % \SetWatermarkFontSize{5cm} | |||
| % % % \SetWatermarkScale{2.5} | |||
| % %\part{Übungsaufgaben ET2} | |||
| % \incrementAufgabennr\input{ET2_Deckblatt_v2} %% vorschlag draft erweiterung | |||
| %\ifthenelse{\equal{\toPrint}{Lösung}}{\input{ET2_Deckblatt}{\input{ET2_Deckblatt_A}} | |||
| \input{ET2_Deckblatt} | |||
| \incrementBlatt | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 12 -- Aufgabe 1}} %% Blatt 12 komplett Revision 19. 05. 2013 | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{Blatt 12 -- Aufgabe 1 -- 2}}{\chead{Blatt 12 -- Aufgabe 1}} | |||
| \incrementAufgabennr\input{ET2_L_B12_A1} % Blitzableiter | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 12 -- Aufgabe 2}} | |||
| \incrementAufgabennr\input{ET2_L_B12_A2} % Drahtschleife | |||
| %\chead{Blatt 12 -- Aufgabe 3} | |||
| % \ifthenelse{\equal{\toPrint}{Lösung}}{\chead{Blatt 12 -- Aufgabe 3}}{} | |||
| \incrementAufgabennr\input{ET2_L_B12_A3} % Metallstab | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{ \chead{Blatt 12 -- Aufgabe 4}} | |||
| % \ifthenelse{\equal{\toPrint}{Lösung}}{}{\chead{Blatt 12 -- Aufgabe 4 -- 5}} | |||
| \incrementAufgabennr\input{ET2_L_B12_A4} % Spannungsverlauf | |||
| %\chead{Blatt 12 -- Aufgabe 5} | |||
| \incrementAufgabennr\input{ET2_L_B12_A5} % Phasenanschnitt | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 12 -- Aufgabe 6}} | |||
| % \ifthenelse{\equal{\toPrint}{Lösung}}{xxxx}{\chead{Blatt 12 -- Aufgabe 6}} | |||
| \incrementAufgabennr\input{ET2_L_B12_A6} % Rechteckspannung | |||
| \incrementBlatt | |||
| %\chead{Blatt 13 -- Aufgabe 1} %% Blatt 16 komplett Revision 19. 05. 2013 | |||
| \incrementAufgabennr\input{ET2_L_B13_A1} % Scheinersatzwiderstände | |||
| %\chead{Blatt 13 -- Aufgabe 2} | |||
| \incrementAufgabennr\input{ET2_L_B13_A2} % Verbraucherleistung | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 13 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B13_A3} % Blindleistungskompensation | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 13 -- Aufgabe 4}} | |||
| \incrementAufgabennr\input{ET2_L_B13_A4} % Energieübertragung | |||
| %\chead{Blatt 13 -- Aufgabe 5} | |||
| \incrementAufgabennr\input{ET2_L_B13_A5} % Wechselstrommotor | |||
| %\chead{Blatt 13 -- Aufgabe 6} | |||
| \incrementAufgabennr\input{ET2_L_B13_A6} % Parallelschaltung von $L$ und $C$ | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 13 -- Aufgabe 7}} | |||
| \incrementAufgabennr\input{ET2_L_B13_A7} % Werte $R_L$ und $L$ einer Spule | |||
| %\chead{Blatt 14 -- Aufgabe 1} | |||
| \incrementBlatt | |||
| \incrementAufgabennr\input{ET2_L_B14_A1} % Zeigerdiagramm | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 14 -- Aufgabe 2}} | |||
| \incrementAufgabennr\input{ET2_L_B14_A2} % Gesamtwiderstand | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 14 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B14_A3} % Brückenschaltung | |||
| %\chead{Blatt 14 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B14_A4} % Zeigerdiagramm Netzwerk | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 14 -- Aufgabe 5}} | |||
| \incrementAufgabennr\input{ET2_L_B14_A5} % Blind- Wirk- und Scheinleistung | |||
| \incrementBlatt | |||
| %\chead{Blatt 15 -- Aufgabe 1} | |||
| \incrementAufgabennr\input{ET2_L_B15_A1} % Komplexe Wechselstromrechnung Netzwerk Strom | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 15 -- Aufgabe 2}} | |||
| \incrementAufgabennr\input{ET2_L_B15_A2} % Übergang Zeitabhängige zu Komplexe Größen | |||
| %\chead{Blatt 15 -- Rekapitulieren} | |||
| \input{ET2_L_B15_A2rekap} % An Tafel rekapitulieren | |||
| %\chead{Blatt 15 -- Aufgabe 3} | |||
| \incrementAufgabennr\input{ET2_L_B15_A3} % Leitwert | |||
| %\chead{Blatt 15 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B15_A4} % Strom L-R-C | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| %\chead{Blatt 15 -- Aufgabe 5} | |||
| \incrementAufgabennr\input{ET2_L_B15_A5} % Überlagerungsmethode | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 15 -- Aufgabe 6}} | |||
| \incrementAufgabennr\input{ET2_L_B15_A6} % Momentan Leistung | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 16 -- Aufgabe 1}} %% Blatt 16 komplett Revision 20. 05. 2013 | |||
| \incrementBlatt | |||
| \incrementAufgabennr\input{ET2_L_B16_A1} % CLR Netzwerk | |||
| \input{ET2_L_B16_A1a} | |||
| %\chead{Blatt 16 -- Aufgabe 2} | |||
| \incrementAufgabennr\input{ET2_L_B16_A2} % Wirkleistung | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{ \chead{Blatt 16 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B16_A3} % Abgebbare Wirkleistung | |||
| %\chead{Blatt 16 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B16_A4} % Wirkleistung der Spannungsquelle | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 16 -- Aufgabe 5}} | |||
| \incrementAufgabennr\input{ET2_L_B16_A5} % Dualitatskonstanten | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| %\chead{Blatt 16 -- Aufgabe 6} | |||
| \incrementAufgabennr\input{ET2_L_B16_A6} % Dualitätskonstante verlustbehaftete Bauelemente | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 16 -- Aufgabe 7}} | |||
| \incrementAufgabennr\input{ET2_L_B16_A7} % Vierpol Y-Parameter | |||
| %\chead{Blatt 16 -- Aufgabe 8} | |||
| \incrementAufgabennr\input{ET2_L_B16_A8} % Spannung Vierpol | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 17 -- Aufgabe 1}} | |||
| \incrementBlatt | |||
| \incrementAufgabennr\input{ET2_L_B17_A1} % | |||
| %\chead{Blatt 17 -- Aufgabe 2} | |||
| \incrementAufgabennr\input{ET2_L_B17_A2} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 17 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B17_A3} % | |||
| %\chead{Blatt 17 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B17_A4} % | |||
| %\chead{Blatt 17 -- Aufgabe 5} | |||
| \incrementAufgabennr\input{ET2_L_B17_A5} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 17 -- Aufgabe 6}} %% Blatt 17 ab A6 komplett Revision 20. 05. 2013 | |||
| \incrementAufgabennr\input{ET2_L_B17_A6} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 17 -- Aufgabe 7}} | |||
| \incrementAufgabennr\input{ET2_L_B17_A7} % | |||
| %\chead{Blatt 17 -- Aufgabe 8} | |||
| \incrementAufgabennr\input{ET2_L_B17_A8} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{ \chead{Blatt 18 -- Aufgabe 1}} %% Blatt 18 alle komplett Revision 20. 05. 2013 | |||
| \incrementBlatt | |||
| \incrementAufgabennr\input{ET2_L_B18_A1} % | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| %\chead{Blatt 18 -- Aufgabe 2} | |||
| \incrementAufgabennr\input{ET2_L_B18_A2} % | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 18 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B18_A3} % | |||
| %\chead{Blatt 18 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B18_A4} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 18 -- Aufgabe 5}} | |||
| \incrementAufgabennr\input{ET2_L_B18_A5} % | |||
| %\chead{Blatt 18 -- Aufgabe 6} | |||
| \incrementAufgabennr\input{ET2_L_B18_A6} % | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 18 -- Aufgabe 7}} | |||
| \incrementAufgabennr\input{ET2_L_B18_A7} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 18 -- Aufgabe 8}} | |||
| \incrementAufgabennr\input{ET2_L_B18_A8} % | |||
| \incrementBlatt | |||
| %\chead{Blatt 19 -- Aufgabe 1} %% Blatt 19 komplett Revision 19. 05. 2013 | |||
| \incrementAufgabennr\input{ET2_L_B19_A1} % | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 19 -- Aufgabe 2}} | |||
| \incrementAufgabennr\input{ET2_L_B19_A2} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 19 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B19_A3} % | |||
| %\chead{Blatt 19 -- Aufgabe 4} | |||
| \incrementAufgabennr\input{ET2_L_B19_A4} % | |||
| %\chead{Blatt 19 -- Aufgabe 5} | |||
| \incrementAufgabennr\input{ET2_L_B19_A5} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 19 -- Aufgabe 6}} | |||
| \incrementAufgabennr\input{ET2_L_B19_A6} % | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| }{}% | |||
| %\chead{Blatt 19 -- Aufgabe 7} | |||
| \incrementAufgabennr\input{ET2_L_B19_A7} % | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 20 -- Aufgabe 1}} %% Blatt 20 komplett Revision 19. 05. 2013 | |||
| \incrementBlatt | |||
| \incrementAufgabennr\input{ET2_L_B20_A1} % Ringspule | |||
| %\chead{Blatt 20 -- Aufgabe 2} | |||
| \incrementAufgabennr\input{ET2_L_B20_A2} % Netzwerk | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| %\chead{Leerseite} | |||
| \input{Leerblatt} % Leerseite | |||
| }{}% | |||
| % \ifthenelse{\equal{\toPrint}{xLösung}}{\chead{}}{\chead{Blatt 20 -- Aufgabe 3}} | |||
| \incrementAufgabennr\input{ET2_L_B20_A3} % Knotenpotential | |||
| % \SetWatermarkText{ } | |||
| \clearpage | |||
| %\chead{Ergebnisse} | |||
| \renewcommand{\thesection}{} | |||
| \input{ET2_L_Ergebnisse} | |||
| \ifthenelse{\equal{\toPrint}{xLösung}}{} | |||
| {\chead{Anhang} | |||
| \incrementAufgabennr\input{ZeichenHilfe} | |||
| }% | |||
| % % % Diverses | |||
| % %\incrementAufgabennr\input{NONsmoker} | |||
| % %\incrementAufgabennr\input{NI} | |||
| % %\incrementAufgabennr\input{RLC} | |||
| % %\incrementAufgabennr\input{Specials} % Besondere Formatierungen | |||
| % %\incrementAufgabennr\input{Test_A3} | |||
| % %\incrementAufgabennr\input{LinLin}% Lineares Koordinatensystem | |||
| % %\incrementAufgabennr\input{LogLog} % Doppelt Logarithmisches Koordinatensystem | |||
| % %\incrementAufgabennr\input{LogLin} % Einfach y Logarithmisches Koordinatensystem | |||
| % %\incrementAufgabennr\input{LinLog} % Einfach x Logarithmisches Koordinatensystem | |||
| % %\incrementAufgabennr\input{LogLinPhase} % Einfach Logarithmisches Koordinatensystem | |||
| % %\input {ppr} | |||
| % %\input {tikz_axes} | |||
| % %\input {intro} | |||
| % %\input {tikz} | |||
| % %\input {X_3D} | |||
| % %\input {pgf_anleitung} | |||
| % %\incrementAufgabennr\input{ET2_A_B2} | |||
| \end{document} | |||
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Main_ET2_V2.toc
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| \select@language {ngerman} | |||
| \contentsline {section}{\numberline {12/1{}}Blitzableiter}{3}{section.1} | |||
| \contentsline {section}{\numberline {12/2{}}Drahtschleife}{5}{section.2} | |||
| \contentsline {section}{\numberline {12/3{}}Metallstab}{7}{section.3} | |||
| \contentsline {section}{\numberline {12/4{}}Spannungsverlauf}{11}{section.4} | |||
| \contentsline {section}{\numberline {12/5{}}Phasenanschnitt}{13}{section.5} | |||
| \contentsline {section}{\numberline {12/6{}}Rechteckspannung}{15}{section.6} | |||
| \contentsline {section}{\numberline {13/1{}}Scheinersatzwiderst\"ande}{17}{section.7} | |||
| \contentsline {section}{\numberline {13/2{}}Verbraucherleistung}{19}{section.8} | |||
| \contentsline {section}{\numberline {13/3{}}Blindleistungskompensation}{21}{section.9} | |||
| \contentsline {section}{\numberline {13/4{}}Energie\"ubertragung}{23}{section.10} | |||
| \contentsline {section}{\numberline {13/5{}}Wechselstrommotor}{25}{section.11} | |||
| \contentsline {section}{\numberline {13/6{}}Parallelschaltung von L und C}{27}{section.12} | |||
| \contentsline {section}{\numberline {13/7{}}Werte $R_L$ und $L$ einer Spule}{29}{section.13} | |||
| \contentsline {section}{\numberline {14/1{}}Zeigerdiagramm}{31}{section.14} | |||
| \contentsline {section}{\numberline {14/2{}}Gesamtwiderstand}{33}{section.15} | |||
| \contentsline {section}{\numberline {14/3{}}Br\"uckenschaltung}{35}{section.16} | |||
| \contentsline {section}{\numberline {14/4{}}Zeigerdiagramm Netzwerk}{37}{section.17} | |||
| \contentsline {section}{\numberline {14/5{}}Blind- Wirk- und Scheinleistung}{39}{section.18} | |||
| \contentsline {section}{\numberline {15/1{}}Komplexe Wechselstromrechnung Netzwerk Strom}{43}{section.19} | |||
| \contentsline {section}{\numberline {15/2{}}\"Ubergang Zeitabh\"angige zu Komplexen Gr\"o\IeC {\ss }en}{45}{section.20} | |||
| \contentsline {section}{\numberline {15/3{}}Leitwert}{48}{section.21} | |||
| \contentsline {section}{\numberline {15/4{}}Strom L-R-C}{49}{section.22} | |||
| \contentsline {section}{\numberline {15/5{}}\"Uberlagerungsmethode}{51}{section.23} | |||
| \contentsline {section}{\numberline {15/6{}}Momentan Leistung}{53}{section.24} | |||
| \contentsline {section}{\numberline {16/1{}}CLR Netzwerk}{55}{section.25} | |||
| \contentsline {section}{\numberline {16/1{}}Wirkleistung vs. Blindleistung}{57}{section.26} | |||
| \contentsline {section}{\numberline {16/2{}}Wirkleistung}{58}{section.27} | |||
| \contentsline {section}{\numberline {16/3{}}Abgebbare Wirkleistung}{60}{section.28} | |||
| \contentsline {section}{\numberline {16/4{}}Wirkleistung Spannungsquelle}{62}{section.29} | |||
| \contentsline {section}{\numberline {16/5{}}Dualit\"atskonstante}{64}{section.30} | |||
| \contentsline {section}{\numberline {16/6{}}Dualit\"atskonstante verlustbehaftete Bauelemente}{65}{section.31} | |||
| \contentsline {section}{\numberline {16/7{}}Vierpol Y-Parameter}{67}{section.32} | |||
| \contentsline {section}{\numberline {16/8{}}Spannung Vierpol}{69}{section.33} | |||
| \contentsline {section}{\numberline {17/1{}}Stromortskurve}{73}{section.34} | |||
| \contentsline {section}{\numberline {17/2{}}Leitwerts-, Widerstandsortskurve}{77}{section.35} | |||
| \contentsline {section}{\numberline {17/3{}}Ortskurve}{79}{section.36} | |||
| \contentsline {section}{\numberline {17/4{}}Stromortskurve}{81}{section.37} | |||
| \contentsline {section}{\numberline {17/5{}}Widerstandstransformation}{83}{section.38} | |||
| \contentsline {section}{\numberline {17/6{}}Br\"uckenschaltung}{85}{section.39} | |||
| \contentsline {section}{\numberline {17/7{}}Wechselstrombr\"ucke}{87}{section.40} | |||
| \contentsline {section}{\numberline {17/8{}}Wechselstrombr\"ucke}{89}{section.41} | |||
| \contentsline {section}{\numberline {18/1{}}\"Ubertrager im Leerlauf}{91}{section.42} | |||
| \contentsline {section}{\numberline {18/2{}}\"Ubertrager mit kapazitiver Last}{93}{section.43} | |||
| \contentsline {section}{\numberline {18/3{}}\"Ubertrager mit Verbindung zum Eingang}{97}{section.44} | |||
| \contentsline {section}{\numberline {18/4{}}Impedanzmatrix}{99}{section.45} | |||
| \contentsline {section}{\numberline {18/5{}}Netztransformator}{101}{section.46} | |||
| \contentsline {section}{\numberline {18/6{}}3-Phasen Spannungssystem}{103}{section.47} | |||
| \contentsline {section}{\numberline {18/7{}}3-Phasen System mit unsymmetrischem Verbraucher}{105}{section.48} | |||
| \contentsline {section}{\numberline {18/8{}}Strangstr\"ome 3-Phasen System mit unsymmetrischem Verbraucher}{107}{section.49} | |||
| \contentsline {section}{\numberline {19/1{}}Resonanzfrequenz Zweipol}{109}{section.50} | |||
| \contentsline {section}{\numberline {19/2{}}RLC-Reihenschwingkreis}{111}{section.51} | |||
| \contentsline {section}{\numberline {19/3{}}Effektivwert und Klirrfaktor}{113}{section.52} | |||
| \contentsline {section}{\numberline {19/4{}}Klirrfaktor}{115}{section.53} | |||
| \contentsline {section}{\numberline {19/5{}}Momentanspannung}{117}{section.54} | |||
| \contentsline {section}{\numberline {19/6{}}Nichtlinears Bauelement}{119}{section.55} | |||
| \contentsline {section}{\numberline {19/7{}}Wirkleistung Zweipol}{120}{section.56} | |||
| \contentsline {section}{\numberline {20/1{}}Ringspule}{122}{section.57} | |||
| \contentsline {section}{\numberline {20/2{}}Netzwerk Wirk- und Blindanteil}{124}{section.58} | |||
| \contentsline {section}{\numberline {20/3{}}Gleichungen in Matrizenschreibweise}{127}{section.59} | |||
| \contentsline {section}{\numberline {}Ergebnisse}{129}{section.60} | |||
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Main_ET2_V2.u.gnuplot
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| set table "Main_ET2_V2.u.table"; set format "%.5f" | |||
| set samples 100; plot [x=0:5] .62*sin(.5*3.14*x+.96-1.57) | |||
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ZeichenHilfe.tex
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| \clearpage | |||
| \pagestyle{empty} | |||
| Zeichenhilfe:\\ | |||
| \begin{tikzpicture}[thin,scale=1] | |||
| \draw[step=0.5cm,ultra thin,black!50!](-8,-8)grid(8,8); | |||
| \draw[thick](0,-8)--(0,8)(-8,0)--(8,0); | |||
| \fill[blue](0,0)circle(0.075cm); | |||
| \foreach \r in {.25,.75,...,7.75} | |||
| \draw[orange](0,0)circle(\r cm); | |||
| \foreach \r in {.5,1.5,...,7.5} | |||
| \draw[blue](0,0)circle(\r cm); | |||
| \foreach \r in {1,3,...,7} | |||
| \draw[red](0,0)circle(\r cm); | |||
| \foreach \r in {2,4,...,8} | |||
| \draw[red!50!blue](0,0)circle(\r cm); | |||
| \foreach \w in {0,10,...,350} | |||
| \draw[orange](\w:1)--(\w:8); | |||
| \end{tikzpicture} | |||
+ 18
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ohmlogoTH.tex
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| @@ -0,0 +1,18 @@ | |||
| % Reduzierte Papier Größe 6.7*1cm | |||
| \setlength{\paperwidth}{6.7cm} | |||
| \setlength{\paperheight}{1cm} | |||
| \hoffset-81pt | |||
| \voffset-70pt | |||
| \oddsidemargin=0pt | |||
| \topmargin=0pt | |||
| \headheight=0pt | |||
| \headsep = 0pt | |||
| % \fontfamily{phv}\selectfont\tiny % Helvetica Font | |||
| % \fontfamily{phv}\selectfont\footnotesize{ % Helvetica Font | |||
| \xdefinecolor{ohmblue}{rgb}{0,.33,.98} % Logo Farbe | |||
| \begin{tikzpicture}[scale=.33] | |||
| \filldraw[ohmblue](0,0)--(0,.65)--(.45,.65)--(0,1.15)--(0,2.55)--(2.55,2.55)--(2.55,1.15) | |||
| --(2.1,.65)--(2.55,.65)--(2.55,0)--(1.4,0)--(1.4,2)--(1.17,2)--(1.17,0)--(0,0); | |||
| \draw node at(2.7,1.2)[right]{\fontfamily{phv}\fontseries{b}\fontshape{n}\selectfont\tiny\textbf{TECHNISCHE HOCHSCHULE NÜRNBERG}}; | |||
| \draw node at(2.7,.38)[right]{\fontfamily{phv}\fontseries{m}\fontshape{n}\selectfont\tiny {GEORG SIMON OHM}}; | |||
| \end{tikzpicture} | |||
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