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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| @@ -0,0 +1,180 @@ | |||
| \section{Blind- Wirk- und Scheinleistung} | |||
| Von dem untenstehenden Schaltbild ist gegeben:\\ | |||
| $R_L = X_L = 20\,\ohm \quad R_C=200\,\ohm \quad X_C=-100\,\ohm\quad\uline{U}=230\,\volt\cdot e^{j0\,\degree}$\\ | |||
| \renewcommand{\labelenumi}{\alph{enumi})} | |||
| \begin{enumerate} | |||
| \item Der Eingangswiderstand Z der Schaltung nach Betrag und Phasenwinkel | |||
| \item Sämtliche Ströme und Spannungen nach Betrag und Phasenwinkel | |||
| \item Wirk- Blind- und Scheinleistungsaufnahme der Schaltung | |||
| \item Qualitatives Zeigerdiagramm aller Ströme und Spannungen unter der Annahme, daß sich | |||
| die Gesamtschaltung induktiv verhält. | |||
| \end{enumerate} | |||
| \begin{align*} | |||
| \begin{tikzpicture}[very thick,scale=2] | |||
| \begin{scope}[>=latex,very thick,xshift=0cm,yshift=1cm]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,.0667) [above] {$R_L$}; | |||
| % \draw [->,blue] (.3,-.2)--(.7,-.2)node at(.5,-.2)[below]{\footnotesize$U_{R}$}; | |||
| \end{scope} | |||
| % \begin{scope}[>=latex,very thick,xshift=0cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| % \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_1$}; | |||
| % \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=1cm,yshift=1cm]%Spule - | |||
| \draw (0,0)--(.3,0) (.7,0)--(1,0)node at (.5,.0667) [above] {$X_L$}; | |||
| \fill (.3,-0.0667)rectangle(.7,0.0667); | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=2cm,yshift=0cm,rotate=90]%Widerstand - nach EN 60617 | |||
| \draw (0,0)--(.3,0) (.3,-0.0667)rectangle(.7,0.0667) (.7,0)--(1,0)node at (.5,-.0667) [right] {$R_C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick,xshift=3cm,yshift=0cm,rotate=90]%Kondensator | | |||
| \draw (0,0)--(.475,0) (.475,-.125)--(.475,.125) (.525,-.125)--(.525,.125) (.525,0)--(1,0)node at (.5,-.133) [right] {$X_C$}; | |||
| \end{scope} | |||
| \begin{scope}[>=latex,very thick]%Knotenpunkte | |||
| \draw (-.5,0)--(0,0) (-.5,1)--(0,1) (-.,0)--(3,0)--(3,.2) (.8,1)--(3,1)--(3,.8); | |||
| \draw [->,blue] (-.5,.9)--(-.5,.1)node at(-.5,.5)[right]{$\underline{U}$}; | |||
| \fill (-.5,0)circle(.025) (-.5,1)circle(.025); | |||
| \draw [->,red] (-.4,1.1)--(-.1,1.1) node at (-.25,1.1)[above]{$\underline{I}$}; | |||
| \draw [->,red] (2.1,.9)--(2.1,.7) node at (2.1,.8)[right]{$\underline{I}_{R_C}$}; | |||
| \draw [->,red] (3.1,.9)--(3.1,.7) node at (3.1,.8)[right]{$\underline{I}_C$}; | |||
| \draw [->,black!50!] (.3,.8)--(.7,.8) node at (.5,.8)[below]{$\underline{U}_{RL}$}; | |||
| \draw [->,black!50!] (1.3,.8)--(1.7,.8) node at (1.5,.8)[below]{$\underline{U}_L$}; | |||
| \draw [->,black!50!] node at (-.5,.5)[left]{$\uline{Z} =>$}; | |||
| \draw [|-|,black!50!] (2.,-.2)--(3,-.2) node at (2.5,-.2)[below]{$\uline{Z}_{||}$}; | |||
| \end{scope} | |||
| \end{tikzpicture} | |||
| \end{align*} | |||
| \ifthenelse{\equal{\toPrint}{Lösung}}{% | |||
| \begin{align} | |||
| \intertext{Formeln:} | |||
| e^{j\varphi}&=\cos\varphi+j\sin\varphi \quad\text{Eulersche Formel}\\ | |||
| \uline{Z}&=Z\cdot e^{\pm j\varphi}=R\pm jX\\ | |||
| \cos\varphi&=\frac{R}{Z}\\ | |||
| \sin\varphi&=\frac{X}{Z} | |||
| \end{align} | |||