I need to draw this TL equivalent circuit in LaTeX using circuitikz, but I don't know how to draw the ellipses. Does anybody know how to do it?
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5Welcome to TeX.SX! Please add a minimal working example (MWE) that shows the code you have so far. Providing the code for just the circuit without the suspension points saves us needing to create the part of the diagram that you've already done. – Paul Gessler Mar 13 '15 at 15:11
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No comments again.
\documentclass{article}
\usepackage{circuitikz}
\ctikzset{bipoles/thickness =1}
\begin{document}
\begin{circuitikz}[line width=1pt]
\draw (0,2) to[L,l=$L'$,*-*] (2,2)
(2,0) to[C,l=$C'$,-*] (2,2)
(2,0) to[short,-*] (0,0)
;
\begin{scope}[xshift=2cm]
\draw (0,2) to[L,l=$L'$,*-*] (2,2)
(2,0) to[C,l=$C'$,-*] (2,2)
(2,0) to[short,-*] (0,0)
;
\end{scope}
\begin{scope}[xshift=4cm]
\draw (0,2) to[L,l=$L'$,*-*] (2,2)
(2,0) to[C,l=$C'$,-*] (2,2)
(2,0) to[short,-*] (0,0)
;
\end{scope}
\begin{scope}[xshift=6cm]
\draw (0,2) -- (2,2)node[midway,scale=2,fill=white]{$\cdots$};
\draw (0,0) -- (2,0)node[midway,scale=2,fill=white]{$\cdots$};
\end{scope}
\draw (8,2) to[L,l=$L'$,-*] (10,2) to[short,-*] (11,2)
(10,0) to[C,l=$C'$,-*] (10,2)
(11,0) to[short,*-](10,0) to[short,*-] (8,0)
;
\end{circuitikz}
\end{document}
You can do away with scopes too by using proper coordinates.
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+1, but too obvious with
\cdots.. what about\draw[loosely dotted](0,2) -- (2,2);? – Claudio Fiandrino Mar 13 '15 at 16:01 -
@ClaudioFiandrino Fair point. I was trying to reproduce the OP's picture almost as it is. It has solid lines at left and right edges and I thought by
\cdots, I can dispense with at least one coordinate in the middle. – Mar 13 '15 at 22:18 -