I need to write this TL equivalent circuit in Latex using circuitikz but i don't know how to implement the suspension points. Does anybody know how to do it?

enter image description here

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}

enter image description here

You can do away with scopes too by using proper coordinates.

  • +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. – user11232 Mar 13 '15 at 22:18
  • Indeed, that's the perfect solution... – Claudio Fiandrino Mar 14 '15 at 9:09

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