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Hello friends, could you helpme? I need to draw this circuits but I dont know, thanks for your help.

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  • 2
    Well what have you tried? Maybe provide some MWE? Jul 3 at 3:13
  • My personal suggestion is to go through (in the sense: read and try) the first tutorial in the TikZ manual and then the tutorials in circuitikz manual. You need to dedicate a few hours to it, but it's absolutely usefuls if you are serious about using circuitikz.
    – Rmano
    Jul 3 at 8:31
  • Except possibly the ocirc node, you don't need circuitiks for this. Jul 3 at 14:32
  • Any news? You got three answers ... It is time to decide which one of them the best fulfil your expectation and accept it then (by clicking check mark at top left side of selected answer). As I see, so far you have not accepted any of the answers to your questions. By this you tell us, that no one your question is not solved. Is this true?
    – Zarko
    Jul 6 at 19:20
  • For all, thanks!! Jul 8 at 17:15
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As starting point:

\documentclass{article}
\usepackage{circuitikz}

\begin{document}

\begin{tikzpicture}
\draw   (0,0)       to [short,o-] ++ (1,0) node[right] (p) {$p$}
        (p.east) -- ++ (1,0) node[right] (q) {$q$}
        (q.east)    to [short,-o] ++ (1,0);
\end{tikzpicture}


\begin{tikzpicture}
\draw   (0,0)   to [short,o-] ++ (1,0) coordinate (a)
        (a) |- ++ (1, 0.5) node[right] (p) {$p$}
        (p.east) -| ++ (1,-0.5) coordinate (b) 
        (b) to [short,-o] ++ (1,0)
        (a) |- ++ (1,-0.5) node[right] (q) {$q$}
        (q) -| (b);
\end{tikzpicture}

\end{document}

enter image description here

Edit: Considered is @rmano comment.

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  • NIce, +1, --- and I think you found a bug in circuitikz, because your code (which is ok) gives a warning in newer TikZ
    – Rmano
    Jul 3 at 10:50
  • @Rmano, you are right. Since warning is somewhere on the middle of ˙.log file, I didn't notice it.. It complain about node coordinates when construct to [short,-o] is used. bu not at -- : Package pgf Warning: Returning node center instead of a point on node border. D id you specify a point identical to the center of node ``q''? on input line 9. Thank you very much!
    – Zarko
    Jul 3 at 11:20
  • Yes, that's the bug ;-) see github.com/circuitikz/circuitikz/pull/562.
    – Rmano
    Jul 3 at 12:03
  • Anyway, if you want that the branch of the second diagram become symmetrical, you should use ` (p.east) -| ++ (1,-0.5) coordinate (b)` (relative coordinates are computed from the center of the node even if then the border anchor is used).
    – Rmano
    Jul 3 at 12:04
  • @Rmano, you are right. This should be used at relative coordinates. Using it, also warning disappear. So there is no bug?
    – Zarko
    Jul 3 at 12:14
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An alternate way of doing the same. It's mostly tikz. The node shape node[ocirc]{} from circuitikz is used.

\documentclass[border=3mm]{standalone}
\usepackage{circuitikz}
\begin{document}
\begin{circuitikz}
\draw
(0,0)node[ocirc]{} 
-- node[fill=white]{$p$} ++(2,0)
-- node[fill=white]{$q$} ++(2,0)node[ocirc]{}

(0,-2)node[ocirc]{} 
-| ++(1,0.5) 
-- node[fill=white]{$p$} ++(2,0) 
|- ++(1,-0.5)node[ocirc]{}
-| ++(-1,-0.5) 
-- node[fill=white]{$q$} ++(-2,0)
-- ++(0,0.5)
;
\end{circuitikz}
\end{document}

enter image description here

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