Take the 2-minute tour ×
TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. It's 100% free, no registration required.

When using circuitikz to draw some circuit pictures, I noticed that the terminal polarity labels are incorrect (I believe) for polar capacitors. Consider the following:

\documentclass{article}
\usepackage[american voltages,siunitx]{circuitikz}

\begin{document}
\begin{circuitikz}
   \draw (0,0) to [battery=\SI{9}{V}] (3,0)
               to [pC,v=$v_C$] (6,0);
\end{circuitikz}
\end{document}

When I compile this (circuitikz v2.4) I get the following picture:

polar capacitor

I believe that the rounded terminal should be the negative one and the straight terminal is the positive one. Am I missing anything?

Please forgive me if this is not the right place to post this. I noticed that the package author sometimes posts here, and so I thought that this might be appropriate, and also perhaps I'm not understanding something!

share|improve this question
    
With to [pC,v>=$v_C$] (6,0) I get the inversion. –  egreg Mar 12 '12 at 23:28
    
@egreg: you should post this as an answer ;-) –  Gonzalo Medina Mar 13 '12 at 1:25
add comment

2 Answers

I am not quite sure what you are asking.

If you want to change rounded terminal's polarity always:

 \draw (0,0) to [battery=\SI{9}{V}] (3,0)
               to [pC,v^>=$v_C$] (6,0); %<--- change v= to v^>=

enter image description here

But if you want to change the polarity, then reverse the co-ordinates like this:

\documentclass{article}
\usepackage[american voltages,siunitx]{circuitikz}

\begin{document}
\begin{circuitikz}
  \draw (0,0) to  [battery=\SI{9}{V}] (3,0) ;
  \draw (6,0) to  [pC, v=$v_C$] (3,0) ; % <-----------co-ordinates reversed here
\end{circuitikz}
\end{document}

enter image description here

share|improve this answer
    
Thank you for the suggestions. I guess what I was trying to say is that the default polarity is usually correct in circuitikz (for example, the negative terminal on the battery symbol is the short one), however this seems not to be the case for the polar capacitor. The standard notation is to have the rounded terminal be the positive one, whereas circuitikz seems to do the opposite (unless manually specified as the reverse, as suggested by you, or egreg). –  quickk Mar 14 '12 at 18:15
    
@quickk I did not mean like that. What I was saying is the (marked) positive terminal of a device is connected to the positive terminal of a battery. :) –  Harish Kumar Mar 14 '12 at 21:30
add comment

I believe that it has everything to do with the direction of the greater than(>) or less than (<) symbol. In the case of CircuiTikZ, these inequalities are essentially the head of the arrow, pointing in the direction of the positive terminal—at least for bipolar elements, I'm uncertain about tripolar elements.

Notice that the only [non-textual] difference in the following code occurs in that single character:

\begin{circuitikz}[american voltages]
   \draw (0,0) to [battery, i=$$, v_>=$Potential\;increases$] (2.25,0);
\end{circuitikz}

\begin{circuitikz}[american voltages]
   \draw (0,0) to [battery, i=$$, v_<=$Potential\;decreases$] (2.25,0);
\end{circuitikz}

The workaround you discovered with the coordinates may work, I would suggest avoiding it—especially as your circuits increase in complexity.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.