7

circuit

`\documentclass{article}
  \usepackage{graphicx}
  \usepackage[american]{circuitikz}
  \begin{document}
     \begin{figure}
       \centering  
       \begin{circuitikz}
      \draw (0,0)
    to[V,v=$V$] (0,3)        
    to[C=$C_1$] (5,3) 
    to [L=$L_1$] (5,0)
    to[short] (0,0);
   \end{circuitikz}
  \caption{}
   \end{figure}
  \end{document}
7

You can invert the voltage source by changing the counting direction of the voltage using v<=$V$ Your full example will look like:

\documentclass{article}
  \usepackage{graphicx}
  \usepackage[american]{circuitikz}
  \begin{document}
     \begin{figure}
       \centering  
       \begin{circuitikz}
      \draw (0,0)
    to[V,v<=$V$] (0,3)        
    to[C=$C_1$] (5,3) 
    to [L=$L_1$] (5,0)
    to[short] (0,0);
   \end{circuitikz}
  \caption{}
   \end{figure}
  \end{document}

enter image description here

  • this doesn't work for me, still it appears the same way, weather I put the < or not, I'm using texlive – Andrés Alcarraz Dec 12 '17 at 3:17
  • Ditto, this doesn't work for me, even though the docs say it should (and this does). Using the answer above (drawing the other way around) puts the V on the wrong side of the circuit. – aha Jan 25 '18 at 18:41
4

Since version 0.8.3, using v<=$V$ has been removed as a way to change the polarity of a source. The manual recommends using the invert tag on the element:

\begin{center}
    \begin{circuitikz}
        \draw (0,0)
        to[V, v=$V$, invert] (0,3) %Invert the element to achieve required polarity.       
        to[C=$C_1$] (5,3) 
        to[L=$L_1$] (5,0)
        to[short] (0,0);
    \end{circuitikz}
\end{center}

Inverting also keeps the correct voltage/current direction.

PDFLaTeX Rendering

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