Why does the resistor have a straight line?

Question

I'm making the next circuit:

based on the next one

but the resistor has a straight line across. I understand that I am giving the command but I don't understand where.

And by the way, can someone please explain how to work with the nodes, the lines, the cursors, etc? It takes me a lot of work to have done codes like this and perhaps I don't fully grasp if this is working because it's correct or by luck.

\documentclass[tikz,border=3.14]{standalone}
\usepackage{pgf,tikz,pgfplots}
\usepackage[american]{circuitikz}
\usepgfplotslibrary{fillbetween}

\begin{document}
\begin{tikzpicture}

\begin{circuitikz} \draw
 (0,0) node[anchor=east]{$v^{-}_{N}$}
  to[short, o-*] (1,0)
  to[R, l=$R_I$, *-*] (1,2)
  %to [R, v=$v_x$, l=$10\ohm$] (3,2)
 % to[short] (4,2) to[cV, i=$\frac{\siemens}{5}v_x$, *-*] (4,0)
 %  to[short] (3,0) to[R, l=$5\ohm$, *-*] (3,2)

% (3,0) -- (1,0)
 (1,2) to[short, o-*] (0,2)
  node[anchor=east]{$v^{+}_{P}$}
 (3,1) node[anchor=east,xshift=3cm]{$v_O$}  to[R,l=$R_O$, ] (3.5,1) 
  to[short,] (2,1) to[cV, i=$\frac{}{}v_x$, o-] (2,-1) -- (2,-1) node [] {}
 
%  to[short, -] (2,-1)
% (2,0) node [-] {}
  to node[ground]{}(2,-0.5)







;
%    \draw  (2,1) -- node[sloped,currarrow,pos=1] {}  (3,1) ;
\end{circuitikz}



\end{tikzpicture}
\end{document}

Answer 1

For the first image:

\documentclass[border=3.14]{standalone}
\usepackage[american]{circuitikz}

\begin{document}
    \begin{circuitikz} 
\draw   (0,0)   node[left]  {$v^{-}_{N}$}
                to [short,  o-*] ++ ( 1,0) coordinate (aux)
                to [R=$R_I$, -*] ++ ( 0,2)
                to [short,   -*] ++ (-1,0)
                node[left]  {$v^{+}_{P}$}
        (aux)   ++ (3,1)  node[right] {$v_o$}
                to [R,a=$R_0$,o-]++ (-2,0)
                to [cV, l=$v_x$] ++ (0,-2) ++ (0,0.4)
                node[ground]{};
\end{circuitikz}
\end{document}

Note: I didn't investigate, why you have line over resistor. Apparently you draw it. Your code is unnecessary complicated (to me), so it was simpler that I redraw circuit from scratch.

It is not clear, if you like to get a solution for the second image too. From above solution to it is quite simple way: by use of fit library you can draw triangle in the background of above image.

Edit: Code for image should be self-explanatory :-).

Circuit is drawn in two loops. Coordinate for dipoles are determined relatively to beginning of loops ((0,0) for input and (aux) for output) using syntax determined in tikz (on which circuitikz is based) ++. For example \draw (0,0) to [short, o-*] ++ (1,0) means draw line form (0,0) to the right for 1cm with circle on beginning of line and black circle on the end, and (aux) ++ (3,1) is determined a coordinate, which is apart from aux for 3cm to left and 1cm above.

For more details see `TikZ & PGF manual (v3.1.9a), section 13.4 Relative and Incremental Coordinates, page 146.

Edit (1): Improved MWE: added output voltage $v_o$. Edit (2): Considered @Rmano comment.