# Circuitikz: Electrical diagram

Below is what I have so far

\documentclass[convert = false]{standalone}
\usepackage[american]{circuitikz}
\usepackage[utf8]{inputenx}%  http://ctan.org/pkg/inputenx
% Euler for math | Palatino for rm | Helvetica for ss | Courier for tt
\renewcommand{\rmdefault}{ppl}% rm
\usepackage[scaled]{helvet}% ss //  http://ctan.org/pkg/helvet
\usepackage{courier}% tt // http://ctan.org/pkg/courier
\usepackage{eulervm}  %  http://ctan.org/pkg/eulervm
% a better implementation of the euler package (not in gwTeX)
\normalfont%
\usepackage[T1]{fontenc}%  http://ctan.org/pkg/fontenc
\usepackage{textcomp}%  http://ctan.org/pkg/textcomp

\begin{document}
\begin{circuitikz}[scale = 2]
\draw                                    (1, 1.5)
to[short]                          (2, 1.5)
to[open, v^ = $y(t)$, o-o]         (2, 0)
to[short]                          (1, 0);
%
\draw                                    (-2, 0)
to[short]                          (1, 0)
to[C, l_ = $C$, *-*]               (1, 1.5)
to[L = $L$]                        (-1, 1.5)
to[R = $R$]                        (-2, 1.5);

\end{circuitikz}
\end{document}


Produces

I can't figure out how to add the C = 1<\farad> as well as for R and L. Additionally, I don't know how to complete the circuit. Below you will see the circuit I am trying to create.

Edit 1:

I have been able to complete the circuit diagram.

1. How do I switch the american voltage source + and -?

2. Also, how do I move x(t) to the left hand side?

3. Can I move L and R so that they are above? Apparently L, l_ moves it above. I don't know why since to me l should be left and underscore should mean below.

\documentclass[convert = false]{standalone}
\usepackage[american]{circuitikz}
\usepackage[utf8]{inputenx}%  http://ctan.org/pkg/inputenx
% Euler for math | Palatino for rm | Helvetica for ss | Courier for tt
\renewcommand{\rmdefault}{ppl}% rm
\usepackage[scaled]{helvet}% ss //  http://ctan.org/pkg/helvet
\usepackage{courier}% tt // http://ctan.org/pkg/courier
\usepackage{eulervm}  %  http://ctan.org/pkg/eulervm
% a better implementation of the euler package (not in gwTeX)
\normalfont%
\usepackage[T1]{fontenc}%  http://ctan.org/pkg/fontenc
\usepackage{textcomp}%  http://ctan.org/pkg/textcomp

\begin{document}
\begin{circuitikz}[scale = 2]
\draw                                           (1, 1.5)
to[short]                                 (2, 1.5)
to[open, v^ = $y(t)$, *-*]                (2, 0)
to[short]                                 (1, 0);
%
\draw                                           (-1.5, 0)
to[short]                                 (1, 0)
to[C, l_ = $C$]                           (1, 1.5)
to[L = $L$]                               (-.5, 1.5)
to[R = $R$]                               (-1.5, 1.5)
to[american voltage source, v^ = $x(t)$]  (-1.5, 0);
\end{circuitikz}
\end{document}


Edit 2:

So I added \usepackage{siunitx} but the below doesn't work

• to[C, l_ = $C = \SI{1}{\farad}$]
• to[C, l_ = $C$= 1<\farad>]

1. I defined a new style for a component similar to american voltage source, but having the + and - inverted. This style is called american voltage source inv.

2. Instead of v^= $x(t)$, use v_= $x(t)$.

3. You've answered it yourself: l_=<label>.

4. Use an extra pair de braces to enclose the inner equal sign, as in l_=${R=\SI{1}{\ohm}}. The code: \documentclass{article} \usepackage[american]{circuitikz} \usepackage[utf8]{inputenx}% http://ctan.org/pkg/inputenx % Euler for math | Palatino for rm | Helvetica for ss | Courier for tt \renewcommand{\rmdefault}{ppl}% rm \linespread{1.05}% Palatino needs more leading \usepackage[scaled]{helvet}% ss // http://ctan.org/pkg/helvet \usepackage{courier}% tt // http://ctan.org/pkg/courier \usepackage{eulervm} % http://ctan.org/pkg/eulervm % a better implementation of the euler package (not in gwTeX) \normalfont% \usepackage[T1]{fontenc}% http://ctan.org/pkg/fontenc \usepackage{textcomp}% http://ctan.org/pkg/textcomp \usepackage{siunitx} \makeatletter \def\pgf@circ@vsourceaminv@path#1{\pgf@circ@bipole@path{vsourceAMinv}{#1}} \tikzset{american voltage source inv/.style = {\circuitikzbasekey, /tikz/to path=\pgf@circ@vsourceaminv@path, \circuitikzbasekey/bipole/is voltage=true, v=#1}} \pgfcircdeclarebipole{}{\ctikzvalof{bipoles/vsourceam/height}}{vsourceAMinv}{\ctikzvalof{bipoles/vsourceam/height}}{\ctikzvalof{bipoles/vsourceam/width}}{ \pgfsetlinewidth{\pgfkeysvalueof{/tikz/circuitikz/bipoles/thickness}\pgfstartlinewidth} \pgfpathellipse{\pgfpointorigin}{\pgfpoint{0}{\pgf@circ@res@up}}{\pgfpoint{\pgf@circ@res@left}{0}} \pgftext[bottom,rotate=90,y=\ctikzvalof{bipoles/vsourceam/margin}\pgf@circ@res@down]{$-$} \pgftext[top,rotate=90,y=\ctikzvalof{bipoles/vsourceam/margin}\pgf@circ@res@up]{$+$} \pgfusepath{draw} } \makeatother \begin{document} \begin{circuitikz}[scale = 2] \draw (1, 1.5) to[short] (2, 1.5) to[open, v^ =$y(t)$, *-*] (2, 0) to[short] (1, 0); % \draw (-1.5, 0) to[short] (1, 0) to[C, l_ =${C=\SI{1}{\farad}}$] (1, 1.5) to[L, l_=${L=\SI{1}{\henry}}$] (-.5, 1.5) to[R ,l_=${R=\SI{1}{\ohm}}$] (-1.5, 1.5) to[american voltage source inv,v_=$x(t)$] (-1.5, 0); \end{circuitikz} \end{document}  I am sorry, I wanted to provide a very short solution to the figure that you originally wanted using circuitikz. Hence the code below. Plus the cute inductors that really make it good. \documentclass[border=1mm]{standalone} \usepackage[american,cuteinductors]{circuitikz} \begin{document} \begin{circuitikz} \draw (0,0) to [V,l=$x(t)$](0,2) to [R,l=R{=}$1\Omega$](2,2) (2,2) to [L,l=L{=}$1H$](5,2) (5,2) to [C,l=C{=}$1F$](5,0) (5,2) to [short,-o] (7,2) node[anchor=west] {+} (0,0) to [short,-o] (7,0) node[anchor=west] {-}; \node at (7.5,1) {$y(t)$}; \end{circuitikz} \end{document}  This gives me this :- Just for typing exercise with PSTricks. \documentclass[pstricks,border=12pt]{standalone} \usepackage{pst-circ,siunitx} \begin{document} \begin{pspicture}(-1,0)(10,6) \pnodes(0,0){O}(0,5){A}(3,5){B}(6,5){C}(6,0){D}(9,5){E}(9,0){F} \Ucc[labelInside=2,labeloffset=1](O)(A){$x(t)$} \resistor[dipolestyle=zigzag](A)(B){$R=\SI{1}{\ohm}$} \coil[labeloffset=-.5](C)(B){$L=\SI{1}{\henry}$} \capacitor[labeloffset=-1.5](C)(D){$C=\SI{1}{\farad}$} %\tension[tensionstyle=pm](E)(F){$y(t)$} unavailable \uput[0](E){$+$}\uput[0](F){$-$}\pcline[linestyle=none](E)(F)\naput{$y(t)$} \psset{arrows=-*}\wire(C)(E)\wire(O)(F) \end{pspicture} \end{document}  ## Notes Unfortunately tensionstyle=pm is not available for \tension so we have to draw the tension with \uput[0](E){$+$}\uput[0](F){$-$}\pcline[linestyle=none](E)(F)\naput{$y(t)$}. I don't know why my request (several decades ago in PSTricks mailing list) to provide tensionstyle=pm for tension was ignored. A PSTricks solution: \documentclass{article} \usepackage{pst-circ} \usepackage{siunitx} \begin{document} \begin{pspicture}(-1.05,-0.15)(8.1,3.65) \psset{unit = 0.5} \pnodes(14,0){A}(9,0){B}(0,0){C}(0,3){D}(0,6){E}(4.5,6){F}(9,6){G}(14,6){H} \rput(16,0){$I$} \wire[arrows = *-](A)(C) \uput[0](A){$-$} \Ucc[labeloffset = 1.5](C)(E){$x(t)$} \rput(-0.7,2.3){$-$} \rput(-0.7,3.7){$+$} \resistor[dipolestyle = zigzag](E)(F){$R = \SI{1}{\ohm}$} \coil[labeloffset = 1](F)(G){$L = \SI{1}{\henry}$} \wire[arrows = *-](H)(G) \uput[0](H){$+$} \capacitor[labeloffset = -2](B)(G){$C = \SI{1}{\farad}$} \rput(14,3){$y(t)\$}
\end{pspicture}

\end{document}


Note that pst-circ is used to draw the electrical diagram and siunitx is used to typeset the physical quantities.