# Problem with relative coordinates within \draw command

I've got a problem with the following code. I'm trying to draw a circuit using circuitikz. However, the current source should be shifted by 1 unit in the x-direction, so that it is placed parallel to the first inductor of the transformer. I figured it should work with this relative coordinate positioning but there seems to be a problem.

\begin{figure}[h!]
\begin{center}
\begin{circuitikz}
\draw (0,0)
node[transformer core, yscale=1.25] (T) {};
\draw (T.A2)+(-1,0)
to [sI] (T.A1)+(-1,0);
\end{circuitikz}
\caption{My first circuit.}
\end{center}
\end{figure}


You should use something like

\draw (T.A2) -- +(-1,0)
to [sI] ($(T.A1)+(-1,0)$) -- (T.A1);


Start with T.A2, move the pen to left by 1 unit, then draw the current source up to the point ($(T.A1)+(-1,0)$) (this needs calc library) and then connect to T.A1.

Full code:

\documentclass{article}
\usepackage{circuitikz}
\usetikzlibrary{calc}
\begin{document}
\begin{circuitikz}
\draw (0,0)
node[transformer core, yscale=1.25] (T) {};
\draw (T.A2) -- +(-1,0)
to [sI] ($(T.A1)+(-1,0)$) -- (T.A1);
\end{circuitikz}
\end{document} A PSTricks solution using the pst-circ package:

\documentclass{article}

\usepackage{pst-circ}

\begin{document}

\begin{pspicture}(4.5,4)
\pnodes(0.5,0){A}(0.5,4){B}(2.5,4){C}(4.5,4){D}(4.5,0){E}(2.5,0){F}
\transformer(C)(F)(D)(E){}
\wire(A)(F)
\vac(A)(B){}
\wire(B)(C)
\end{pspicture}

\end{document} 