How can I draw the switch in circuitikz?

Like this?

with use of the switch symbol spdt (see circuitikz package documentation, page 108) and use default (american) style of circuits drawing :

\documentclass[border=3.141592mm]{standalone}
\usepackage{circuitikz}

\begin{document}
    \begin{circuitikz}
\node[spdt, rotate=90] (sw) {};
\draw   (sw.in)     to [L={$(L,r)$}] ++ (0,-2)
                    to [R=$R$] ++ (0,-2) coordinate (aux1)
        (sw.out 2)  node[above] {(2)}   to [short] ++ (+1,0) |- (aux1)
        (sw.out 1)  node[above] {(1)}   to [short] ++ (-1,0) coordinate (aux2)
                    to [vsource, a=$E$]    (aux2 |- aux1)
                    to [short] (aux1);
    \end{circuitikz}
\end{document}

For european style, you need to add circuitikz option european:

\begin{circuitikz}[european]

In this case result is the following:

However, if you like to have mix og the both styles, then MWE is:

\documentclass[border=3.141592mm]{standalone}
\usepackage{circuitikz}

\begin{document}
    \begin{circuitikz}
\node[spdt, rotate=90] (sw) {};
\draw   (sw.in)     to [L={$(L,r)$}] ++ (0,-2)
                    to [R=$R$, european] ++ (0,-2) coordinate (aux1)
        (sw.out 2)  node[above] {(2)}   to [short] ++ (+1,0) |- (aux1)
        (sw.out 1)  node[above] {(1)}   to [short] ++ (-1,0) coordinate (aux2)
                    to [vsource, a=$E$]    (aux2 |- aux1)
                    to [short] (aux1);
    \end{circuitikz}
\end{document}

which produce:

And for the label, you have to add curly brackets INSIDE the $: ${(L,r)}$, otherwise the parser mess up. It gives you finally:

\documentclass[border=3.141592mm]{standalone}
\usepackage{circuitikz}

\begin{document}
    \begin{circuitikz}
\node[spdt, rotate=90] (sw) {};
\draw   (sw.in)     to [L=${(L,r)}$] ++ (0,-2)
                    to [R=$R$, european] ++ (0,-2) coordinate (aux1)
        (sw.out 2)  node[above] {(2)}   to [short] ++ (+1,0) |- (aux1)
        (sw.out 1)  node[above] {(1)}   to [short] ++ (-1,0) coordinate (aux2)
                    to [vsource, a=$E$]    (aux2 |- aux1)
                    to [short] (aux1);
    \end{circuitikz}
\end{document}```

here is my contribution

\documentclass[border=2mm]{standalone}
\usepackage[european, cute inductors]{circuitikz}

\begin{document}
    \begin{circuitikz}
\node[spdt, rotate=90] (inter) {};
\draw   (inter.in)
    to [L, l=\mbox{$(L,r)$}] ++ (0,-2)
    to [R, l=$R$] ++ (0,-2) coordinate (aux1)
    (inter.out 2)  node[above] {(2)}
    to [short] ++ (+1,0) |- (aux1)
    (inter.out 1)  node[above] {(1)}   
    to [short] ++ (-1,0) coordinate (aux2)
    to [vsource, a=$E$]    (aux2 |- aux1)
    to [short] (aux1);
    \end{circuitikz}
\end{document}

I used \mbox to write the detail of the dipole L, r.

Being European, I insert the option european into the circuitikz package directly. As Zarko indicates, on the other hand, you must also add the option cute inductors to avoid obtaining a black rectangle as the inductor.