如何将 circuitkz 中的两幅图像之间的箭头置于中心？

Question

搜索该问题我只找到图片的答案......

我想将这些箭头移到这里：

这里我不仅想移动箭头，还想移动方程式。可以吗？

这里有两个代码：

\documentclass[11pt]{article}

\usepackage[margin=1in]{geometry}
\usepackage{graphicx}
\usepackage[english]{babel}
\usepackage{circuitikz}
\usepackage{tikz}                     
\usetikzlibrary{shapes.geometric, arrows}

\begin{center}
        \begin{circuitikz}[american voltages]
        \ctikzset { label/align = straight }
        \draw (0,0) 
        to[R, l=$R_c$, o-o] (4,0)
        to[R, l=$R_a$, o-o] (2,-3)
        to[R, l=$R_b$, o-o] (0,0)
        {[anchor=east] (0,0) node {a} [anchor=west] (4,0) node {b} [anchor = north] (2,-3) node {c}};
        \end{circuitikz}
        $\hspace{0.25in} \displaystyle \Leftrightarrow \hspace{0.25in}$
        \begin{circuitikz}[american voltages]
        \ctikzset { label/align = straight }
        \draw (0,0) 
        to[R, l=$R_1$, o-*] (2,-1)
        to[R, l=$R_2$, *-o] (4,0) (2,-1)
        to[R, l=$R_3$, *-o] (2,-3)
        {[anchor=east] (0,0) node {a} [anchor=west] (4,0) node {b} [anchor = north] (2,-3) node {c}};
        \end{circuitikz}
        \end{center}

\begin{figure}[h!]
        \begin{center}
        \begin{circuitikz}[american voltages]
        \draw (0,0) 
        to[V,v=$V$] (2,0) % The voltage source
        to[short,-o](2.75,0);
        \draw[thick](2.78,0)-- +(30:0.46);
        \draw(3.25,0)to[short,o-](4,0)
        to[short, i=$i$] (4,-2)
        to[L,l=$L$] (2,-2)
        to[R,l=$R$] (0,-2)
        to[short] (0,0) ;
        \end{circuitikz}
        $\hspace{0.25in} \displaystyle \Leftrightarrow \hspace{0.25in}
        L \frac{di}{dt} + Ri = V$ (switch closed)
        \end{center}
        \end{figure}

Answer 1

,baseline=(current bounding box.center)在选项中使用circuitikz

\documentclass[11pt]{article}

\usepackage[margin=1in]{geometry}
\usepackage{graphicx}
\usepackage[english]{babel}
\usepackage{circuitikz}
\usepackage{tikz}
\usetikzlibrary{shapes.geometric, arrows}
\begin{document}
\begin{center}
        \begin{circuitikz}[american voltages,baseline=(current bounding box.center)]
        \ctikzset { label/align = straight }
        \draw (0,0)
        to[R, l=$R_c$, o-o] (4,0)
        to[R, l=$R_a$, o-o] (2,-3)
        to[R, l=$R_b$, o-o] (0,0)
        {[anchor=east] (0,0) node {a} [anchor=west] (4,0) node {b} [anchor = north] (2,-3) node {c}};
        \end{circuitikz}
        $\hspace{0.25in} \displaystyle \Leftrightarrow \hspace{0.25in}$
        \begin{circuitikz}[american voltages,baseline=(current bounding box.center)]
        \ctikzset { label/align = straight }
        \draw (0,0)
        to[R, l=$R_1$, o-*] (2,-1)
        to[R, l=$R_2$, *-o] (4,0) (2,-1)
        to[R, l=$R_3$, *-o] (2,-3)
        {[anchor=east] (0,0) node {a} [anchor=west] (4,0) node {b} [anchor = north] (2,-3) node {c}};
        \end{circuitikz}
        \end{center}

\begin{figure}[h!]
        \begin{center}
        \begin{circuitikz}[american voltages,baseline=(current bounding box.center)]
        \draw (0,0)
        to[V,v=$V$] (2,0) % The voltage source
        to[short,-o](2.75,0);
        \draw[thick](2.78,0)-- +(30:0.46);
        \draw(3.25,0)to[short,o-](4,0)
        to[short, i=$i$] (4,-2)
        to[L,l=$L$] (2,-2)
        to[R,l=$R$] (0,-2)
        to[short] (0,0) ;
        \end{circuitikz}
        $\hspace{0.25in} \displaystyle \Leftrightarrow \hspace{0.25in}
        L \frac{di}{dt} + Ri = V$ (switch closed)
        \end{center}
        \end{figure}
        \end{document}