Circuitikz - 对齐节点

Question

我在 tikz 中绘制电路并发现了问题。

第一个问题：我无法将垂直打开的节点对齐到左侧。有没有简单的解决方案？

第二个问题：有没有办法获取 T.A1 的 Y 坐标并将其放入另一个坐标中？例如：\draw (0,T.A1) -- (3,T.A1);

\begin{tikzpicture}%[european]
  \ctikzset{transformer L2/.style={inductors/coils=8,inductors/width=1.2}}
  % \ctikzset{quadpoles/transformer/width=1, quadpoles/transformer/height=6}
  \ctikzset{bipoles/capacitor/width=.1}
  \draw (0,0) to [short,o-] (0.5,0) to [/tikz/circuitikz/bipoles/length=1cm,R=$R_z$] (2.5,0) node[transformer core, anchor=A1](T){};
  \draw (T.B1) to [/tikz/circuitikz/bipoles/length=1cm,R=$R_t$] (6.5,0) to [/tikz/circuitikz/bipoles/length=1cm,C=$C_1$] (8,0) to [/tikz/circuitikz/bipoles/length=1cm,D*=$P_1$] (10,0) to [short,-o] (11,0) node[anchor=south west] {A};
  \draw (8,-2) to [/tikz/circuitikz/bipoles/length=1cm,D*=$P_2$] (8,0);
  \draw (10,-2) to [/tikz/circuitikz/bipoles/length=1cm,C=$C_2$] (10,0);
  \draw (6.5,0) to [/tikz/circuitikz/bipoles/length=0.75cm,C=$C'_d$] (6.5,-1) to [/tikz/circuitikz/bipoles/length=.75cm,C=$C''_d$] (6.5,-2);
  \draw (T.B2) to [-] ($(T.B2)+(2,0)$) -| (6.5,-2);
  \draw (T.B2) to [-] ($(T.B2)+(2,0)$) -| (8,-2);
  \draw (T.B2) to [-] ($(T.B2)+(2,0)$) -| (10,-2);
  \draw (10,-2) node[ground](GND){};
  % \draw (and_2.in 2) to[short, -o] ++ (-60pt, 0) coordinate(x3) node[left]{$x_3$};
  \draw (T.A2) to [short,-o] ++(-60pt,0) coordinate(x3);% -| (0,-2);
 
\end{tikzpicture}

Answer 1

有一种语法允许您选择一个坐标，该坐标采用X另一个坐标的值和是另一个坐标的值。语法的(A |- B)含义是X的坐标值A和是的坐标值B。还有，(A -| B)然后使用X的坐标值B。

因此，基于此，我可能会执行以下操作：

\documentclass[border=10pt]{standalone}
\usepackage{circuitikz}

\ctikzset{
    transformer L2/.style={
        inductors/coils=8, 
        inductors/width=1.2,
    },
    quadpoles/transformer core/height=2.5,
    bipoles/capacitor/width=.1,
    bipoles/length=1cm,
    /tikz/short bipole/.style={
        /tikz/circuitikz/bipoles/length=.75cm,
    }
}

\begin{document}
\begin{circuitikz}

  \draw (0,0) to [short, o-] (0.5,0) 
    to[R=$R_z$] (2.5,0) 
    node[transformer core, anchor=A1] (T) {};
  \draw (T.A2) 
    to[short, -o] (T.A2 -| 0,0);
  
  \draw (T.B1) 
    to[R=$R_t$] (6.5,0) 
    to[C=$C_1$] (8,0) 
    to[D*=$P_1$] (10,0) 
    to[short, -o] (11,0) node[anchor=south west] {A};
  \draw (8,0 |- T.A2) 
    to[D*=$P_2$] (8,0);
  \draw (10,0 |- T.A2) 
    node[ground] (GND) {}
    to[C=$C_2$] (10,0);
  \draw (6.5,0) 
    to[short bipole, C=$C'_d$] (6.5,0 |- T.center) 
    to[short bipole, C=$C''_d$] (6.5,0 |- T.A2);
  \draw (T.B2) to (GND.center);
 
\end{circuitikz}
\end{document}