绘制具有等效阻抗的电路

Question

您能帮我将下面的代码改为我所附的图吗？

\documentclass{article}
\usepackage{graphicx}
\usepackage{siunitx}
\usepackage[american]{circuitikz}

\begin{document}
\begin{figure}
\centering  
\begin{circuitikz}
\draw (0,3)
to[V,v<=$V_s$] (0,0)
(0,3) to[ R, R=$ \SI{4.5}{\ohm}$] (3,3) 
(3,3) to[ L=$j\SI{1.5}{\ohm}$] (6,3) 
(0,2.2) to[ open, v=$V$] (6.5,2.2) 
(6,3) to [R=$ \SI{25}{\ohm}$] (6,0)
(6,3) to[short] (8,3)
(8,3) to [L=$ j\SI{20}{\ohm}$] (8,0)
(8,3) to[short, -*] (10,3)
(10,0) to[short, *-] (0,0)
(10.2,3) to[ open, v=$V_o$] (10.2,0) 
(6,0) to[short] (0,0);
\end{circuitikz}
\end{figure}

\end{document}

Answer 1

\documentclass[a4paper]{article}
\usepackage[siunitx ]{circuitikz}
\usetikzlibrary{positioning, arrows.meta, calc}
\usetikzlibrary{decorations.pathreplacing,calligraphy}

\begin{document}
\begin{circuitikz}
    \draw 
    (0,3)               to[american voltages, V, v_=$V_s$, ]      (0,0)
    (0,3) coordinate(c) to[ R, R=$ {4.5}$]                              (2,3) 
                        to[ L=$j{1.5}$]                                 (4,3) coordinate(d) 
                        to[short, -*]                                   (4,3)
                        to [R=$ {25}$]                                  (4,0) coordinate(e)
                        to[short, -*]                                   (4,0)
    (4,3)               to[short, -*]                                   (6,3)
                        to [L=$ j{20}$]                                 (6,0) coordinate(f)
                        to[short, -*]                                   (6,0)
    (6,3)               to[short, -o]                                   (8,3) coordinate(g);
    \draw
    (10,1.5) coordinate(a) to [C=$-j{5}$]                           ++ (0,0.3) coordinate(b);
    \draw[dashed]
    (8,3)               to[short]                                       (10,3)
                        to                                              (b);
    \draw[dashed]   
    (a)                 to[short]                                       (10,0)
                        to                                              (8,0);
    \draw   
    (8,0) coordinate(h) to[short, o-]                                   (0,0);
\draw [decorate, decoration={calligraphic brace, mirror,amplitude=8pt}] ($(c.south)+(0.5,-4pt)$) -- node[below=2ex] {$Z_1$} ($(d.south)+(-0.5,-4pt)$);
\draw [decorate, decoration={calligraphic brace, mirror,amplitude=8pt}] ($(e.south)+(0,-4pt)$) -- node[below=2ex] {$Z_2$} ($(f.south)+(0,-4pt)$);
\node[below=1ex] at (8,3) (){+};
\node[above=1ex] at (8,0) (){--};
\node at( $(g)!0.5!(h) $)(){$V_0$};
\end{circuitikz}
\end{document}

Answer 2

对于锻炼：

\documentclass[border=3.141592]{standalone}
\usepackage{siunitx}
\usepackage[american]{circuitikz}
\usetikzlibrary{decorations.pathreplacing,
                calligraphy}
\tikzset{BC/.style args = {#1/#2}{decorate,
        decoration={calligraphic brace, amplitude=2mm,
        pre =moveto, pre  length=2mm,
        post=moveto, post length=2mm,
        raise=#1,
              #2},% for mirroring of brace
        thick,
        pen colour=red}
        }
\ctikzset{bipoles/capacitor/height=0.5,
          bipoles/capacitor/width=0.1,
          }

\begin{document}
    \begin{circuitikz}
\draw   (0,3)   to[V, a=$V_s$]  (0,0)            coordinate (aux0)
        (0,3)   to[R=$\SI{4.5}{\ohm}$]  ++ (2,0)
                to[L=$j\SI{1.5}{\ohm}$] ++ (2,0) coordinate (aux1)
                to[R=$\SI{25}{\ohm}$,*-*]   ++ (0,-3)
        (aux1)  -- ++ (1.5,0)                    coordinate (aux2)
                to[L=$j\SI{20}{\ohm}$,*-*]  ++ (0,-3)
        (aux2)  to[short,-o]            ++ (2, 0) coordinate[label=$a$] (aux3)
                to[open, v=$V_o$]       ++ (0,-3) node[below] {$b$}
                to[short,o-]            (aux0);
\draw[densely dashed]   
        (aux3)  to [short,o-] ++ (1,0) 
                to[C=$-j\SI{5}{\ohm}$]  ++ (0,-3)
                to [short,-o] (aux0 -| aux3);
\draw[BC=3mm/mirror]   
        (0,3) -- node[below=5mm] {$Z_1$}  (aux1); 
\draw[BC=1mm/mirror]   
        ([xshift=-4mm] aux0 -| aux1) -- node[below=3mm] {$Z_2$}  ([xshift=4mm] aux0 -| aux2);
    \end{circuitikz}
\end{document}

附录： 如果您喜欢cute inductance固体电容器的条形图，那么通过考虑下面的@Rmano 评论（@Rmano 非常感谢您的帮助），MWE 可以是：

\documentclass[border=3.141592]{standalone}
\usepackage{siunitx}
\usepackage[american, cute inductors]{circuitikz} % <---
\usetikzlibrary{decorations.pathreplacing,
                calligraphy}
\tikzset{BC/.style args = {#1/#2}{decorate,
        decoration={calligraphic brace, amplitude=2mm,
        pre =moveto, pre  length=2mm,
        post=moveto, post length=2mm,
        raise=#1,
              #2},% for mirroring of brace
        thick,
        pen colour=red},
        }
\ctikzset{bipoles/capacitor/height=0.5,
          bipoles/capacitor/width=0.1,
          }

\begin{document}
    \begin{circuitikz}
\draw   (0,3)   to[V, a=$V_s$]  (0,0)            coordinate (aux0)
        (0,3)   to[R=$\SI{4.5}{\ohm}$]  ++ (2,0)
                to[L=$j\SI{1.5}{\ohm}$] ++ (2,0) coordinate (aux1)
                to[R=$\SI{25}{\ohm}$,*-*]   ++ (0,-3)
        (aux1)  -- ++ (1.5,0)                    coordinate (aux2)
                to[L=$j\SI{20}{\ohm}$,*-*]  ++ (0,-3)
        (aux2)  to[short,-o]            ++ (2, 0) coordinate[label=$a$] (aux3)
                to[open, v=$V_o$]       ++ (0,-3) node[below] {$b$}
                to[short,o-]            (aux0);
% capacitor as capacitorshape node with dashed connections
\path   (aux3)  -- node (c) [right=10mm,
                             capacitorshape, fill=white, solid, rotate=90,
                             label=below:$-j\SI{5}{\ohm}$] {} (aux0 -| aux3);
\draw[densely dashed] 
        (aux3)  to [short,o-] ++ (1,0) to (c.right) 
        (c.left)to (aux0 -| c) to [short,-o] (aux0 -| aux3);
% 
\draw[BC=3mm/mirror]
        (0,3) -- node[below=5mm] {$Z_1$}  (aux1);
\draw[BC=1mm/mirror]
        ([xshift=-4mm] aux0 -| aux1) -- node[below=3mm] {$Z_2$}  ([xshift=4mm] aux0 -| aux2);
    \end{circuitikz}
\end{document}