我是这个网站的新手,也是 TeX 的新手。我已经设法绘制了基本的电路,但是附件中的电路超出了我目前的技能水平。希望有人能帮忙。
干杯
答案1
这是其中一种方法。解释分散在代码中。许多线路连接/电路元件分配都使用正交坐标。
代码
\documentclass[border=20pt]{standalone}
\usepackage[american,siunitx]{circuitikz}
\usetikzlibrary{arrows,shapes,calc,positioning}
\begin{document}
\begin{circuitikz}[american]
\draw (0,0) node [scale=1.5,transformer core
] (T){}
(T.A1) node[above] {A1}
(T.A2) node[below] {A2}
(T.B1) node[above left] {B1}
(T.B2) node[below left] {B2}
(T.base) node{};
\path (T.B1) -| ++ (2,1) coordinate (tb1){};% define a coordinate at top right relative to B1
\draw (T.B1) -- ++ (0,1) to[D*,-*] (tb1);
\draw (T.B2) to[D*] ++(2,0)coordinate(tba){} -| (tba |- tb1);
\path (T.B2) |- ++(3,-1) coordinate(tb2){}; % define a coordinate at bottom right relative to B2
\draw (tb2) to[D*,*-] ++(-3,0) -- (T.B2);
\draw (tb2) --(tb2 |- T.B1) to[D*] (T.B1);
% place voltage labels
\draw(T.A1) to[open,v<={$240V_{rms}$,o-o}](T.A2);
\draw($(T.B1)!0.3!(T.B2)$) node[]{$12V_{rms,AC}$}(T.B1);
\draw ($(T.B1)!0.8!(T.B2)$)node[]{$12V_{rms,AC}$}(T.B2);
\draw[thick] ($(T.B1)!0.51!(T.B2)-(1cm,0)$)coordinate[](c){} -- ++ (12.44,0) -| ++ (1,-1) node[ground]{}; % add a gorund to neutral line
% place capacitors and resistor (upper branches)
\draw(6,1) node[](d1){} to [C,l_=$C_1$, *-*] (c -| d1);
\draw(10,1)node[](d2){} to [C,l_=$C_3$, *-*] (c -| d2);
\draw(13,1)node[](d3){} to [R,l=$R_L$] (c -| d3);
% place rectangles
\draw (7.5,0.5) rectangle (8.5,1.5) node[below left= 0.25cm and 0.1cm]{780s};
\draw (7.5,-3.6) rectangle (8.5,-4.6)node[above left=0.25cm and 0.1cm]{790s};
% place capacitors and resistors (lower branches)
\draw(6,-4.15) node[](d4){} to [C,l_=$C_2$, *-*] (c -| d4);
\draw(10,-4.15)node[](d5){} to [C,l_=$C_4$, *-*] (c -| d5);
\draw(13,-4.15)node[](d6){} to [R,l=$R_L$] (c -| d6);
\draw (8,0.5) node[](d7){} to [short,-*] (c -| d7) --(8,-3.6);
% place top and bottom lines
\draw(2,1) -- (7.5,1) (8.5,1) --(11,1) to[short,i^={$I_L$}] (13,1);
\draw(2,-4.15) -- (7.5,-4.15) (8.5,-4.15) --(13,-4.15);
\end{circuitikz}
\end{document}