答案1
作为起点。
\documentclass[margin=3mm]{standalone}
\usepackage[RPvoltages]{circuitikz}
\usepackage{amsmath}
\usepackage{siunitx}
\begin{document}
\begin{circuitikz}[scale=0.8,transform shape]
\draw (0,0) to[sV,l_=$240 \angle{0^\circ} \si{\volt}$,name=vf](0,1.5)to [L,l=$\mathrm{j}0.12\ \si{\ohm}$](0,2.5) to [R=$0.08\ \si{\ohm}$](0,4)to [short,-o](2,4);
\node at (2,4) [above]{A};
\draw (2,4)to [short](3,4) to [R,l=$0.2\ \si{\ohm}$](4.5,4)to[L,l=$\mathrm{j}0.5\ \si{\ohm}$](6,4)to [short,i=$I_A$,-o](6.5,4)to [short](7.5,4) to [short,i=$I_P$](7.5,3.5)to[R,l=$8\ \si{\ohm}$](7.5,2.25)to[L,l=$\mathrm{j}5\ \si{\ohm}$](7.5,1)to [short](7.5,0)to[short](0,0);
\draw[red,thick,rounded corners,-latex] (2,0.5)--(2,3.5)--++(0:4.5)--++(-90:2.9)--++(180:3.75)node[above]{KVL};
\node at (vf.45){$+$};
\draw (8.35,4) to [open,v^=$V_P$] (8.35,0);
\end{circuitikz}
\end{document}
答案2
作为 @ferahfeza 好答案的补充 (+1)。使用的是
siunitx
语法选项- 重新定义
siunitx
复数符号 - 相对坐标
- 重新定义双极子尺寸
- 电气方案略作修改
- 使用的是 CTAN 上的最新版
circuitikz
(版本 1.0.0-pre1(2019/12/22))
\documentclass[margin=3mm]{standalone}
\usepackage[siunitx]{circuitikz}
\sisetup{complex-root-position=before-number,
output-complex-root=j}
\begin{document}
\begin{circuitikz}
\ctikzset{bipoles/length=11mm}
\draw (0,0) to[sV, l_=$240\angle\ang{0}$\,V, name=V] ++ (0,1.5)
node[below left] {$+$}
to[L=j 0.12<\ohm>] ++ (0,1.5)
to[R=0.08<\ohm>] ++ (0,1.5)
to[short,-o] ++ (1,0) coordinate[label=A] (auxA)
%
to[R=0.08<\ohm>] ++ (1.5,0)
to[short,i=$I_A$] ++ (1,0)
to[L=j0.12<\ohm>,-o]++ (1.5,0) coordinate[label=B] (auxB)
%
to[short] ++ (1.0,0)
to[R=8<\ohm>] ++ (0,-2)
to[short, i=$I_p$] ++ (0,-0.5)
to[L=j5<\ohm>] ++ (0,-2)
to[short,-o] ++ (-1.0,0) coordinate[label=below:B'] (auxBB)
%
to[short,-o] ++ (-4,0) coordinate[label=below:A'] (auxAA)
to[short] (0,0);
\draw[<->, shorten >=1mm, shorten <=1mm]
(auxB) -- node[left] {$V_p$} (auxBB);
%
\draw[red,->, thick]
(V.south east) |- ([shift={(0.5,-0.4)}] auxB)
|- ([xshift=9mm] V.south west) node[pos=0.75,above,text=black] {KVL};
\end{circuitikz}
\end{document}