答案1
解决方案的一个可能示例(再次检查电路---它现在没有什么意义!)
\documentclass[border=10pt]{standalone}
\usepackage[siunitx, RPvoltages]{circuitikz}
\begin{document}
\begin{circuitikz}[european]
\draw (0,0) node[en amp] (opamp){}
(opamp.+) -- ++(0,-1) -| (opamp.out)
to[R=\SI{10}{\ohm}, *-] ++(3,0) node[right]{$v_{\mathit{out}}$}
(opamp.-) to[R, l_=\SI{9.87}{\kohm}, *-] ++(0,3) node[cground, rotate=180]{}
(opamp.-) to[short,-*] ++(-1,0) coordinate(divid);
\draw (divid) to[R, l=\SI{988}{\ohm}] ++(0,2) -- ++(-2,0)
to[battery, l_=\SI{5}{V}, invert] ++(0,-4) node[cground](GND){};
\draw (divid) to[R, l_=\SI{364.1}{\ohm}] (divid|-GND) node[cground]{};
\end{circuitikz}
\end{document}
答案2
\documentclass[margin=3mm]{standalone}
\usepackage{siunitx}
\usepackage[european, straightvoltages]{circuitikz}
\usetikzlibrary{positioning}
\ctikzset{
amplifiers/fill=cyan,
resistors/fill=violet,
}
\begin{document}
\begin{circuitikz}[
line width = 0.8pt,
]
\node[op amp] (A1) {\texttt{}};
\draw(A1.out) to [R,l^=10\si{\ohm},*-]++(4,0)node[right]{$v_{out}$};
\draw(A1.out) to++(0,-2)-|(A1.+);
\draw(A1.-)to [R,l_=987k\si{\ohm},*-]++(0,2)to node[cground, rotate=180]{}++(0,1);
\draw(A1.-)to [short,-*]++(-1,0)to [R,l_=3641\si{\ohm}]++(0,-2)to node[cground](cgnd){}++(0,-1);
\draw(A1.-)to [short,-*]++(-1,0)to [R,l^=988\si{\ohm}]++(0,2) to [short]++(-2,0)coordinate(tmp) to [battery, l_=15V] (tmp |-cgnd) node[cground]{};
\end{circuitikz}
\end{document}