如何对公式进行垂直排序？

Question

平均能量损失

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{array}
I_1\cdot(R_1 + R_4)+I_2\cdot R_1 +I_3\cdot 0 =\varepsilon_1\\
I_1\cdot R_1 + I_2\cdot(R_1+R_2+R_5) +I_3\cdot R_5 =\varepsilon_1\\
I_1\cdot 0 + I_2\cdot R_5+I_3\cdot(R_3 + R_5)=\varepsilon_2
\end{array}


\begin{array}
I_3=I_1+I_2\\
\downarrow  \downarrow&  \downarrow \\
\dfrac{U_1}{R_4}=\dfrac{\varepsilon_1-U_1}{R_1}+\dfrac{U_2-U_1}{R_2}
\end{array}

\end{document}

Answer 1

在环境的帮助下array：

\documentclass{article}
\usepackage{amsmath}

\begin{document}
\renewcommand{\arraystretch}{1.5}
\[
\begin{array}{c@{~\cdot~}c@{~+~}c@{~\cdot~}c@{~+~}c@{~\cdot~}c@{~=~}c}
R_1 + R_4 & I_1 & R_1         & I_2 & 0         & I_3 & \varepsilon_1\\
R_1       & I_1 & R_1+R_2+R_5 & I_2 & R_5       & I_3 & \varepsilon_1\\
0         & I_1 & R_5         & I_2 & R_3 + R_5 & I_3 & \varepsilon_2
\end{array}
\]

\[
\begin{array}{ccccc}
I_3             & = & I_1                            & + & I_2\\
\downarrow      &   & \downarrow                     &   &  \downarrow \\
\dfrac{U_1}{R_4}& = & \dfrac{\varepsilon_1-U_1}{R_1} & + & \dfrac{U_2-U_1}{R_2}
\end{array}
\]

\end{document}

Answer 2

我忍不住要为通过 获得的第一组三个方程式提出另一种布局pstricks：由于组与组之间或多或少有较大的空白处，我将符号放在+每个空白处的中心：

\documentclass{article}
\usepackage{amsmath}
\usepackage{pst-node, multido, auto-pst-pdf}

\begin{document}

\begin{postscript}
\[ \begin{array}{lrr}
I_1 \cdot (R_1 + R_4)\pnode{A1} & \pnode{B1} R_1 \cdot I_2 \pnode{C1}& \pnode{D1}0 \cdot I_3 =\varepsilon_1\\[1ex]
I_1 \cdot R_1 \pnode{A2} & \pnode{B2} (R_1+R_2+R_5) \cdot I_2 \pnode{C2} & \pnode{D2} R_5 \cdot I_3 =\varepsilon_1\\[1ex]
I_1 \cdot 0 \pnode{A3}&\pnode{B3} R_5 \cdot I_2\pnode{C3} &\enspace \pnode{D3} (R_3 + R_5) \cdot I_3 =\varepsilon_2
\end{array} \]
\psset{offset =\fontdimen22\textfont2,linestyle=none}
\multido{\i =1+1}{3}{\ncline{A\i}{B\i}\ncput*{+}\ncline{C\i}{D\i}\ncput*{+}}
\end{postscript}

\end{document}