\documentclass[10pt,a4paper,final]{report}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{fontspec}%使用xetex
\setmainfont[BoldFont=黑体]{宋体}               % 使用系统默认字体
\XeTeXlinebreaklocale "zh"                      % 针对中文进行断行
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\begin{document}
%\begin{equation}
%\begin{aligned}
%\frac{\partial P_{ij}}{\partial V_i}
%&=
%-V_{ij}(g_{ij}cos\theta_{ij}+b_{ij}sin\theta_{ij})+
%2(g_{ij}+g_{si})V_i
%\end{aligned}
%\end{equation}
线路功率（不考虑接地导纳）
\newline
\begin{equation}
\begin{aligned}
\dot{I}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
(G+jB) \\
&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)](G+jB) \\
&=(V_1 cos\theta_1+ jV_1 sin\theta_1- V_2 cos \theta_2 -j V_2 sin \theta_2)(G+jB) \\
&=(V_1 cos \theta_1 -V_2 cos \theta_2)G + j(V_1 sin \theta_1- V_2 sin \theta_2)G
\\&+j( V_1 cos \theta_1 -V_2 cos \theta_2)B + j(jV_1 sin \theta_1-jV_2 sin \theta_2)B \\
&=G(V_1 cos \theta_1- V_2 cos \theta_2) - B(V_1 sin \theta_1 -V_2 sin \theta_2) \\
&+jB(V_1 cos \theta_1 - V_2 cos \theta_2) + jG(V_1 sin \theta_1 - V_2 sin \theta_2) \\
&= V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
&+jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)
\end{aligned}
\end{equation}
电流的共轭复数
\begin{equation}
\begin{aligned}
\dot{I}^*_{12}&=V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) \\
&-jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)
\end{aligned}
\end{equation}
线路复功率
\begin{equation}
\begin{aligned}
\dot{S}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} ) 
\dot{I}^*_{12} \\
&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)]\dot{I}^*_{12} \\
&=[V_1 (cos\theta_1+jsin\theta_1) - V_2 (cos\theta_2+jsin\theta_2)] \\
&[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2) 
-jV_1(Bcos \theta_1+Gsin\theta_1)+jV_2(Bcos\theta_2+Gsin\theta_2)]\\
&=V_1( G cos \theta_1-B sin \theta_1) 
[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
&-V_1( G cos \theta_1-B sin \theta_1)
[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)] \\
&- V_2 (cos\theta_2+jsin\theta_2)
[V_1( G cos \theta_1-B sin \theta_1) -V_2 (G cos \theta_2 - B sin \theta_2)]\\
&V_2 (cos\theta_2+jsin\theta_2)
[jV_1(Bcos \theta_1+Gsin\theta_1)-jV_2(Bcos\theta_2+Gsin\theta_2)]
\end{aligned}
\end{equation}
另一种推导方式
\begin{equation}
\begin{aligned}
\dot{I}_{12}=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
(G+jB)
\end{aligned}
\end{equation}
电流的共轭复数
\begin{equation}
\begin{aligned}
\dot{I}^*_{12}=(V_1 e^{-j\theta_1} -  V_2 e^{-j\theta_2} )
(G-jB)
\end{aligned}
\end{equation}
线路复功率
\begin{equation}
\begin{aligned}
\dot{S}_{12}&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )\dot{I}^*_{12} \\
&=(V_1 e^{j\theta_1} -  V_2 e^{j\theta_2} )
(V_1 e^{-j\theta_1} -  V_2 e^{-j\theta_2} )
(G-jB) \\
&=(V_1 e^{j\theta_1}V_1 e^{-j\theta_1} - V_1 e^{j\theta_1}V_2 e^{-j\theta_2} 
-  V_2 e^{j\theta_2}V_1 e^{-j\theta_1}+  V_2 e^{j\theta_2}V_2 e^{-j\theta_2}
)
(G-jB) \\
&=(V_1^2+V_2^2-V_1V_2e^{j(\theta_1-\theta_2)} -V_1V_2e^{j(\theta_2-\theta_1)} )
(G-jB) \\
&=[V_1^2+V_2^2
-V_1V_2(e^{j(\theta_1-\theta_2)}+e^{j(\theta_2-\theta_1)})
](G-jB) \\
&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)+jsin(\theta_1-\theta_2)
+cos(\theta_2-\theta_1)+jsin(\theta_2-\theta_1)]\}(G-jB) \\
&=\{V_1^2+V_2^2-V_1V_2[cos(\theta_1-\theta_2)
+cos(\theta_1-\theta_2)]\}(G-jB) \\
&=[V_1^2+V_2^2-2V_1V_2cos(\theta_1-\theta_2)](G-jB) \\
&=(V_1^2+V_2^2)G-j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G-j2V_1V_2cos(\theta_1-\theta_2)(-jB) \\
&=(V_1^2+V_2^2)G-2V_1V_2cos(\theta_1-\theta_2)B
-j(V_1^2+V_2^2)B-j2V_1V_2cos(\theta_1-\theta_2)G
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
P_{12}=(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
Q_{12}=-(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
\end{aligned}
\end{equation}
考虑接地支路,接地电流为
\begin{equation}
\label{接地电流}
\dot{I}_{1}=V_1e^{j\theta_1}(G_d+jB_d)
\end{equation}
(\ref{接地电流})共轭为
\begin{equation}
\dot{I}_{1}^{*}=V_1e^{-j\theta_1}(G_d-jB_d)
\end{equation}
%Sd
\begin{equation}
\begin{aligned}
\dot{S}_{1}&=V_1e^{j\theta_1}\dot{I}_{1}^{*} \\
&=V_1e^{j\theta_1} 
V_1e^{-j\theta_1}(G_d-jB_d) \\
&=V_1^2(G_d-jB_d)
\end{aligned}
\end{equation}
因而
\begin{equation}
\begin{aligned}
P_{1}&=V_1^2 G_d \\
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
Q_{1}&=-V_1^2 B_d \\
\end{aligned}
\end{equation}
考虑节点支路后的注入电流
\begin{equation}
\begin{aligned}
\tilde{P}_{12}&=V_1^2 G_d 
+(V_1^2+V_2^2)G+2V_1V_2sin(\theta_1-\theta_2)B
\end{aligned}
\end{equation}
\begin{equation}
\begin{aligned}
\tilde{Q}_{12}&=-V_1^2 B_d 
-(V_1^2+V_2^2)B+2V_1V_2sin(\theta_1-\theta_2)G
\end{aligned}
\end{equation}
\end{document}
