此时功角特性方程
'E0U1.321?1sin??sin??0.550sin? P2?X22.402（3）切除故障后
此时的系统电抗为
'?XT1?Xl?XT2?0.820 X3?Xd功角特性方程为
'E0U1.321?1sin??sin??1.611sin? P3?X30.82（4）计算极限切除角?cm
根据等面积定则的基本原理，求?cm。先求?h
?h???arcsin?cm?arccosPT1.0?180?arcsin?141.63P31.611
?arccos?79.2PT(?h??0)?P3cos?h?P2sin?0P3?P21.0??180
(141.63?30.74)?1611cos141.63?0.550cos30.741.611?0.550由已知，?c?50得，?c??cm，所以该简单系统能保持暂态稳定