由式(2)解得: R u12Y R rR2+r2R3+RR 1△=W12△/R12-31△k31 uoaur-uuR 2△A=23△R23-u12△/R12(1 Y (3) rR2+R2R3+RBR 3A=l3131-L23△/23 R 31Y 23Y R Y RR+retro 根据等效条件,比较式(3与式(1),得由Y接→△接的变换结果: RR R12=R1+R,+ 12 G,G R 12 3 G1+G2+G3 RR GG R,=R,+R,+ 或 23 G,++g GG R,,=R,+R,+ 31 R 2 G1+G2+G3由式(2)解得： i3 =u31 /R31 – u23 /R23 i2 =u23 /R23 – u12 /R12 i1 =u12 /R12 – u31 /R31 (1) 1 2 2 3 3 1 12Y 3 31Y 2 1Y R R R R R R u R u R i + + − = 1 2 2 3 3 1 23Y 1 12Y 3 2Y R R R R R R u R u R i + + − = 1 2 2 3 3 1 31Y 2 23Y 1 3Y R R R R R R u R u R i + + − = (3) 根据等效条件，比较式(3)与式(1)，得由Y接→接的变换结果： 2 3 1 31 3 1 1 2 3 23 2 3 3 1 2 12 1 2 R R R R R R R R R R R R R R R R R R = + + = + + = + + 1 2 3 3 1 31 1 2 3 2 3 23 1 2 3 1 2 12 G G G G G G G G G G G G G G G G G G + + = + + = + + = 或