2-19解 S=75 3×10 U11000/33 133.349 43.3 1、空载实验 低压方:Z 29_400/3 =3.859 R≈Rn P03800 =0.3592xn=239389 x0≈xm=VZ。-R=3832 由短路实验:20C时;Zs IKφ 440/ 5.879 43.3 10900 R P 1949 3l。3×43.3 Rs =00145 xs=√Z-R=5.542 xs=0.0415 所以r1=2=2R =0979 xn=xn=2x3=272 CoO,=0.8(滞后)且β=1 △M=B(RCo2+xSm)=00145×08+00415×06=00365 U2=U2(1-△)=400×09635=3854V βP+P 7 BS. o B PE+Po 10900+3800 =97.6% 750000×0.8+10900+3800 当 Cos p2=08(超前)时: Au= B(Rs Cos -xs Sin )=-0.0133 U2=U2(1-△)=400×10133=40532V2－19 解： 43.3 3 10 750 3 1 1 =  = = U S I N N N A = = = 133.34 43.3 10000 / 3 1 1 1 I U Z N N N   1、空载实验： 低压方： = = = 3.85 60 400 / 3 20 20 0 I U Z   =    = = 0.35 3 3800 3 60 2 2 0 0 0 I P R Rm  xm = 2393.8  = − = 3.83 2 0 2 x0 xm Z0 R 由短路实验：200C 时； = = = 5.87 43.3 440 / 3 1 1 I U Z K K S   =   = = 1.94 3 10900 3 43.3 2 2 I P R k k S  = 0.0145  RS = − = 5.54 2 2 xS ZS RS = 0.0415  xS 所以 = = = 0.97 2 ' 1 r1 r 2 RS = = = 2.77 2 ' 1 x1 x2 xS 2、 0.8 2 Cos = （滞后）且  = 1 ( )  2  2 u  R Cos x Sin S S    = + = 0.01450.8+ 0.04150.6 = 0.0365 (1 ) 400 0.9635 385.4 2 2 U =U − u =  = N V S P P P P N kN kN Cos 0 2 2 0 2 1 + + + = −      97.6% 750000 0.8 10900 3800 10900 3800 1 =  + + + = − 当 0.8 2 Cos = （超前）时： ( ) 0.0133 2 2  = − = −   u  RS Cos xS Sin (1 ) 400 1.0133 405.32 2 2 U =U − u =  = N V