(3) Z1=10+j50Z2=400+1000 Z2 B B=?,I2 leads v90° 12+B1 Z111+Z212=Z1(12+BI2)+ s=Z1+BZ1+22 2 leads vs 90 Re[Z1+B21+22=0 10+B10+400=0B=-41 B I2 leads v。60 S=(1+B)Z1+ 50(+B)+1000 10(1+B)+400tan(-60°) B=-2615=  = + = + • • ?, 90 10 50 400 1000 2 1 2 Vs I leads Z j Z j  • • • • • • • • • 1 = 2 + 2 = 1 1 + 2 2 = 1 2 + 2 + 2 2 I I I V Z I Z I Z (I I ) Z I  s  1 1 2 2 Z Z Z I Vs = + + • •  2 90 Re[ 1 + 1 + 2 ] = 0 • • I leadsVs Z  Z Z 10 +  10 + 400 = 0  = −41 (3) • 1 I =  • •  ?, I 2 leadsVs 60 1 2 2 (1 ) Z Z I Vs = + + • •  tan( 60 )  = −26.15 10(1 ) 400 50(1 ) 1000 = −  + + + +  