Example 2 P47 Exercise 2.3.1 (a)(c) (a利用虚短与虚断的概念 R,20K2 虚短：yw=Vp=0→y2V 虚断：p=iw=0→ii2 =i2→(0-v)R1=(y-vo)/R2 可求得：vo=6V R,20K2 (c)虚短：yw=vp=0→y2V 虚断：ip=iw0→iw=0 i=0→y,=y=2V (a) 利用虚短与虚断的概念 ◆Example 2 P47 Exercise 2.3.1 (a) (c) vo - + R2 R1 10KΩ 20KΩ vN vP 2V vo - + R2 20KΩ vN vP 2V v1 i1 i2 iN iP 虚短: vN =vP =0  v1 =2V 虚断: iP = iN =0  i1= i2 i1 = i2  (0-v1 )/R1 = (v1 -vO )/R2 可求得：vO＝6V (c) 虚短: vN =vP =0  v1 =2V v1 i iN iP 虚断: iP = iN =0  i= iN =0 i=0  vo =v1 =2V