Example: What is the maximum power that can be delivered to an external load ri by the network 200 36 0.2A R Oc Solution:。=20×+u1=1.51u1=30(+0.2)1=24 40 36 40+0.2×30 20 +0.12 =0.2 ×30=2.4 20+3040 20+30 s2.4 0+0.12=0.18 Rm=36/0.18=200c2 36 162Ⅳ na 4×200Example: What is the maximum power that can be delivered to an external load RL by the network. 1 1 1 1.5 40 20    oc =  + = 0.2) 24V 40 30( 1 1 1 = +  =   oc = 36V 0 12 20 30 40 30 0 2 40 1 1 . = + . + = +    sc i 30 2.4 20 30 20 1 0.2  = +  =  = + 0.12 = 0.18 = 36/ 0.18 = 200 40 2.4 s c Rt h i P 1.62W 4 200 362 max =  = Solution: sc i  oc