(B): We apply a la source externally, measure the resultant voltage, and then set r=v/1(独立源为零) U=3k×1+(1+ 2k=5k+ 2k 3k十 4000 2 D=lOkv 4000 h 10k9 (C): We could also apply a iv source externally, determine i=1/ Rth, and then set rth=1/(独立源为零) 2k 3k 1=3+(i+)2k=5ki+ +D 400 4000 U=1∴i=1/10kR==10 10 The Thevenin-equivalent circuit 下8（ B): We apply a 1A source externally, measure the resultant voltage, and then set .Rth =v/1(独立源为零) kV k k k 10 2 )2 5 4000 3 1 (1 = =  + + = +     Rth = = 10k 1  （C）:We could also apply a 1V source externally, determine i=1/Rth, and then set Rth =1/i.(独立源为零) i k ki i k ki x x x 1 1/10 2 )2 5 4000 1 3 ( =  = = + + = +    = = k i Rth 10 1 The Thevenin-equivalent circuit: 4000  x − +  x 2k 3k 1A = 4000  x − +  x 2k 3k 1V i − + 8V − + 10k