3)U, R1 B3 EE R1+R2 C 20 ×15=-10V c2 20+10 -IcrC 15-1×6=9V U 2 U R B 10V<U.<9V U, EE 图P4-6B UEE R R R U 1 2 1 3 + (3) = 15 10V 20 10 20  = − + = − C CC CQ RC U =U − I 1 =15−16 = 9V −10V Uic  9V