(1+B)R101×10 174 R r+ b be 2+3.8 R c2 Q0 bI b2 (3)R d R R B B 2(re +RB) =2×(3.8+2) 11.6k R L R (单) b2 2 □。+=10k2 R B RR U B ic2 2RE 2R E174 2 3.8 (1 ) 101 10 = +   + +  B b e E R r  R ＋ － Uod2 R L ＋ V1 V2 Uid1 Uid2 － － ＋ RC Ib1 Ib2 Ic2 R B R B ＋ － Uoc2 R L ＋ V1 V2 Uic1 Uic2 － － ＋ RC Ib1 Ib2 Ic2 2RE 2RE RB RB id id id I U R = 1 1 2 b id I U = 2( ) be RB = r + = 2(3.8 + 2) =11.6k R oc(单）= RC =10k (3)