R2(β+1)b3 6-10 ff: U=IcBR, Uo=(B+D Ib3R2+Ib3 be3=IE3R2+Ib3 be xlc3R2    = = = = + + = +  − i f u f f C b b b e E b b e C UU UU A U I R U I R I r I R I r I R 0 0 3 1 0 1 3 2 3 3 3 2 3 3 3 2 , 6 10 解： （ ） u0 R2 rbe3 Ib3 βIb3 (β+1)Ib3