一个白炽灯泡L上标有“4v 4w”的字样,将它与一个定值电阻R串联后接入电源电压恒定为6V的电路中

小灯泡：P = UU/R0 ,R0 = UU/P=4*4/4=4欧 ,
I = U'/(R+R0) =6/(R+4),
I*I*R =2
联立这2个方程,得R1 = 8 ,R2=2,
1、根据题意,如果R=2,R0=4,U'=6 ,串联分压后,灯泡的电压正好是4V,应该正常发光,所以R不是2欧,R应该是8欧.
2、 I = U'/(R+R0) =6/(R+4)= 6/12 = 0.5A
灯泡功率p' = I * I *R0 = 0.25*4=1WI = U'/(R+R0) =6/(R+4), I*I*R =2怎么解啊？？？I=6/(R+4), I*I*R =2II = 2/R II = 36/(RR+8R+16)； 2/R=36/(RR+8R+16) 整理得RR - 10R +16 = 0解得：R1 = 8 , R2=2