title: 一道物理（电）1请分析并解答，谢谢~~ 
desc: 请并解答，谢谢~~
answer: ∵R1、R2、R3并联 ==> U=I1R1=I2R2=I3R3 ==>I1:I2=R2:R1, I2:I3=R3:R2
  ∵I2=1A ==>I1=R2:R1(A), I3=R2:R3(A)
  ∵R1=4Ω ==>I1=U/R1=U/4
  ∵P3=12W ==>U×I3=12W ==>I3=12/U
  ∵I=4.5A ==>I1+I2+I3=4.5 ==>U/4+1+12/U=4.5 ==>U=6V or U=8V
  当U=6V时
  I1=1.5A (由I1:I2=R2:R1) ==>R2=6Ω
  根据P=U^2/R ==>R3=3Ω
  当U=8V时
  I1=2A ==>R2=8Ω
  根据P=U^2/R ==>R3=16/3Ω
  
  (注：因为P=U×I,所以I3=P3/U，再根据I2:I3=R3:R2，可以算出R3，可用来做题目的验证)