分类：教育/科学-学习帮助
问题：一道物理（电）1请分析并解答，谢谢~~ 
请并解答，谢谢~~
回答：∵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)