由已知条件得知P1=I1^2R1,P0=I2^2R0,P2=I2^2R2,R1/R2=1/2P1/P2=I1^2R1/I2^2R2=9/8所以I1/I2=3/2,I2=0.2A,因为P0=0.8,R0=20Ω解方程组:U=0.3(R1+20) U=0.2(R2+20) R2=2R1得:R1=20Ω,R2=40ΩU=12V,R0先后两次消耗的功率之比=I1^2/I2^2=9/4选A