当两开关均闭全时,R1与R2并联后与R3串联;外部总电阻R= 4 2 +2=4Ω;由闭合电路欧姆定律可知,总电流I= 3.5 4+1 =0.7A;流过R1中的电流I1= 0.7 2 =0.35A;则R1的功率P=I12R1=(0.35)2×4=0.49;当K2闭合,K1断开时,R1与R3串联;则R1的功率P'=( E R1+R3+r )2R1=( 3.5 4+2+1 )2×4=1W;故功率之比为:P:P′=0.49:1=49:100;故答案为:49:100.