(1)根据P=
可得,L1的电阻:U2 R
R1=
=U2 P1
=6Ω;(6V)2
6W
(2)两灯泡的电阻之比:
=R1 R2
=
U2 P1
U2 P2
=P2 P1
=3W 6W
,1 2
∵串联电路总各处的电流相等,
∴根据欧姆定律可得:
=U1 U2
=IR1
IR2
=R1 R2
,1 2
∵串联电路中总电压等于各分电压之和,
∴
=U甲 U乙
=U2
U1+U2
=2 1+2
.2 3
故答案为:6;2:3.