(1)因这盏灯标有“220V 100W”字样,则R L=
U 2额
P 额 =
(220V) 2
100W =484Ω.
答:灯泡的电阻为484Ω.
(2)白炽灯的实际电功率为81W时,
电路中电流: I 1 =
P 实
R L =
81W
484Ω =0.41A ,
U 1=I 1R L=0.41A×484Ω=198.44V,
所以U 2=U-U 1=220V-198.44V=21.56V,
功率为P 2=U 2I 1=21.56V×0.41A≈9W.
故答案为:(1)484;(2)9.
(1)因这盏灯标有“220V 100W”字样,则R L=
U 2额
P 额 =
(220V) 2
100W =484Ω.
答:灯泡的电阻为484Ω.
(2)白炽灯的实际电功率为81W时,
电路中电流: I 1 =
P 实
R L =
81W
484Ω =0.41A ,
U 1=I 1R L=0.41A×484Ω=198.44V,
所以U 2=U-U 1=220V-198.44V=21.56V,
功率为P 2=U 2I 1=21.56V×0.41A≈9W.
故答案为:(1)484;(2)9.