(1)由P=UI得 I=
P 1
U =
30
220 A=0.14A .
答:保温时通过电热水瓶的电流是0.14A.
(2) R=
U 2
P 2 =
220 2
1200 Ω=40.33Ω .
答:加热时电热水瓶的电阻为40.33Ω.
(3)消耗的电能W=P 2t=1200W×5×60s=3.6×10 5J.
答:加热时电热水瓶的消耗的电能3.6×10 5J.
(1)由P=UI得 I=
P 1
U =
30
220 A=0.14A .
答:保温时通过电热水瓶的电流是0.14A.
(2) R=
U 2
P 2 =
220 2
1200 Ω=40.33Ω .
答:加热时电热水瓶的电阻为40.33Ω.
(3)消耗的电能W=P 2t=1200W×5×60s=3.6×10 5J.
答:加热时电热水瓶的消耗的电能3.6×10 5J.