(1)由P 额=
,得:电水壶的电阻R=
=
=40Ω;
(2)t=3min=180s,
t时间内电水壶消耗的电能:W=
kW·h=0.05kW·h=1.8×10 5J,
电水壶的实际电功率:P 实=
=
=1000W;
(3)由P=
,即 U=
,
得:该同学家的实际电压:U 实=
=
=200V
(1)由P 额=
,得:电水壶的电阻R=
=
=40Ω;
(2)t=3min=180s,
t时间内电水壶消耗的电能:W=
kW·h=0.05kW·h=1.8×10 5J,
电水壶的实际电功率:P 实=
=
=1000W;
(3)由P=
,即 U=
,
得:该同学家的实际电压:U 实=
=
=200V