(1)该灯泡的额定电流是I=
=P U
=0.02A;4.4W 220V
(2)消耗的电能W=Pt=0.0044kW×5h×30=0.66kW?h
(3)设两种灯泡正常发光的时间为t,
∵P=
,W t
∴LED消耗的电能为W1=P1t=0.0044kW?t,白炽灯消耗的电能为W2=P2t,
∵在达到相同亮度的条件下,LED灯可以节约90%的电能,
∴
=90%,
P2t?0.0044kWt
P2 t
解得:P2=0.044kW=44W.
故答案为:0.02;0.66;44