问题
填空题
一只标有“3.8V 0.3A”的小灯泡,若加在两端的电压是1.9伏,则灯泡的额定功率是______瓦,实际功率是______瓦.
答案
灯泡的额定电压是3.8V,额定电流是0.3A,故额定功率P=UI=3.8V×0.3A=1.14W,
灯泡的电阻R=
=U I
=12.7Ω,3.8V 0.3A
当电压是1.9V时,实际功率P′=
=U′2 R
=0.28W.(1.9V)2 12.7Ω
故本题答案为:1.14;0.28.