额定电压与额定功率分别是220V,40W和220V,100W的两盏电灯A和B,把它们串联起来接到110V的电压上,它们的实际功率Pa= Pb=
问题描述：
额定电压与额定功率分别是220V,40W和220V,100W的两盏电灯A和B,把它们串联起来接到110V的电压上,它们的实际功率Pa= Pb=
答
RA = U*U / pA = 220v*220v / 40 w= 1210Ω
RB = U*U / pB= 220v*220v / 100w = 484Ω
串联后，R 总= RA+RB = 1694Ω
I = u/R= 110v /1694Ω = 0。065A
Pa = I*I*RA = 0.065A*0.065A*1210Ω= 5.1 W
Pb = I*I*RB = 0.065A*0.065A*484Ω= 2.05 W
答
RA = U*U / RA = 220*220 / RA = 40 RA = 1210RB = U*U / RB= 220*220 / RB = 100 RA = 484串联后,R = RA+RB = 1694 I = 110/R= 110 /1694 = 0.065Pa = I*I*RA = 0.065*0.065*1210= 5.1 WPb = I*I*RB = 0.065*0.065*...