U2 |
P额 |
(220V)2 |
110W |
(2)电热毯处于高温加热状态时,10min消耗的电能:W=Pt=110W×10×60s=6.6×104J;
(3)当电热毯处于低温加热状态时,R1、R2串联,电路中的电流:I=
U |
R1+R2 |
220V |
440Ω+660Ω |
P实=I2R1=(0.2A)2×440Ω=17.6W.
答:(1)R1的阻值为440Ω;
(2)该电热毯处于高温加热状态时,10min消耗的电能为6.6×104J;
(3)已知R2的阻值为660Ω,当电热毯处于低温加热状态时,R1的实际功率为17.6W.