答案
解:由U-I图象可知,当电流表的示数I=0.2A时,U
a=2V U
b=1V,
(1)电阻b的电功率P
b=U
bI=1V×0.2A=0.2W,
答:电阻b的电功率是0.2W.
(2)两电阻串联在1min内消耗的电能
W=UIt=(U
a+U
b)It=(2V+1V)×0.2A×60s=36J,
答:电阻a、b在lmin内消耗电能之和是36J.
解:由U-I图象可知,当电流表的示数I=0.2A时,U
a=2V U
b=1V,
(1)电阻b的电功率P
b=U
bI=1V×0.2A=0.2W,
答:电阻b的电功率是0.2W.
(2)两电阻串联在1min内消耗的电能
W=UIt=(U
a+U
b)It=(2V+1V)×0.2A×60s=36J,
答:电阻a、b在lmin内消耗电能之和是36J.