一灯泡电压3v,电流0.3A.工作0.5min消耗的电功为多少J
问题描述：
一灯泡电压3v,电流0.3A.工作0.5min消耗的电功为多少J
答
P=UIT=3×0.3×(0.5÷60)=0.0075W
答
t=0.5min=30s
消耗的电能=电流的功=Pt=UIt=3V*0.3A*30s=27J
答
W=UIt=3×0.3×0.5×60=27(J)
答
27J