(1)
电流做功这个说法不确切吧,如果说电机的额定功率就是电流做功的话,
W(电流做功)= P(额定功率)X 时间 = 4.4KW X 5min = 1,320,000J
如果电流做功是指电动机的电磁功率所做功,那么还需考虑机械损耗,附加损耗等.
I(额定电流)=P/U =20A
W(线圈产热)=I X I X R X t= 20A^2 X 0.5 ohm X 5min=60,000J
(2)
R1上流过的电流 I1=24/8 = 3A,故R2上流过的电流 I2=9A - 3A = 6A
所以R2=24/6 = 4 Ohm
P(R2)=UI = 24*6=144W
(3)
R2两端的电压为12-3=9V,故P(R2消耗的功率)=UxU/R = 9*9/12 = 6.75 W
流过R2的电流为9/12 = 0.75A 故R1的阻值为 R1=3/0.75 = 4 ohm
(4)串电阻两端的电压应为 9-6 = 3V,而流过灯泡的电流 I = P/U = 0.667A
故,串电阻阻值R=U/I = 4.5 Ohm
P(R)=UxU/R = 3X3/4.5 = 2W