小明注意到家中灯泡比平时暗,他做如下实验,关闭家中气压用电器,只开一盏220v,

1个回答

  • 1、这盏灯的电阻是 U^2/P = 220^2/100 = 484Ω;

    2、这盏灯消耗的电能是 81/3000 x 60min/20min =0.081kWh = 0.081 x1000w x3600s = 291600J = 291.6kJ.

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