求什么?
1）R2=U/I=24V/0.2A=120Ω
2)R1电流=24V/80Ω=0.3A
3)耗能=24V*(0.2+0.3)A*10S=120J
电阻R1与R2并联接入电路中,两端所加电压为24V,如果R1为80Ω,通过R2的电流为0.2A,若通电时间为10s,求这
电阻R1与R2并联接入电路中,两端所加电压为24V,如果R1为80Ω,通过R2的电流为0.2A,若通电时间为10s,求这
物理人气：789 ℃时间：2019-08-21 10:01:32
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