某工地因施工需要,要安装44盏“220V 100W”的灯泡,该工地距离发电站550m,现采用每米电阻0.005欧的单芯导
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1.R灯=U²/P=(220v)²/100W=484ΩR总=0.005Ω*550+484Ω=486.75ΩI=U/R=220V/486.75Ω=80/177AP灯实际=I²R=(80/177A)²*484Ω=98.87W2.I正常=P/U=100W/220V=5/11AU总=IR=(5/11A)*486.75Ω=221....