116问答网 > 某水电站，用总电阻为2.5Ω的输电线输电给500km外的用户，其输出电功率是3×106 kW．现用500kV电压输电，

某水电站，用总电阻为2.5Ω的输电线输电给500km外的用户，其输出电功率是3×106 kW．现用500kV电压输电，

2025-04-07 18:48:58
推荐回答（1个）
回答1：

A、根据P=UI知，输电线上的输送电流I=

P
U
=
3×109
5×105
=6000A．故A错误．
B、输电线上损失的电压△U=IR=6000×2.5V=15000V=15kV．故B正确．
C、若改用5kV电压输电，则输送电流I=
P
U
=
3×109
5000
=6×105A，则损失的功率P损=I2R=36×1010×2.5=9×1011W=9×108 kW．故C正确．
D、输电线上损失的功率为△P＝I2r＝(
P
U
)2rP为输电功率，U为输电电压，r为输电线上的电阻；故D正确．
故选：BCD．

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