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58问答库 > 把标有6伏 3瓦的灯泡L1和标有12伏3瓦的灯泡L2串联在电压为6伏的电源两端，通过它的电流为（

把标有6伏 3瓦的灯泡L1和标有12伏3瓦的灯泡L2串联在电压为6伏的电源两端，通过它的电流为（

2025-03-30 21:03:56

推荐回答（3个）

回答（1）：

先算出R1=U1^2/P1=6×6/3=12Ω R2=U2^2/P2=12×12/3=48Ω 串联总电阻是60Ω 电流是0.1A
因为电流一样，所以消耗功率之比等于电阻之比等于1:4
并联到电源上后，R1两端电压和额定电压一样，那功率就和额定功率一样是3w，R2的功率等于6×6/48=0.75 并联后总功率等于3.75w
希望能够采纳，谢谢

回答（2）：

求出这俩灯泡的电阻就行了。

回答（3）：

R1=6^2/3=12，R2=12^2/3=48，R总=60，I=6/60=0.1A，P1：P2=I^2R1：I^2R2=R1：R2=1：4

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