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230问答网 > 【一道物理电功率题，高手快快来，要超超超级具体】

【一道物理电功率题，高手快快来，要超超超级具体】

2024-12-19 20:01:31

推荐回答（4个）

回答1：

甲灯“6V 6W”，正常发光的电流是1A，电阻是6Ω；乙灯“6V 3W”，正常发光的电流是0.5A，电阻是12Ω。电压表甲测的是乙灯的电压，电压表乙测的是电源电压，由串联电路的分压原则可知，V甲/V乙=R乙/R总=12Ω/18Ω=2/3 所以选C

回答2：

选C
先求出R1=6欧,R2=12欧去掉电压表观察电路是串联电路，所以电压比等于电阻比，所以V甲测L2的电压，V乙测电路总电压，V甲/V乙=R2/R1+R2=12欧/18欧=2/3

回答3：

选c，电阻R=V的平方/W，可求得L1、L2的电阻分别为6和12欧，则L2分得的电压就为2/3，故选C

回答4：

C

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