初中物理题(高手速进)1如图,电源电压保持不变,已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,在游端时,电压表8V,求总电压,R0,RP
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![初中物理题(高手速进)1如图,电源电压保持不变,已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,在游端时,电压表8V,求总电压,R0,RP](/uploads/image/z/11427436-28-6.jpg?t=%E5%88%9D%E4%B8%AD%E7%89%A9%E7%90%86%E9%A2%98%EF%BC%88%E9%AB%98%E6%89%8B%E9%80%9F%E8%BF%9B%EF%BC%891%E5%A6%82%E5%9B%BE%2C%E7%94%B5%E6%BA%90%E7%94%B5%E5%8E%8B%E4%BF%9D%E6%8C%81%E4%B8%8D%E5%8F%98%2C%E5%B7%B2%E7%9F%A5%E6%BB%91%E7%89%87P%E6%BB%91%E7%89%87P%E4%BB%8E%E5%B7%A6%E5%88%B0%E5%8F%B3%2CR0%E7%9A%84%E5%8A%9F%E7%8E%87%E4%B9%8B%E6%AF%94%E4%B8%BA9%EF%BC%9A1%2C%E5%BD%93%E6%BB%91%E7%89%87%E5%9C%A8%E5%B7%A6%E7%AB%AF%E6%97%B6%2C%E7%94%B5%E6%B5%816A%2C%E5%9C%A8%E6%B8%B8%E7%AB%AF%E6%97%B6%2C%E7%94%B5%E5%8E%8B%E8%A1%A88V%2C%E6%B1%82%E6%80%BB%E7%94%B5%E5%8E%8B%2CR0%2CRP)
初中物理题(高手速进)1如图,电源电压保持不变,已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,在游端时,电压表8V,求总电压,R0,RP
初中物理题(高手速进)
1如图,电源电压保持不变,已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,在游端时,电压表8V,求总电压,R0,RP
初中物理题(高手速进)1如图,电源电压保持不变,已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,在游端时,电压表8V,求总电压,R0,RP
通过:已知滑片P滑片P从左到右,R0的功率之比为9:1,当滑片在左端时,电流6A,
可以求出滑片在右边时的电流为=2A ( 6*6*R0/(I*I*R0)=9 ) .
6*R0=2*(R0+RP) 可得出RP=2R0 2*RP=8 RP=4Ω R0=2Ω
参考公式:P=UI=UU/R 。电源电压保持不变即U不变。
当滑片在左端时,由R0组成的串联电路:I(R0)=6A,P(R0)=UI(R0)=UU/R0 ``````````````````````````````````````````````````````````````````````````````````````````````1
当滑片在右端时,由R0,RP组成的串联...
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参考公式:P=UI=UU/R 。电源电压保持不变即U不变。
当滑片在左端时,由R0组成的串联电路:I(R0)=6A,P(R0)=UI(R0)=UU/R0 ``````````````````````````````````````````````````````````````````````````````````````````````1
当滑片在右端时,由R0,RP组成的串联电路:U(RP)=8V,P‘(R0)=U'(R0)U'(R0)/R0,U=U(RP)+U'(R0)```````````````````````````````````````````````````````````````````````````````2
P(R0)/P‘(R0)=9:1``````````````````````````````````````````````````````````````````````3
由1,2,3可得:U=12伏,R0=2欧,RP=4欧.
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由P0=I2R0知,R0的电功率之比为9:1,那么,P在右端时的电流为在左端电流的1/3,因此P在右端时电流为2A。然后根据电压相等列方程:6A×R0=2A×R0+8V,可解出R0=2Ω
U=12V R0=2Ω Rp=4Ω
看公式 自己推 又不难 善思考 能考好 知道不?