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
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?