（2008•昌平区二模）如图所示的电路中，R为滑动变阻器，R1、R2为定值电阻，且R1＞R2，E为电压恒定的电源，当滑动变阻器的滑片滑动时，通过R、R1、R2的电流将发生变化，电流变化值分别为

Question

（2008•昌平区二模）如图所示的电路中，R为滑动变阻器，R₁、R₂为定值电阻，且R₁＞R₂，E为电压恒定的电源，当滑动变阻器的滑片滑动时，通过R、R₁、R₂的电流将发生变化，电流变化值分别为△I、△I₁、△I₂表示，则（　　）

A．当滑动片向右滑动时，有△I₁＜△I＜△I₂
B．当滑动片向左滑动时，有△I＜△I₁＜△I₂
C．无论滑动片向左还是向右滑动，总有△I=△I₁=△I₂
D．无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I<sub>112121212

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Answer 1

由电路图可知，R与R₂2并联后与R₁1串联，且R₁1＞R₂2，
设R₁1=2Ω，R₂2=1Ω，U=1V，
电路中的总电阻R_总总=R₁1+(RR2)/(R+R2)=(RR2+RR1+R1R2)/(R+R2)=(3R+2)/(R+1)，
电路中的电流I₁=(U)/(R总)=(R+1)/(3R+2)，
并联部分得的电压U_并=I₁×R_并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (RR2)/(R+R2)RR2RR2RR22R+R2R+R2R+R22=(RR2+RR1+R1R2)/(R+R2)=(3R+2)/(R+1)，
电路中的电流I₁=(U)/(R总)=(R+1)/(3R+2)，
并联部分得的电压U_并=I₁×R_并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (RR2+RR1+R1R2)/(R+R2)RR2+RR1+R1R2RR2+RR1+R1R2RR2+RR1+R1R22+RR1+R1R21+R1R21R22R+R2R+R2R+R22=(3R+2)/(R+1)，
电路中的电流I₁=(U)/(R总)=(R+1)/(3R+2)，
并联部分得的电压U_并=I₁×R_并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (3R+2)/(R+1)3R+23R+23R+2R+1R+1R+1，
电路中的电流I₁1=(U)/(R总)=(R+1)/(3R+2)，
并联部分得的电压U_并=I₁×R_并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (U)/(R总)UUUR总R总R总总=(R+1)/(3R+2)，
并联部分得的电压U_并=I₁×R_并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R+1)/(3R+2)R+1R+1R+13R+23R+23R+2，
并联部分得的电压U_并并=I₁1×R_并并=(R+1)/(3R+2)×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R+1)/(3R+2)R+1R+1R+13R+23R+23R+2×(R)/(R+1)=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R)/(R+1)RRRR+1R+1R+1=(R)/(3R+2)，
因R与R₂并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R)/(3R+2)RRR3R+23R+23R+2，
因R与R₂2并联，
所以I=(U并)/(R)=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (U并)/(R)U并U并U并并RRR=(1)/(3R+2)，
I₂=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (1)/(3R+2)1113R+23R+23R+2，
I₂2=(U并)/(R2)=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (U并)/(R2)U并U并U并并R2R2R22=(R)/(3R+2)；
当滑动变阻器接入电路的电阻变为R′时
△I₁=|I₁-I₁′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R)/(3R+2)RRR3R+23R+23R+2；
当滑动变阻器接入电路的电阻变为R′时
△I₁1=|I₁1-I₁1′|=|(R+1)/(3R+2)-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R+1)/(3R+2)R+1R+1R+13R+23R+23R+2-(R′+1)/(3R′+2)|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R′+1)/(3R′+2)R′+1R′+1R′+1′+13R′+23R′+23R′+2′+2|=|(R′−R)/((3R+2)×(3R′+2))|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R′−R)/((3R+2)×(3R′+2))R′−RR′−RR′−R′−R(3R+2)×(3R′+2)(3R+2)×(3R′+2)(3R+2)×(3R′+2)′+2)|，
△I=|I-I′|=|(1)/(3R+2)-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (1)/(3R+2)1113R+23R+23R+2-(1)/(3R′+2)|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (1)/(3R′+2)1113R′+23R′+23R′+2′+2|=|(3(R′−R))/((3R+2)×(3R′+2))|，
△I₂=|I₂-I₂′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (3(R′−R))/((3R+2)×(3R′+2))3(R′−R)3(R′−R)3(R′−R)′−R)(3R+2)×(3R′+2)(3R+2)×(3R′+2)(3R+2)×(3R′+2)′+2)|，
△I₂2=|I₂2-I₂2′|=|(R)/(3R+2)-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R)/(3R+2)RRR3R+23R+23R+2-(R′)/(3R′+2)|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (R′)/(3R′+2)R′R′R′′3R′+23R′+23R′+2′+2|=|(2(R′−R))/((3R+2)×(3R′+2))|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂＞△I₁．
故选D． (2(R′−R))/((3R+2)×(3R′+2))2(R′−R)2(R′−R)2(R′−R)′−R)(3R+2)×(3R′+2)(3R+2)×(3R′+2)(3R+2)×(3R′+2)′+2)|；
所以无论滑动片向左还是向右滑动，总有△I＞△I₂2＞△I₁1．
故选D．