(1)当变阻器滑片置于N端时,只有R1连入电路,电路中的电流I=
=U R1
=0.75A.6V 8Ω
(2)电压表测的R1电压,电流I′=
=U1 R1
=0.5A,滑动变阻器的电压U2=U-U1=6V-4V=2V,4V 8Ω
滑动变阻器的电阻R2=
=U2 I′
=4Ω.2V 0.5A
(3)R1与R3并联后有:
=1 R并
+1 R1
,1 R3
解得:R并=
Ω,它们两端的电压U′=I″R并=0.6A=×16 3
Ω=3.2V,16 3
R1中的电流I1=
=U′ R1
=0.4A.3.2V 8Ω
答:(1)当变阻器滑片置于N端时,电流表的示数为0.75A;
(2)当电压表为4伏时,变阻器R2连入电路的电阻为4Ω;
(3)通过R1的电流为0.4A.