某同学利用电压表和电流表测量定值电阻Rx的阻值(约9Ω左右),电源选用两节干电池.(1)请将图中电流表 某同学利用电压表和电流表测量电阻R1的阻值(约9Ω左右),电...

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(1)电路中的最大电流I=
U
R
=
3V
≈0.33A,故电流表的量程选择0~0.6A;将开关、电流表以及电阻串联起来;如下图所示:

(2)在闭合开关之前,滑动变阻器应处于最大阻值的右端,这样做的目的是为了防止电流过大,保护电路;
(3)移动滑片,电压表示数总为3V,说明电压表测的是电源电压,即定值电阻断路或滑动变阻器短路;
(4)清除故障后,将滑片P向左滑动时,其电阻减小,整个电路的电流增大,电压表示数将增大;
由图2可知:U=1.4V,I=0.28A,所以R=
U
I
=
1.4V
0.28A
=5Ω;这个结果不可靠,应多次测量求平均值,以减小实验的误差.
(5)(a)闭合开关S1,测出通过已知电阻R0的电流I0
(b)闭合开关S1、S2,测出干路电流I.
(c)∵并联电路中干路电流等于各支路电流之和,
∴通过Rx的电流为Ix=I-I0
∵并联电路中各支路两端的电压相等,
∴I0R0=IxRx
I0R0=(I-I0)Rx
Rx=
I0R0
I?I0

故答案为:(2)右;(3)定值电阻断路;滑动变阻器短路;(4)增大;5;没有多次测量求平均值,误差太大;(5)(a)S1;(b)闭合开关S1、S2,测出干路电流I;(c)
I0R0
I?I0


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