在探究“电流与电阻的关系”实验过程中,老师给同学们准备了以下器材:蓄电池(6V)、电流表(0-0.6A,0-

2025-03-13 22:32:35
推荐回答(1个)
回答1:

(1)连接电路时,开关应处于断开状态;
(2)将滑动变阻器与电阻串联,滑动变阻器已接了上面一个接线柱,应再接下面一个接线柱,如图所示:

(3)由图丙知,电流表的量程为0~0.6A,示数为0.4A;
将5Ω的电阻更换为10Ω电阻,根据串分压的知识,电压表的示数将变大,为使电压表的示数不变,应增大滑动变阻器的阻值,将滑片向右移动;
(4)换接15Ω电阻时,根据串分压的知识,电压表的示数变大,为使电压表的示数不变,所需滑动变阻器的阻值更大,电压表的示数无法达到2V,可能是滑动变阻器的阻值太小或是定值电阻两端的电压保持太低或是电源电压太高造成;
(5)由表格中数据知,电阻与电流的乘积不变,可得电压不变,导体中的电流与导体的电阻成反比;
(6)该实验中始终保持电阻两端的电压不变,用到的是控制变量法,根据此电路还可保持电阻不变,通过变阻器改变电阻两端的电压,探究电流和电压的关系;也可测量电阻的阻值及电功率.
故答案为:(1)断开;
(2)见上图;
(3)0.4;右;
(4)滑动变阻器最大阻值太小(或定值电阻两端电压太低或电源电压太高);
(5)电压一定时,导体中的电流与导体的电阻成反比;
(6)控制变量法;探究电流与电压的关系.

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