为了探究并联电路的电流规律,小薇设计了如图1所示的电路进行实验.(1)小薇先将电流表接在L1所在的支路

2025-04-07 09:37:15
推荐回答(1个)
回答1:

(1)灯L2发光,但灯L1不发光,电流表的示数为零,说明L1支路不可能短路,只能是断路,即电路可能存在的故障是L1或电流表开路.
(2)乙电流表的量程为0~0.6A,分度值为0.02A,所以I2=0.52A;
丙电流表的量程为0~3A,分度值0.1A,所以I=1A;
在误差范围内I=I1+I2
(3)本实验多次测量的目的是为零避免实验现象的偶然性,从而能够得出普遍规律.
(4)从实物图中可以看出,电流从正极流出,经过开关、电流表进行分支,分别经过两个灯泡回到电源负极,因此电流表测量干路电流.
(5)要测灯L2的电流,将L1与电流表负接线柱相连的一端改接电流表的0.6A的接线柱,如下图所示:

故答案为:(1)L1断路;
(2)0.52;1;I1+I2=I;
(3)找到普遍的实验规律;
(4)干路;
(5)如上图.

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