在“描绘小电珠的伏安特性曲线”的实验中,用导线a、b、c、d、e、f、g和h按图1所示方式连接好电路,电路

2025-04-07 10:49:42
推荐回答(1个)
回答1:

(1)调节滑动变阻器,小电珠的亮度变化,但电压表、电流表示数总不能为零,说明滑动变阻器不起分压作用,滑动变阻器实际上接成了限流接法,由图甲所示电路图可知,可能是g导线断路.
(2)某同学测出一个电源和小电珠的U-I图线如图乙所示,则小灯泡的电阻值随工作电压的增大而增大.
由图乙所示电源U-I线可知,电源电动势E=3.0V,电源内阻为:r=

△U
△I
=
3
3
=1.0Ω;
由图乙所示小电珠的U-I图象可知,当电流为1.0A时,灯泡两端电压为1.2V,此时灯泡电阻为:RL=
1.2
1.0
=1.2Ω,灯泡与电阻串联后接在电源两端,
由闭合电路欧姆定律得:I=
E
r+R+RL

解得R=0.8Ω;
(3)因a元件的I-U图线是直线,说明其电阻不随电压的变化而变化,故可作为标准电阻,故正确对,B错误.
b的阻值随电压升高而增大,c的阻值随电压升高而降低,故C正确,D错误.
故选:AC.
(4)该小组用多用表的“×10”欧姆挡试测这段材料在常温下的电阻,操作步骤正确,发现表头指针偏转的角度很大.为了准确地进行测量,应换到×1档.
如果换档后立即用表笔连接待测电阻进行读数,那么欠缺的步骤是:两表笔直接接触,调节欧姆调零旋钮,使指针指在“0Ω”处,补上该步骤后,表盘的示数如图所示,则该材料的电阻是11.0Ω.
故答案为:(1)g;(2)3.0,1.0,0.8.(3)AC
(4)×1,欧姆调零旋钮,11.0

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