[题目]有一灯泡上标有“6 V 0．1 A 字样.现要测绘该灯泡的伏安特性曲线.有下列器材可供选用A．电压表(0-5 V.内阻2．0 kΩ)B．电压表(0-10 V.内阻3．0 kΩ)C．电流表(0-100m A.内阻2．0 Ω)D．电流表(0-6 A.内阻1．5 Ω)E．滑动变阻器(30 Ω.2 A)F．滑动变阻器(100 Ω.0．5 A)G．学生题目和参考答案——青夏教育精英家教网—

题目内容

【题目】有一灯泡上标有“6 V 0．1 A”字样，现要测绘该灯泡的伏安特性曲线，有下列器材可供选用

A．电压表（0～5 V，内阻2．0 kΩ）

B．电压表（0～10 V，内阻3．0 kΩ）

C．电流表（0～100m A，内阻2．0 Ω）

D．电流表（0～6 A，内阻1．5 Ω）

E．滑动变阻器（30 Ω，2 A）

F．滑动变阻器（100 Ω，0．5 A）

G．学生电源（直流9 V）及开关、导线等

（1）实验中所用的电压表应选________，电流表应选________，滑动变阻器应选________。

（2）画出实验电路图，要求电压从零开始测量。

（3）根据你的实验电路，将图中所示的器材连成实验电路。

（4）该同学描绘出的IU图像应是图中的________。

【答案】（1）B；C；E （2）如图（3）如图（4）B

【解析】（1）电压表的量程应大于小灯泡的额定电压6 V，故电压表应选B。小灯泡的额定电流是0．1 A，故电流表应选C。由题目要求可知滑动变阻器采用分压式，故应选电阻较小的E。

（2）小灯泡的电阻R＝＝Ω＝60 Ω，电压表的内阻与待测电阻的比为=50，待测电阻与电流表的内阻的比为＝30，＞，故电流表应采用外接法，滑动变阻器接成分压式，电路图见答案。

（3）电路图如图所示

（4）小灯泡的电阻随着温度的升高而变大，所以B正确。

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