问题 实验题

(10分)在“描绘小灯泡的伏安特性曲线”实验中,待测小灯泡上标有“3V,1.5W”的字样。测量小灯泡两端的电压和小灯泡的电流时,备有如下器材:

A.直流电源3V(内阻可不计)

B.直流电流表0 -3A(电阻约0.1Ω)

C.直流电流表O- 600mA(内阻约0.5Ω)

D.直流电压表0-3V(内阻约3kΩ,)E.直流电压表O -15V(内阻约200kΩ)      F.滑动变阻器10Ω、1A

G.滑动变阻器lkΩ、300mA

①除开关、导线外,为了完成该实验,需要从上述器材中选用的器材有      .(用序号字母表示)

②利用给出的器材通过实验描绘出小灯泡的伏安特性曲线,要求测量多组实验数据,请你在答卷虚线框内画出实验原理电路图;并将图中实物连接成实验电路图.

③下表中的各组数据是此学习小组在实验中测得的,根据表格中的数据在方格纸上作出该小灯泡的伏安特性曲线.

④试估算当小灯泡两端电压为2.2V时,该灯泡的实际功率为         W(结果保留两位有效数字)

答案

①     ACDF     (2分)

③  

④      1.0      (2分)

题目分析:①根据小灯泡额定电压3v判断电压表选择量程即D,“3V,1.5W”可得灯泡额定电流,所以电流表选择C。由于要描绘伏安特性曲线,电压从0开始调节,滑动变阻器选择小的即F,滑动变阻器G的最大电流小于灯泡额定电流而排除,电源必选。

②由于要描绘伏安特性曲线,电压从0开始调节,滑动变阻器选择分压式,灯泡正常发光电阻根据,依据”大内偏大小外偏小”,而,电流表选择外接法。具体电路见答案。

③描点作图,而电压与电流并不成正比例关系,图像是平滑的曲线。随电压升高,电阻变大。

④根据上题目中描点作图,当小灯泡两端电压为2.2V时,电流,消耗的电功率

单项选择题
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