某同学利用电压表和电阻箱测定一种特殊电池的电动势（电动势E大_爱下问搜题

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问题问答题

某同学利用电压表和电阻箱测定一种特殊电池的电动势（电动势E大约在9V左右，内阻r约为50Ω），已知该电池允许输出的最大电流为l50mA．该同学利用如图（a）所示的电路进行实验，图（a）中电压表的内阻约为2KΩ，R为电阻箱．阻值范围0～9999Ω，R₀是定值电阻，起保护电路的作用．

（1）实验室备有的定值电阻尺．有以下几种规格：

A．2Ω B．20Ω C．200Ω D．2000Ω，本实验应选______（填入相应的字母）作为保护电阻．

（2）在图（b）的实物图中，已正确地连接了部分电路，请完成余下电路的连接．

（3）该同学完成电路的连接后，闭合开关s，调节电阻箱的阻值；读取电压表的示数，其中电压表的某一次偏转如图（c）所示，其读数为______．

（4）改变电阻箱阻值，取得多组数据，作出了如图（d）所示的图线，则根据该同学所作出的图线可求得该电池的电动势E为______V，内阻r为______Ω．（结果保留两位有效数字）（5）用该电路测电动势与内阻，测量和真实值的关系E_测______E_真，r_测______r_真（填“大于”、“小于”或“等于”）

答案

（1）当电阻箱的电阻调为零时，电路中电流最大，根据闭合电路欧姆定律得

I_m=

E

R0+r

得R₀=10Ω

所以实验应选B．

（2）根据电路图连接实物图，电动势E大约在9V左右，电压表应选择15V的量程，如图：

（3）电压表应选择15V的量程，所以读数为6.5V，

（4）闭合开关，调整电阻箱的阻值，读出电压表的示数，再改变电阻箱的电阻，得出多组数据．根据

1

U

=

1

E

+

r

E

1

R0+R

，知图线的纵轴截距表示电动势的倒数，图线的斜率等于

r

E

．

有：

1

E

=0.1，解得E=10V．

r

E

=5，解得r=50Ω．

（5）如果考虑电压表的内阻，根据实验的原理E=U+（

U

R0+R

+

U

Rv

）r，得：

1

U

=

1

E

（1+

r

Rv

）+

r

E

?

1

R0+R

，

考虑电压表的内阻，此时图线的纵轴截距表示

1

E

（1+

r

Rv

），所以E_测小于E_真，r_测小于r_真．

故答案为：（1）B．（2）（3）6.5 v （4）10，50Ω （5）小于，小于

填空题

下表是元素周期表的一部分，针对所给的9种元素，完成下列各小题．

周期

（1）非金属性最强的元素是：______；

（2）H₂O的电子式：______；

（3）Ar原子结构示意图为：______；

（4）NaOH与Al（OH）₃中，碱性较强的是：______；

（5）制作光导纤维的主要材料是：______；

（6）N与C中，原子半径较小的是：______；

（7）Al₂O₃与盐酸反应的化学方程式：______．

阅读理解

阅读理解。

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