一道普通高中物理题 帮忙解一下 谢谢!!!!!!!!

2024-11-19 12:33:30
推荐回答(1个)
回答1:

精锐教育温馨解读:
实物图自己画吧
a状态Rp与R串联,b状态是Rp与Rx并联后再与R串联
要解决此题首先要了解此实验的实验原理:Rp虽然是电阻箱,然而电阻箱是粗略的测量电阻工具,本实验是要精确测量Rx,因此a状态的作用其实就是利用伏安法测出电阻箱的精确电阻,Rp=U1/I1。
当状态b时,Rx两端的电压就是U2,因此只要找出流过Rx的电流,即可求出Rx,而流过Rx的电流等于干路电流I2减去流过Rp的电流,流过Rp的电流可比较容易求出:此时状态的Rp电压是U2,电阻Rp通过a状态已经求出,因此流过Rp的电流可以求出后求流过Rx的电流,最后算出Rx。
Rp=U1/I1 (1)式 b状态时流过Rp的电流为 U2/Rp,将(1)式带入后得电流为U2I1/U1 (2)式
干路电流I2减去(2)式就是Rx电流,经过计算得(U1I2-U2I1)/U1 (3)式
最后Rx用U2除以(3)式可得阻值为U1U2/(U1I2-U2I1)即为第二问答案

误差分析:Rx两端的电压U2无误差,误差只能出现在通过Rx的电流,干路电流无误差,所以出现误差的一定是流过电阻箱Rp的电流,而b状态中的Rp两端电压U2无误差,因此Rp的电流误差肯定是因为Rp电阻计算出的误差,很显然,Rp的计算误差是因为a状态下伏安法外界法引起的误差,所以此题就转化为伏安法外接法误差问题,很显然外接法计算得出的Rp值偏小
Rp偏小,那么流过Rp的电流U2/Rp就偏大,那么流过Rx的电流就偏小,最后计算Rx的阻值就偏大了,所以最后结果为偏大。
点评:此题的关键是误差的来源分析,最根本的还是要摸透实验原理,通过逆推法一步一步得出误差的最原始来源,然后具体的偏大还是偏小再反过来顺推得出最后结果。

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