物理的電路問題

Let R be the resistance of the voltmeter.
Hence, when the voltmeter is connected to a 100-ohm resistor, the equivalent resistance = 100R/(100+R)
Hence, total resistance of the circuit = 100 + [100R/(100+R)]
= (200R+10000)/(100+R)

Since voltage across the voltmeter = 2.9 v, we have
2.9 = 6 x [100R/(100+R) / (200R+10000)/(100+R)]
after simplification,
2.9 = 6 x (R/(2R+100))
6R = 5.8R + 290
R = 1450 ohms

The internal resistance of the voltmeter is 1450 ohms

我首先採用方法也是天同的那方法, 也計到1450歐。
至於類似no.回答者, 主要透過電流方法, 我也試過, 但計到的仍是1450歐。
主要考慮電流個個方法, 涉及基爾霍夫第一定律, 會麻煩點。
所以, 我鼓勵用"天同"那個方法。

先計總電流i=v/r=6/200=3ma

當伏特計並聯電阻時，讀數係2.9v，計100歐電阻電流i=v/r=2.9/100==2.9ma

換言之通過伏特計電流係0.1ma，計伏特計電阻r=v/i=2.9/0.01=290歐