question_answer

Two resistances of \[400\Omega \]and \[800\Omega \] are connected in series with 6 volt battery of negligible internal resistance. A voltmeter of resistance \[10,000\Omega \] is used to measure the potential difference across\[400\Omega \]. The error in the measurement of potential difference in volt approximately is

A) 0.01     

B)        \[0.02\]     

C) \[0.03\]                         

D) \[0.05\]

Correct Answer: D

Solution :

Before connecting voltmeter potential difference across \[400\,\Omega \] resistance is \[{{V}_{i}}=\frac{400}{(400+800)}\times 6=2V\]

After connecting voltmeter equivalent resistance between A and B \[=\frac{400\times 10,000}{(400+10,000)}=384.6\Omega \]

Hence, potential difference measured by voltmeter \[{{V}_{f}}=\frac{384.6}{(384.6+800)}\times 6=1.95\,V\]

Error in measurement \[={{V}_{i}}-{{V}_{f}}=2-1.95=0.05V\]