question_answer

Two resistances of 400 W and 800 W are connected in series with 6 volt battery of negligible internal resistance.  A voltmeter of resistance 10,000 W is used to measure the potential difference across 400 W.  The error in the measurement of potential difference in volts approximately is                                                  [EAMCET 2003]

A)         0.01                                  

B)         0.02

C)         0.03                                  

D)         0.05

Correct Answer: D

Solution :

                                         Before connecting voltmeter potential difference across 400W 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 \[{{V}_{f}}=\frac{384.6}{(384.6+800)}\times 6=1.95V\] Error in measurement = \[{{V}_{i}}-{{V}_{f}}=2-1.95\] = 0.05V.