Current Electricity Charging Discharging Of Capacitors Question 297
Question: 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]
Options:
A) 0.01
B) 0.02
C) 0.03
D) 0.05
Show Answer
Answer:
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.
BETA
