Q. 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
Explanation
(d) 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
$V_{f} = \frac{384.6}{(384.6 + 800)} \times 6 = 1.95V$
Error in measurement $= V_{i} - V_{f} = 2 - 1.95 = 0.05V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.