Q. A galvanometer of resistance 50$\ \Omega$ is connected to a battery of 3V along with a resistance of 2950$\ \Omega$ in series. A full scale deflection of 30 divisions is obtained in the galvanometer. In order to reduce this deflection to 20 divisions, the resistance in series should be
A) 5050$\ \Omega$
B) 5550$\ \Omega$
C) 6050$\ \Omega$
D) 4450$\ \Omega$
Explanation
(d) Current through the galvanometer
$I = \frac{3}{(50 + 2950)} = 10^{- 3}A$
Current for 30 divisions $= 10^{- 3}A$
Current for 20 divisions
$= \frac{10^{- 3}}{30} \times 20 = \frac{2}{3} \times 10^{- 3}A$
For the same deflection to obtain for 20 divisions, let resistance added be R
$\therefore\ \frac{2}{3} \times 10^{- 3} = \frac{3}{(50 + 1R)}\ $
$or\ \ \ R = 4450\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.