Question

A galvanometer of resistance $50 \, \Omega $ is connected to a battery of $3 \, V$ 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 $

Answer 1

Answer: (D)

Current through the galvanometer
$I=\frac{3}{\left(\right. 50 + 2950 \left.\right)}=\left(10\right)^{- 3 \, }\text{A}$
Current for 30 divisions $=10^{- 3 \, }\text{A}$
Current for 20 divisions
$=\frac{10^{- 3}}{30}\times 20=\frac{2}{3}\times 10^{- 3 \, }\text{A}$

For the same deflection to obtain for 20 divisions, let resistance added be $R$
$\frac{2}{3} \times 10^{-3}=\frac{3}{(50+R)}$ or $\quad R=4450 \Omega$