Question

A galvanometer of resistance $50\, Ω$ is connected to a battery of $3V$ along with a resistance of $2950\, Ω$ in series shows full-scale deflection of $30$ divisions. The additional series resistance required to reduce the deflection to $20$ divisions is

• A. $1500 \, \Omega$
• B. $4450 \, \Omega$
• C. $7400 \, \Omega$
• D. $2950 \, \Omega$

Answer 1

Answer: (B)

Current flowing in galvanometer ,
$I=\frac{3}{(50+2950)}$
$I=10^{-3} \,A$

Current for 30 division $=10^{-3} \,A$
Current for 20 division $=\frac{10^{-3}}{30} \times 20=\frac{2}{3} \times 10^{-3} \,A$
Let the series resistance $=R$
$ \therefore \frac{2}{3} \times 10^{-3} =\frac{3}{(50+R)}$
$ R =4450 \,\Omega$