Question

A galvanometer of resistance $50\Omega$ is connected to a battery of $3V$ along with a resistance of $2950\Omega$ 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 $=R$
$\therefore \frac{2}{3}\times 10^{-3}=\frac{3}{(50+R)}$
$R=4450\Omega$