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)

$4450\,\Omega$

Answer 2

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 $R$
$\therefore \frac{2}{3} \times 10^{-3} =\frac{3}{(50+1 R)}$
or $R =4450 \Omega$