Question

A galvanometer of resistance $50\Omega$ is connected to a battery of $3V$ along with a resistance $2950\Omega$ 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. $4450\Omega$
• C. $6050\Omega$
• D. $5550\Omega$

Answer 1

Answer: (B)

$R^{\prime }=(n-1)(G+R)$
$=\left(\frac{30}{20}-1\right)3000=1500\Omega$
Total resistance $=2950+1500=4450\Omega$