Question

A galvanometer of resistance $50\Omega$ is connected to a battery of $3V$ 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
$I=\frac{3}{(50+2950)}={\left(10\right)}^{-3}\text{A}$
Current for $30$ divisions $=10^{-3}\text{A}$
Current for $20$ divisions
$=\frac{10^{-3}}{30}\times 20=\frac{2}{3}\times 10^{-3}\text{A}$

To obtain a deflection of $20$ divisions, let resistance added be $R$
$\therefore \frac{2}{3}\times 10^{-3}=\frac{3}{(50+R)}$
$orR=4450\Omega$