Question

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

• A. $4440\Omega$
• B. $1500\Omega$
• C. $7400\Omega$
• D. $2950\Omega$

Answer 1

Answer: (B)

$1500\Omega$

The current flowing through the galvanometer in series with $2950\Omega$ is $I=\frac{V}{R}=\frac{3}{2950+50}=0.001A$

Thus, the galvanometer deflects by 30 divisions for a current of $0.001A$.

Let the added resistance in series be $R$.

The current flowing through the galvanometer now is $\frac{3}{2950+50+R}$

The galvanometer deflection is 20 divisions.

We know that, deflection is proportional to the current in the galvanometer.

Thus, $\frac{20}{30}=\frac{\frac{3}{3000+R}}{0.001}$

$\Rightarrow \frac{3000+R}{3000}=\frac{3}{2}\Rightarrow R=1500\Omega$