Question

The resistance of the coil of a tangent galvanometer is $60\Omega$ . It is connected to a battery of negligible internal resistance. The deflection is found to be ${60}^o$. Now a shunt resistance of $30\Omega$ is connected across the coil of the tangent galvanometer. The deflection produced will be :

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${37}^o$

Answer 1

The correct option is C ${60}^o$

Let the potential across the galvanometer be $V$ volts.

Initially current flowing be $i_i=\frac{V}{60}$

Now after the shunt is connected, $R_{eff}=\frac{30\times 60}{30+60}=20\Omega$

So, $i=\frac{V}{20}$

Also from $60i_g=30(i-i_g)⟹i_g=\frac{i}{3}=\frac{\left(\frac{V}{20}\right)}{3}$

Thus, $i_g=\frac{V}{60}$ i-e, current flowing through the galvanometer remains same although after the shunt is connected.

Now $\theta =\frac{nAB}{k}I$

i.e., $\theta \propto I$

Hence deflection$\left(\theta \right)$ will same as $I$ is same.