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 ${60}^{\circ }$. Now a shunt resistance of $30\Omega$ is connected across the coil of the tangent galvanometer. The deflection produced will be________${}^{\circ }$

Answer 1

For a tangent galvanometer , $i=k\tan \theta$

$\because i_1=\frac{V}{G}=k\tan {\theta }_1........\left(1\right)$

After shunting , $i_2=\frac{V}{\left(\frac{GR}{R+G}\right)}=k\tan {\theta }_2....\left(2\right)$

Dividing $\left(1\right)$ and $\left(2\right)$ gives $\frac{R+G}{R}=\frac{\tan {\theta }_1}{\tan {\theta }_2};{\theta }_1={60}^{\circ }{\theta }_2=?$

$\frac{30+60}{30}=\frac{\tan {60}^{\circ }}{\tan {\theta }_2}$

$\Rightarrow \tan {\theta }_2=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}\Rightarrow {\theta }_2={30}^{\circ }$