Question

When $2$ ampere current is passed through a tangent galvanometer, it gives a deflection of ${30}^o$. For ${60}^o$ deflection, the current must be

• A. $1$ amp
• B. $2\sqrt{3}$amp
• C. $4$amp
• D. $6$amp

Answer 1

The correct option is D $6$amp

From the above concept $I\propto \tan \theta$

$\frac{I_1}{I_2}=\frac{\tan {\theta }_1}{\tan {\theta }_2}$

$\frac{2}{I_2}=\frac{\tan 30}{\tan 60}$

$\frac{2}{I_2}=\frac{\frac{1}{\sqrt{3}}}{\sqrt{3}}=\frac{1}{3}$

$\Rightarrow I_2=2\times 3=6A$