Question

When 2 Amperes current is passed through a tangent galvanometer, it gives a deflection of ${30}^0$. For ${60}^0$ deflection, the current must be

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

Answer 1

The correct option is B 6 A

Tangent galvanometer

For a tangent galvanometer, if deflection is $\theta$, then current reading is given by

$i=ktan\theta$

Where, reduction factor galvanometer

$k=\frac{2r{B}_H}{{\mu }_{0}n}$

(r= radius of coil, n= no. of turns)

$\Rightarrow i\propto tan\theta$

$\therefore \frac{i_1}{i_2}=\frac{tan{\theta }_1}{tan{\theta }_2}$

$\Rightarrow \frac{2A}{i_2}=\frac{tan{30}^0}{tan{60}^0}$

$\Rightarrow i_2=(2\times 3)=6A$

Hence, the correct option is (D)