Question

The deflection angle of the pointer of an ideal moving iron ammeter is 20° for 1.0 ampere dc current. If a current of 3 sin (314 t) amperes is passed through the ammeter then the deflection angle is

• A. 0°
• B. 42°
• C. 60°
• D. 90°

Answer 1

The correct option is D 90°

In the moving iron ammeter,

$\theta \propto i_{rms}^2$

$\frac{20}{\theta }=\frac{(1{)}^2}{{\left(\frac{3}{\sqrt{2}}\right)}^2}=\frac{2}{9}$

$\therefore \theta ={90}^{\circ }$