Question

The change of inductance for a moving-iron ammeter is 2μH/degree. The control spring constant is 5 × 10⁻⁷ N-m/degree. The maximum deflection of the pointer is 100º, what is the current corresponding to maximum deflection ?

Answer 1

As seen from the deflecting torque is given by

T_d = 1/2I²(dL/dθ) N-m

Control spring constant = 5 × 10⁻⁷ N-m/degree 

Deflection torque for 100° deflection = 5 × 10⁻⁷ × 100 = 5 × 10⁻⁵ N-m ; dL/dθ = 2 μH/degree = 2 × 10⁻⁶ H/degree.

∴ 5 × 10⁻⁵ = 1/2 I² × 2 × 10⁻⁶

∴ I² = 50 and I = 7.07 A