Question

The inductance of attraction type instrument is given by L = (10 + 5θ − θ²)μH where θ is the deflection in radian from zero position. The spring constant is 12 × 10⁻⁶ N-m/rad. Find out the deflection for a current of 5A.

Answer 1

L = (10 + 5θ − θ²) × 10⁻⁶ H

∴ dL/dθ = (0 + 5 − 2 × θ) × 10^–6

= (5 − 2θ) × 10⁻⁶ H/rad

Let the deflection be θ radians for a current of 5A, then deflecting torque, 

T_d = 12 × 10⁻⁶ × θ N-m

Also, T_d = 1/2 I² dL/dθ

Equating the two torques, we get

12 × 10⁻⁶ × θ = 1/2 × 5² × (5 − 2θ) × 10⁻⁶

∴ θ = 1.689 radian