Page 864 - Basic Electrical Engineering

Solution:

Area of the coil = 0.04 × 0.03 m 2
–4
= 12 × 10 m 2

Deflecting torque, T = ϕ NI = BAN I
d
–4
= 0.2 × 12 × 10 × 100 × I Nm

–2
or Td = 2.4 × 10 × I Nm

–6
Control torque produced by the spring when elongated at 48°, T = 1 × 10 ×
c
48 Nm.

For steady deflection, T = T .
d
c
–2
Equating, 2.4 × 10 × I = 48 × 10 –6

Example 11.8 In a moving iron instrument a current of 5 A produces a

deflection of 60°. What will be the deflection when a current of 2 A is
flowing through the coil of the instrument? Assume that the instrument is

spring controlled.

Solution:

We know that for a moving iron instrument deflecting torque is proportional

to the square of the current flowing through the coil.

Therefore, T α I 2
c

When spring controlled, the control torque is proportional to the angle of
deflection.