If the current through a moving iron instrument is increased by 20%, what is the percentage increase in the deflection torque?

This question was previously asked in

ISRO (VSSC) Technical Assistant Electrical 2017 Official Paper

Option 4 : 44

CT 1: Network Theory 1

11532

10 Questions
10 Marks
10 Mins

**Concept:**

In moving iron instruments, the deflecting torque is unidirectional (acts in the same direction) whatever may be the polarity of the current.

The deflecting torque is given by

**\({T_d} = {I^2}\frac{{dL}}{{dθ }}\)**

Where,

I is operating current

L is self-inductance

θ is deflection

\({T_d}\propto {I^2}\)

So that the deflection torque of the moving iron instrument is proportional to the square of the RMS value of the operating current and change in self-inductance.

**Calculation:**

When the current through a moving iron instrument is increased by 20%, then deflection torque is

\({T'_d}\propto {{(1.2I)}^2}\)

∴ T'_{d} = 1.44 T_{d}

Hence **deflection torque is increased by 44%**