Question:

Published on: 27 June, 2022

** How that driving torque in a moving iron instrument is given by \(\theta = \frac{1}{2K}i^2\frac{\text{d}L}{\text{d}\theta}\) . Where the symbols have their usual meaning.**

Answer:

Let ‘\(\theta\)’ be the deflection corresponding to a current of ‘I’ amp

Let the current increases by di, the corresponding deflection is ‘\(\theta+d\theta\)’

There is change in inductance since the position of moving iron change w.r.t the fixed electromagnets.

Let the new inductance value be ‘L+dL’. the current change by ‘di’ is dt seconds.

Let the emf induced in the coil be ‘e’ volt.

\(e = \frac{d}{dt}(Li) = L\frac{di}{dt}+i\frac{dL}{dt}\)

Mutiplying by ‘idt’ in equation (1.22)

\(e × idt =L\frac{di}{dt}×idt +i\frac{dL}{dt}×idt\)

\(e×idt =Lidi+i^2dL\)

Equation (1.24) gives the energy is used in two forms. part of energy is stored in the inductance. Remaining energy is converted in to mechanical energy which produces deflection.

Change in energy stored=final energy-initial energy stored

\(=\frac{1}{2} (L+dL)(i+di)^2-\ \frac{1}{2}Li^2\)

\(=\frac{1}{2}{(L+dL) (i^2 +di^2 +2idi) - Li^2}\)

\(= \frac{1}{2}\ {(L+dL) (i^2 + 2idi) - Li^2}\)

\(= \frac{1}{2} {Li^2 +2Ldi+ i^2dl +2ididL - Li^2}\)

\(= \frac{1}{2} {2Lidi + i^2dL}\)

\(= Lidi + \frac{1}{2}\ i^2dL\)

Mechanical work to move the pointer by d\(\theta\)

\(= T_dd\theta\)

By law of conservation of energy,

Electrical energy supplied=Increase in stored energy + Mechanical work done

Input energy = Energy stored + mechanical energy

\(Lidi + i^2dL= Lidi+ \frac{1}{2} i^2dL+T_dd\theta\)

\(\frac{1}{2} i^2dL= T_dd\theta\) ...(1.28)

\(T_d= \frac{1}{2}\ i^2\frac{\text{d}L}{\text{d}\theta}\)

At steady condition \(T_d=T_C\)

\(\frac{1}{2} i^2\frac{\text{d}L}{\text{d}\theta} =K\theta\)

\(\theta= \frac{1}{2K}\ i^2\frac{\text{d}L}{\text{d}\theta}\)

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