# The inductance of a certain moving-iron ammeter is expressed as $L = 10 + 3\theta – \dfrac{{{\theta ^2}}}{4}\mu H$

The inductance of a certain moving-iron ammeter is expressed as $L = 10 + 3\theta - \dfrac{{{\theta ^2}}}{4}\mu H$ where θ is the deflection in radians from the zero position. The control spring torque is 25 × 106 Nm/radian. The deflection of the pointer in radians, when the meter carries a current of 5 A RMS is

#### SOLUTION

Given inductance

$L = 10 + 3\theta – \dfrac{{{\theta ^2}}}{4}\mu H$

Control spring torque (K) = 25 × 106 Nm/radian

Current I = 5A

The pointer will come to rest at a position where controlling torque TC, is equal to the deflecting torque Td

TC = Td

$\begin{array}{l}25{\rm{ }} \times {\rm{ }}{10^{ – 6}} \times \theta = \dfrac{1}{2}{(5)^2}\dfrac{{d\left( {10 + 3\theta – \dfrac{{{\theta ^2}}}{4}} \right)}}{{d\theta }}\\\\25{\rm{ }} \times {\rm{ }}{10^{ – 6}} \times \theta = \dfrac{1}{2}{(5)^2} \times \left( {3 – \dfrac{\theta }{2}} \right) \times {10^{ – 6}}\\\\\theta = \dfrac{1}{2}\left( {3 – \dfrac{\theta }{2}} \right)\end{array}$

5θ/2 =3