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
Right Answer is:
1.2
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
θ = 1.2 rad