Answer to Question #238201 in Mechanical Engineering for DILLIP

Question

The inductance of moving iron ammeter in micro henry is given by expression: L=20+5θ-2θ² Where θ is deflection in radians from the zero position. Determine the angular in radians for a current of 10A if the for a current of 5A is 30⁰.Also determine the spring constant

Answer 1

$$L=20+5\theta-2\theta^2, \mu H$$ $$\dfrac{dL}{d\theta}=5-4\theta , \mu H/rad$$ $$\theta=\dfrac{I^2}{2k}\cdot\dfrac{dL}{d\theta}$$ $$k=\dfrac{I^2}{2\theta}(5-4 \theta )\times10^{-6}$$

For a current $I=10\ A,$ $\theta=30\degree=\dfrac{\pi}{6}\ rad.$

$$k=\dfrac{(10)^2}{2\big(\dfrac{\pi}{6}\big)}\bigg(5-4 \big(\dfrac{\pi}{6}\big)\bigg)\times10^{-6}$$ $$=\bigg(\dfrac{15}{\pi}-2\bigg)\times10^{-4} Nm/rad\approx2.775\times10^{-4}\ Nm/rad$$ $$2k\theta\times10^{6}=5I^2-2I^2\theta$$ $$\theta=\dfrac{2.5I^2}{10^{6}k+I^2}$$ $$\theta=\dfrac{2.5(10)^2}{10^{6}\cdot\big(\dfrac{15}{\pi}-2\big)\times10^{-4}+(10)^2}\ rad$$ $$\approx0.6623\ rad\approx38\degree$$ $$\theta=38\degree$$ $$k=2.775\times10^{-4}\ Nm/rad$$