Question

The inductance of a moving iron instrument is given by $L=(10+5\theta -{\theta }^2)$ $\mu H$, where $\theta$ is the deflection in degree from zero position. The spring constant is $3\times {10}^{-6}Nm/degree$ then the deflection for a currrent of 5 A.

• A. 2.23 rad
• B. 2.23 degree
• C. 127.76 degree
• D. 8.03 rad

Answer 1

The correct option is B 2.23 degree

Rate of change of inductance with deflection,

$\frac{dL}{d\theta }=\frac{d}{d\theta }(10+5\theta -{\theta }^2)=(5-2\theta )\mu H/degree$

The deflection is,

$\theta =\frac{1}{2}\cdot \frac{I^2}{K}\cdot \frac{dL}{d\theta }$

$\theta =\frac{1}{2}\times \frac{\left(5^2\right)}{3\times {10}^{-6}}\times (5-2\theta )\times {10}^{-6}$

$\theta =\frac{1}{2}\times \frac{25}{3}\times (5-2\theta )$

$6\theta =25(5-2\theta )$

$\Rightarrow \theta =\frac{125}{56}=2.23degree$