Question

The inductance of a moving iron instrument is given by $L=(18+5\theta -4{\theta }^2)\mu H$, (where is the deflection in radian from zero position). If the spring constant is ${10}^{-5}$ Nm/rad, the deflection for 5 A current will be

• A. 1.69 radian
• B. 0.568 radian
• C. 32.54 degree
• D. 96.82 degree

Answer 1

Answer: (B), (C)

32.54 degree

For a moving iron instrument, the deflection $\left(\theta \right)$ is given by,

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

$L=18+5\theta -4{\theta }^2$,

$So,\frac{dL}{d\theta }=5-8\theta$

$\therefore$ $\theta =\frac{1}{2}\times \frac{25}{{10}^{-5}}(5-8\theta )\times {10}^{-6}$

$as,I=5A,K={10}^{-5}Nm/rad)$

$\Rightarrow$ $\theta =1.25(5-8\theta )$

$\Rightarrow$ $\theta =6.25-10\theta$

$\Rightarrow$ $11\theta =6.25$

$\Rightarrow$ $\theta =0.568radian={32.54}^o$