Question

A moving iron ammeter produces a full scale torque of $240\mu Nm$ with deflections of ${120}^{\circ }$ at a current of 10 A. The rate of change of self inductance $\left(\mu H/radian\right)$ of the instrument at full scale is

• A. $2.0\mu H/radian$
• B. $4.8\mu H/radian$
• C. $12.0\mu H/radian$
• D. $114.6\mu H/radian$

Answer 1

The correct option is B $4.8\mu H/radian$

Torque produced,

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

where,
$I=10A$

$andT=240\mu N-m$

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

$240\times {10}^{-6}=\frac{1}{2}\times {10}^2\times \frac{dL}{d\theta }$

Rate of change of self-inductance,

$\frac{dL}{d\theta }=4.8{\times }^{-6}H/radian$

$\Rightarrow \frac{dL}{d\theta }=4.8\mu H/radian$