Question

The full-sclae deflecting torque of a 10 ampere moving iron ammeter is $4\times {10}^{-5}N-m.$ The rate of changes of self inductance of instrument at full scale is

• A. $0.8\mu H/rad$
• B. $0.2\mu H/rad$
• C. $0.4\mu H/rad$
• D. $0.8mH/rad$

Answer 1

The correct option is A $0.8\mu H/rad$

Full scale deflection torque,
$T_d=4\times {10}^{-5}N-m$

Full scale current, I = 10 A

Since,
$T_d=\frac{1}{2}.\frac{I^{2}dL}{d\theta }$

$\frac{dL}{d\theta }=\frac{2T_d}{I^2}=\frac{2\times 4\times {10}^{-5}}{(10{)}^2}=0.8\mu H/rad$