Question

The inductance of a certain moving-iron ammeter is expressed as:
$L=10+3\theta -\left({\theta }^2/4\right)\mu H$, where $\theta$ is the deflection in radians from the zero position. The control spring torque in $25\times {10}^{-6}Nm/radian.$ The deflection of the pointer in radian when the meter carries a current of 5 A, is

• A. 2.4
• B. 2.0
• C. 1.2
• D. 1.0

Answer 1

The correct option is C 1.2

Deflecting torque in moving-iron ammeter,
$T_d=\frac{1}{2}{I}^2\frac{dL}{d\theta }$

Inductance,
$L=(10+3\theta -\frac{{\theta }^2}{4})\mu H$

Rate of change of inductance with deflection,
$\frac{dL}{d\theta }=\frac{d}{d\theta }(10+3\theta -\frac{{\theta }^2}{4})$

$Current,I=5A$

Deflecting torque,
$T_d=\frac{1}{2}{I}^2\frac{dL}{d\theta }$

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

$=\frac{25}{2}(3-\frac{\theta }{2})\times {10}^{-6}Nm$

Controlling torque,
$T_c=k\theta =25\times {10}^{-6}\theta$

At equilibrium,

$T_c=T_d$

$25\times {10}^{-6}\theta =\frac{25}{2}(3-\frac{\theta }{2})\times {10}^{-6}$

$\frac{5\theta }{2}=3$

$\theta =1.2rad$