Question

The inductance of a moving iron ammeter is given by the expression $L=(10+5\theta -2{\theta }^2)\mu H$, where $\theta$ is the angular deflection in radians from zero position. Angular deflection (in degrees) for a current of 10 A if the deflection for a current of 5 A is ${45}^o$ is

• A. ${54.7}^o$
• B. ${56.7}^o$
• C. ${62.41}^o$
• D. ${73.5}^o$

Answer 1

The correct option is C ${62.41}^o$

$L=(10+5\theta -2{\theta }^2)\mu H$

$\frac{dL}{d\theta }=(5-4\theta )\mu H/radian$

and also,

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

$\therefore$ $(5-4\theta )\times {10}^{-6}=\frac{2k\theta }{I^2}$ ...(i)

Substituting, $\theta =\frac{\pi }{4}$ and I = 5 A in above expression, we get

$[5-4\left(\frac{\pi }{4}\right)]\times {10}^{-6}=\frac{2k\times \frac{\pi }{4}}{(5{)}^2}$

$[5-\pi ]\times {10}^{-6}=\frac{\pi }{2\times 25}k$

$\frac{50}{\pi }[5-\pi ]\times {10}^{-6}=k$

$k=2.95\times {10}^{-5}$ Nm/radian

Substituting, I = 10 A and k = 2.95 $\times {10}^{-5}$ in equation (i), we get

$(5-4\theta )\times {10}^{-6}=\frac{2\times 2.95\times {10}^{-5}\times \theta }{{10}^2}$

$(5-4\theta )\times {10}^{-6}=5.9\times {10}^{-7}\theta$

$5-4\theta =0.59\theta$

$5=4.59\theta$

$\theta =\frac{5}{4.59}=1.089radian\left(or\right){62.41}^o$