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The inductance of a 25 A electrodynamic ammeter changes uniformly at the rate of 0.0035 μH / degree. The spring constant is 10-6 Nm/degree. The angle of deflection at full scale will be
1. 135°
2. 125°
3. 115°
4. 105°

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Correct Answer - Option 2 : 125°

Concept:

Deflecting torque:

It is proportional to quantity under measurement, this torque deflects the pointer away from initial or zero position.

TD Measurable quantity.

For electrodynamometer, deflecting Torque is defined as:

\({T_D} = {I^2}\frac{{dm}}{{d\theta }}\)      ---(1)

Controlling Torque:

It is opposite to the deflecting Torque when deflecting Torque equals to the controlling Torque Pointer comes to a final steady-state position.

For electrodynamometer, controlling Torque is defined as:

Tc = k θ      ----(2)

At equilibrium,

 Td = Tc       -----(3)

Calculation:

Given:

\(\frac{{dm}}{{d\theta }} = 0.0035\mu H/degree\)

K = 10-6 Nm/degree

From equation (1), (2), (3), we get:

\(K\theta = {I^2}\frac{{dm}}{{d\theta }}\)

\(\theta = \frac{{{I^2}}}{K}\frac{{dm}}{{d\theta }}\)

\(\theta = \frac{{{{\left( {25} \right)}^2}}}{{{{10}^{ - 6}}\frac{{Nm}}{{degree}}}} \times 0.0035 \times {10^{ - 6}}\frac{H}{{degree}}\)

= 2.18 rad

Converting radian into the degree, we get:

\(\theta = 2.18 \times \frac{{180^\circ }}{\pi } = 125^\circ \)

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