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The inductance of a certain moving- iron ammeter is expressed as L = 10 + 3θ – \(\frac{θ^2}{4}\) μH, where θ is the deflection in radian from the zero position. The control spring torque is 25 × 10^-6 Nm/rad. The meter carries a current of 5 A. What is the deflection?(a) 2.4(b) 2.0(c) 1.2(d) 1.0I got this question in an interview.My question is based upon Advanced Problems on Indicating Instruments topic in chapter Extension of Instrument Ranges of Electrical Measurements

Answer» RIGHT answer is (C) 1.2

The explanation is: At EQUILIBRIUM,

Kθ = \(\FRAC{1}{2} I^2 \frac{dl}{dθ} \)

(25 × 10^-6) θ = \(\frac{1}{2} I^2 (3 – \frac{θ}{2}) \) × 10^-6

∴ 2 θ + \(\frac{θ}{2}\) = 3

Or, θ = 1.2.


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