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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. If the meter is carrying a current of 5 A, the deflection is ____________(a) 2.4(b) 2.0(c) 1.2(d) 1.0This question was posed to me by my college director while I was bunking the class.This intriguing question comes from Advanced Problems on Magnetically Coupled Circuits topic in chapter Resonance & Magnetically Coupled Circuit of Network Theory

Answer»

The CORRECT choice is (c) 1.2

To explain: 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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