Question

# The inductance of a certain moving-iron ammeter is expressed as: L=10+3θ−(θ2/4)μH, where θ is the deflection in radians from the zero position. The control spring torque in 25×10−6Nm/radian. The deflection of the pointer in radian when the meter carries a current of 5 A, is

A
2.4
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B
2.0
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C
1.2
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D
1.0
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Solution

## The correct option is C 1.2Deflecting torque in moving-iron ammeter, Td=12I2dLdθ Inductance, L=(10+3θ−θ24)μH Rate of change of inductance with deflection, dLdθ=ddθ(10+3θ−θ24) Current, I=5A Deflecting torque, Td=12I2dLdθ Td=12×52×(3−θ2)×10−6 =252(3−θ2)×10−6Nm Controlling torque, Tc=kθ=25×10−6θ At equilibrium, Tc=Td 25×10−6θ=252(3−θ2)×10−6 5θ2=3 θ=1.2rad

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