Q1Electronic Measurement & Instrumentation

Question

The inductance of a moving iron ammeter in micro-henry is given by the expression. $L = 20 + 10 θ - 3 θ^{2}$ Where $θ$ is the deflection in radians from zero position. Determine the deflection of ammeter for a current of $10 A$ if the spring constant is $8 \times 1 0^{- 6} Nm/radian$ .

Answer 1

Given parameters for the moving iron ammeter: - Inductance $L = (20 + 10 θ - 3 θ^{2}) \times 1 0^{- 6} H$ - Current $I = 10 A$ - Spring constant $K = 8 \times 1 0^{- 6} Nm/rad$

The deflecting torque $T_{d}$ for a moving iron instrument is given by: $T_{d} = \frac{1}{2} I^{2} \frac{d L}{d θ}$

First, differentiate $L$ with respect to $θ$ : $\frac{d L}{d θ} = \frac{d}{d θ} (20 + 10 θ - 3 θ^{2}) \times 1 0^{- 6} = (10 - 6 θ) \times 1 0^{- 6} H/rad$

Substitute $I$ and $\frac{d L}{d θ}$ into the torque equation: $T_{d} = \frac{1}{2} (10)^{2} (10 - 6 θ) \times 1 0^{- 6} = 50 (10 - 6 θ) \times 1 0^{- 6} Nm$

At steady-state deflection, Deflecting Torque = Controlling Torque ( $T_{d} = T_{c} = K θ$ ): $50 (10 - 6 θ) \times 1 0^{- 6} = 8 \times 1 0^{- 6} θ$

Cancel $1 0^{- 6}$ from both sides and solve for $θ$ : $500 - 300 θ = 8 θ$ $308 θ = 500$ $θ = \frac{500}{308} \approx 1.623 radians$

Q1Electronic Measurement & Instrumentation

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