An LVDT has mutual inductances M₁,M₂ to its two secondary coils. Their self-inductances remain constant with displacement x. At x = 0, M₁ = M₂ = M₀. Over ±10 mm, M₁,M₂ change linearly and oppositely; each changes in magnitude by 0.25M₀ at either endpoint. The secondary excitation phasors are V₁ at 0° and V₂ at 180°. At x = D the detector voltage is zero when |V₂| = 1.25|V₁|. Find D in mm, to one decimal place.
For the symmetric LVDT shown, the opposing induced primary voltages null when $M_1|V_1|=M_2|V_2|$. Thus $M_1/M_2=1.25$. The linear changes give $M_1=M_0(1+0.025D)$ and $M_2=M_0(1-0.025D)$ with D in mm. Solve $1+0.025D=1.25(1-0.025D)$ to obtain $0.05625D=0.25$, so D = 4.444 mm, rounding to 4.4 mm.
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