GATE 2020 IN Q.53: solution and answer

GATE 2020 · INQ.53 · 2 marks · NAT

The mutual inductances between the primary coil and the secondary coils of a linear variable differential transformer (LVDT) shown in the figure are M1 and M2. Assume that the self-inductances Ls1 and Ls2 remain constant and are independent of x. When x = 0, M1 = M2 = M0. When x is in the range ±10 mm, M1 and M2 change linearly with x. At x = +10 mm or −10 mm, the change in the magnitudes of M1 and M2 is 0.25 M0. For a particular displacement x = D, the voltage across the detector becomes zero when |V2| = 1.25|V1|. The value of D (in mm, rounded off to one decimal place) is ___

GATE 2020 IN question 53: LVDT schematic. — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

The detector voltage is zero at $D \approx 4.4 mm$ , with the core displaced towards the upper secondary $L_{s 1}$ .

Official answer: 4.3 to 4.6Source: Official answer key, GATE 2020 IN (IIT Delhi)

Linear variation with a change of $0.25 M_{0}$ at $∣ x ∣ = 10 mm$ gives the slope $k = \frac{0.25}{10} = 0.025 mm^{- 1}$ , as a fraction of $M_{0}$ per mm.
In the figure $+ x$ points upward, towards the upper secondary $L_{s 1}$ . So $M_{1} = M_{0} (1 + k x)$ and $M_{2} = M_{0} (1 - k x)$ .
Each secondary carries a current set by its own source and self-inductance: $I_{1} = \frac{V _{1}}{j ω L _{s 1}}$ and $I_{2} = \frac{V _{2}}{j ω L _{s 2}}$ . Each current induces a voltage in the primary: $M_{1} I_{1}$ from the first and $M_{2} I_{2}$ from the second, times $j ω$ . Both secondaries have the same dot position, and the sources are $18 0^{\circ}$ apart, so the two induced voltages oppose.
The detector voltage is zero when the magnitudes are equal: $\frac{M _{1} ∣ V _{1} ∣}{L _{s 1}} = \frac{M _{2} ∣ V _{2} ∣}{L _{s 2}}$ . The two secondaries are identical coils in the symmetric figure, so take $L_{s 1} = L_{s 2}$ . Then $M_{1} ∣ V_{1} ∣ = M_{2} ∣ V_{2} ∣$ .
With $∣ V_{2} ∣ = 1.25∣ V_{1} ∣$ this gives $\frac{M _{1}}{M _{2}} = 1.25$ .
Substitute: $\frac{1 + k x}{1 - k x} = 1.25 \Rightarrow 1 + k x = 1.25 - 1.25 k x \Rightarrow 2.25 k x = 0.25 \Rightarrow k x = \frac{1}{9}$ .
$D = \frac{1}{9 \times 0.025} mm = 4.44 mm \approx 4.4 mm$ This lies inside the $\pm 10 mm$ linear range and inside the key range 4.3 to 4.6.

KEY IDEA

The detector reads zero when the two opposing induced voltages are equal in magnitude. With equal self-inductances this means $M_{1} ∣ V_{1} ∣ = M_{2} ∣ V_{2} ∣$ . Since $M_{1}$ and $M_{2}$ vary linearly and oppositely with $x$ , the ratio $M_{1} / M_{2} = 1.25$ fixes the displacement.

WATCH OUT FOR

Using $M_{1} / M_{2} = ∣ V_{1} ∣/∣ V_{2} ∣ = 0.8$ (the inverted ratio) gives a negative displacement. Another slip is to take the slope as $0.25$ per $10 mm$ of $M_{0}$ but apply it to $M$ rather than to the fraction of $M_{0}$ .

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 2 Motion & force questions, with notes