GATE 2025 IN Q.53: solution and answer

GATE 2025 · INQ.53 · 2 marks · NAT

A metallic strain-gauge (SG) with resistance $R_{S G}$ is connected as shown in the figure, where $R_{L 1}$ , $R_{L 2}$ , $R_{L 3}$ represent the lead wire resistances. The SG has a gauge factor of 2 and nominal resistance $R_{N}$ of 125 $Ω$ . When the SG is subjected to a tensile strain of $2 \times 1 0^{- 3}$ , the resulting change in $R_{S G}$ is $Δ R$ . The $Δ R$ value is measured as $Δ R_{M E A S} = R_{E Q 2} - R_{E Q 1}$ . The $R_{E Q 1}$ and $R_{E Q 2}$ are the equivalent resistances measured between the terminals 1 and 2, and terminals 2 and 3, respectively. If $R_{L 1} = R_{L 2} = 5 Ω$ and $R_{L 3} = 4.95 Ω$ , the measured value of tensile strain is ___ $\times 1 0^{- 3}$ (rounded off to two decimal places).

GATE 2025 IN question 53: Three-wire strain gauge connection. — Figure from the GATE 2025 IN paper. It is part of the question.

Answer and solution

ANSWER

The measured tensile strain is $1.80 \times 1 0^{- 3}$ , against the true value of $2 \times 1 0^{- 3}$ .

Official answer: 1.75 to 1.85Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Read the figure: $R_{N}$ and $R_{S G} = R_{N} + Δ R$ are in series on the left. Terminal 1 connects through $R_{L 1}$ to the top of $R_{N}$ , terminal 2 through $R_{L 2}$ to the junction of $R_{N}$ and $R_{S G}$ , and terminal 3 through $R_{L 3}$ to the bottom of $R_{S G}$ .
Between terminals 1 and 2 the path is $R_{L 1}$ , then $R_{N}$ , then $R_{L 2}$ : $R_{E Q 1} = R_{L 1} + R_{N} + R_{L 2}$
Between terminals 2 and 3 the path is $R_{L 2}$ , then $R_{S G}$ , then $R_{L 3}$ : $R_{E Q 2} = R_{L 2} + (R_{N} + Δ R) + R_{L 3}$
Subtract. $R_{N}$ and $R_{L 2}$ cancel: $Δ R_{M E A S} = R_{E Q 2} - R_{E Q 1} = Δ R + R_{L 3} - R_{L 1}$
The true change in the gauge follows from $G = \frac{Δ R / R _{N}}{ε}$ : $Δ R = G R_{N} ε = 2 \times 125 \times 2 \times 1 0^{- 3} = 0.5 Ω$
The lead mismatch is $R_{L 3} - R_{L 1} = 4.95 - 5 = - 0.05 Ω$ , so $Δ R_{M E A S} = 0.5 - 0.05 = 0.45 Ω$
The instrument converts the measured change to strain assuming no lead error: $ε_{m e a s} = \frac{Δ R _{M E A S}}{G R _{N}} = \frac{0.45}{2 \times 125} = 1.8 \times 1 0^{- 3}$

KEY IDEA

In the three-wire connection, $R_{L 2}$ appears in both $R_{E Q 1}$ and $R_{E Q 2}$ and cancels in the difference. Only the mismatch $R_{L 3} - R_{L 1}$ is left as an error added to $Δ R$ . Strain is then $Δ R_{M E A S} / (G R_{N})$ .

WATCH OUT FOR

Using $R_{E Q 2} - R_{E Q 1} = Δ R + R_{L 1} - R_{L 3}$ with the sign reversed gives 0.55 and a strain of 2.2. The path between terminals 2 and 3 contains $R_{L 3}$ , and the path between 1 and 2 contains $R_{L 1}$ . Another slip is reporting the true strain of 2.00 and ignoring the lead mismatch.

From the GATE 2025 IN paper. Every question of this paper · Download the full paper · All 19 Sensing principles questions, with notes

GATE 2025 IN Q.53: solution and answer

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