GATE 2023 IN Q.54: solution and answer

GATE 2023 · INQ.54 · 2 marks · NAT

Four strain gauges R_A, R_B, R_C and R_D, each with nominal resistance R, are connected in a bridge configuration. When a force is applied, R_A and R_D increase by ΔR and R_B and R_C decrease by ΔR as shown. A potentiometer with total resistance R_v is connected as shown. If R = 100 Ω, and ΔR = 1 Ω, the minimum value of resistance R_v required to balance the bridge is ______ Ω (rounded off to two decimal places).

GATE 2023 IN question 54: Wheatstone bridge excited by source V_B (plus terminal up) across the top node and ground. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The minimum pot resistance needed to balance the bridge is $R_{v} \approx 4.04 Ω$ .

Official answer: 4.00 to 4.10Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Circuit: the sensors are in all four arms of a full bridge. The supply $V_{B}$ drives the top node. Left arm: $R_{A}$ then $R_{C}$ then the left part $R_{1}$ of the potentiometer, down to the grounded wiper. Right arm: $R_{B}$ then $R_{D}$ then the right part $R_{2}$ of the potentiometer, down to the same ground. The output $V_{o}$ is taken between the left node (between $R_{A}$ and $R_{C}$ ) and the right node (between $R_{B}$ and $R_{D}$ ). With $R = 100 Ω$ and $Δ R = 1 Ω$ : $R_{A} = R_{D} = 101 Ω$ and $R_{B} = R_{C} = 99 Ω$ . The wiper splits the pot as $R_{1} + R_{2} = R_{v}$ , with $R_{1}, R_{2} \geq 0$ .
Left node voltage (potential divider): $V_{L} = V_{B} \frac{99 + R _{1}}{101 + 99 + R _{1}} = V_{B} \frac{99 + R _{1}}{200 + R _{1}}$
Right node voltage: $V_{R} = V_{B} \frac{101 + R _{2}}{99 + 101 + R _{2}} = V_{B} \frac{101 + R _{2}}{200 + R _{2}}$
Balance requires $V_{o} = V_{L} - V_{R} = 0$ , so $(99 + R_{1}) (200 + R_{2}) = (101 + R_{2}) (200 + R_{1})$
Expand. The $R_{1} R_{2}$ terms cancel: $19800 + 99 R_{2} + 200 R_{1} = 20200 + 101 R_{1} + 200 R_{2}$ which gives $99 R_{1} - 101 R_{2} = 400$
Put $R_{1} = R_{v} - R_{2}$ : $99 R_{v} - 200 R_{2} = 400$ , so $R_{v} = \frac{400 + 200 R _{2}}{99}$
Since $R_{2} \geq 0$ (the wiper cannot go past the end of the pot), $R_{v}$ is smallest when the wiper sits at the end stop, $R_{2} = 0$ .
Then $99 R_{v} = 400$ , so $R_{v} = \frac{400}{99} \approx 4.04 Ω$

KEY IDEA

Write the two arm voltages as potential dividers to the grounded wiper. Balance gives the linear condition $99 R_{1} - 101 R_{2} = 400$ . The pot's total resistance is smallest when the wiper is at one end stop, so $R_{2} = 0$ .

WATCH OUT FOR

Treating the grounded wiper as a simple series trim in one arm, or ignoring that the pot is shared between both arms. Another slip is balancing with $R_{1} = R_{2}$ (wiper at the centre) and getting a larger $R_{v}$ , instead of finding the smallest pot that can reach balance, which puts the wiper at the end stop.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 19 Sensing principles questions, with notes

GATE 2023 IN Q.54: solution and answer

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