GATE 2022 IN Q.65: solution and answer

GATE 2022 · INQ.65 · 2 marks · NAT

In the Wheatstone bridge circuit shown, $R_{1} = 1.5 k Ω$ and $R_{2} = R_{3} = R_{4} = 1 k Ω$ . The switch is initially open and the voltage between the points C and D is $V_{C D}$ . Upon closing the switch at $t = 0$ , the resistance in the arm AD changes by an amount $δ R_{1}$ , and the voltage between C and D changes by $δ V_{C D}$ . The sensitivity of the bridge (in volt/kiloohm), defined as $\frac{δ V _{C D}}{δ R _{1}}$ , is ________ (round off to two decimal places)

GATE 2022 IN question 65: A diamond-shaped Wheatstone bridge with node A at the top, B at the bottom, D on the left and C on the right. — Figure from the GATE 2022 IN paper. It is part of the question.

Answer and solution

ANSWER

The sensitivity is $\frac{δ V _{C D}}{δ R _{1}} = \frac{28}{17} \approx 1.65 V/k Ω$ .

Official answer: 1.62 to 1.72Source: Official answer key, GATE 2022 IN (IIT Kharagpur)

Take node B as the $0 V$ reference, so node A is at $10 V$ . Each half of the bridge is a voltage divider. For the right half: $V_{C} = 10 \cdot \frac{R _{4}}{R _{2} + R _{4}} = 10 \cdot \frac{1}{1 + 1} = 5 V$ . The switch does not touch this half, so $V_{C}$ stays the same after it closes.
Switch open: $V_{D} = 10 \cdot \frac{R _{3}}{R _{1} + R _{3}} = 10 \cdot \frac{1}{1.5 + 1} = 4 V$ , so $V_{C D} = V_{C} - V_{D} = 5 - 4 = 1 V$ .
Switch closed: the $30 k Ω$ branch is connected across arm AD, so it sits in parallel with $R_{1}$ : $R_{1}^{'} = \frac{1.5 \times 30}{1.5 + 30} = \frac{45}{31.5} = \frac{10}{7} k Ω \approx 1.4286 k Ω$ .
Change in arm AD: $δ R_{1} = \frac{10}{7} - \frac{3}{2} = - \frac{1}{14} k Ω \approx - 0.0714 k Ω$ .
New potential at D: $V_{D}^{'} = 10 \cdot \frac{R _{3}}{R _{1}^{'} + R _{3}} = 10 \cdot \frac{1}{1 + 10/7} = \frac{70}{17} \approx 4.1176 V$ , so $V_{C D}^{'} = 5 - \frac{70}{17} = \frac{15}{17} \approx 0.8824 V$ .
Change in output: $δ V_{C D} = \frac{15}{17} - 1 = - \frac{2}{17} V \approx - 0.1176 V$ .
The question defines sensitivity as the ratio of these actual (finite) changes: $\frac{δ V _{C D}}{δ R _{1}} = \frac{2/17}{1/14} = \frac{28}{17} \approx 1.65 V/k Ω$

KEY IDEA

Closing the switch puts the 30 kΩ branch in parallel with $R_{1}$ , which lowers arm AD by a finite $δ R_{1} = - \frac{1}{14} k Ω$ . Only the divider potential at D changes; $V_{C}$ stays at 5 V. Work out $V_{C D}$ before and after, and divide the finite change in output by the finite change in resistance, exactly as the question defines it.

WATCH OUT FOR

Using the small-signal derivative $\frac{d V _{C D}}{d R _{1}} = \frac{10 R _{3}}{( R _{1} + R _{3} ) ^{2}} = \frac{10}{2. 5 ^{2}} = 1.60 V/k Ω$ at $R_{1} = 1.5 k Ω$ . That value falls outside the key, because the change of $- 0.0714 k Ω$ is too large to treat as infinitesimal. The question asks for the ratio of the actual changes. Another slip is to add 30 kΩ in series with $R_{1}$ . In the figure, the switch and the 30 kΩ resistor join both ends of arm AD, so they form a parallel path.

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 14 Bridges questions, with notes

GATE 2022 IN Q.65: solution and answer

Answer and solution

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