GATE 2024 IN Q.51: solution and answer

Official answer: 4Source: Official answer key, GATE 2024 IN (IISc Bengaluru)

1. An ideal INA has infinite input resistance, so it draws no current and does not load terminals A-B. The equivalent resistance is therefore that of the bridge alone, seen between A (the right node) and B (the left node).

2. To find the equivalent resistance, deactivate the independent source. The ideal voltage source $V_{in}$ is replaced by a short circuit, which joins the top node T and the bottom node Bot into a single node.

3. From B (left node) to the merged T/Bot node, the two left arms are in parallel: $$(R+\Delta R)\parallel (R-\Delta R)=\frac{(R+\Delta R)(R-\Delta R)}{(R+\Delta R)+(R-\Delta R)}=\frac{R^2-\Delta R^2}{2R}$$ Numerically this is $6\text{k}\Omega \parallel 3\text{k}\Omega =2\text{k}\Omega$.

4. From A (right node) to the merged T/Bot node, the two right arms are also in parallel: $$(R-\Delta R)\parallel (R+\Delta R)=\frac{R^2-\Delta R^2}{2R}$$ Numerically this is $3\text{k}\Omega \parallel 6\text{k}\Omega =2\text{k}\Omega$.

5. The two parallel pairs are in series through the merged node: $$R_{AB}=2\cdot \frac{R^2-\Delta R^2}{2R}=\frac{R^2-\Delta R^2}{R}$$

6. Substitute $R=4.5\text{k}\Omega$ and $\Delta R=1.5\text{k}\Omega$: $$R_{AB}=\frac{4.5^2-1.5^2}{4.5}\text{k}\Omega =\frac{20.25-2.25}{4.5}\text{k}\Omega =\frac{18}{4.5}\text{k}\Omega =4\text{k}\Omega$$ This agrees with the numerical check $2+2=4\text{k}\Omega$.

7. The result is already an integer, so rounding changes nothing: $R_{AB}=4\text{k}\Omega$.