GATE 2022 IN – Question 65
The Wheatstone bridge has $R_1=1.5$ k$\Omega$ and $R_2=R_3=R_4=1$ k$\Omega$. Initially the switch is open and the voltage between C and D is $V_{CD}$. Closing it at $t=0$ changes the AD-arm resistance by $\delta R_1$ and C–D voltage by $\delta V_{CD}$. Its sensitivity $|\delta V_{CD}/\delta R_1|$ is ______ V/k$\Omega$ (two decimal places).
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Correct answer: 1.62 to 1.72
Explanation
Closing adds 30 kΩ parallel to 1.5 kΩ, giving R1′=10/7 kΩ and δR1=−1/14 kΩ. VD changes from 10/2.5=4 V to 10/(17/7)=70/17 V; VC stays 5 V.
Thus |δVCD/δR1|=(2/17)/(1/14)=**28/17=1.65 V/kΩ**.
This uses the finite switched difference specified in the question. The infinitesimal derivative at the initial resistance would be 1.60 V/kΩ.