GATE 2018 IN Q.51: solution and answer

Official answer: 0.48 to 0.51Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. Gage factor: $G=\frac{\Delta R/R}{\epsilon }$, so $\Delta R=G\epsilon R=2.5\times 400\times 10^{-6}\times 1000=1\Omega$.

2. The strain is positive, so the gage resistance rises: $R_g=1000+1=1001\Omega$.

3. The gage $R_g$ is the bottom-left arm of the bridge; the other three arms are $1000\Omega$. The $2\text{V}$ source is across the top and bottom nodes, with its $+$ terminal at the top node. Take the bottom node as $0\text{V}$. $V_o$ is read from the left middle node ($+$) to the right middle node ($-$).

4. Left divider, measured from the bottom node: $V_L=2\times \frac{R_g}{1000+R_g}=2\times \frac{1001}{2001}\text{V}$.

5. Right divider: $V_R=2\times \frac{1000}{1000+1000}=1\text{V}$.

6. $$V_o=V_L-V_R=\frac{2002}{2001}-1=\frac{1}{2001}\text{V}\approx 0.4998\text{mV}\approx 0.50\text{mV}$$

7. Check with the quarter-bridge approximation: $V_o\approx \frac{V_s}{4}\frac{\Delta R}{R}=\frac{2}{4}\times 10^{-3}\text{V}=0.5\text{mV}$, which agrees.