GATE 2018 IN Q.47: solution and answer

Official answer: 2.49 to 2.53Source: Official answer key, GATE 2018 IN (IIT Guwahati)

1. A full-scale accuracy fixes the same absolute error at every point on the scale. For the voltmeter: $\delta V=\pm 1\%\times 300\text{V}=\pm 3\text{V}$.

2. At the reading of $200\text{V}$, that is a relative error of $\frac{3}{200}\times 100=\pm 1.5\%$.

3. For the ammeter: $\delta I=\pm 0.5\%\times 5\text{A}=\pm 0.025\text{A}$.

4. At the reading of $2.5\text{A}$, that is a relative error of $\frac{0.025}{2.5}\times 100=\pm 1\%$.

5. The resistance is a quotient: $R=\frac{V}{I}=\frac{200}{2.5}=80\Omega$. A limiting error is the worst case, so the relative errors of a quotient add; they do not subtract: $\frac{\delta R}{R}=\pm \left(\frac{\delta V}{V}+\frac{\delta I}{I}\right)$.

6. $$\frac{\delta R}{R}=\pm (1.5\%+1\%)=\pm 2.50\%$$

7. So the limiting error in the computed load resistance is $2.50\%$, which is $\pm 2\Omega$ on $80\Omega$.