In a linear element, the measured current is \(100\) mA \(\pm 2.5\%\) and the measured voltage is \(5\) V \(\pm 5\%\). Which of the following options show(s) the consumed power?
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For a product of two measured quantities, \(P=IV\), the percentage errors add: \(\%\Delta P = \%\Delta I + \%\Delta V\). Check whether each option's uncertainty accounts for both the current's and the voltage's tolerance.
Step 1: Start from the logarithmic form of the error propagation rule.
For $P = IV$, take the natural log of both sides: $\ln P = \ln I + \ln V$. Differentiating gives
\[ \frac{dP}{P} = \frac{dI}{I} + \frac{dV}{V} \]
For the worst-case (maximum) uncertainty, the magnitudes of the two fractional errors are added rather than allowed to cancel:
\[ \left|\frac{dP}{P}\right|_{max} = \left|\frac{dI}{I}\right| + \left|\frac{dV}{V}\right| = 2.5\% + 5\% = 7.5\% \]
Step 2: Get the nominal power.
\[ P = (0.1 \text{ A})(5 \text{ V}) = 0.5 \text{ W} = 500 \text{ mW} \]
Step 3: Apply the 7.5% uncertainty to get the absolute bound.
\[ dP = 0.075 \times 500 \text{ mW} = 37.5 \text{ mW} \]
so $P = (500 \pm 37.5)$ mW, the same as writing $P = 500 \text{ mW} \pm 7.5\%$.
Step 4: Rule out the incomplete options by working backward.
If only the current's error were used, $dP = 2.5\% \times 500 = 12.5$ mW, giving option (B) and, in percentage form, option (D). Since the voltage also carries a $5\%$ tolerance and $P$ depends on it just as directly as on $I$, dropping that term understates the true uncertainty, so (B) and (D) do not represent the full error in the consumed power.
\[ \boxed{P = 500\text{ mW} \pm 7.5\% = (500 \pm 37.5)\text{ mW}} \]
