Question

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?

• A. \((500 \pm 37.5)\) mW
• B. \((500 \pm 12.5)\) mW
• C. \(500\) mW \(\pm 7.5\%\)
• D. \(500\) mW \(\pm 2.5\%\)

Hint:

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.

Answer 1

The Correct Option is A, C

Step 1: Recall how errors combine in a product.
The consumed power in a linear element is \(P=IV\). When two measured quantities are multiplied, the maximum possible relative (percentage) error in the product is the sum of the individual relative errors:
\[ \%\Delta P = \%\Delta I + \%\Delta V \]
This worst-case addition is used because the two uncertainties can, in the least favourable case, both push the result in the same direction at once.

Step 2: Compute the nominal power.
\[ P = I \times V = 100 \text{ mA} \times 5 \text{ V} = 500 \text{ mW} \]

Step 3: Add the percentage errors.
\[ \%\Delta P = 2.5\% + 5\% = 7.5\% \]

Step 4: Convert the percentage error into an absolute value.
\[ 7.5\% \text{ of } 500 \text{ mW} = 0.075 \times 500 = 37.5 \text{ mW} \]
So the power can be written either as \((500 \pm 37.5)\) mW in absolute form, or as \(500\) mW \(\pm 7.5\%\) in percentage form; these two are the same result written two ways.

Step 5: Check each option.
Option (A), \((500 \pm 37.5)\) mW, matches Step 4 exactly, so it is correct. Option (C), \(500\) mW \(\pm 7.5\%\), is the same value in percentage form, so it is also correct. Option (B), \((500 \pm 12.5)\) mW, corresponds to only a \(2.5\%\) error (\(12.5/500=2.5\%\)); this is the current's tolerance alone with the voltage's \(5\%\) tolerance dropped, so it is wrong. Option (D), \(500\) mW \(\pm 2.5\%\), makes the same mistake in percentage form and is also wrong, for the same reason.

Final Answer:
Both (A) and (C) correctly state the consumed power; (B) and (D) understate the uncertainty by ignoring the voltage's error contribution. \[ \boxed{P = (500 \pm 37.5)\text{ mW} = 500\text{ mW} \pm 7.5\%} \]