GATE 2026 IN Q.29: solution and answer

GATE 2026 · INQ.29 · 1 mark · MSQ

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)
$500 mW \pm 2.5%$

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Answer and solution

ANSWER

Options (A) and (C) are correct. The consumed power is $500 mW \pm 7.5%$ , which is the same as $(500 \pm 37.5) mW$ .

Official answer: A, CSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Nominal power: $P = V I = 5 V \times 100 mA = 500 mW$ .
Rule for a product: taking logarithms, $ln P = ln V + ln I$ , so $\frac{d P}{P} = \frac{d V}{V} + \frac{d I}{I}$ . A limiting (guaranteed) error assumes the worst case, where both errors push the same way, so the relative limiting errors add in magnitude: $\frac{δ P}{P} = \pm (\frac{δ V}{V} + \frac{δ I}{I})$ .
Substituting: $\frac{δ P}{P} = \pm (5% + 2.5%) = \pm 7.5%$ , so $P = 500 mW \pm 7.5%$ . Option (C) is correct.
In absolute terms: $δ P = 0.075 \times 500 mW = 37.5 mW$ , so $P = (500 \pm 37.5) mW$ . This is the same result as (C) written as an absolute error, so option (A) is also correct.
Check by extremes: $P_{m a x} = 5.25 V \times 102.5 mA = 538.125 mW$ (that is $+ 7.625%$ ) and $P_{m i n} = 4.75 V \times 97.5 mA = 463.125 mW$ (that is $- 7.375%$ ). The exact extremes differ from $\pm 7.5%$ only by the cross-product term $(0.05) (0.025) = 0.125%$ of $P$ , which is $0.625 mW$ . For small errors this second-order term is neglected, which is why the standard limiting error is $\pm 7.5%$ , i.e. $\pm 37.5 mW$ .
Options (B) and (D) say the same thing: $\frac{12.5}{500} = 2.5%$ . That equals the current's error alone, or the difference $5% - 2.5%$ . It is not the sum of the errors, so both options are wrong. (The root-sum-square value, $5^{2} + 2. 5^{2} % \approx 5.59%$ , is not offered either.)

KEY IDEA

When quantities are multiplied or divided, their limiting (worst-case) relative errors add in magnitude. For $P = V I$ : $\pm (5% + 2.5%) = \pm 7.5%$ of $500 mW$ , which is $\pm 37.5 mW$ . An MSQ can list the same result twice, once as a percentage and once as an absolute error, and then both options are correct.

WATCH OUT FOR

Picking only one of (A) and (C) when they state the same error in two forms. Another mistake is subtracting the relative errors ( $5% - 2.5% = 2.5%$ , giving (B)/(D)) or taking the root-sum-square ( $\approx 5.59%$ ). A limiting error is worst-case, so the magnitudes add.

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GATE 2026 IN Q.29: solution and answer

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