GATE 2020 IN Q.25: solution and answer

GATE 2020 · INQ.25 · 1 mark · NAT

Two $100 Ω$ resistors having tolerance $3 %$ and $4 %$ are connected in series. The effective tolerance of the series combination (in %, rounded off to one decimal place) is ____

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Answer and solution

ANSWER

The effective tolerance of the series combination is 3.5 % (worst-case limiting error: 200 Ω ± 7 Ω). The official key accepts 2.5 or 3.5; 2.5 % is the root-sum-square figure.

Official answer: 2.5, or 3.5Source: Official answer key, GATE 2020 IN (IIT Delhi)

A tolerance is a guaranteed (limiting) error, so it is a worst-case bound. First turn each percentage into an absolute error in ohms: $δ R_{1} = 0.03 \times 100 = 3 Ω$ and $δ R_{2} = 0.04 \times 100 = 4 Ω$ .
Rule for a sum: in a series combination $R = R_{1} + R_{2}$ , the worst-case absolute errors add. $δ R = δ R_{1} + δ R_{2} = 3 + 4 = 7 Ω$
Nominal value of the combination: $R = 100 + 100 = 200 Ω$ .
Effective tolerance of the combination: $\frac{δ R}{R} \times 100 = \frac{7}{200} \times 100 = 3.5 %$
Cross-check with the weighted-average form: $\frac{R _{1}}{R} ϵ_{1} + \frac{R _{2}}{R} ϵ_{2} = \frac{1}{2} (3) + \frac{1}{2} (4) = 3.5 %$ . With equal resistors the result is simply the average of the two tolerances, so the combination is $200 Ω \pm 7 Ω$ , that is $\pm 3.5 %$ .
Why the official key also accepts $2.5$ : if the two errors are treated as independent random errors and combined by root-sum-square instead of worst case, $δ R = 3^{2} + 4^{2} = 5 Ω, \frac{5}{200} \times 100 = 2.5 %$ The question does not say which convention to use, so both answers are accepted. For a tolerance (a guaranteed limit), the worst-case value $3.5 %$ is the standard answer.

KEY IDEA

For a sum of quantities, the absolute limiting errors add: 3 Ω + 4 Ω = 7 Ω on a 200 Ω total. So the relative error of a series combination is the resistance-weighted average of the individual tolerances, not their sum. Treating the errors as independent and combining them by root-sum-square (5 Ω) gives 2.5 % instead, and the key accepts both.

WATCH OUT FOR

Adding the percentages directly (3 % + 4 % = 7 %). Percentage errors add only for a product or a quotient. For a sum you add the absolute errors in ohms and then divide by the total resistance, which gives 7/200 = 3.5 %.

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GATE 2020 IN Q.25: solution and answer

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