GATE 2025 EE Q.31: solution and answer

GATE 2025 · EEQ.31 · 1 mark · NAT

The maximum percentage error in the equivalent resistance of two parallel-connected resistors of $100 Ω$ and $900 Ω$ with each having a maximum $5%$ error is ________ $%$ (round off to nearest integer value).

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Answer and solution

ANSWER

The maximum percentage error in the equivalent resistance is 5%.

Official answer: 5Source: Official answer key, GATE 2025 EE (IIT Roorkee)

Nominal equivalent resistance: $R = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{100 \times 900}{100 + 900} = 90 Ω$ .
Find the limiting error from the total differential, treating $R_{1}$ and $R_{2}$ as independent variables: $\frac{\partial R}{\partial R _{1}} = \frac{R _{2}^{2}}{( R _{1} + R _{2} ) ^{2}}, \frac{\partial R}{\partial R _{2}} = \frac{R _{1}^{2}}{( R _{1} + R _{2} ) ^{2}}$
Divide by $R = \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ to get the relative form: $\frac{δ R}{R} = \frac{R _{2}}{R _{1} + R _{2}} \frac{δ R _{1}}{R _{1}} + \frac{R _{1}}{R _{1} + R _{2}} \frac{δ R _{2}}{R _{2}}$ Both weights are positive and add up to 1, so the worst case is when both resistors are off in the same direction.
Substitute the values: $\frac{δ R}{R}_{m a x} = \frac{900}{1000} (5%) + \frac{100}{1000} (5%) = 4.5% + 0.5% = 5%$
Check with the extreme values. At $+ 5%$ the resistors are $105 Ω$ and $945 Ω$ , so $R = \frac{105 \times 945}{1050} = 94.5 Ω$ , which is $+ 5%$ of $90 Ω$ . At $- 5%$ they are $95 Ω$ and $855 Ω$ , so $R = \frac{95 \times 855}{950} = 85.5 Ω$ , which is $- 5%$ . If both resistors change by the same percentage, the parallel value changes by that same percentage.
Maximum percentage error $= 5%$ .

KEY IDEA

For resistors in parallel, the relative error in the result is a weighted average of the individual relative errors, with weights $R_{2} / (R_{1} + R_{2})$ and $R_{1} / (R_{1} + R_{2})$ . The weights add up to 1, so when every resistor has the same maximum percentage error, the combination has that same maximum percentage error. Another way to see it: conductances add, so $G = G_{1} + G_{2}$ , and a sum of terms that each carry 5% error also carries at most 5%.

WATCH OUT FOR

Using the product/quotient rule on $R_{1} R_{2} / (R_{1} + R_{2})$ as if the numerator and denominator were independent: 5% + 5% for the product plus 5% for the sum gives 15%. The same resistors appear in both, so their errors partly cancel. Simply adding the two errors to get 10% is also wrong. A parallel combination never has a larger percentage error than the worst of its resistors.

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GATE 2025 EE Q.31: solution and answer

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