Two resistances are given as R₁ = (10 ± 0.5) Ω and R₂ = (15 ± 0.5) Ω… | JEE Main 2023 (6 Apr, Shift 1)

Q67JEE Main 2023 · 6 APR · SHIFT 1Easy

Errors in Measurement

Two resistances are given as $R_1 = (10 \pm 0.5)\ \Omega$ and $R_2 = (15 \pm 0.5)\ \Omega$ . The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

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Answer: (c)

$R_p = \frac{R_1R_2}{R_1 + R_2} = \frac{150}{25} = 6\ \Omega$
From $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$ : $\frac{\Delta R_p}{R_p^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$
$\Delta R_p = 36\left(\frac{0.5}{100} + \frac{0.5}{225}\right) = 0.26\ \Omega$
Percentage error $= \frac{0.26}{6} \times 100 \approx 4.33\%$
So the answer is (c).

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