Two resistors R1 = (4 ± 0.8) Ω and R2 = (4 ± 0.4) Ω are connected in… | JEE Main 2021 (1 Sep, Shift 2)

Q47JEE Main 2021 · 1 SEP · SHIFT 2Medium

Combination of Resistors

Two resistors R1 = (4 $\pm$ 0.8) $\Omega$ and R2 = (4 $\pm$ 0.4) $\Omega$ are connected in parallel. The equivalent resistance of their parallel combination will be:

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Answer: (c)

$R_p = \dfrac{R_1 R_2}{R_1 + R_2} = \dfrac{16}{8} = 2\ \Omega$ .
From $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ : $\dfrac{\Delta R_p}{R_p^2} = \dfrac{\Delta R_1}{R_1^2} + \dfrac{\Delta R_2}{R_2^2}$
$\Delta R_p = 4\left(\dfrac{0.8}{16} + \dfrac{0.4}{16}\right) = 0.3\ \Omega$
So the answer is (c) $(2 \pm 0.3)\ \Omega$ .

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