Q80 / 90 · JEE Main 2021 (01 Sep Shift 2) · Physics

Two resistors $R_{1} = (4 \pm 0.8) Ω$ and $R_{2} = (4 \pm 0.4) Ω$ are connected in parallel. The equivalent resistance of their parallel combination will be :

A
$(4 \pm 0.4) Ω$
B
$(2 \pm 0.4) Ω$
C
$(4 \pm 0.3) Ω$
Correct answer
D
$(2 \pm 0.3) Ω$

Solution

$\frac{Δ R}{R^{2}} = \frac{Δ R_{1}}{R_{1}^{2}} + \frac{Δ R_{2}}{R_{2}^{2}}$
$Δ R = {(R)}^{2} [\frac{Δ R_{1}}{R_{1}^{2}} + \frac{Δ R_{2}}{R_{2}^{2}}]$
$= {(2)}^{2} [\frac{0.8}{4^{2}} + \frac{0.4}{4^{2}}]$
$= [0.2 + 0.1] = 0.3$
$R_{net} = R + Δ R = 2 \pm 0.3$

Hence, the correct answer is option: 'D'.

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

Solution

Relevant Questions

Two resistors $R_{1} = (4 \pm 0.8) Ω$ and $R_{2} = (4 \pm 0.4) Ω$ are connected in parallel. The equivalent resistance of their parallel combination will be :