Two resistors of resistances R1=(100±3)Ω and R2=(200±4)Ω are connected in parallel. The equivalent resistance of the combination

1R=1R1+1R2 & ∆RR2=∆R1R12+∆R2R22⇒R=2003=66.7 Ω , ∆R=169≈1.8 Ω

Two resistors of resistances R1=(100±3)Ω and R2=(200±4)Ω are connected in parallel. The equivalent resistance of the combination

1R=1R1+1R2 & ∆RR2=∆R1R12+∆R2R22⇒R=2003=66.7 Ω , ∆R=169≈1.8 Ω

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Two resistors of resistances $R_{1} = (100 \pm 3) Ω$ and $R_{2} = (200 \pm 4) Ω$ are connected in parallel. The equivalent resistance of the combination

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$(66.7 \pm 7) Ω$

$(300 \pm 1.8) Ω$

$(66.7 \pm 1.8) Ω$

$(300 \pm 7) Ω$

answer is A.

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$\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ & $\frac{∆ R}{R^{2}} = \frac{∆ R_{1}}{R_{1}^{2}} + \frac{∆ R_{2}}{R_{2}^{2}}$

$\Rightarrow R = \frac{200}{3} = 66.7 Ω$ , $∆ R = \frac{16}{9} \approx 1.8 Ω$

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Two resistors of resistances R1=(100±3)Ω and R2=(200±4)Ω are connected in parallel. The equivalent resistance of the combination

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Two resistors of resistances $R_{1} = (100 \pm 3) Ω$ and $R_{2} = (200 \pm 4) Ω$ are connected in parallel. The equivalent resistance of the combination