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Two resistors of resistances $R_{1} = 100 \pm 3 Ω a n d R_{2} = 200 \pm 4 Ω$ are connected in parallel, then the equivalent resistance in parallel is (in ohm) $(U s e \frac{1}{R'} = \frac{1}{R_{1}} + \frac{1}{R_{2}} a n d \frac{Δ R'}{Δ R'^{2}} = \frac{Δ R_{1}}{R_{1}^{2}} + \frac{Δ R_{2}}{R_{2}^{2}})$

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66.7±1.8

300±7

150.8±2

92.3±3

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detailed solution

Correct option is A

The equivalent resistance of parallel combination is R'=R1R2R1+R2=2003=66.7 ΩFrom, 1R'=1R1+1R2, we getΔR'R'2=ΔR1R12+ΔR2R22ΔR'=(R'2)ΔR1R12+(R'2)ΔR2R22=(66.7100)23+(66.7200)24=1.8 ΩHence, R'=(66.7±1.8)Ω

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Two resistors of resistances R1=100±3Ω and R2=200±4Ω are connected in parallel, then the equivalent resistance in parallel is (in ohm) (Use 1R'=1R1+1R2 and ΔR'ΔR'2=ΔR1R12+ΔR2R22)

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