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Two resistances are given as R₁=(10±0.5) Ω and R₂=(15±0.5) Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

Asked in JEE Main 6th April 1st Shift 2023 · Error in a sum or difference

Answer: (3) 4.33

Step-by-step solution

Given: R₁=10±0.5 Ω, R₂=15±0.5 Ω.

Equivalent: Rₚ=(R₁R₂)/(R₁+R₂)=(150)/(25)=6 Ω.

Idea: differentiating 1/(Rₚ)=1/(R₁)+1/(R₂) gives (Δ Rₚ)/(Rₚ²)=(Δ R₁)/(R₁²)+(Δ R₂)/(R₂²).

Δ Rₚ=36((0.5)/(100)+(0.5)/(225))=36(0.005+0.00222)=0.26 Ω.

Percentage: (0.26)/6×100=4.33%.

Why the other options are wrong

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