Two resistors are measured as R₁ = (100 ± 2) Ω and R₂ = (200 ± 4) Ω and are connected in parallel. The maximum percentage error in the equivalent resistance R is:
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Two resistors are measured as R₁ = (100 ± 2) Ω and R₂ = (200 ± 4) Ω and are connected in parallel. The maximum percentage error in the equivalent resistance R is:

G1 Physics · Accepted Answer

For parallel combination, 1/R = 1/R₁ + 1/R₂. Differentiating: -dR/R² = -dR₁/R₁² - dR₂/R₂². For maximum error, ΔR/R² = ΔR₁/R₁² + ΔR₂/R₂². Hence ΔR/R = R(ΔR₁/R₁² + ΔR₂/R₂²). Here R = (100×200)/(100+200) = 200/3 Ω. Therefore ΔR/R = (200/3)[2/100² + 4/200²] = (200/3)(0.0002 + 0.0001) = 0.02. Thus maximum percentage error = 2%.