Two resistors R₁ = (6.0 ± 0.1) Ω and R₂ = (3.0 ± 0.1) Ω are connected in parallel across a voltage V = (12.0 ± 0.2) V. Total power is calculated as P = V²/Rₑq. The maximum percentage error in P is approximately:
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Question

Two resistors R₁ = (6.0 ± 0.1) Ω and R₂ = (3.0 ± 0.1) Ω are connected in parallel across a voltage V = (12.0 ± 0.2) V. Total power is calculated as P = V²/Rₑq. The maximum percentage error in P is approximately:

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G1 Physics · Accepted Answer

For the parallel combination, 1/Rₑq = 1/R₁ + 1/R₂. Differentiating: ΔRₑq/Rₑq² = ΔR₁/R₁² + ΔR₂/R₂². Here Rₑq = (6×3)/(6+3) = 2 Ω. Hence ΔRₑq/Rₑq = 2[0.1/6² + 0.1/3²] = 2[0.1/36 + 0.1/9] = 0.02778 = 2.778%. Voltage percentage error = (0.2/12.0)×100 = 1.667%. Since P = V²/Rₑq, maximum percentage error in P = 2(1.667%) + 2.778% = 6.112% ≈ 6.11%.