Two resistors R₁ = (4 ± 0.8) Ω and R₂ = (4 ± 0.4) Ω are connected in parallel. The equivalent resistance of their parallel combination will be :

Solution

Physics sir · 4 steps

Work with reciprocals: each error in 1/R gets divided by its own R².

1. Central value. Two equal 4 Ω resistors in parallel give 1/R = 1/4 + 1/4 = 1/2, so R = 2 Ω.
2. Differentiate the relation. Take the differential of 1/R = 1/R₁ + 1/R₂. Each term gives -(Δ R)/R². Errors always add, so drop the minus signs.

   (Δ R)/R² = (Δ R₁)/R₁² + (Δ R₂)/R₂²

3. Put in numbers. (Δ R)/2² = 0.8/16 + 0.4/16 = 1.2/16 = 0.075. So Δ R = 0.075 × 4 = 0.3 Ω.
4. So. The parallel combination is (2 ± 0.3) Ω, which is option (c).

The idea to keep. For a parallel combination, do not use the product-over-sum form for errors. Differentiate 1/R directly: (Δ R)/R² is the sum of (Δ Rᵢ)/Rᵢ².

Answer from the NTA final answer key. Solution by PrepWiser.