A metallic wire having resistance of 20 Ω is cut into two equal parts

A metallic wire having resistance of 20 Ω is cut into two equal parts in length. These parts are then connected in parallel. The resistance of this parallel combination is equal to

20 Ω
10 Ω
5 Ω
15 Ω
This question was previously asked in
UPSC NDA-2 – 2020
The original metallic wire has a resistance R = 20 Ω. When it is cut into two equal parts in length, the resistance of each part is halved, assuming the material and cross-sectional area remain uniform. So, each part has a resistance R₁ = R₂ = R/2 = 20 Ω / 2 = 10 Ω. When these two parts are connected in parallel, the equivalent resistance (Rₚ) is given by the formula 1/Rₚ = 1/R₁ + 1/R₂. Substituting the values, 1/Rₚ = 1/10 Ω + 1/10 Ω = 2/10 Ω = 1/5 Ω. Therefore, the equivalent resistance Rₚ = 5 Ω.
The resistance of a wire is directly proportional to its length. For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
For two resistors R₁ and R₂ in parallel, the equivalent resistance can also be calculated as Rₚ = (R₁ * R₂) / (R₁ + R₂). In this case, Rₚ = (10 Ω * 10 Ω) / (10 Ω + 10 Ω) = 100 / 20 = 5 Ω.
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