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

[amp_mcq option1=”20 Ω” option2=”10 Ω” option3=”5 Ω” option4=”15 Ω” correct=”option3″]

This question was previously asked in

UPSC NDA-2 – 2020

Download PDF Attempt Online

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.

More similar MCQ questions for Exam preparation:-