Question

A piece of wire of resistance 20 Ω is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.

Answer 1

We know that : The Longer the Length of the Wire, The Greater the Resistance it offers to the Electric Current Flow.

Let the Original Length of the Wire be : L₁

Given : This Piece of Wire offers a Resistance of 20Ω

⇒ R₁ = 20Ω

Given : The Length of this Wire is increased to Twice it Original Length

⇒ New Length (L₂) = 2L₁

As : Length of the Wire and Resistance are Directly Proportional

$\bold{\implies \frac{R_1}{R_2} = \frac{L_1}{L_2}}$

$\bold{\implies \frac{20}{R_2} = \frac{L_1}{2L_1}}$

⇒ R₂ = 40Ω

⇒ The Resistance of the Wire in the New Situation is 40Ω