A 4Ω resistance wire doubled on it. Calculate the new resistance of the wire.

CONCEPT:

• Resistance (R): The resistance offered to the flow of current is known as the resistance.
• SI unit of resistance is the ohm (Ω).
• Mathematically resistance can be written as

\(⇒ R = \frac{{\rho l}}{A}\)

Where R = resistance, l = length, A = area of cross-section and ρ = resistivity

EXPLANATION:

• The wire is doubled on it means to fold the length of wire. That is its length will get half and the area of the cross-section will get doubled.

Given: R₁ = R = 4 Ω 

As we know that the resistance of the wire is,

\(\Rightarrow R = \frac{{\rho l}}{A}\)

When the wire is doubled then, then its area will increase automatically. But the volume of the wire will be the same.

∴ The volume of original wire = volume of new wire

⇒ A1l1 = A2l2

⇒ A1l1 = A2 l₁/2

⇒ A2 = 2A1

• The resistance of the wire in 1st case

\(\Rightarrow {R_1} = R = \frac{{\rho {l_1}}}{{{A_1}}}\)       ----(1)

• The resistance of the wire in 2nd case

\(\Rightarrow {R_2} = \frac{{\rho {l_2}}}{{{A_2}}} = \frac{{\rho {\frac{l_1}{2}}}}{{{2A_1}}} = \frac{{\rho {l_1}}}{{{4A_1}}}\)         -----(2)

Divide equation 1 and 2, we get

\(\Rightarrow \frac{R_1}{{{R_2}}} = \frac{{\frac{{\rho {l_1}}}{{{A_1}}}}}{ \frac{{\rho {l_1}}}{{{4A_1}}}} = 4\)

⇒ R2 = 1 Ω