The resistance of a wire is defined as
R = (ρL)/A
where
ρ = resistivity
L = length
A = cross-sectional area
Before the wire is drawn, let
L₁ = the length
A₁ = the cross-sectional area.
The volume of the wire and its resistivity remain unchanged after it is drawn.
After it is drawn, the new length is
L₂ = 4L₁
To preserve the volume, the new cross-sectional area is
A₂ = (L₁A₁)/L₂ = (L₁A₁)/(4L₁) = A₁/4
The initial resistance is given as
R₁ = (ρL₁)/A₁ = 5.3 Ω
The final resistance is
[tex]R_{2}= \frac{\rho L_{2}}{A_{2}} = \frac{\rho (4L_{1})}{A_{1}/4} = 16 (\frac{\rho L_{1}}{A_{1}}) = 16*5.3 \, \Omega = 84.8 \, \Omega[/tex]
Answer: 84.8 Ω
