Question

A copper wire has diameter 0.5 mm and resistivity of $1.6×10^{–8} \ Ω m$. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer 1

Resistance (R) of a copper wire of length l and cross-section A is given by the expression,
$R = ρ\frac {l}{A}$
Where,
ρ is resistivity of copper = 1.6 × 10–8 Ω m
R = 10 Ω,
radius of wire r = $\frac {0.5}{2}$mm = 0.25 mm = 0.00025 m
𝐴 = 𝜋𝑟2
A = 3.14×(0.00025)2
A = 0.000000019625 m2
⟹𝑙 = $\frac {𝑅𝐴}{ρ}$

⟹𝑙 = $\frac {10×0.000000019625}{1.6×10^{−8}}$
⟹𝑙 = 122.72 𝑚
If the diameter (radius) is doubled, the new radius r = 0.5 mm = 0.0005 m
𝐴 = 𝜋𝑟2
A = 3.14×(0.0005)2
A = 0.000000785 m2
So, the new resistance will be
$𝑅^′=ρ\frac {𝑙}{𝐴}$

$R' = \frac {1.6×10^{−8} \times 122.72}{0.000000785 }$
$R'=2.5\ Ω$
Now,
$\frac {𝑅^′}{𝑅}=\frac {2.5}{10}$

⟹ $\frac {𝑅^′}{𝑅}=\frac {1}{4}$

⟹ $𝑅^′=\frac 14𝑅$

Hence, the new resistance will become $\frac 14$ times the original resistance.