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?

Given,

diameter = 0.5 mm

area = $\dfrac{πd^2}{4} = \dfrac{3.14 \times (5 \times 10^{-4})^2}{4}$ = 19.625 x 10^–8 m

resistivity = 1.6 × 10^–8 Ω m

R = 10 Ω

l = ?

We know, R = ρ $\dfrac{l}{A}$

Substituting we get,

10 = 1.6 × 10^–8 x $\dfrac{l}{19.625 \times 10^{-8}}$

l = $\dfrac{10 \times 19.625}{1.6 }$ = 122.7 m

The length of the wire is 122.72 m.

When the diameter is doubled

According to the formula,

R = ρ $\dfrac{l}{A}$ where A = $\dfrac{πd^2}{4}$

Hence, R ∝ $\dfrac{1}{d^2}$

When diameter is double,

Resistance will change by $\dfrac{1}{(2d)^2}$ = $\dfrac{1}{4d^2}$

∴ Resistance will become $\dfrac{1}{4}^{\text{th}}$ of its original value.

Hence, new resistance = $\dfrac{1}{4} \times 10$ = 2.5 Ω