A copper wire has diameter 0.5 mm and resistivity of 1.6 x 108 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: Radius, r = 0.25 mm = 0.25 x 10-3 m
Length of the wire, l = 1
Resistance, R = 10 Ω
Resistivity, ρ = 1.6 x 108 Ω m
Applying the formula R = \(\frac{\rho l}{A}\)
= \(\frac{\rho l}{\pi r^2}\), we get
l = \(\frac{R\pi r^2}{\rho l}\)
= \(\frac{10 \times 3.14 \times 0.25 \times 10^{-3}\times 0.25 \times 10^{-3}}{1.6 \times 10^{-8}}\)
= \(\frac{10 \times 3.14 \times 0.0625 \times 100}{1.6}\)
= 122.6 m
Again R = \(\frac{\rho l}{A}\) = \(\frac{\rho l}{\frac{\pi d^2}{4}}\) [where d is diameter of wire]
R ∝ \(\frac{1}{d}\)
Thus on doubling the diameter, the resistance is reduced to \(\frac{1}{4}\)th of the original value.
New resistance = \(\frac{10}{4}\) = 2.5Ω
Change in resistance = 10 - 2.5 = 7.5 Ω