Question

A given wire having resistance R is stretched so as to reduce its diameter to half of its previous value. What will be its new resistance?

Answer 1

If l is the length of a wire and A is its area of cross-section, then resistance of the wire.

R = ρ\(\frac{l}{A}\)

When its diameter is reduced to half, its area of cross-section becomes \(\frac{A}{4}\)

i.e. A₁ = \(\frac{A}{4}\)

If l₁ is the new length, then

l₁A₁ = lA

or l₁ \(\frac{A}{4}\) = lA

or l₁ = 4l

∴ New resistance,

R₁ = \(\frac{\rho4l}{A/4}\) = 16 (\(\rho\)\(\frac{l}{A}\)) = 46R

or R₁ = 16R.