Question

A copper wire has diameter 0.5 mm and resistivity of $1.6\times {10}^{-8}\Omega m$. What will be the length of this wire to make its resistance $10\Omega$?

Answer 1

Given:

$d=0.5mm=0.5\times {10}^{-3}mr=\frac{d}{2}=\frac{0.5\times {10}^{-3}}{2}=0.25\times {10}^{-3}mA=\Pi r^2=3.14\times (0.25\times {10}^{-3}{)}^2\rho =1.6\times {10}^{-8}\Omega mR=10\Omega WeknowR=\frac{\rho l}{A}orl=\frac{RA}{\rho }=\frac{10\times 3.14\times (0.25{)}^2\times {10}^{-6}}{1.6\times {10}^{-8}}=1.226\times {10}^2=122.6m$