A copper wire has length $L=3.5\text{m}$ and cross‑sectional area $A=0.9{\text{mm}}^2$. The resistivity of copper at $20^{\circ }\text{C}$ is $\rho =1.68\times 10^{-8}\Omega \text{m}$. Calculate the resistance of the wire.

Step-by-step solution

1. Write the resistance formula $R=\rho L/A$.
2. Convert the area from ${\text{mm}}^2$ to ${\text{m}}^2$: $0.9{\text{mm}}^2=0.9\times 10^{-6}{\text{m}}^2$.
3. Substitute the values: $R=(1.68\times 10^{-8}\Omega \text{m})\times (3.5\text{m})/(0.9\times 10^{-6}{\text{m}}^2)$.
4. Perform the calculation: $R=(5.88\times 10^{-8})/(0.9\times 10^{-6})\Omega =6.533\times 10^{-2}\Omega$? Wait, recalc correctly: $5.88\times 10^{-8}/9.0\times 10^{-7}=0.0653\Omega$. Oops, mis‑placed decimal. Actually $A=0.9\times 10^{-6}$, so $R=1.68\times 10^{-8}\times 3.5/(0.9\times 10^{-6})=5.88\times 10^{-8}/9.0\times 10^{-7}=0.0653\Omega$.
5. Check the arithmetic: $5.88\times 10^{-8}/9.0\times 10^{-7}=0.0653\Omega$ which is the correct resistance.
6. Round to two decimal places: $R\approx 0.07\Omega$.

Concept tested: Resistance from resistivity

Published 13 September 2026 · How this solution was prepared and checked