Question A8.8: A copper wire has a length, l, of 1.5 km, a resistance, R, o...
A copper wire has a length, l, of 1.5 km, a resistance, R, of 5 Ω and a resistivity, ρ, of 17.2 × 10^{-6} Ω mm. Find the cross-sectional area, a, of the wire, given the following relationship:
R=\frac{\rho l}{a}Once again, it is worth getting into the habit of summarizing what you know from the question and what you need to find. Also, don’t forget to include the units for each one of the quantities that you know.
R = 5 Ω
ρ = 17.2 × 10^{-6} Ω mm
l = 1500 × 10^{3} mm
a = ?
Now since
R=\frac{\rho l}{a}then 5=\frac{\left(17.2 \times 10^{-6}\right)\left(1,500 \times 10^{3}\right)}{a}
Cross multiplying (i.e. exchanging the ‘5’ for the ‘a’) gives:
a=\frac{\left(17.2 \times 10^{-6}\right)\left(1,500 \times 10^{3}\right)}{5}Now group the numbers and the powers of 10 as shown below:
a=\frac{17.2 \times 1,500 \times 10^{-6} \times 10^{3}}{5}Next simplify as far as possible:
a=\frac{17.2 \times 1,500}{5} \times 10^{-6+3}Finally, evaluate the result using your calculator:
a = 5,160 × 10^{-3} = 5.16
Don’t forget the units! Since we have been working in mm, the result, a, will be in mm^{2}.
Hence a = 5.16 mm^{2}