If the resistivity of #12 copper wire (diameter of 80.81 miles) is 10.4 ohms-cm/ft), what is the resistivity of the #18 gaauge copper wire (diameter of 40.30 miles)?
get the question correct so it can be answered. diameter 80 miles, you probably mean 80 mils 10.4 ohms-cm/ft !! that is not the units of resistivity. Resistivity of copper is 17.2 nΩ-m .
There is something wrong with the question. Resistivity is measured in ohm-cm (or ohm-m), not ohm-cm/ft. It is an intrinsic property of the copper and does not depend upon the diameter or length of the wire. The resistivity of copper at 20°C is: rho = 1.72x10^(-8) ohm-m, or 1.72x10^(-6) ohm-cm. It is the same for #12 and #18 copper wire. The resistance R (not resistivity rho) of a wire varies according to the formula: R = rho x L / A ----- (Equation 1) where rho = resistivity (ohm-cm) L = length of wire (cm) A = cross sectional area of wire (cm^2) R = resistance (ohm) The diameters of the two wires are: D12 = 80.81 mils = 0.08081 inch D18 = 40.30 mils = 0.04030 inch where mil is 1/1000 inch or 0.001 inch The resistance of the #18 wire can be calculated relative to the #12 wire as follows: Use Equation 1 twice: R18 = rho x L / A18 --- Eqn 2 R12 = rho x L / A12 --- Eqn 3 Divide Eqn 2 by Eqn 3, sustitute in D, and solve: R18/R12 = A12/A18 = (pi x D12)^2 / (pi x D18)^2 = (D12/D18)^12 R18/R12 = (80.81mil / 40.30mil)^2 = 4.021 or R18 = 4.021 x R12 The #18 wire will have a resistance that is 4.021 times greater than the #12 wire. I hope that helps. The problem needs to be clarified.
