Here is my attempt.A copper cable and an aluminium cable are connected in parallel to supply a 230 A load. Both have a length of 200 metres and cross-sectional areas of 40 square millimeters. At the normal working temperatures of the cables, the resistivity of aluminium is 28 nohms/metre and copper's is 18 nohms/metre. Calculate the voltage drop, and the currents carried by each of the cables.

Given that the resistance of each of the cables will be represented by the following equation:

R=roh * (l/a)

roh = resistivity of material

l = length of material

a = cross sectional area

Rcopper = 18x10^-9 * (200/4x10^-5) = 2.42 x 10^-12 Ohms

Raluminium=28x10^-9 * (200/4x10^-5) = 0.14 Ohms

Because they're connected in parallel, the equivalent resistance would be 2.42x10^-12 Ohms.

Then, using Ohm's law to get the voltage drop

V = 240 * 2.42x10^-12 = 5.8x10^-10 V

My book says the answer should be 12.4 V. I'm not sure whether what I've done is just a Maths error, or my thinking is totally off. Either way, I'd appreciate some input because I've gone over my numbers several times and can't find the answer. Thanks.