Here is the question:
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.
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.