Calculating voltage drop across connected cables.

Thread Starter


Joined Mar 21, 2009
Here is the question:

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.
Here is my attempt.

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.


Joined Jul 23, 2009
Hi Zaraphrax,

Although you had a real good start, you have a real bad math error. Note that the resistivity of copper compared to aluminum is less than a 1:2 ratio. Since the lengths and cross-sectional areas are equal, the resistances had better be in a less than 1:2 ratio of copper to Aluminum and your's is way off.

After you get that fixed, the way you went about calculating the voltage drop is fine, but for the currents in each, you need to use the current dividing principle (the dual of the voltage dividing principle). Here's how it goes:

If you have 2 resistors in parallel, the current will divide this way:
Icopper = Itotal(RAluminum)/(Rcopper + RAluminum).

Notice it's almost the same as the voltage divider formula, except that for the resistor that you want the current of, the numerator gets multiplied by the other resistor (in this case, the Aluminum resistor), then divided by the sum. This is for two resistors in parallel only.

Good luck,
Kamran Kazem


Joined Jul 7, 2009
Zaraphrax, I'm not trying to pick on you, but I suspect your arithmetic error came about because you blindly trusted your calculator and didn't look at the resistance numbers. How could one of them be 12 orders of magnitude different than the other? Before calculators, one had to make approximations to the answers in one's head to make sure the order of magnitude of the answer was about right. This was because the intermediate calculations would be done on a slide rule which knew nothing about exponents.

I maintain it's still important for students to learn to do this order of magnitude estimating, even with a calculator. The reason is, as your example shows, that you had no way to check your answer. I've seen this happen time and again with folks using calculators. And of course I've made the same mistakes myself, many times.

Here's one way to do the approximate calculation:

Leave the resistivity in nohms. Then (good, you remembered to convert square mm to square m)

Rcopper = 18 * (200/(40(1e-6))

Now, round 18 to 20 to get 20*200*1e6/40 = (4000/40)*1e6 = 400/4*1e6 = 100*1e6 = 1e8 nohms. Since you rounded 18 up to 20, you know the answer is about about 10% too high. If you wanted a closer answer, subtract 1e7 from 10e7 to get 9e7 nohms or 9e-2 ohms or, finally, 90 mΩ.

Does what I did make sense? I did it without a calculator -- just simple arithmetic and the ability to add and subtract exponents. It's one of the most powerful techniques an engineer/scientist can learn because it gives you a way of checking your work. The basic technique is simply rounding things off to 1 significant figure and doing the resultant arithmetic. Physicists call it an order of magnitude calculation because it's purpose is to get you about within a power of 10 of the right answer. Of course, with not much care, it can do better than that.

Once you get the voltage drop, it's easy to figure the current in each wire -- the current is the voltage drop divided by the resistance of the wire. No need to use the formula kkazem gave unless you want to.

Again, I'm not trying to make you feel bad. But if there is one thing I would try to teach to all students in the world when doing arithmetic, it's this technique of checking your work by checking the order of magnitude of the answer. You'll really appreciate it when you're doing design work on real problems and a mistake can cost money, time, or injury.

Thread Starter


Joined Mar 21, 2009
Thanks guys - that's good advice. I obviously wrote it down wrong when I was doing the calculation (I have a really bad habit of doing that), becasue now that I've read this and walked through the problem again, I can see what I did wrong.

Thanks again. :)