## How many amps is this cable good for?

The answer to that, I’m afraid, is another question. How many volts can you afford to lose?

Voltage drop is inevitable, unfortunately. Ohm’s Law says that the voltage dropped across a cable will be equal to the current flowing in it, times its resistance. Resistance, in turn, is proportional to length, and inversely proportional to cross-sectional area.

Note: Although cable cross-sectional areas are often given in square millimetres, here I will be talking about areas in square metres. Trust me: the maths is much easier if you use “whole” units throughout, because you don’t have to worry about the prefices later. The answer will always come out in “whole” units, and *then* you can add a prefix to make it more manageable.

Cable sizes may be given in one of three ways:

- Square millimetres. This is the easiest to deal with; there are 1 000 000 of these in a square metre, so just use the exponent button on your calculator. For instance, if the area is 2.5 sq. mm., enter it as 2.5 [EXP] 6 [+/-].
- Strand count and diameter: for instance, 16/0.2. This means 16 strands of 0.2 mm. diameter. To get the area of each strand, use π / 4 * D * D. Then multiply this by the number of strands. The area of a 16/0.2 cable would be 16 * π / 4 * 0.2 * 0.2 = 0.503 sq. mm. = 0.503 E-6 sq. m. (Attention, pure maths pedants: Before you point out that that should be written 5.03 E-7, which it would be if we were doing pure maths, look up “engineers’ notation”. Sometimes it’s just more useful to have the exponent be a multiple of three, and therefore correspond with a named prefix, than to have the mantissa lie between 1 and 10. That’s how your calculator will display it by default, but the [ENG] button adjusts the exponent downwards to the nearest multiple of three. Or [Inv] [ENG] adjusts it upwards.)
- AWG number. Don’t bother converting this. Just look it up on the Internet.

Now you know the cross-sectional area of the conductor, you just need to divide the resistivity of the material by the cross-sectional area to find the resistance per unit length. Resistivity is measured in ohm-metres (that’s a product, not a ratio) and represents the resistance between opposite faces of a one-metre cube of the material in question. (It will be a small number, because a square metre is very big compared to the cross-sectional area of a real-life cable: a millimetre is a thousandth of a square metre, so a square millimetre is a millionth of a square metre. A lump with a cross-sectional area of one square metre is equivalent to a million cables of one square millimetre in parallel.) Divide that by square metres, and you get ohms __per__ metre. Multiply that by the length of the cable, double it (to allow for the return conductor), and you have the resistance. Finally, multiply that by amperes to get the voltage drop.

**Example 1**: The resistivity of copper at 20°C is 1.68 E-8 Ωm. Our cable has a cross-sectional area of 1.5mm². This gives a resistance of 1.68 E-8 / 1.5 E-6 = 0.0112 ohms per metre. Which means that each metre of this cable will drop 0.0112 volts per ampere.

If we were using this for a mains extension lead carrying 13 amps, then at full load the voltage drop would be 0.1456 V/m., and the power dissipation would be 1.9 watts per metre. But both the live and neutral conductors are dropping this voltage, so the total voltage drop is 0.2912 V/m., and the power dissipated will be 3.8 W/m. A metre of 1.5 mm² three-core flex has a much bigger surface area than a 4 watt resistor, so it should easily be able to dissipate that heat to its surroundings. (*Unless it’s coiled up in a confined space! How hot do you think a 40 watt light bulb is going to get in a container the size of a ten-metre cable drum?*)

**Example 2**: What size of copper cable would be adequate for a lamp drawing 800mA from a 12V supply, if the maximum acceptable voltage drop in both conductors combined is 0.5V along the 5m. length?

We want to drop no more than 0.05V per metre per conductor at 0.8A, which corresponds to a resistance of 0.0625 ohms / metre. So the cross-sectional area will be 1.68 E-8 / 0.0625 = 2.688 E-7 m² = 0.2688 mm². The next commonly-available size up is 0.5mm².