Wire Resistance, Ampacity and the Skin Effect

Calculating the resistance of an electrical cable can be a real pain in the arse, of course you could just measure it but not everyone has a low Ohm meter laying around and sometimes you just want a rough figure for a given cable length, this can also be of great value when checking a cable since you'll have a figure for what is to be expected.

Wire measures systems are frankly a total mess with wire commonly sold in AWG (American Wire Gauge), SWT (Standard Wire Gauge) and more recently IEC 60228 which is a move in the right direction.

Cross Sectional Area

For most cases the only measurements you need are the Cross Sectional Area and length, and diameter CSA is important since it's used to calculate current capacity and resistance.

In the case of stranded or braided wires the total CSA is the sum of each individual strand so for example if you have a cable with 7 wires each with a CSA of 0.4mm2 the total is 2.8mm2

The CSA is fairly easy to calculate, if it's not available, although trying to measure a very small wire is a challenge, in that case it may be easier just to treat it as a solid core wire, first take the diameter and calculate the radius, from there you can calculate the CSA:

RadiusCross Sectional Area
`r=d/2`
`A=r^2*pi`


Resistivity and Resistance

The resistivity of materials is specified in Ohm's per meter (ρ), for common metals this is:

Copper (annealed)1.724 x10-8
Aluminium2.82 x10-8
Gold2.44 x10-8
Iron9.71 x10-8
Nichrome1.10 x10-6

Since the above table and practically all tables on material resistivity are in Ohm's per meter you need to convert it to Ohm's per mm2.

`rho_{mm^2} = rho / {1*10^-3}`

This equation in full is known as Pouillet's law which can be used to calculate the resistance for any length or area.

`R = {rhoL}/A`

So for copper this comes to 0.01724 Ohms per mm2
For aluminium 0.0282 Ohm's per mm2
You can then calculate the resistance for your length and CSA, for example 10 meters of 2.5mm2 copper wire:

`{0.01724 * 10} / {2.5} = 68.96mOmega`

or if you just want to do it all in one go:

`({1.724*10^-8}/{1*10^-3}) \/ 2.5 * 10 = 68.96mOmega`

Another example with a 7 strand aluminium wire with each strand 0.2mm2, 1m length

Cross Sectional AreaResistance (1 meter)
`0.2 * 7 = 1.4mm^2`
`({2.82*10^-8}/{1*10^-3}) \/ 1.4 = 0.504 Omega`
`rho = 2.82^-8` for aluminium

Temperature Coefficient

All metals have a temperature coefficient which means the resistance is going to vary over temperature, for copper and aluminium these are:

Copper = 0.00393
Aluminium = 0.0039
Being almost identical using value 0.039 is usually fine.
To calculate the resistance we use the following formula:

`R = R_0 * ( 1 + k * (T - T_0))`

R0 being the resistance measured at T0
T0 being the ambient temperature
T being the maximum acceptable temperature
k being the resistance coefficient as specified above

So for a copper wire we might have the following:
Resistance at 20c = 0.1 Ohm
Maximum temperature = 70c

`0.1 * ( 1 + 0.00393 * (70 - 20)) = 0.1965 Omega`

This is an change of +96.5 mOhm or +96.5% which is very significant since higher resistance means more power dissipation, this in turn leads to more heat, typically a 20c rise is fine, but more than 50c can be a cause for concern, if your wiring is in a hot ambient environment you also need to factor that in, the actual value will be less then this since the temperature coefficient reduces with increasing temperature.

Skin Depth

The Skin Effect causes AC current to concentrate towards the outside of the conductor which increases the effective resistance of the wire, this only applies to AC and increases with frequency, at power line frequencies this is only an issue in very thick cables.

To see if the skin effect may be a problem in your application you can calculate the skin depth, if your solid core wire radius is greater than the skin depth you need to switch to a stranded cable, with the per strand radius smaller than the skin depth or for very high frequencies a hollow conductor (pipe).

`delta = sqrt( {2rho} / {(2pif)* (mu_0mu_r)})`

`delta` is the skin depth in meters
`rho` is the resistivity
f is the frequency
`mu_0` is the permeability of free space (1.2566 x10-6)
`mu_r` is the relative permeability (1 for copper and aluminium)

For example if you was designing an audio cable the skin depth at 20kHz for aluminium is ~0.6mm, a diameter of ~1.2mm, which is a CSA of 1.13mm^2 or about 17/18 in AWG/SWG, if you needed lower resistance which for a high power amplifier might be the case stranded is the preferable option, you can also calculate the AC resistance:

`R = {Lrho} / {pi(D - delta)delta}`

L is the length in meters
D is the diameter of the wire in meters
`delta` is the skin depth in meters

Current Capacity (Ampacity)

This is more difficult to calculate since you need to consider many factors for your specific case.

This list goes on and on, in general you should never design anything near the calculated maximum ampacity.
The Neher–McGrath formula can be used to calculate ampacity with a reasonable degree of accuracy:

`I = sqrt( {T-T_0} / {R * (1+Y_c) * R_{ca} * 10^3}`

T is the maximum temperature
T0 is the ambient temperature
R is the resistance measured at T0
Yc is the AC resistance from skin and proximity effect
Rca is the thermal resistance

The skin effect is not really of any importance at low frequencies in copper or aluminium conductors, for mains power frequencies you'd need a solid conductor greater than 1cm in diameter for the skin effect to become apparent so it's only a concern with high current solid core cables above 78mm2 or bus bars.

Proximity effect can usually be ignored, it only becomes really significant at RF frequencies and in the design of inductors and transformers.
Thermal resistance is the thermal resistance to the surrounding enviroment, some examples are:

Soil90
Concrete55
Damp Soil60
Paper550
Polyethylene450
PVC (PolyVinylChloride)650
Dry Soil / Sand120

So for example a cable with 0.2 Ohm resistance with a PVC insulation and 70c max and 20c ambient and assuming no AC loss would give:

`sqrt( {70-20} / {0.2 * 650 * 10^3}) = 62A`

This seems to be a reasonable value, keep in mind thermal resistance is difficult to calculate, for such a cable I would consider 25A to 30A to be acceptable, of course the only true way to find out is to test it but this provides a reasonable estimate.

Conclusion

Doing these calculations in advance can save you a lot of trouble, in particular if you're dealing with high current and or high frequency, as always these are not perfect and do not take in to account things like contact resistance, corrosion, aging, etc which you need to think about when making a long term installation, derating and testing is ultimately what makes a good design.


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