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Author Topic: Wire Resistance without Wire Tables  (Read 2053 times)

Frank Koenig

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Wire Resistance without Wire Tables
« on: October 12, 2014, 04:13:27 pm »

Here's a way to remember wire resistance without having to refer to any tables, at least in the US where we use American Wire Gauge (AWG). It's plenty close enough for routine voltage drop calculations in small power systems or dB loss and damping factor calcs for speaker cables. Do this in your head and impress your friends.

The key: Remember that 10 AWG copper is 1 Ohm per 1000 ft.

10-1-1000, even I can remember that. (This is for 25 deg C and therefore a little lower than the number in the NEC tables, which are for elevated temperature, but who cares?)

Cross-sectional area of wire halves every three gauge numbers, so

R = 2^((AWG - 10) / 3)

where R is the resistance in Ohms per 1000 ft. and AWG is the gauge number. ("^" represents exponentiation so, for example, 2^3 = 8.)

If the difference between the gauge and 10 is a multiple of 3 you're done since the ratio of resistance is a power of 2. For example, 4 AWG is .25 Ohms per 1000 ft, 1 AWG is .125 Ohms per 1000 ft, etc.

If the gauge differs from 10 by other than a multiple of 3 you have to know your 1/3 octave bands. But, luckily, we're audio engineers. Just picture the front of a 1/3 octave graphic equalizer, which has bands at 1000, 1250, and 1600 Hz. These are the required ratios. So, for example, 8 AWG is 1 / 1.6 = .63 Ohms per 1000 ft. and 11 AWG is 1 * 1.25 = 1.25 Ohms per 1000 ft. (The actual ratios are closer to 1.26 and 1.59 but, again, close enough.)

To take the GEQ model a bit further, make 1000 Hz correspond to 10 AWG. Then just count off one band for each gauge number from 10 and the frequency divided by 1000 is the Ohms per 1000 ft. 

Aluminum wire, for the same resistance as copper, is very close to 2 gauge numbers larger. So for aluminum just add 2 to the AWG and figure for copper.

[I posted something along these lines on the other forum a while back and it didn't get any traction. I try again here.]

--Frank
« Last Edit: October 12, 2014, 11:11:38 pm by Frank Koenig »
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Lyle Williams

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Re: Wire Resistance without Wire Tables
« Reply #1 on: October 13, 2014, 01:39:41 pm »

That's pretty cool.  :-) 
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Jonathan Johnson

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Re: Wire Resistance without Wire Tables
« Reply #2 on: October 13, 2014, 06:06:52 pm »

But if you have the Intertubes handy, you could just go here to calculate voltage drop:

http://www.calculator.net/voltage-drop-calculator.html
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Mike Sokol

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Re: Wire Resistance without Wire Tables
« Reply #3 on: October 13, 2014, 07:34:14 pm »

But if you have the Intertubes handy, you could just go here to calculate voltage drop:

Are you implying that the "Intertube" has made us all mentally lazy???  ;D
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Frank Koenig

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Re: Wire Resistance without Wire Tables
« Reply #4 on: October 13, 2014, 09:42:48 pm »

But if you have the Intertubes handy, you could just go here to calculate voltage drop:

http://www.calculator.net/voltage-drop-calculator.html

I do like that the calculator site refers to V = I*R as the voltage drop law rather than "Ohm's law", as folks too often do.

Ohm's law is very similar in form to Hook's law for Young's modulus of materials. I'll take a crack at the parallelism:

Hook's Law

For many materials (the ones for which Hook's law is useful) the modulus of elasticity (stress / strain) is essentially constant for small values of strain.

Ohm's Law

For many conductors (the ones for which Ohm's law is useful) the resistance (voltage / current) is essentially constant for small values of current.

The above assumes a chunk of conductor of fixed length and cross-sectional area. Alternatively, and perhaps more usefully, Ohm's law can be stated in intrinsic terms as follows:

For many conductors (the ones for which Ohm's law is useful) the resistivity (electric field / current density) is essentially constant for small values of current density.

Let's try it for sound (Oh no!)

For an acoustic wave propagating (adiabatically, dunno if this constraint matters) in many (isotropic, homogenous, elastic) media (the ones for which this law, whose name I don't know, is useful) the acoustic impedance (sound pressure / volume velocity) is essentially constant for small values of volume velocity.

I'm not a real physicist, although I have stayed in a Holiday Inn Express, and it's past Happy Hour, so someone please correct me if I've bugered this up.

PS: I learned R = V/I, and its permutations, as the "definition of resistance".

--Frank

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Stephen Swaffer

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Re: Wire Resistance without Wire Tables
« Reply #5 on: October 13, 2014, 09:59:55 pm »

Are you implying that the "Intertube" has made us all mentally lazy???  ;D

Don't blame it all on the intertube-if they are stranded without wifi, they can still use the calculator on their smartphone to do the math with Frank's method. No sense in overworking the mental muscle.
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Steve Swaffer

ProSoundWeb Community

Re: Wire Resistance without Wire Tables
« Reply #5 on: October 13, 2014, 09:59:55 pm »


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