I can't find this anywhere else on the internet so I thought I'd post it here. There is an old rule of thumb for calculating the voltage drop of copper wire.
The Rule of Fourteen
This scales so 14 AWG, 14 Amps 1.4 feet has about a 0.1 volt drop.
For each increase in AWG the current drops by 1dB (10log) or about
so for 16 AWG, 14 feet, (0.8)*(0.8)*14Amps = 8.96 has a 1 volt drop.
Try it
I did. It is off by about 100%.
14 AWG, 28 Amps, 14 feet has about a 1 volt drop. OR 14 AWG, 14 Amps, 28 feet has about a 1 volt drop.Check Ohm's Law (using your link) to verify.
Well -50.5% anyway.
0.002525 ohms per foot x 14 ft = 0.03535 ohms per length x 14 A = 0.4949 V of IR drop across the length 0.4949-1.0 =-0.5051 V difference between actual and predicted drop
-0.5051/1.0 =
-0.5051 or -50.51 %
+-100% difference would've been 2 V or 0 V of IR drop, yes?--Winston
Just kidding. 0.4949 V plus 102% really is ~1.0 V
--Winston
I find it much easier to remember starting with AWG 10 is 1 ohm per 1000 feet and from there it is like decibels: adding 3 to a wire gauge is multiplying the resistance by 2; adding 4, by 2.5; 10, by 10; and so on. So AWG 24 should be about 25 ohms per 1000 feet; AWG 4, 0.25 per 1000.
16 AWG would be about 0.004 ohms per foot, so 14 amps through it would drop about 0.056 V/ft, so a little less than 20 ft (about 18, really) will yield a 1 V drop.
the internet so I thought I'd post
Your right. It is round trip. So you're cutting 2 wires. Power and Ground. Both 14 feet.
10AWG is 1ohm per 1000ft sounds easier to remember.
0 .
Those are truly useful rules of thumb, but all this AWG stuff has always annoyed me. Why not just spec wire by diameter, rather than a lookup table based on circular millicubits?
Cheers, James Arthur
. . .
Two similar rules of thumb: On a 60 volt full wave power supply, 8300 microfarads makes the amps of supply equal the volts of ripple.
And, of course, in any hazmat scene, you hold your thumb up at arm's length and close one eye. If you can still see the scene, you are too close.
What does "60 volt" have to do with it?
Oops. 60 Hertz, full wave.
Approximating the waveform as a triangular wave, 8300 microfarads drops you one volt per amp in 8.3 milliseconds.
Since you have somewhat less than a triangle wave, the rule is conservative.
Right, that's the lookup table.
-- Cheers, James Arthur
.
ve.
one: (928)428-4073
The St. Louis and Suburban Radio Club Handy Reference says its 120Hz or maybe I'm not following what you're saying.
60 Hz into a full-wave bridge rectifier = ?
ost
rop.
00 sative.
Ahh. So 60 Hz into a full-wave bridge rectifier with C of capacitance gives a ripple voltage equal to 8333*I/C at 120Hz.
Right, but, as the thumb rules imply, the rating by "gauge" allows an estimate of the unit resistance without needing to reference a table, whereas the IEC 60228 sizes, which are indeed in easy to remember millimeter increments, really don't give much of a clue.
Have something to add? Share your thoughts — no account required.
Ask the community — no account required