It seems I may have been working under a false understanding of AC vs. DC wire derating? Going back to the Edison vs Tesla "electrical war" on comparison given was:

I'll admit I don't understand the physics behind it but by comparing the DC derating chart from the American Boating and Yacht Council which I have used for many years now.

Versus the National Electrical Code

where

So let's limit this to single phase, no temperature derating, no derating for multiple conductors. Using 120VAC, 100', 20A,

From the AWG

So 100' AWG 12 wire with 10.15Ω/1000 ft. on a 20A circuit breaker

Vd1ϕ=IL(2Zc)/1000 = 20A*100ft(2*10.15Ω/1000ft)/1000= 40.3mV

Which is far less than the derating for DC?

So I am obviously missing something here... I understand that @ Vrms, volts are volts no matter whether AC or DC, and amps are amps. What am I missing here?

I'll admit I don't understand the physics behind it but by comparing the DC derating chart from the American Boating and Yacht Council which I have used for many years now.

Versus the National Electrical Code

**Derating:**- No derating is currently applied to the current ratings tables 310.15(B)(16) and 310.15(B)(17)..
- It is assumed that the maximum ambient temperature is 30°C and the maximum ground temperature id 20°C. For higher temperatures, a derating will have to be applied according to NEC.
- It is assumed that there is only one three current carrying conductors in a raceway, cable or buried. And that the raceways, cables and buried conductors are spaced according to NEC to prenet derating.
- To apply a manual derating, divide the load by the derating factor from NEC, and enter the new load value in the calculator.

**Voltage drop calculation:**- The single phase AC voltage drop is calculated as:

where

**I**is the load current,**L**is the distance, and Zc is the cable impedance in Ω/km.So let's limit this to single phase, no temperature derating, no derating for multiple conductors. Using 120VAC, 100', 20A,

From the AWG

So 100' AWG 12 wire with 10.15Ω/1000 ft. on a 20A circuit breaker

Vd1ϕ=IL(2Zc)/1000 = 20A*100ft(2*10.15Ω/1000ft)/1000= 40.3mV

Which is far less than the derating for DC?

So I am obviously missing something here... I understand that @ Vrms, volts are volts no matter whether AC or DC, and amps are amps. What am I missing here?

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