Originally Posted by

**Carultch**
Bear in mind that when you get questions like this, the voltage drop information could be a red herring. You could go through your voltage drop calculation and determine a size of conductor, that is insufficient for the circuit amps. Sometimes the voltage drop might be the factor that governs the answer, other times ampacity may govern it instead. In this case, all sizes indicated are sufficient for 30A noncontinuous, so voltage drop is the only calculation that will matter.

For lack of additional information, we assume it is a single phase circuit, and we assume copper wires. Which means the total round trip length of the circuit, is twice the distance given. Call the given length L. The round trip length is 2*L.

The wire has a resistance per unit length. Call it r. The total resistance (R) can be calculated with R = 2*L*r. This is a value in the lookup tables in units of thousand feet (kft) and kilometers, so you will often divide by 1000 to take care of this.

Call the operating current, in units of amps, I.

Construct Ohm's Law for voltage drop (dV):

dV = I*R

dV = I*2*r*L

Solve for r:

r = dV/(2*I*L)

The NEC recommends (not requires, but recommends) not exceeding 3% voltage drop. For dV, plug in 3% of 230V. Plug in 30A for I, and plug in 100 ft for L. **Our result is 0.0015 Ohms/ft, which translates to 1.15 Ohms/kft.**

We need a wire resistance per unit length that is does not exceed 1.15 Ohms/kft, which is #8 at 0.778 Ohms/kft.

0.0015 Ohms/ft = 1.5 Ohms per kft, not 1.15. #10 has .999 Ohms/1kft

Electricians do it until it Hertz!

## Bookmarks