Voltage drop is a function of voltage (regardless of DC or AC), amperage, distance, and the resistance of the wire.
Is there a different voltage drop calculation for direct current (DC) as opposed to alternating current (AC)? I have a project that will take my solar-electric module string wires about 300 to 400 feet, carrying 9 amps at 480 volts DC. What size of wire do you recommend, and how do I calculate that?
There is not a different calculation for AC versus DC voltage drop. Voltage drop is a function of voltage (regardless of DC or AC), amperage, distance, and the resistance of the wire.
This equation calculates voltage drop by percentage:
By rearranging the voltage drop equation, you can solve for a specific ohms/Kft. value. Then, you can use the National Electrical Code (NEC) Chapter 9, Table 8, to find the wire size having an ohm/Kft. value that does not exceed the calculated ohms/Kft. value.
Let’s say you are aiming for a maximum voltage drop of 2%. Assuming 480 V, 9 A, and 400 ft., you have:
On the NEC table, we find that #10 AWG stranded (uncoated) copper yields 1.24 Ohms/Kft., less than 1.33. This means we can use #10 AWG or larger diameter wire and not exceed a 2% voltage drop.