Voltage drop is a basic skill required for every electrical engineer in the nuclear power industry. Here’s a summary of how it’s calculated. There are also references to relevant National Electric Code articles . . . .

There are functional purposes that drive voltage-drop design. Those are addressed elsewhere. In this article, we’ll briefly discuss the application and design as it is typically seen in common circuits.

The most common voltage drop criteria engineers deal with are limits of 3% for branch circuits and 5% for feeder-and-branches combined. Some plants are more or less restrictive.

When calculating voltage drop, there are two primary equations you need to keep in mind: the single-phase equation and the three-phase equation. *Remembering their differences is critical.*

Calculating voltage drop the NEC way and using the resistance values provided in Chapter 9 (given in ohms-per-thousand feet), the formula for a single-phase circuit is simple once the one-way circuit length (in feet) and load current (in amps) are known:

where “L” is the one-way circuit length in feet, “R” is the impedance value of a particular cable size per-thousand-feet, and “I” is the full-load current being designed to. The factor of 2 in the numerator indicates that the actual circuit length is double the one-way length.

The voltage drop formula changes somewhat for a three-phase circuit:

Again, *notice the differences*: the factor of two in the first equation and the square-root of three in the second.

*Keep this sanity check in the back of your mind*: voltage drop in a three-phase circuit, assuming everything else is the same, will be lower than in a single-phase circuit. The 2011 NEC Handbook provides a helpful example in Article 215.2(4) that helps to solidify the concept in your mind.

[Not be confused with the “Codebook.” The NEC Codebook contains just the Code. The NEC Handbook contains the Code plus helpful examples and illustrations that help interpret the code requirements.]

To determine if the calculated voltage drop is acceptable, you would divide your result, VD (in volts) by your base voltage. A voltage drop of 7 volts in a 120-volt circuit would produce a voltage drop of 7 ÷ 120 = 5.8%.