Voltage drop gets tested twice over, once as a question about status and once as arithmetic

How the exam asks about voltage drop

Twenty-five of the 500 questions in our bank touch voltage drop, and 13 of those sit in the conductor sizing area. The remainder are spread across theory, motors, load calculations, conduit and the administrative area, which tells you something useful about the subject: it reaches into most of the paper without owning a section of it.

Two question shapes carry almost all of them. The first asks about status, and it is a knowledge-portion item with a percentage sitting in the options waiting to be picked. The second is arithmetic, and it lives on the calculations portion where PSI gives you 110 minutes for 26 items. A drop calculation with a multiplier in it is comfortably a four minute item when you are sure of the method and a lost item when you are reconstructing it.

Both shapes reward the same preparation, because both turn on knowing what the code does and stops doing here. Candidates who learn only the arithmetic answer the status question with a percentage. Candidates who learn only the slogan answer a fire pump item as though nothing were enforceable.

The status question, and its answer

For an ordinary branch circuit or feeder the code sets no percentage at all. The 3 percent and 5 percent figures everybody quotes sit in informational notes, one attached to the branch circuit conductor rule in Article 210 and one attached to the feeder rule at 215.4(A)(2). They describe a conductor sized so the drop at the farthest outlet stays within 3 percent, with the total across feeder and branch circuit staying within 5 percent, as providing reasonable efficiency of operation.

NEC 90.5 sorts every line in the book into mandatory rules, permissive rules and explanatory material, and it puts informational notes in the third group, where they carry no enforceable requirement. So an item asking what the code requires for voltage drop on an ordinary branch circuit has an answer, and the answer is that the code requires no percentage. Any percentage in that option list is there to be left alone.

Our bank runs the working version of it. A specification calls for branch circuit voltage drop to stay under 3 percent, and the question asks whether that figure is a code requirement. The answer is that the figure reaches you through the specification. Two of the wrong options put the number into Article 210 as a requirement, and one denies that voltage drop appears in the code at all, which overcorrects in the other direction.

The two places the limit is real

Both sit outside general wiring and both are written as hard numbers in enforceable text.

NEC 647.5(D) sets 1.5 percent as the ceiling on any branch circuit and 2.5 percent on feeder and branch circuit combined. Article 647 covers separately derived systems supplying sensitive electronic equipment, and those tight figures are the reason the article exists.

NEC 695.8 covers fire pumps and sets two limits under two different conditions. Under 695.8(A) the voltage at the controller line terminals may drop no more than 15 percent below normal under motor starting conditions. Under 695.8(D) the voltage at the contactor load terminals may drop no more than 5 percent below the motor voltage rating with the motor running at 115 percent of its full load current rating. Look at the shape of that second one: the calculation is run at 115 percent of full load, so the arithmetic the stem appears to be asking for is a different calculation from the one it wants. That detail is the item.

A candidate carrying the slogan that voltage drop is only a recommendation will lose a fire pump item, which is why the statement needs both halves of itself.

The multiplier, and the round trip

Voltage drop is current times the resistance of the path, and the path is longer than the run. Current goes out on one conductor and comes back on the other, so a load 180 feet away sits at the end of 360 feet of conductor. One of our theory items asks for that number alone, because it is the single fact the rest of the subject is built on.

The two formula versions differ only in how much conductor the current sees. Single phase drop is 2 times the one-way length times the current times the resistance per unit length. Three phase drop uses 1.732 in place of the 2, because the three phase currents sit 120 degrees apart and the square root of three falls out of that phase relationship. Use the one-way length in both, since the multiplier already carries the return path.

A bank item works the distance version: a 120 volt single phase lighting circuit drawing 16 amperes on a conductor of 1.24 ohms per 1000 feet, with 3.6 volts of drop allowed, may sit 91 feet from the panel. The wrong options are the three ways the multiplier gets mishandled. Dropping the 2 gives 181 feet, applying it twice gives 45 feet, and using 1.732 on a single phase circuit gives 105 feet, and every one of those numbers looks like a distance somebody measured.

The five ways this goes wrong

Each row comes out of the trap lines attached to our bank questions, which record what the candidate who picked that option did.

The first three are one error wearing different clothes, and all three come from writing before picturing the path the current takes. The fourth has its own bank item: a 480 volt three phase feeder dropping 10.6 volts is at 2.2 percent, because the percentage is taken against the voltage the circuit runs at. Dividing by 277 turns a comfortable design into one that appears to fail.

Working backward, and the K method

The harder direction gives you a drop you are willing to accept and asks what conductor delivers it. Rearrange the same formula so the unknown stands alone, and round toward less resistance, which means toward the larger conductor.

The approximate method puts 2 times K times current times one-way length over the circular mil area, with 1.732 in place of the 2 on three phase. K stands in for the resistance of one circular mil foot of the material, 12.9 for copper and 21.2 for aluminum. A bank item runs it backward: a 208 volt single phase circuit carrying 34 amperes 210 feet at 3 percent needs 29,521 circular mils, found by turning the 3 percent into 6.24 volts first and then dividing the rest of the formula by it. The wrong options drop the 2, divide by the full 208 volts, and reach for 1.732.

The ratio of the two K values is worth carrying as a fact. 21.2 divided by 12.9 is 1.64, so an aluminum conductor of the same circular mil area drops about 64 percent more than copper, which is the same physical fact that gives it the lower ampacity. The two methods land close together without matching exactly, because the tabulated resistance is a measured property while K smooths over stranding and coating.

What upsizing moves, and what it leaves

Drop is inversely proportional to conductor area, so doubling the area halves the drop, and a bank item takes a circuit at 4 percent to 2 percent that way. Twice the area and two sizes up are different things, because the circular mil steps between sizes run unevenly.

Three things do move. The conductors take more area, so the fill has to be reworked. The circular mil area in the drop calculation is the whole point of the change. And the equipment grounding conductor is increased in proportion under 250.122(B), where proportion is measured in circular mils. Matching the number of AWG sizes is the intuitive move and it is the one the word proportional rules out, since two sizes near the small end of the table is a very different ratio from two sizes near the large end.

Paralleling gets you to the same place. Adding a second identical set halves the current in each set while its own resistance stays put, so a feeder dropping 8.4 volts drops 4.2 volts, and the parallel rules then govern the installation.

What it costs, and what to fix

Motors are where the money is. Torque follows the square of the applied voltage, so a motor at the end of a run with 8 percent drop starts on 92 percent of nominal and develops about 85 percent of its rated starting torque. A motor that cannot accelerate its load keeps drawing starting current, which is a failure the ampacity calculation cannot see, because derating describes the conductor and this describes the load. Motor circuit items lean on that relationship.

On a job, two things turn guidance into obligation without changing a word of the code. An authority having jurisdiction can amend the code it adopts, and Texas leaves that power intact: Tex. Occ. Code 1305.201(c) lets a municipality adopt procedures for local amendments and 1305.201(d) requires work inside the corporate limits to follow applicable local ordinances, so what a city can require is worth asking before you price anything. A specification does the same thing privately, and failing it is a commercial problem of its own size.

Before test day, do three things. Learn the sentence in place of the formula, because current times the resistance of a path longer than the run rebuilds both versions and both directions without recalling which letter stood for what. Tab Chapter 9 Table 8 and Table 9, since the first carries direct-current resistance by material and construction and the second carries alternating-current resistance and reactance for conductors in a raceway. And answer the status question out loud once, so a percentage in an option list stops looking like an answer. The worked arithmetic for every calculation above sits in the question set.

Questions people ask

Does the NEC require 3 percent voltage drop in Texas?

For an ordinary branch circuit or feeder it requires no percentage. The 3 percent and 5 percent figures sit in informational notes, one attached to the branch circuit conductor rule in Article 210 and one at 215.4(A)(2) for feeders, and NEC 90.5 puts informational notes in the explanatory group where they carry no enforceable requirement. What can bind you is a local amendment, since Tex. Occ. Code 1305.201(c) and (d) leave a Texas municipality free to amend the adopted code and require compliance with local ordinances, or a specification you signed.

Where does the code set a real voltage drop limit?

In two places, both in enforceable text. NEC 647.5(D) sets 1.5 percent on any branch circuit and 2.5 percent on feeder and branch circuit combined, for separately derived systems supplying sensitive electronic equipment. NEC 695.8 covers fire pumps with two limits under two conditions: 15 percent below normal at the controller line terminals under motor starting conditions at 695.8(A), and 5 percent below the motor voltage rating at the contactor load terminals with the motor at 115 percent of its full load current rating at 695.8(D). That 115 percent is the detail the item is built on.

What length goes into a voltage drop calculation?

The one-way distance, with the multiplier carrying the return trip. Current goes out on one conductor and comes back on the other, so a load 180 feet from the panel sits at the end of 360 feet of conductor, and the formula accounts for that with a 2 on single phase and 1.732 on three phase. Doubling the length and also multiplying by 2 doubles the answer, and using the one-way distance with no multiplier halves it. Those two slips produce most of the wrong answers our bank records on this subject.

Why is the three phase multiplier 1.732?

Because the three phase currents sit 120 degrees apart, and the square root of three falls out of that phase relationship, the same term that appears in a three phase power calculation. Use the one-way length with it, exactly as on single phase. A bank item gives a 480 volt three phase feeder carrying 62 amperes on a conductor of 0.308 ohms per 1000 feet with a 2 percent budget, and the permitted run is 290 feet. Using the single phase multiplier of 2 shortens that to 251 feet and rejects a design that complies.

Does upsizing for voltage drop change the breaker?

The device stays where it was. It was chosen to protect the conductor the load required and to serve that load, and neither of those moved when the conductors grew. A branch circuit takes its rating from the device that protects it, so heavier wire pulled to hold the drop down leaves the circuit the same circuit. Three things do move: the raceway fill has to be reworked, the circular mil area in the drop calculation is the point of the exercise, and the equipment grounding conductor is increased in proportion under 250.122(B).

How is the grounding conductor increased when conductors are upsized?

In proportion to the increase, measured in circular mils. NEC 250.122(B) asks for a proportional increase, and the steps between adjacent AWG sizes run unevenly, so moving two sizes near the small end of the table is a very different ratio from moving two sizes near the large end. The method that works is to find the ratio by which the ungrounded conductors grew, apply that ratio to the area of the grounding conductor the table called for, then take the first size at or above the result. Matching the number of AWG sizes is the move the word proportional rules out.