Voltage Drop Calculations Practice

The circular-mil method, single-phase against three-phase, and sizing a conductor to hold the drop within a limit.

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What it is, and why electricians need it

Voltage drop is the voltage consumed by the conductors themselves rather than delivered to the load, and it is proportional to current and to length and inversely proportional to conductor area. A conductor with ample ampacity can still be the wrong size, because ampacity asks whether the conductor will overheat and voltage drop asks whether the load will get enough voltage to work properly.

The distinction matters in practice. A 20-ampere circuit on 12 AWG copper is perfectly sized for ampacity, but run it 100 feet at 120 volts and the drop is about 7.9 volts, or 6.6 percent — more than double the commonly specified 3 percent. Nothing overheats; the equipment at the far end simply underperforms, lamps run dim, and motors run hot and start poorly.

It is worth being precise about the Code status of the familiar percentages. The 3 percent branch-circuit and 5 percent combined figures sit in Informational Notes, which are explanatory and not enforceable as requirements for general branch circuits and feeders. Particular articles do impose mandatory limits for specific applications, and project specifications almost always require 3 percent contractually — so you size for it, but you should know why if an exam question asks whether it is a requirement.

Higher system voltages tolerate the same absolute drop far better, which is the other reason long runs are made at the highest practical voltage. One volt is 0.83 percent of 120 volts but only 0.21 percent of 480 volts.

Formulas and principles

VD = 2 × K × I × L ÷ CM
Single phase. K = 12.9 copper or 21.2 aluminium, I in amperes, L the one-way length in feet, CM the circular-mil area.
The 2 accounts for the return conductor, so use the one-way length.
VD = 1.732 × K × I × L ÷ CM
Three phase. Everything else is the same.
Produces about 87 percent of the single-phase drop for the same current and length.
CM = 2 × K × I × L ÷ VD
Rearranged to size a conductor. Use 1.732 in place of 2 for three phase.
Gives the minimum circular mils; then pick the next standard size up.
Percentage drop = VD ÷ Vsystem × 100
Convert the volt figure into a percentage of the supply voltage.
Or go the other way first: 3 percent of 120 V is 3.6 V.

Circular-mil areas you will need

The circular-mil area is what goes in the denominator. These values come from the conductor properties table in Chapter 9 of the 2017 NEC. For kcmil sizes the circular-mil area is simply the kcmil figure times 1,000.

Circular-mil areas by conductor size
Size	Circular mils	Size	Circular mils
14 AWG	4,110	1 AWG	83,690
12 AWG	6,530	1/0 AWG	105,600
10 AWG	10,380	2/0 AWG	133,100
8 AWG	16,510	3/0 AWG	167,800
6 AWG	26,240	4/0 AWG	211,600
4 AWG	41,740	250 kcmil	250,000
3 AWG	52,620	350 kcmil	350,000
2 AWG	66,360	500 kcmil	500,000

Source: NEC 2017 Chapter 9, Table 8, from Public.Resource.Org's publication of incorporated standards.

Now try it on 14 questions

One question at a time, marked the moment you answer, with the reasoning and the source behind it. Anything you miss is saved so you can come back to it.

Start the Voltage Drop drill

Worked examples

Scenario: A single-phase 120-volt circuit carries 20 amperes 100 feet to the load on 12 AWG copper. Find the drop, and whether it meets a 3 percent limit.

VD = 2 × K × I × L ÷ CM.
VD = 2 × 12.9 × 20 × 100 ÷ 6,530 = 51,600 ÷ 6,530.
VD = 7.9 volts.
As a percentage: 7.9 ÷ 120 × 100 = 6.6 percent.
That is more than double the 3 percent recommendation, so the conductor needs upsizing even though 12 AWG has ample ampacity for 20 amperes.

Answer: 7.9 volts, or 6.6 percent — well outside 3 percent

Scenario: Same circuit, but now size the conductor to hold the drop to 3 percent of 120 volts.

Convert the limit to volts first: 120 × 0.03 = 3.6 volts.
Rearrange: CM = 2 × K × I × L ÷ VD.
CM = 2 × 12.9 × 20 × 100 ÷ 3.6 = 51,600 ÷ 3.6 = 14,333 circular mils minimum.
From the table, 10 AWG is 10,380 CM — too small. 8 AWG is 16,510 CM.
Select 8 AWG copper, two sizes larger than ampacity alone would have required.

Answer: 8 AWG copper, from a minimum of 14,333 circular mils

Scenario: A balanced three-phase feeder carries 50 amperes 200 feet on 4/0 copper at 480 volts. Find the drop and the percentage.

Three phase, so VD = 1.732 × K × I × L ÷ CM.
VD = 1.732 × 12.9 × 50 × 200 ÷ 211,600 = 223,428 ÷ 211,600.
VD = 1.06 volts.
As a percentage: 1.06 ÷ 480 × 100 = 0.22 percent.
Negligible. Here ampacity, not voltage drop, is clearly the governing factor.

Answer: 1.06 volts, or 0.22 percent

Common mistakes

Doubling the length as well as using the factor of 2. The 2 already accounts for the return path. Use the one-way distance. This error gives an answer exactly twice the correct one, and it is the most common mistake on these questions.
Using 2 for a three-phase circuit. Three phase uses 1.732. Using 2 overstates the drop by about 15 percent, which can push a compliant design into an apparent violation.
Putting circular mils in the numerator. CM goes in the denominator. If your arrangement makes a bigger conductor produce more drop, you have inverted it.
Comparing a volt figure against a percentage. Convert one into the other first. Three percent of 120 volts is 3.6 volts, and 3.6 is what you compare your calculated drop against.
Stopping once the voltage-drop size is found. Voltage drop and ampacity are separate tests. A conductor sized for drop may still need checking against ampacity, termination temperature limits and the small-conductor overcurrent rules.
Treating the 3 percent figure as an enforceable requirement everywhere. For general branch circuits and feeders it lives in an Informational Note and is a recommendation. Know this, because exams do ask.

Questions about this topic

Is the 3 percent voltage drop limit actually a Code requirement?

Not for general branch circuits and feeders. The 3 percent branch-circuit figure appears in an Informational Note to 210.19(A), and the matching feeder note is at 215.2(A)(1); both describe drops that provide reasonable efficiency of operation, and Informational Notes are explanatory rather than enforceable. Specific articles elsewhere do impose mandatory limits for particular applications, and designers and project specifications routinely make 3 percent a contractual requirement, so in practice you size for it.

Do I use the one-way or round-trip length in the formula?

The one-way length. The factor of 2 in the single-phase formula already accounts for the return conductor. Doubling the length yourself as well is the single most common error on these questions, and it produces an answer exactly twice the correct one.

Why does three-phase use 1.732 instead of 2?

Because of how the currents and voltages relate in a balanced three-phase system. The practical consequence is that the same current over the same distance in the same conductor produces about 87 percent of the single-phase drop, which is one of the reasons three-phase is preferred for long runs.

What are the K values and where do they come from?

K is a resistivity constant in ohm-circular-mils per foot: 12.9 for copper and 21.2 for aluminium. These are the industry-standard approximations used throughout exam preparation, not values published in a Code table. Aluminium being 64 percent higher is why an aluminium run of the same size and length drops substantially more voltage.

How do I size a conductor for a voltage drop limit?

Rearrange the formula to solve for circular mils: CM = 2 × K × I × L ÷ VD for single phase. Convert the percentage into volts first, so you are comparing volts with volts. Then pick the smallest conductor whose circular-mil area meets or exceeds the result — and remember to re-check ampacity, because the voltage-drop answer does not guarantee the conductor is large enough to carry the load.

Why does excessive voltage drop matter for motors?

A motor must deliver the mechanical load it is coupled to, so when voltage sags it draws more current to compensate, and that extra current heats the windings. Starting torque suffers worse still, because it varies roughly with the square of the applied voltage — an undersized feeder can leave a motor unable to start its load at all.

Does voltage drop or ampacity usually govern conductor size?

It depends entirely on the length. On a short run ampacity governs and voltage drop is negligible. On a long run voltage drop takes over, often demanding a conductor several sizes larger than ampacity alone would require. Always check both, because satisfying one says nothing about the other.

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