How to Calculate Voltage Drop for the Electrician Exam

Voltage drop is one of the calculation types you should be able to do quickly and without second-guessing on exam day. The math is short. The mistakes come from the setup: wrong length, wrong multiplier, wrong table value. This guide walks through the formula, where every number comes from, and four worked problems you can check against the free voltage drop calculator.

What voltage drop actually is

Every conductor has resistance. When current flows through that resistance, some voltage is used up in the wire itself instead of reaching the load. A long run, a heavy load, or a small conductor all make the drop bigger. The load then sees less than the source voltage: a motor runs hotter, a heater puts out less heat, and electronics can misbehave.

On the exam you will usually be asked one of four things: the drop in volts, the drop as a percent, the voltage at the load, or the minimum conductor size that keeps the drop under a stated limit. Read the question twice and circle which one it wants before you start.

The two formulas

• VD is the voltage drop in volts.
• K is the resistance of a circular-mil-foot of conductor. Use 12.9 for copper and 21.2 for aluminum (ohm-cmil per foot at 75 °C).
• I is the load current in amperes.
• L is the one-way length of the circuit in feet, from the source to the load.
• CM is the conductor area in circular mils, from NEC Chapter 9, Table 8.

The 2 in the single-phase formula accounts for the current going out on one conductor and coming back on the other. That is why L is the one-way length. If you plug in the round-trip length and also keep the 2, you have doubled the answer. In the three-phase formula the 1.732 (the square root of 3) replaces the 2 because of how the phase currents combine.

Where K = 12.9 and 21.2 come from

K is not a magic number. You can rebuild it from Chapter 9, Table 8, which lists the DC resistance of conductors at 75 °C in ohms per 1,000 feet. Take 12 AWG uncoated copper: about 1.98 ohms per 1,000 ft, and an area of 6,530 circular mils.

Do the same for 12 AWG aluminum (3.25 ohms per 1,000 ft) and you get 21.2. Knowing this matters because some exam questions give you a resistance value and skip K entirely. In that case use VD = 2 × I × R for single-phase, where R is the resistance of one conductor for the one-way length.

Worked problem 1: single-phase branch circuit

A 120 V single-phase circuit feeds a 20 A load 100 ft away using 12 AWG copper. Find the voltage drop in volts and percent.

That is more than double the 3% recommendation. Try the next sizes up, keeping everything else the same:

8 AWG is the first size at or under 3% for this run. Notice the numerator (51,600) never changed. Only CM changed, so once you have the top of the fraction you can test sizes fast.

Worked problem 2: three-phase feeder

A 480 V three-phase feeder carries 100 A over 250 ft on 1/0 AWG copper.

Higher voltage helps a lot: the same drop in volts is a much smaller share of 480 V than of 120 V. That is one reason long feeders run at higher voltage.

Worked problem 3: find the minimum size

When the question gives you a maximum percent and asks for the conductor size, flip the formula to solve for CM.

Example: 240 V single-phase, 50 A load, 200 ft one way, copper, keep the drop at 3% or less.

Now go to Chapter 9, Table 8 and pick the first size with at least 35,833 circular mils. 6 AWG is 26,240 (too small). 4 AWG is 41,740, so the answer is 4 AWG copper. Never round down here: a conductor with fewer circular mils than you calculated will exceed the limit.

The voltage drop answer is not the whole job. The conductor still has to have enough ampacity for the load under NEC 310.16 and the termination rules in 110.14(C). On the exam, if a question asks only about voltage drop, answer only that. On the job, check both and use the larger conductor.

Worked problem 4: find the maximum length

Solve for L when the question asks how far you can run a given conductor.

Example: 12 AWG copper, 16 A, 120 V single-phase, 3% maximum.

Is 3% a code requirement?

For general branch circuits and feeders, no. In NEC 2020 and 2023, the 3% and 5% figures appear in informational notes under 210.19 (branch circuits) and 215.2 (feeders). Informational notes explain and recommend; they are not enforceable requirements. The usual reading is 3% maximum on a branch circuit or feeder, and 5% maximum combined from the service to the farthest outlet.

A few specific articles do set mandatory voltage limits, for example the fire pump rules in 695.7. If a question names a special system, look in that article rather than assuming the 3% note.

Fast checklist for exam day

1. Single-phase or three-phase? That decides 2 or 1.732.
2. Copper or aluminum? That decides 12.9 or 21.2.
3. Is the length one-way? If the question gives total conductor length, divide by 2 for single-phase.
4. Look up CM in Chapter 9, Table 8. Do not guess it from memory under pressure.
5. Keep full precision until the last step, then round.
6. Answer what was asked: volts, percent, voltage at the load, size, or length.

Run your own numbers through the voltage drop calculator. It shows every step, so you can compare it to your work line by line instead of only checking the final answer.