AmpSizerHow voltage drop is calculated
Voltage drop is 2 Γ amps Γ ohms per 1,000 ft Γ one-way feet Γ· 1,000 on a single-phase or DC circuit, and
the same expression with 1.732 in place of the 2 on a balanced three-phase circuit.
The factor of two is where most errors come from. Current leaves on one conductor and returns on the other, so the copper it travels through is twice the distance you measured β and the distance you measure is one way.
A worked example: 20 amps down 100 feet of 12 AWG copper, which has 1.98 ohms per 1,000 ft.
2 Γ 20 Γ 1.98 Γ 100 Γ· 1,000 = 7.92 volts. On a 120 V circuit that is 6.6%; on 240 V, the same 7.92 volts is
3.3%.
Working it through, line by line
Take a detached garage: a 20 amp circuit, 100 feet of 12 AWG copper, 120 volts.
Step 1 β find the resistance. Chapter 9, Table 8 gives 12 AWG stranded copper as 1.98 ohms per 1,000 ft.
Step 2 β account for both conductors. The current goes out and comes back, so it passes through 200 feet of copper. That is the factor of two.
Step 3 β multiply out. 2 Γ 20 A Γ 1.98 Ξ©/kft Γ 100 ft Γ· 1,000 = 7.92 volts.
Step 4 β turn it into a percentage. 7.92 Γ· 120 Γ 100 = 6.6%.
Step 5 β compare with the target. 6.6% is over both 3% and 5%, so this conductor is too small for this run even though it carries 20 amps comfortably. Stepping to 8 AWG at 0.778 ohms gives 3.11 volts, which is 2.6%.
Where the numbers come from
The factor of two. A circuit is a loop. Nothing physical happens at the load to send the current home by a shorter route.
The 1.732 on three phase. The three phases are 120 degrees apart, so the currents do not add arithmetically. The vector sum works out to the square root of three, and in a balanced system there is no neutral current to account for at all.
The 1,000 in the divisor. Resistance is tabulated per thousand feet because that is a convenient size for the numbers. It cancels the thousands in the length.
Why the percentage matters differently to different loads
| Load type | What 5% low does |
|---|---|
| Resistive heater | Delivers about 10% less heat β power falls with the square of voltage |
| Incandescent lamp | Noticeably dimmer, and lives much longer |
| LED driver | Almost nothing; it draws more current and keeps output constant |
| Switch-mode power supply | Almost nothing, for the same reason |
| Induction motor | Torque falls about 10%, current rises, windings run hotter |
| Motor starting under load | Worst case β it may not start at all, and stalling is what burns motors |
The pattern is that anything electronic shrugs at voltage drop by drawing more current, which moves the problem into the conductor. Anything with a motor in it suffers directly. Anything resistive simply does less work.
The 120 volt penalty
The volts lost in a conductor do not care what the system voltage is. Only the percentage does.
That means the same load in watts, at 240 volts instead of 120, halves the current and doubles the number the loss is measured against β a four-times improvement. It is the single biggest lever available on a long run, and it is why a detached building gets a 240 volt feed and a small panel rather than a long 120 volt circuit.
Try it on the voltage drop calculator: switch the same load between 120 and 240 volts and watch the longest-run figure quadruple.
Frequently asked questions
Is voltage drop a code requirement?
Not in the NEC for ordinary branch circuits and feeders. The 3% and 5% figures appear in informational notes to 210.19(A) and 215.2(A), and informational notes are explicitly not enforceable. There are places where the code does mandate a limit β sensitive electronic equipment, fire pumps, some PV and EV provisions β and other codes and standards do treat it as a requirement.
Should I use 3% or 5%?
The convention is 3% on the branch circuit and 5% total from the service to the load, which leaves 2% for the feeder. Designing to 3% end to end costs a little more copper and buys margin for the day somebody extends the circuit. Designing past 5% is where equipment starts behaving differently rather than just less efficiently.
Do I use AC or DC resistance?
Chapter 9, Table 8 gives direct-current resistance, and it is what nearly every calculator including this one uses. Table 9 gives alternating-current impedance, which is a few percent higher for large conductors in steel raceway because of skin effect and reactance. Below about 1/0 the difference is inside the rounding; above 250 kcmil in steel conduit it is worth doing properly.
Does a longer run need a bigger breaker?
No. The breaker protects the conductor, so it is sized from the load and the conductor, and distance is not part of that. What a longer run changes is the conductor β and once you upsize the conductor for voltage drop, 250.122(B) makes you upsize the equipment grounding conductor in the same proportion.
How does temperature affect voltage drop?
Copper’s resistance rises about 0.4% per degree Celsius, so a conductor running hot drops more than the table figure. Table 8 is quoted at 75 Β°C, which is roughly a fully loaded conductor, so it is already the pessimistic case. A lightly loaded conductor at room temperature drops slightly less than the calculation says.
How we calculate this
k Γ I Γ R Γ L Γ· 1,000, where k is 2 for single-phase and direct-current
circuits and the square root of three (1.732) for balanced three-phase, I is the load current in amps, R
is the conductor’s resistance in ohms per 1,000 ft, and L is the one-way length of the run in feet.
Resistance figures are the direct-current values for stranded uncoated conductors in NEC Chapter 9, Table 8,
quoted at 75 Β°C. Percentage drop is volts lost divided by nominal system voltage. The 3% branch-circuit and
5% total figures are from the informational notes to 210.19(A) and 215.2(A) and are recommendations, not
enforceable requirements.Sources: