Voltage Drop Explained: The 3% Rule and How to Calculate It

When current flows through a wire, some voltage is lost to the resistance of the conductor itself. The load at the end of the run doesn't see the full supply voltage, it sees the supply voltage minus that drop. This is voltage drop, and it's the reason you can't just pick a wire by ampacity alone.

A motor rated for 240 V that's actually receiving 222 V draws more current to compensate. That extra current generates extra heat. The NEC doesn't set a mandatory maximum for most circuits, but the 3% guideline in NEC 210.19(A) informational note exists because problems become predictable above that threshold.

Why it matters in practice

For lighting, the effect is visible: a 5% drop on a 120 V circuit means the fixture sees 114 V, and incandescent bulbs dim noticeably. For motors, it's less visible but more damaging. Motors are designed to operate within roughly ±10% of rated voltage. Operating at the low end of that range consistently shortens insulation life.

The 3% rule is a rule of thumb for branch circuits. NEC 210.19(A) suggests keeping it under 3% for branch circuits and under 5% total (feeder plus branch combined). These aren't hard requirements in most cases, but inspectors and engineers treat them as best practice.

Long runs are where this bites. A 20 A circuit from a panel 10 feet away will almost never have a voltage drop problem on 12 AWG wire. The same circuit at 150 feet very often will.

The formula

The standard voltage drop formula for DC and single-phase AC circuits is:

V_drop = (2 × K × I × L) / A

Where:

• V_drop is the voltage drop in volts
• K is the resistivity constant (12.9 for copper, 21.2 for aluminum, in ohm-cmil/ft)
• I is the current in amperes
• L is the one-way length of the run in feet
• A is the conductor cross-section in circular mils

The factor of 2 accounts for the round trip: current goes out on one conductor and returns on another.

For three-phase circuits, the formula changes slightly:

V_drop = (1.732 × K × I × L) / A

The 1.732 (square root of 3) comes from the geometry of three-phase power.

You can also work in ohms per 1,000 feet instead of the K constant. NEC Chapter 9, Table 9 lists AC resistance values for conductors in conduit. For 12 AWG copper at 75°C in PVC conduit, that's about 2.0 ohms per 1,000 feet. Multiply by the run length in thousands of feet, multiply by current, multiply by 2 for single-phase, same result.

Worked example

You're running a 20 A, 120 V circuit to a workshop 60 feet from the panel. You're planning to use 12 AWG copper.

12 AWG copper has a cross-section of 6,530 circular mils.

V_drop = (2 × 12.9 × 20 × 60) / 6,530
V_drop = 30,960 / 6,530
V_drop ≈ 4.74 V

As a percentage of 120 V: 4.74 / 120 = 3.95%

That's above 3%. To get under, you could step up to 10 AWG (10,380 circular mils):

V_drop = (2 × 12.9 × 20 × 60) / 10,380
V_drop = 30,960 / 10,380
V_drop ≈ 2.98 V → 2.48%

10 AWG keeps you under the guideline with a little margin. For a workshop with power tools, that's the right call anyway.

This is the calculation that a cable size calculator handles automatically when you enter your voltage, current, run length, and conductor material.

When to use a larger wire than ampacity requires

Ampacity is the minimum safe conductor size for the load current. Voltage drop sometimes forces you higher. The rule is: always use the larger of the two results.

A 30 A circuit on a 200-foot run is a common example. 10 AWG is rated for 30 A (at 75°C, copper). But at 30 A over 200 feet on 240 V single-phase:

V_drop = (2 × 12.9 × 30 × 200) / 10,380 ≈ 14.9 V → 6.2%

That's well over 3%. You'd need to step up to 6 AWG (26,240 circular mils) to get the drop under 3%, even though 6 AWG is rated for 65 A, far more than the 30 A load.

The wire is sized for voltage drop, not ampacity, in that case.

What the calculation doesn't cover

This formula gives you steady-state voltage drop at the design current. It doesn't account for:

• Motor starting currents, which can be 6–8x running current for a fraction of a second
• Harmonic distortion in circuits feeding variable-frequency drives
• Temperature effects on conductor resistance (resistance increases with temperature)

For most residential and light commercial work, the steady-state formula is sufficient. For industrial motor circuits, consult the motor nameplate and NEC Article 430.

The 3% guideline comes from NEC 210.19(A) informational notes and is widely used but not universally mandatory. Verify requirements with the adopted code in your jurisdiction and consult a licensed electrician for permanent installations.