Voltage Drop: What It Is and How to Fix It

Every electrical conductor has inherent resistance. When current flows through that resistance, some voltage is lost as heat — this loss is called voltage drop. A 120V circuit with 5% voltage drop delivers only 114V at the load, which can cause motors to overheat, lights to dim, and sensitive electronics to malfunction.

The NEC does not legally mandate maximum voltage drop — it is a recommendation in NEC 210.19(A) Informational Note No. 4. However, most AHJs (Authorities Having Jurisdiction) enforce the 3%/5% guideline as a practical standard, and it is required on most engineering specifications.

Single-Phase: VD = (2 × L × I × R) / 1000 — where L is one-way conductor length in feet, I is load current in amps, and R is conductor resistance in ohms per 1000 feet from NEC Chapter 9, Table 8.

Three-Phase: VD = (√3 × L × I × R) / 1000 — the √3 factor (1.732) replaces the factor of 2 because current returns through the other two phases rather than a dedicated neutral.

Percentage: VD% = (VD / Vsource) × 100. For a 240V circuit with 6V drop: 6/240 × 100 = 2.5% — within the 3% recommendation.

Resistance values (R) come from NEC Chapter 9, Table 8 for DC resistance or Table 9 for AC impedance. For most branch circuits, Table 8 values are sufficient. For larger conductors (≥ 4/0 AWG) or long runs, use Table 9 AC impedance to account for skin effect and reactance.

Problem: A 120V single-phase branch circuit supplies a 16A continuous load. The run length is 150 feet. Using 12 AWG copper THHN. Does it pass the 3% recommendation?

Step 1: Look up resistance — NEC Chapter 9, Table 8: 12 AWG uncoated copper = 1.98 Ω/1000 ft.

Step 2: Calculate voltage drop — VD = (2 × 150 × 16 × 1.98) / 1000 = 9.50V.

Step 3: Calculate percentage — VD% = 9.50 / 120 × 100 = 7.92%. This exceeds 3% — FAIL.

Step 4: Solution options — (a) Upsize to 10 AWG (1.24 Ω/1000 ft): VD = (2 × 150 × 16 × 1.24) / 1000 = 5.95V = 4.96%. Still fails. (b) Upsize to 8 AWG (0.778 Ω/1000 ft): VD = (2 × 150 × 16 × 0.778) / 1000 = 3.73V = 3.11%. Still borderline. (c) Upsize to 6 AWG or shorten circuit length.

This example illustrates why long runs almost always require conductor upsizing beyond the ampacity minimum.

Branch Circuits: NEC 210.19(A) Informational Note No. 4 recommends maximum 3% voltage drop at the farthest outlet on the branch circuit.

Feeders: NEC 215.2(A) Informational Note No. 2 recommends maximum 3% voltage drop for the feeder, with a combined feeder + branch circuit total not exceeding 5%.

Sensitive Equipment: NEC 647.4(D) for sensitive electronic equipment requires maximum 1.5% drop on the branch circuit, for a total of 2.5%.

Fire Pump Circuits: NEC 695.7 requires voltage at the fire pump motor terminals to be not less than 115% of the motor nameplate minimum operating voltage at rated load — effectively limiting voltage drop to approximately 5%.

Increase conductor size — The most common solution. Each step up in AWG roughly halves the resistance. Going from 12 AWG to 10 AWG reduces resistance from 1.98 to 1.24 Ω/1000 ft (37% reduction).

Shorten the circuit — Reduce the physical distance between the panel and the load. This is often impractical after construction but should be considered during design.

Increase voltage — A 240V circuit has half the current of a 120V circuit for the same wattage. Converting from 120V to 240V cuts voltage drop percentage in half because both current and percentage base change.

Use parallel conductors — For large feeders, running two sets of conductors in parallel (each carrying half the current) reduces voltage drop by approximately 50%. NEC 310.10(G) permits parallel conductors for 1/0 AWG and larger.

Relocate the panel — Installing a subpanel closer to the load centers can dramatically reduce conductor lengths for branch circuits.

Using nameplate amps instead of design current — Always include the 125% continuous load factor before calculating voltage drop. A 16A continuous load should be calculated at 20A for the conductor, but the voltage drop calculation uses actual load current (16A), not the adjusted value.

Forgetting the round-trip distance — The factor of 2 in the single-phase formula accounts for current traveling to the load AND back. Omitting it gives exactly half the actual voltage drop.

Ignoring conduit material — Steel conduit increases AC impedance due to magnetic effects. For large conductors in steel conduit, use NEC Table 9 rather than Table 8.

Not considering motor starting current — Motors draw 6-8× FLC during starting. While brief, this inrush causes significant momentary voltage drop that can trip sag-sensitive controls or cause visible light flicker.