How to Calculate Voltage Drop

Every conductor has resistance, so current flowing through it loses some voltage on the way to the load. On a short run you never notice. On a long run the load sees a lower voltage than the panel supplies: lights dim, motors run hot, and electronics can drop out. Calculating the drop before you pull wire is a few lines of arithmetic.

The voltage drop formula

Two formulas cover most runs. Use the first for single-phase and DC circuits, and the second for balanced three-phase circuits. Length is the one-way distance from the source to the load, in feet, and resistance is in ohms per 1000 ft.

Single-phase or DC:  drop (V) = 2 x length x R x amps / 1000
Three-phase:         drop (V) = sqrt(3) x length x R x amps / 1000
Percent drop = drop (V) / supply voltage x 100
Voltage at the load = supply voltage - drop (V)

Single-phase counts the length twice because the current goes out on one conductor and comes back on the other. Three-phase uses the square root of 3 instead because the three line currents are 120 degrees apart. If you run parallel sets, divide the result by the number of sets, since the current splits between them.

Where the resistance number comes from

The NEC gives you the resistance in Chapter 9. Table 8 lists the direct-current resistance of each conductor size. Table 9 lists the effective impedance for alternating current at the power factor the table assumes, which includes the conductor reactance and changes with the raceway: PVC, aluminum or steel conduit. Table 9 Note 2 says the Effective Z column is a line-to-neutral value, so the same length factors above apply with either table.

Table 8 is slightly conservative on the smaller sizes shown here: its figure is higher than Table 9 down through 4 AWG. From 2 AWG up in this table the Table 9 figure is the higher one, because reactance becomes a real share of the impedance, so it is the more realistic number for AC circuits on large conductors. Table 9 assumes a single power factor, so for a motor or other load with a different power factor, a full impedance calculation with the actual power factor is the better tool.

What the NEC says about voltage drop

The 3% and 5% numbers most people quote are not mandatory rules. They appear in Informational Notes: 3% for a branch circuit to the farthest outlet, and 5% for the feeder and branch circuit together. The notes are in 210.19 and in 215.2(A)(2) Informational Note No. 2 in NEC 2023, 215.4(A)(2) Informational Note No. 2 in NEC 2026. The 2023 and 2026 values are the same; only the feeder section number moved. Informational Notes are recommendations, so check whether the specification, another code or the equipment maker sets a hard limit on your job.

Worked examples

Example 1: long 120 V branch circuit

A 16 A load sits 100 ft from the panel on 12 AWG copper at 120 V. Using the Table 8 resistance of 1.98 ohms per 1000 ft:

Drop = 2 x 100 ft x 1.98 x 16 A / 1000 = 6.34 V
Percent = 6.34 V / 120 V x 100 = 5.28 %
Voltage at the load = 113.66 V

That is above the 3% figure recommended in the Informational Note. The smallest size that stays within 3% for these inputs is 8 AWG, which drops 2.49 V (2.07 %).

Example 2: three-phase feeder, DC against AC method

A 100 A load on a 480 V three-phase feeder, 250 ft one way, 1/0 AWG copper in steel conduit. With Table 8 (0.122 ohms per 1000 ft) the drop is 5.28 V, or 1.1 %. With Table 9 Effective Z (0.13 ohms per 1000 ft):

Drop = sqrt(3) x 250 ft x 0.13 x 100 A / 1000 = 5.63 V
Percent = 5.63 V / 480 V x 100 = 1.17 %
Voltage at the load = 474.37 V

Here the AC method gives the higher drop, which is why it is the better choice on larger conductors. Judged against the 5% combined feeder and branch circuit figure from the Informational Notes (the 3% figure is for the branch circuit, not a feeder figure), both results are within it, and that leaves room for the branch circuit drop. Any branch circuit drop downstream adds to the feeder figure.

Example 3: 240 V single-phase subpanel feed

A 50 A load 150 ft away on 6 AWG copper at 240 V (Table 8: 0.491 ohms per 1000 ft) drops 7.37 V, which is 3.07 % of 240 V and above the 3% figure. The smallest size within 3% is 4 AWG, dropping 4.62 V (1.93 %).

How to fix excessive voltage drop

• Upsize the conductor. Resistance falls as the size goes up, and this is the usual fix. A bigger wire is not free: check that the lugs and terminals accept it.
• Use parallel sets on larger feeders. The current splits between the sets, so the drop divides by the number of sets.
• Shorten the run, for example by moving the panel closer to the load.
• Use copper instead of aluminum at the same size. Aluminum has higher resistance per 1000 ft.
• Raise the supply voltage where the equipment allows it. The same load at 240 V draws half the current it does at 120 V, so the percent drop falls sharply.

A size picked for voltage drop must still meet ampacity, termination and overcurrent protection rules. Use the wire size calculator for those, and keep the larger of the two answers.