Voltage drop

Why voltage drops

A conductor is a long, thin resistor. Copper is a very good conductor, but a 150 ft run of 6 AWG still has 0.074 Ω in each leg, and current has to flow out on one conductor and back on the other. By Ohm’s law, every ampere through that resistance costs some voltage, and the load gets what is left. Nothing is wrong with the circuit: this is simply what wire does. The engineering question is how much loss you will accept.

Three things set the size of the drop:

• Resistance per foot. It falls as the conductor gets bigger and is higher for aluminum than for copper. NEC Chapter 9 Table 8 lists DC resistance at 75 °C; IEC 60228 lists maximum resistance at 20 °C.
• Length. The drop is proportional to the one-way length, and the return conductor doubles it on a two-wire circuit.
• Current. Twice the load, twice the drop. Motor starting current, which can be several times the running current, makes a brief but much larger dip.

Alternating current adds reactance. The magnetic field around each conductor opposes changes in current, and that opposition grows with conductor spacing and with steel raceways. In Chapter 9 Table 9 the reactance of 10 AWG copper in PVC conduit is only 4 % of its resistance; at 1/0 it is 37 %, and from 350 kcmil up it is the bigger of the two. How much of it shows up as voltage drop depends on the power factor of the load. Temperature matters too: copper’s resistance rises about 0.32% per °C near operating temperature, so a conductor running at 50 °C has about 8 % less resistance than the 75 °C figure in Table 8. Calculations that use the table are slightly conservative for lightly loaded runs, which is the safe direction.

What the drop looks like along a run

Picture a detached garage fed from the house panel. The current leaves on one conductor and returns on the other, and the voltage between the two shrinks steadily with distance. At the panel the pair carries the full 240 V; at the far end the load sees the supply minus the drop in both conductors.

The same picture applies to a 120 V circuit, where the second conductor is the neutral. It is why the voltage at a receptacle at the end of a long kitchen run reads a few volts low under load and returns to normal when the load is switched off: with no current there is no drop.

Working it out

Two-wire circuits: single-phase and dc

Multiply the current by the resistance of both conductors. With resistance per 1,000 ft from Chapter 9 Table 8 and the one-way length in feet:

For the garage run on 6 AWG copper (0.491 Ω/kft): 2 × 40 × 0.491 × 150 / 1000 = 5.89 V, which is 2.46 % of 240 V. The 2 accounts for the return conductor, so the length in the formula is always the one-way distance.

Balanced three-phase circuits

In a balanced three-phase circuit the return current is shared by the other phases, and the line-to-line drop works out to √3 times the drop in one conductor:

Compared with the two-wire formula for the same conductor and current, the result is √3/2 = 0.866 as large, and it is a fraction of the line-to-line voltage, so a 480 V feeder tolerates far more volts of drop than a 120 V branch circuit for the same percentage.

The K method

Field calculations often use a constant K, the resistance of a conductor one circular mil in area and one foot long. Divide by the actual area and you have the resistance per foot, so the two-wire drop becomes 2 × K × I × L / CM. IAEI’s articles give K as 12.9 for copper and 21.2 for aluminum at 75 °C. Those numbers fall straight out of Table 8 (resistance × area / 1000):

K is not quite constant, because stranding and the table’s rounding vary from size to size, but the spread is small: averaged over these sizes it is 12.81 for copper and 21.08 for aluminum. K is a DC quantity, which is its weakness on large AC conductors.

AC with reactance: Table 9 effective impedance

Chapter 9 Table 9 gives AC resistance and inductive reactance per 1,000 ft for three single conductors in PVC, aluminum or steel conduit, plus an “effective Z” at a power factor of 0.85. The effective impedance is the part of the conductor impedance that lines up with the load current: R × cos θ + X × sin θ. For a different power factor you compute it from R and X rather than using the tabulated column.

Here is how far the DC method can be off for three-phase feeders at 480 V in steel conduit, with the drop computed from Table 9 at 0.85 power factor and from Table 8 DC resistance:

On the 500 kcmil feeder the DC method misses 48 % of the drop: in steel conduit its reactance (0.048 Ω/kft) is larger than its AC resistance (0.029 Ω/kft), and the DC method ignores reactance altogether. Steel matters: for 4/0 conductors, Table 9 puts the reactance in steel conduit 24 % above the value in PVC or aluminum conduit. The power factor matters too. At unity power factor the effective impedance of the 500 kcmil run is just its AC resistance; recomputed at 0.8 from R and X it is 0.0520 Ω/kft, against the tabulated 0.05 at 0.85.

Percent, and the voltage left at the load

Percent drop is the drop divided by the nominal system voltage: line-to-neutral for a 120 V or 277 V circuit, line-to-line for a 240 V, 208 V or 480 V load. The voltage at the load is the supply minus the drop. Remember that the supply itself may already be below nominal at the service, which is why the limits are kept modest.

Run the numbers for your circuit

The calculator below covers the NEC editions from 2017 to 2026 as well as IEC 60364 and BS 7671. Enter the load, the length and the conductor, and it shows each step of the working.

If the tool does not load here, you can calculate voltage drop on its own page. To pick a size that satisfies ampacity and drop together, find the wire size for the load.

What the NEC says

The NEC’s two figures are informational notes. The branch-circuit note says that conductors sized to keep the drop under 3% at the farthest outlet, with no more than 5% for the feeder and branch circuit together, give reasonable efficiency of operation. The feeder note says the same thing from the feeder’s side. Article 90 makes informational notes explanatory, not enforceable, so an inspector cannot fail a general-purpose circuit for a 4% drop on the strength of these notes alone. The notes have moved around between editions:

Voltage drop does become a requirement in a few places:

• Sensitive electronic equipment on the 60 V-to-ground technical power systems of Article 647: for fixed equipment the drop may not exceed 1.5% on a branch circuit or 2.5% for feeder and branch circuit combined (647.4(D)), with lower limits for receptacle circuits. The 2026 edition renumbers the article, which moves the rule to 647.5(D).
• Fire pumps: the voltage at the controller’s line terminals may not dip more than 15% below the controller rating while the motor starts, and with the motor running at 115% of full-load current the voltage at the controller’s motor-side terminals may not fall more than 5% below the motor rating (695.7, which follows NFPA 20). The 2026 edition renumbers Article 695, which moves the rule to 695.8.
• Equipment instructions. Listed equipment has to be installed according to its instructions (110.3(B)). If a pump, charger or HVAC unit states a minimum supply voltage, that number is enforceable.
• Energy codes. Commercial energy codes can add their own voltage-drop rule; ASHRAE 90.1, adopted by many states for commercial buildings, has one in section 8.4.1. We have not checked its current wording against a public copy, so confirm the edition your jurisdiction uses with the plans examiner.

IEC 60364 and BS 7671

IEC 60364-5-52 deals with voltage drop in clause 525 and gives recommended maxima in its Annex G, measured from the origin of the installation to the equipment and expressed as a percentage of nominal voltage:

Final circuits should stay within the first row where possible, even on a private supply. Where the main wiring is longer than 100 m, the limits may grow by 0.005% per metre beyond 100 m, up to 0.5% extra: a 180 m run gains 0.4%, and anything from 200 m up gets the full 0.5%. Motor starting and other inrush may exceed the limits briefly, provided the equipment still gets the voltage its own standard requires.

BS 7671 follows the same pattern. Regulation 525.202 treats the requirement as met when the drop from the origin to a socket-outlet or fixed equipment does not exceed the values in Appendix 4, section 6.4, which are the same 3% and 5%: 6.9 V and 11.5 V on a 230 V supply. You will still see these limits attributed to Appendix 12; NICEIC and NAPIT both place them in Appendix 4, section 6.4 of the current edition. Appendix 4 also gives the drop for each cable type as millivolts per ampere per metre, so the working is mV/A/m × current × length / 1000. For a 6 mm² twin-and-earth radial carrying 32 A over 20 m that is 4.67 V, or 2.03 %.

The UK has a modern reason to stay well inside 5%. Many EV chargers on PME supplies have a built-in open-PEN detection device that disconnects when the voltage at the charger leaves the 207 to 253 V band. The supply itself may legitimately be 6% low, 216.2 V, and a 5% circuit drop then leaves 204.7 V, below the trip point, while 3% leaves 209.3 V. The IET therefore recommends a 3% design limit for these chargers. At 32 A, that sets the longest twin-and-earth run:

The IET’s own figures, worked from the armoured-cable table, come out at the same lengths. Current-carrying capacity is a separate check: see the cable size calculator.

What drop does to motors, LEDs and electronics

• Motors. Torque varies roughly with the square of the voltage, so a motor at 90% voltage has about 81 % of its torque. It slows, draws more current to carry the same load and runs hotter. Contactors and relays may chatter or drop out when the voltage sags during starting; ABB’s handbook puts reliable contactor holding at no lower than 85% of rated voltage.
• Heaters. A resistive heater’s output varies with the square of the voltage, so a 3% drop costs 5.9 % of the output. Water heats more slowly, but nothing is damaged.
• LED lighting. Mains LED drivers regulate their output over a wide input range, so a few percent of drop changes nothing you can see. Low-voltage DC strips are the opposite. A 12 V strip drawing 4 A at the end of a 20 ft pair loses 1.05 V (8.7 %) on 18 AWG and 0.41 V (3.4 %) on 14 AWG: the far end of a long strip visibly dims or shifts colour.
• Electronics. Switch-mode power supplies tolerate a wide input range, but a deep sag can reset controllers, drop network gear or push a UPS onto battery. Incandescent and halogen lamps simply dim and look redder.
• Energy. The lost volts times the current is heat in the cable. The garage feeder on 6 AWG wastes 236 W at full load, about 688 kWh a year if it runs 8 hours a day.

Long runs: a detached garage and a well pump

A detached garage 150 ft away

Take a 40 A load at 240 V single-phase, 150 ft from the house panel. With Table 8 DC resistance at 75 °C:

8 AWG copper misses the 3% suggestion; 6 AWG copper meets it with room for a short branch circuit in the garage, and 4 AWG aluminum lands close to 6 AWG copper at 2.54 %. Whichever you choose still has to have enough ampacity for the load and its overcurrent device, which is a separate check.

A submersible well pump

Pumps are where voltage drop bites hardest, because the cable runs to the well and then down it. Suppose a 240 V pump draws 10 A running, the well is 150 ft from the house and the pump hangs 250 ft down: 400 ft of cable each way. Assume it draws 5 times running current for the moment it starts (check the nameplate or the manufacturer’s data):

Counting only the horizontal distance would suggest a far smaller cable than the pump needs. Pump manufacturers publish maximum cable lengths for each motor and cable size, and those tables, not a general rule, decide the size; the starting dip is often what limits it.

How far can you go?

Because drop is proportional to length, each conductor size has a distance at which it reaches a given percentage. For a 40 A load at 240 V:

Read across at 3% to find the reach of each size, then halve it for a load of twice the current, or double it for half. At 120 V the same current gets half the distance, because the same volts of drop are twice the percentage.

How to reduce voltage drop

1. Use a bigger conductor. Two AWG sizes up cuts resistance by roughly 37 % (6 AWG to 4 AWG copper).
2. Use a higher voltage. A 2,400 W load 100 ft away on 12 AWG drops 6.4 % at 120 V but only 1.6 % at 240 V: half the current gives half the volts, and those are a smaller share of twice the voltage.
3. Shorten the run. Put a subpanel near the loads and feed it with one large feeder rather than running many long branch circuits.
4. Choose copper where size is tight; aluminum needs about two sizes more for the same drop.
5. Keep circuit conductors together in the same raceway, and prefer non-magnetic raceway on very large feeders, to keep reactance down.
6. Improve power factor on large inductive loads, which reduces the current for the same real power.

Upsizing has a knock-on effect in the NEC: if you increase the ungrounded conductors beyond the minimum, the equipment grounding conductor has to grow in proportion to the circular-mil area (250.122(B) through the 2023 edition, 250.122(D) in 2026). See grounding and bonding for a worked example.

Common mistakes

• Doubling the length twice. The 2 in the two-wire formula already counts the return conductor; use the one-way distance.
• The wrong base voltage. A 240 V load’s drop is a share of 240 V, a 120 V load’s of 120 V, and a three-phase drop of the line-to-line voltage.
• Forgetting vertical runs in wells, risers and tall buildings.
• Trusting K on big AC feeders. Above about 1/0, and in steel conduit, use Table 9 impedance or the manufacturer’s data.
• Checking only running current for motors that start often or start under load.
• Treating 3% as a code violation. It is advice in the NEC, unless your jurisdiction, the equipment instructions or an article such as 647 or 695 says otherwise.
• Forgetting the grounding conductor when upsizing for drop.