Voltage Drop Explained: Causes, Calculation, and NEC Limits

Voltage drop is the voltage lost in the cable itself, between the source and the load. It is caused by the resistance of the conductor, it grows with current and with length, and it shrinks as the conductor gets larger. A small drop is unavoidable and harmless. A large drop dims lights, slows motors, overheats equipment, and wastes energy as heat in the cable. This guide explains the calculation, the National Electrical Code recommendations that most electricians design to, and the practical decision — a larger conductor, a shorter run, or a higher supply voltage — when the drop is too large.

What causes voltage drop

Every conductor has resistance. When current flows through it, some voltage is used up pushing the current along the cable, and that voltage never reaches the load. The loss follows Ohm’s law applied to the cable: the drop equals the current times the resistance of the length of cable involved. Copper has lower resistance than aluminium of the same size, a larger conductor has lower resistance than a smaller one, and resistance rises with temperature, so a hot cable drops more voltage than a cool one carrying the same current.

Length is the multiplier most people underestimate. The current travels out and back, so a single-phase circuit’s drop depends on twice the one-way length: the supply conductor and the return conductor both contribute. A load 150 feet from the panel sits at the end of 300 feet of conductor, and the drop is calculated on 300 feet, not 150. Three-phase circuits behave differently — the factor is 1.732, the square root of 3 — but the principle is the same: the whole path counts.

Voltage drop is not the same as a voltage sag from the supply. A sag affects the whole installation and comes from the utility or from a large load starting elsewhere. Voltage drop is local to one circuit and grows with that circuit’s own current. If one circuit’s lights dim when its own load switches on, and the rest of the building is steady, the cause is usually drop in that circuit’s conductors.

Try the method on your own figures with the Voltage Drop Calculator.

The calculation, step by step

For a single-phase circuit, the voltage drop is the current, times twice the one-way length, times the conductor resistance per unit length, divided by the length unit of the resistance table. Using resistance in ohms per 1,000 feet and length in feet, the formula is Vd = (2 × L × I × R) ÷ 1,000. The percentage drop is the drop divided by the supply voltage, times 100.

Worked example: a 120 V circuit, 150 feet one way, carrying 20 A, wired in 12 AWG copper. Table 8 gives 12 AWG copper at 75°C as 1.98 ohms per 1,000 feet. The drop is (2 × 150 × 20 × 1.98) ÷ 1,000, which is 11.88 V, or 9.9% of 120 V — far over the recommended 3%. The load sees only about 108 V, and the cable dissipates the rest as heat.

Fix it by upsizing. The same run in 6 AWG copper, at 0.491 ohms per 1,000 feet, drops (2 × 150 × 20 × 0.491) ÷ 1,000, which is 2.95 V, or 2.5% — inside the recommendation. That is the usual trade: a larger conductor costs more copper and is harder to pull, but it delivers the voltage the load was designed for. A second worked example at 240 V shows why higher voltage helps: 100 feet, 30 A, 10 AWG copper at 1.24 ohms per 1,000 feet drops (2 × 100 × 30 × 1.24) ÷ 1,000, which is 7.44 V, or 3.1% of 240 V — the same cable and current that would be 6.2% at 120 V.

The NEC recommendations: 3% and 5%

The National Electrical Code addresses voltage drop in informational notes, not in enforceable requirements, with one narrow exception for fire pumps. The notes in Articles 210.19(A) and 215.2(A) recommend sizing branch-circuit conductors so the drop does not exceed 3% at the farthest outlet, and sizing feeders so the combined drop of feeder and branch circuit does not exceed 5%. The purpose stated in the notes is reasonable efficiency of operation.

Informational does not mean ignorable. The 3% and 5% figures are the design targets used across the trade, they are adopted as requirements by some local codes and specifications, and equipment manufacturers commonly specify a supply-voltage range that assumes the drop stays within them. A motor that must start under load, a sensitive electronic load, or a long lighting run will all show the effects of excessive drop long before any safety device operates.

Two cautions apply. First, the recommendations are about the farthest outlet on the circuit, at full load, not about the average outlet at average load — design for the worst case, because the worst case is when the equipment notices. Second, voltage drop and ampacity are separate checks. A conductor can be large enough for the current — satisfying Table 310.16 — and still too small for the length. Every long run needs both checks: ampacity first, then drop.

What excessive drop does to equipment

Resistive loads — heaters, incandescent lamps — simply produce less: a heater at 108 V instead of 120 V produces about 81% of its rated heat, because power falls with the square of the voltage. That is wasteful but rarely destructive. Motors are the opposite: an induction motor tries to deliver its rated torque, draws more current as the voltage falls, and overheats. Sustained undervoltage shortens motor life and can trip overloads on starting.

Electronic loads vary. Switch-mode power supplies tolerate a wide input range and often compensate by drawing more current, which increases the drop further — a feedback loop that ends at the supply’s limit or a tripped breaker. Older transformer-based equipment and some lighting ballasts simply misbehave. The symptom that brings most complaints is flicker or dimming when a large load starts, which is a transient drop on top of the steady-state drop, and it points to the same remedy: a larger conductor, a shorter run, or a dedicated circuit.

Energy waste is the quiet cost. The power lost in the cable is the drop times the current, dissipated as heat along the run. In the 12 AWG example above, 11.88 V times 20 A is about 238 watts of heat in the cable — more than two old-style 100-watt lamps, running whenever the circuit is loaded. Over a year of regular use, the wasted energy can cost more than the larger conductor would have.

Fixing a drop that is too large

The first fix is a larger conductor. Going from 12 AWG to 10 AWG cuts resistance by about 37%, and each further size up cuts it by roughly another quarter to a third. The calculator shows the drop for each size, so the smallest size that meets the target can be chosen rather than guessed. Remember to re-check ampacity, termination temperature ratings, and conduit fill after upsizing — a larger conductor changes more than the drop.

The second fix is a shorter run. Moving the panel, the subpanel, or the load closer to the supply cuts the drop in direct proportion: halving the length halves the drop. Where a long run is unavoidable, a subpanel fed by a larger feeder, with short branch circuits from it, often beats one very long branch circuit. The third fix is a higher supply voltage where the equipment allows it: the same power at 240 V draws half the current of 120 V, and the percentage drop halves with it, which is why long runs are commonly wired at 240 V.

When none of those is possible, the honest answer is to accept the drop and choose equipment rated for it, or to reduce the load on the circuit. What is never the answer is ignoring it: a circuit that works on a cool day at half load and fails on a hot day at full load has been designed for the wrong conditions.

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