Voltage drop calculations

Every conductor has resistance, and current through resistance produces a voltage drop. On a short circuit the drop is trivial; on a long run with a heavy load it can leave motors struggling to start, lights dim and electronics misbehaving. The NEC treats voltage drop mostly as a design recommendation rather than a rule, but master exams test it often, because it is a good calculation question and because a handful of articles do make it mandatory. This lesson explains where voltage drop appears in the Code, the formulas exam questions use, how to calculate the drop, the minimum conductor size for a given drop, and the maximum length for a given conductor.

Where the Code talks about voltage drop

The most quoted voltage drop guidance is in informational notes, which, as MELU-101 explained, are explanatory and not enforceable. The notes to 210.19 and 215.2 suggest sizing branch circuit conductors so the voltage drop does not exceed 3 out of 100 at the farthest outlet, and so the total drop on feeders and branch circuits together does not exceed 5 out of 100. Because these are informational notes, an exam answer that says the Code requires a 3 out of 100 limit on a general branch circuit is wrong; a question asking what the Code recommends is looking for those figures.

Some rules are mandatory. Article 695 sets a voltage limit at the controller and motor terminals of a fire pump during starting and running. Article 647, for sensitive electronic equipment, limits voltage drop on certain branch circuits and feeders. Energy codes adopted alongside the NEC in many states also set voltage drop limits, and project specifications often do. On the job, Dana treats the informational note values as the minimum standard of good work.

The formulas

Exam questions use a simple direct-current approximation based on conductor resistance. For single-phase circuits, the voltage drop is 2 times K times I times L divided by the circular mil area of the conductor. For three-phase circuits, replace the 2 with the square root of 3, about 1.732. K is the resistance of a conductor one circular mil in area and one foot long, in ohms: about 12.9 for copper and about 21.2 for aluminum at normal operating temperatures. I is the load current in amperes, L is the one-way length of the circuit in feet, and the circular mil area comes from Chapter 9 Table 8.

The factor of 2 in the single-phase formula accounts for the current traveling out and back. The three-phase formula uses 1.732 because the phase currents are displaced and the line-to-line drop is the square root of 3 times the drop in one conductor. Some questions instead use the resistance per thousand feet directly from Chapter 9 Table 8 or Table 9: voltage drop equals 2 times the length times the resistance per foot times the current for single-phase. Both methods give nearly the same answer for small conductors, and both are acceptable exam methods unless the question specifies one.

Calculating the drop

# Voltage drop, minimum size and maximum length (K = 12.9 copper, 21.2 aluminum)
CM = {"14": 4110, "12": 6530, "10": 10380, "8": 16510, "6": 26240, "4": 41740, "2": 66360, "1/0": 105600}

def vdrop(k, amps, feet, cmil, three_phase=False):
    m = 1.732 if three_phase else 2
    return m * k * amps * feet / cmil

print("Garage feeder:", round(vdrop(12.9, 40, 150, CM["6"]), 2), "V")
print("Delgado CNC feeder:", round(vdrop(12.9, 120, 220, CM["1/0"], True), 2), "V")
print("Same garage feeder in aluminum 6 AWG:", round(vdrop(21.2, 40, 150, CM["6"]), 2), "V")

Garage feeder: 5.9 V
Delgado CNC feeder: 5.59 V
Same garage feeder in aluminum 6 AWG: 9.7 V

Finding the minimum size for an allowed drop

Rearranging the formula gives the minimum circular mil area for a target drop: circular mils equal 2 times K times I times L divided by the allowed voltage drop for single-phase, or 1.732 times K times I times L divided by the allowed drop for three-phase. You then choose the smallest conductor in Chapter 9 Table 8 with at least that area, and finally confirm that its ampacity is also enough. Voltage drop never replaces the ampacity check; it adds to it.

Upsizing branch circuit conductors for voltage drop has a consequence many candidates forget. Section 250.122(B) requires the equipment grounding conductor to be increased in proportion to the increase in the ungrounded conductors' circular mil area. If the 12 AWG conductors the 20-ampere circuit would normally need are increased to 6 AWG, the equipment grounding conductor must grow by the same ratio. MELU-104 covers that rule in detail.

Finding the maximum length

The same formula solved for length gives the longest run for a conductor size and allowed drop: length equals allowed drop times circular mils divided by 2 times K times I, for single-phase. This is a common exam form: 'What is the maximum distance a 20-ampere, 120-volt load can be from the panel on 10 AWG copper with a drop of no more than 3 out of 100?' The answer is 3.6 times 10,380 divided by 2 times 12.9 times 20, which is about 72 feet.

# Maximum one-way length for an allowed drop
def max_length(volts, share, k, amps, cmil, three_phase=False):
    m = 1.732 if three_phase else 2
    return volts * share * cmil / (m * k * amps)

print("20 A at 120 V on 10 AWG copper, 3 out of 100:", round(max_length(120, 0.03, 12.9, 20, 10380)), "ft")
print("30 A at 240 V on 10 AWG copper, 3 out of 100:", round(max_length(240, 0.03, 12.9, 30, 10380)), "ft")
print("50 A at 480 V three-phase on 6 AWG copper:", round(max_length(480, 0.03, 12.9, 50, 26240, True)), "ft")

20 A at 120 V on 10 AWG copper, 3 out of 100: 72 ft
30 A at 240 V on 10 AWG copper, 3 out of 100: 97 ft
50 A at 480 V three-phase on 6 AWG copper: 338 ft

Exam habits for voltage drop

Write the formula first, then plug in values with units. Check whether the length is one-way, which the formula expects, or total. Check whether the system is single-phase or three-phase, and use 1.732 only for three-phase. Use the K value the question gives; if none is given, use 12.9 for copper and 21.2 for aluminum, but some exam bulletins specify slightly different values. And watch for the question that asks for the voltage at the load rather than the drop: subtract the drop from the source voltage.

Voltage at the load and motor starting

Some questions ask for the voltage at the load rather than the drop. The answer is the source voltage minus the calculated drop: a 240-volt circuit with a 7.5-volt drop delivers 232.5 volts at the load. Motors are the loads most sensitive to drop, because a motor's starting current is several times its running current, and the drop during starting is correspondingly larger. A motor that runs well may fail to start, or start slowly and overheat, if its circuit is long and the conductors are sized only for running current. That is why the fire pump rules in Article 695 limit the voltage drop during starting as well as running, and why many specifications add a starting drop limit for large motors.

Voltage drop also wastes energy. The power lost in the conductors is the current squared times the resistance, which shows up on the owner's electric bill as heat in the walls. For a heavily loaded circuit that runs all day, the energy saved by one size larger conductor can repay its cost within a few years, an argument Dana uses with the Copper Kettle's owner when she recommends 8 AWG for the long kitchen exhaust fan circuit.