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Voltage Drop Calculator

DC, single-phase and three-phase. AWG or mm², copper or aluminium. Every figure comes with the formula it came from.

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  • VVoltage
  • ΔVVoltage Drop
  • %VDPercentage
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Voltage drop calculator

Electrical system
V
Load given as
A

Conductor

Size system
Material

Run

One-way is the distance from the source to the load. Round-trip is the total conductor length there and back. The formulas already account for the return path, so choose the one you actually measured.

%
Advanced options

Identical conductors run in parallel. Leave blank for one.

°C

Reference data is at 75 °C.

1 for a purely resistive load. Motors and electronics typically run 0.7 to 0.95.

Stranded conductors read about 2% higher resistance than solid at the same size.

6.60% is above the 3% limit you set. About 6.60% of the 120 V supply is lost in the conductors under these conditions, leaving 112.08 V at the load. That is 7.92 V dropped across the run, out and back.

Your voltage drop appears here

Enter a supply voltage, a load current and a cable length. The result updates as you type.

Voltage dropOver limit

6.60%

7.92 V lost against a3% limit

Supply
120.00V
At the load
112.08V
Lost as heat
158.4W

6.60% is above the 3% limit you set. About 6.60% of the 120 V supply is lost in the conductors under these conditions, leaving 112.08 V at the load. That is 7.92 V dropped across the run, out and back.

To bring it down: increase the conductor size, shorten the run, split the load across parallel conductors, or supply the load at a higher voltage. Doubling the conductor area roughly halves the drop; halving the run length halves it exactly.

Source120.00 V
At load112.08 V
3% limit
Voltage falls from 120.00 volts at the source to 112.08 volts at the load, a drop of 6.60 percent against a limit of 3 percent.
Working, assumptions and sources
Conductor resistance
0.396 Ω
Impedance per 1000 ft
1.98 Ω

Calculation details

  1. Conductor resistance per 1000 ft

    R = table value for size and material12 AWG copper, stranded, at 75 °C

    Result: 1.98 Ω per 1000 ft

    From NEC Chapter 9 Table 8, direct-current resistance at 75 °C.

  2. Voltage drop

    V_drop = 2 × I × R × LV_drop = 2 × 20 A × 1.98 Ω/1000 ft × 0.1000 (1000 ft)

    Result: 7.92 V

    A two-wire circuit carries the full load current out and back, so the current travels twice the one-way run length.

  3. Voltage drop percentage

    %V_drop = (V_drop ÷ V_source) × 100%V_drop = (7.92 V ÷ 120 V) × 100

    Result: 6.60%

  4. Voltage at the load

    V_load = V_source − V_dropV_load = 120 V − 7.92 V

    Result: 112.08 V

  5. Power lost in the conductors

    P_loss = I² × R_path × number of current-carrying pathsP_loss = 20² × 0.198 Ω × 2

    Result: 158.4 W

    Only resistance dissipates power. Reactance stores and returns energy each cycle rather than turning it into heat.

Assumptions and sources

  • Resistance only. Inductive reactance is not included. This matches the widely used simplified method and is exact at unity power factor.

Resistance: NEC Chapter 9, Table 8 (DC resistance at 75 °C)

02Definition

What is voltage drop?

Voltage drop is the voltage a circuit loses on the way to whatever it is supplying. Copper and aluminium both resist the current passing through them, so what arrives at the load is less than what left the source, and the lost share leaves as heat. Longer runs, smaller conductors and higher currents all increase it.

What is voltage drop covers why it happens and what it does to motors, lighting and electronics.

Voltage along a cable run, falling from 120.00 volts at the source to 112.08 volts at the loadTwo drawings sharing one horizontal scale. Above, a plot of voltage along the run: a dashed line marks the 120.00 volt supply level, and a solid line starting at the same point falls steadily to 112.08 volts by the far end, a drop of 7.92 volts or 6.60 per cent. Below, the circuit itself: a source on the left joined to a load on the right by two conductors, one carrying current out and one carrying it back, spanning the same distance as the plot above. The vertical scale of the plot is exaggerated for legibility.Voltage along the run120.00 V112.08 V−7.92 V (6.60%)SourceLoadConductor — current outReturn conductorOne-way run length
A 120 V single-phase circuit feeding a 20 A load through 100 ft of 12 AWG copper. The supply leaves the panel at 120 V and reaches the load at 112.08 V. The 7.92 V difference is the voltage drop, and it leaves the circuit as heat in the two conductors rather than doing any work at the load.

03Method

How to calculate voltage drop

Current and conductor resistance set the drop directly. Length, wire size, material and temperature are what set that resistance. Five steps turn them into a number.

  1. Find the conductor impedance

    Look up resistance per 1000 ft for the size and material. For AC below unity power factor, add inductive reactance and combine the two at the load’s power factor angle.

  2. Reduce the run to a one-way length

    Every formula here takes the distance from source to load. A round-trip measurement is halved first, and metres are converted to feet, because the published resistance data is per 1000 ft.

  3. Apply the system multiplier

    2 for DC and single-phase, because current goes out and comes back. √3 for balanced three-phase, because no conductor carries the full return current.

  4. Multiply by current

    Ohm’s law does the rest. Current times impedance times length times the multiplier gives the volts lost in the conductors.

  5. Express it as a percentage

    Divide the drop by the supply voltage. The percentage is what limits are written against, and what tells you whether the run is workable.

The full method, the unit traps and the other two mistakes that cause most bad results are set out on how to calculate voltage drop.

04Formula

Voltage drop formula

One equation covers every case here: current multiplied by the impedance of the run, scaled by a factor set by the circuit. The 2 counts two conductors, since current leaves along one and returns along the other. The √3 is smaller because a balanced three-phase circuit has no return conductor carrying full load current.

The percentage is %V_drop = (V_drop ÷ V_source) × 100, and the voltage left at the equipment is V_load = V_source − V_drop.

Derivations, the metric ohms-per-kilometre form, the circular-mil K method and the versions rearranged for conductor size or maximum length are on the voltage drop formula page.

DC and single-phase AC
V_drop = 2 × I × Z × L
I
Load current · A
Z
Conductor impedance per unit length · Ω / 1000 ft
L
One-way run length · thousands of feet
Balanced three-phase AC
V_drop = √3 × I × Z × L
√3
Approximately 1.732 · line-to-line
Effective impedance below unity power factor
Z_e = R × cos θ + X_L × sin θ
R
AC resistance · Ω / 1000 ft
X_L
Inductive reactance at 60 Hz · Ω / 1000 ft
θ
Power factor angle · arccos PF

05Worked example

Voltage drop calculation, start to finish

A 120 V branch circuit taken through every step the calculator takes: resistance looked up for the size and material, the run reduced to a one-way length, then drop, percentage and the voltage left at the load.

Given

System
Single-phase AC
Supply
120 V
Load
20 A
Conductor
12 AWG stranded copper
Run
100 ft one-way
Limit
3%

Working

  1. Resistance of 12 AWG stranded copper at 75 °C

    R = 1.98 Ω per 1000 ft

    NEC Chapter 9, Table 8.

  2. Voltage drop

    V_drop = 2 × 20 A × 1.98 Ω/1000 ft × 0.1 (1000 ft)V_drop = 7.92 V
  3. As a percentage

    %V_drop = (7.92 ÷ 120) × 100%V_drop = 6.60%
  4. Voltage at the load

    V_load = 120 − 7.92V_load = 112.08 V

Four more worked examples, including three-phase and a resistor, are on the voltage drop calculation page.

07Questions

Voltage drop questions, answered

What is voltage drop?

Voltage drop is the voltage lost between the source and the load because the conductors carrying the current are not perfect. Every conductor resists the current passing through it, and pushing current through resistance costs voltage, so what arrives at the equipment is always less than what left the supply. The difference does no useful work; it leaves the circuit as heat in the cable. It is normally quoted as a percentage of the supply, because 3 V lost from 120 V matters far more than 3 V lost from 480 V.

How do you calculate voltage drop?

Multiply the load current by the impedance of the run and by the multiplier for the circuit type. In practice that is five steps: look up the conductor resistance per unit length for the size and material, reduce the run to a one-way length, pick the multiplier (2 for DC and single-phase, √3 for balanced three-phase), multiply current × impedance × length × multiplier, then divide by the supply voltage for the percentage. The accuracy of the answer depends entirely on how well you know the current, the length and the conductor temperature.

What is the voltage drop formula?

For DC and single-phase AC it is V_drop = 2 × I × Z × L. For balanced three-phase it is V_drop = √3 × I × Z × L. I is the load current in amperes, Z is the conductor impedance per unit length, and L is the one-way distance from source to load in the units Z is published in. The 2 accounts for current travelling out on one conductor and back on the other; √3 is smaller because a balanced three-phase circuit has no return conductor carrying full load current. Z is plain resistance for DC and for AC at unity power factor, and R × cos θ + X_L × sin θ below it.

How do I calculate voltage drop in a cable?

The same way as any other run, using the resistance of the cable conductors. Working in metric units, V_drop = 2 × I × (R_km ÷ 1000) × L, with R_km the conductor resistance in ohms per kilometre and L the one-way cable run in metres. Substitute √3 for the 2 on a balanced three-phase cable. One point catches people out: for a two-core cable the cable length is the one-way distance, because the second core is the return path the multiplier already accounts for.

How do I calculate voltage drop across a resistor?

Use Ohm’s law on its own: V = I × R. Multiply the current through the resistor in amperes by its resistance in ohms. A 470 Ω resistor carrying 20 mA drops 0.02 × 470 = 9.4 V. There is no length term, no conductor material and no multiplier, because you are looking at one component rather than a run of cable. Rearranged, I = V ÷ R gives the current from a known drop and R = V ÷ I gives the resistance needed to produce one. Power dissipated is P = V × I, which is what sets the wattage rating the part needs.

What causes voltage drop in a wire?

Conductor resistance, and on AC a second contribution from inductive reactance. A wire is a lattice of metal atoms with electrons moving through it; the electrons collide with the lattice as they go and each collision gives up a little energy as heat. The bulk effect is resistance, and Ohm’s law turns it into a voltage difference between the two ends. On AC the alternating current also sets up a changing magnetic field around the conductor which opposes the change, adding to the drop without dissipating energy. Terminations, splices and switch contacts add resistance of their own that no calculation predicts.

Does wire size affect voltage drop?

Strongly, and inversely through cross-sectional area. A larger conductor gives the current more room, so resistance falls roughly in proportion to the increase in area and the drop falls with it. Each AWG step down in gauge number is about 1.26 times the area, and three steps roughly doubles it and halves the drop. A 120 V circuit carrying 20 A over 100 ft of 12 AWG copper drops 6.60%; the same run in 8 AWG copper drops 2.59%, because the conductor resistance falls from 1.98 to 0.778 Ω per 1000 ft.

Does cable length affect voltage drop?

Yes, in direct proportion. Double the run and you double the drop; halve it and you halve it exactly. Resistance accumulates with every metre of conductor, so length is the one factor that scales the answer with no diminishing returns in either direction. This is why shortening a run, or moving a distribution point closer to the load, is often cheaper than upsizing the whole cable. Enter the one-way distance from source to load, not the total length of conductor there and back, or the result comes out doubled.

Does current affect voltage drop?

Yes, and also in direct proportion. Drop and current rise together, so a circuit carrying twice the current on the same conductor drops twice the voltage. Power lost as heat behaves differently: it follows I² × R, so doubling the current quadruples the heating. This is why the drop should be calculated at the actual load current rather than at the rating of the breaker protecting the circuit, which is usually higher and will overstate the result.

What is a DC voltage drop?

It is the voltage lost along the conductors of a direct-current circuit, where resistance is the only thing opposing the current. There is no reactance and no power factor, which makes DC the simplest case and the one where a resistance-only calculation is exactly right rather than an approximation. It matters more than the arithmetic suggests because DC systems usually run at low voltage: losing half a volt on a 120 V circuit is 0.42% and nothing notices, while losing the same half volt on a 12 V battery circuit is 4.2%.

How do you calculate DC voltage drop?

Use V_drop = 2 × I × R × L, with I the load current, R the conductor resistance per unit length and L the one-way run. The 2 is there because current leaves along one conductor and returns along the other, meeting the resistance of the run twice. No power factor term appears, so the calculation is exact for the conditions you give it. Low-voltage DC systems also draw much higher current for the same power, and drop rises with current, so the two effects compound on battery, solar and vehicle circuits.

How do you calculate AC voltage drop?

Start from the same formula but replace resistance with effective impedance: V_drop = 2 × I × Z_e × L for single-phase, or √3 × I × Z_e × L for balanced three-phase. Below unity power factor, Z_e = R × cos θ + X_L × sin θ, where X_L is the inductive reactance of the conductor and θ is the power factor angle. That is the component of the impedance lying along the direction of the load current, not the magnitude √(R² + X²). At unity power factor sin θ is zero, reactance drops out, and the result collapses back to a resistance-only calculation.

How do you calculate voltage drop percentage?

Divide the drop in volts by the supply voltage and multiply by 100: %V_drop = (V_drop ÷ V_source) × 100. A 7.92 V drop on a 120 V supply is (7.92 ÷ 120) × 100 = 6.60%, leaving 112.08 V at the load. The percentage is the figure worth working in, because it is what design targets are written against and what tells you whether a run is usable. The same volts lost mean very different things on a 12 V circuit and a 480 V one.

What is the difference between voltage drop and voltage loss?

In everyday use the two are interchangeable and both mean the volts that fail to reach the load. If you want to be precise, voltage drop is the potential difference measured across the conductors, in volts, while the associated power loss is what that difference costs you, in watts. Drop is measured with a voltmeter; loss is calculated as I² × R over the current-carrying conductors. Some equipment manufacturers also use "voltage loss" for the drop inside a device rather than in the wiring feeding it.

How do you measure voltage drop?

By reading the voltage at two points of an energised circuit while it is carrying its normal load, and taking the difference. A circuit at rest carries no current and therefore drops nothing, which is why a reading taken at an idle outlet tells you nothing useful. Measurement and calculation answer different questions: the calculation models an ideal run, while a reading includes terminations, splices, contact resistance and the real conductor temperature. Live testing carries genuine risk and belongs to people trained and equipped for it, working to the safety procedures that apply where they are.

What is a voltage drop test?

A diagnostic check that looks for excessive resistance in a circuit that is under load, rather than measuring resistance directly. Because the reading is taken with current flowing, it reveals problems a resistance measurement on a dead circuit can miss: a corroded joint or a loose terminal may read near zero ohms yet drop a significant voltage once real current passes through it. Comparing the measured figure against a calculated one is what makes the result meaningful, since a reading far above the calculation points at a connection rather than an undersized conductor.

How can I reduce voltage drop?

Four things work. Increase the conductor size, since drop falls roughly in proportion to the increase in cross-sectional area. Shorten the run, which reduces it exactly in proportion. Split the load across conductors in parallel, which divides the impedance of the path by the number of sets. Or supply the load at a higher voltage, which cuts the current needed for the same power and reduces the drop twice over. Switching from aluminium to copper at the same size is a fifth option and cuts the drop by about 39%.

What wire size should I use to reduce voltage drop?

There is no single answer, because the size depends on the current, the run length and the supply voltage together. Work backwards instead: set the drop limit you want to hit and solve for the smallest conductor that clears it. Bear in mind that the answer has to satisfy two independent checks. Voltage drop is an engineering calculation; ampacity, the current a conductor may carry, is a separate question answered by published tables and then modified for ambient temperature, conductor bundling and terminal ratings. A size that passes one says nothing about the other.

What is an acceptable voltage drop?

3% on a branch circuit and 5% for feeder and branch circuit combined are the figures most often quoted. In the National Electrical Code they appear in Informational Notes to 210.19(A) and 215.2(A), which are explanatory rather than enforceable, so within the code itself they are recommendations. Some local amendments, energy codes and utility requirements make equivalent limits mandatory, and individual equipment may specify something tighter. Treat 3% and 5% as engineering targets, and check what actually applies to your installation and jurisdiction rather than assuming these figures are binding.

Can voltage drop affect electrical equipment?

Yes, though the effects are usually gradual rather than dramatic. Motors draw more current at reduced voltage, start harder and run hotter, which shortens their life. Incandescent and some LED lighting dims. Resistive heating elements take longer to reach temperature because power falls with the square of the voltage. Electronics with a specified minimum input voltage may reset, behave erratically or refuse to start. Separately, the energy lost in the conductors is paid for at the meter and never delivered: a 20 A load on 100 ft of 12 AWG copper turns about 158 W into heat in the wiring.

How does copper vs aluminum affect voltage drop?

Aluminium drops noticeably more at the same conductor size. It has roughly 61% the conductivity of copper, so about 1.6 times the resistance and about 1.6 times the drop. For example, 2 AWG copper is 0.194 Ω per 1000 ft against 0.319 Ω per 1000 ft for 2 AWG aluminium. To match a copper conductor on drop you generally need to go up about two AWG sizes in aluminium, which is why aluminium runs are physically larger for the same electrical result even though the conductor itself is lighter and cheaper.

How does temperature affect voltage drop?

Conductor resistance rises with temperature, so a hot conductor drops more voltage than a cold one carrying the same current. The correction published in NEC Chapter 9, Table 8 is R₂ = R₁ × [1 + α × (T₂ − 75)], with α of 0.00323 for copper and 0.00330 for aluminium against a 75 °C reference. Going from 75 °C to 90 °C raises copper resistance by about 4.8%. The reference data used here is quoted at 75 °C, which represents a loaded conductor; calculating at room temperature because that is what a data sheet quotes will understate a cable that is actually working.

Can I calculate voltage drop for a long cable run?

Yes, and the arithmetic does not change with distance, but long runs are where the result usually decides the design. Because drop scales directly with length, a run several times longer needs a conductor several times larger in area to hold the same percentage, and it is common for voltage drop rather than current-carrying capacity to set the size. Two other things become worth checking on long runs: the conductor temperature you assume, and on AC the reactance term, which is negligible on small conductors but not on large ones at low power factor.

What is the difference between single-phase and three-phase voltage drop?

The multiplier. Single-phase uses 2, because the current goes out along one conductor and returns along the other, meeting the resistance of the run twice. Balanced three-phase uses √3, about 1.732, because no single conductor carries the full return current. For the same current, conductor and distance, the three-phase drop is √3 ÷ 2, or about 86.6%, of the single-phase drop, and the figure is the line-to-line voltage drop. The comparison is a little misleading in practice, since a three-phase circuit delivers considerably more power at that same current.

Can a voltage drop calculator determine the correct wire size?

It can tell you the smallest conductor that keeps the drop inside a limit you set, which is one of the two checks a conductor has to pass. The other is ampacity, and the two are answered from different data and can give different answers; on a long run the drop limit usually decides, while on a short heavily loaded run ampacity does. The tools here report each separately rather than merging them into a single recommendation, because doing so would obscure which constraint is binding. Neither output is a determination of code compliance.