Give this voltage drop calculator the size, length, current and temperature of a DC, single-phase or three-phase cable and it returns the drop, the load voltage and the power lost. Or give it a budget and find the smallest AWG or metric wire that stays within it.
Line-to-line for three-phase.
Source to load. The return path is added for you.
A warm cable has more resistance. NEC tables use 75 °C.
Voltage drop
828.1 mV
Same run with other wire sizes
| Wire | Drop | Percent | At load | Power lost |
|---|---|---|---|---|
| 17 AWG | 1.66 V | 13.8 % | 10.34 V | 16.6 W |
| 16 AWG | 1.32 V | 11 % | 10.68 V | 13.2 W |
| 15 AWG | 1.04 V | 8.7 % | 10.96 V | 10.4 W |
| 14 AWG | 828 mV | 6.9 % | 11.17 V | 8.28 W |
| 13 AWG | 656 mV | 5.47 % | 11.34 V | 6.56 W |
| 12 AWG | 521 mV | 4.34 % | 11.48 V | 5.21 W |
| 11 AWG | 414 mV | 3.45 % | 11.59 V | 4.14 W |
Use Vd = 2 × I × L × R for DC and single-phase circuits, with L the one-way length and R the resistance per unit length. 10 A through 5 m of 14 AWG copper (8.281 Ω/km at 20 °C) drops 828 mV, or 6.9 % of a 12 V supply, and the cable dissipates 8.28 W. For balanced three-phase, replace the 2 with √3.
The drop is the load current times the resistance of the conductors it flows through. The multiplier counts those conductors.
Schneider Electric’s Electrical Installation Guide gives the same expression for single-phase and balanced three-phase cables in its Figure G29. The term R cos φ + X sin φ comes from the phasor sum of source voltage, load voltage and the current through the cable’s R + jX. The part of the drop in quadrature with the load voltage is left out, which changes the magnitude very little while the drop is small. With power factor 1 and no reactance the formula is purely resistive.
The percentage is Vd ÷ source voltage. Power lost is 2 × I² × R × L for two conductors and 3 × I² × R × L for three-phase.
18 AWG copper is 20.95 Ω/km at 20 °C. Vd = 2 × 5 × 0.005 × 20.95 = 1.05 V, or 8.73 %. The strip sees 10.95 V and the cable burns 5.24 W. That much drop shows up on an unregulated strip as dimming at the far end, so a thicker feed or a second feed point is worthwhile.
The budget allows 720 mV, so R must be below 720 mV ÷ (2 × 20 A × 0.01 km) = 1.8 Ω/km. The first gauge under that is 7 AWG, at 1.634 Ω/km, which drops 654 mV (2.72 %) and dissipates 13.1 W.
Stranded copper at 70 °C is 3.507 Ω/km. At power factor 0.85, Vd = √3 × 25 × 0.08 × 3.507 × 0.85 = 10.3 V, which is 2.58 % of 400 V. The three conductors dissipate 526 W.
DC, 10 A over a 10 m one-way run of copper at 20 °C (20 m of conductor). Scale the drop and the loss linearly with length; the drop scales linearly with current and the loss with its square.
| AWG | Area | Resistance | Drop | Cable loss |
|---|---|---|---|---|
| 4/0 AWG | 107.2 mm² | 0.1608 Ω/km | 32.2 mV | 322 mW |
| 2/0 AWG | 67.43 mm² | 0.2557 Ω/km | 51.1 mV | 511 mW |
| 1/0 AWG | 53.49 mm² | 0.3223 Ω/km | 64.5 mV | 645 mW |
| 2 AWG | 33.62 mm² | 0.5128 Ω/km | 103 mV | 1.03 W |
| 4 AWG | 21.15 mm² | 0.8152 Ω/km | 163 mV | 1.63 W |
| 6 AWG | 13.3 mm² | 1.297 Ω/km | 259 mV | 2.59 W |
| 8 AWG | 8.367 mm² | 2.061 Ω/km | 412 mV | 4.12 W |
| 10 AWG | 5.261 mm² | 3.277 Ω/km | 655 mV | 6.55 W |
| 12 AWG | 3.308 mm² | 5.212 Ω/km | 1.04 V | 10.4 W |
| 14 AWG | 2.082 mm² | 8.281 Ω/km | 1.66 V | 16.6 W |
| 16 AWG | 1.308 mm² | 13.19 Ω/km | 2.64 V | 26.4 W |
| 18 AWG | 0.8229 mm² | 20.95 Ω/km | 4.19 V | 41.9 W |
| 20 AWG | 0.5189 mm² | 33.23 Ω/km | 6.65 V | 66.5 W |
| 22 AWG | 0.3243 mm² | 53.16 Ω/km | 10.6 V | 106 W |
| 24 AWG | 0.2047 mm² | 84.22 Ω/km | 16.8 V | 168 W |
| 26 AWG | 0.1281 mm² | 134.6 Ω/km | 26.9 V | 269 W |
One-way length in meters for DC or single-phase at 20 °C on copper. Multiply by 3.281 for feet. Length scales inversely with current.
| AWG | 5 V | 12 V | 24 V | 48 V | 120 V |
|---|---|---|---|---|---|
| 2 AWG | 14.6 m | 35.1 m | 70.2 m | 140 m | 351 m |
| 4 AWG | 9.2 m | 22.1 m | 44.2 m | 88.3 m | 221 m |
| 6 AWG | 5.78 m | 13.9 m | 27.8 m | 55.5 m | 139 m |
| 8 AWG | 3.64 m | 8.74 m | 17.5 m | 34.9 m | 87.4 m |
| 10 AWG | 2.29 m | 5.49 m | 11 m | 22 m | 54.9 m |
| 12 AWG | 1.44 m | 3.45 m | 6.91 m | 13.8 m | 34.5 m |
| 14 AWG | 0.906 m | 2.17 m | 4.35 m | 8.69 m | 21.7 m |
| 16 AWG | 0.569 m | 1.37 m | 2.73 m | 5.46 m | 13.7 m |
| 18 AWG | 0.358 m | 0.859 m | 1.72 m | 3.44 m | 8.59 m |
| 20 AWG | 0.226 m | 0.542 m | 1.08 m | 2.17 m | 5.42 m |
The calculation assumes the current stays fixed. A switching regulator draws more current as its input sags, so near its minimum input a real drop is worse than computed. Connectors, fuses and crimps add resistance that the cable table does not count, and at low voltage they can matter as much as the wire. Use a realistic conductor temperature: a cable near its rating runs far above 20 °C. Voltage drop and heating are separate limits, so a wire that passes this check still needs an ampacity check. Metric sizes use the nominal area: IEC 60228 also sets resistance limits for each size and a real cable can measure higher, so take the resistance from the data sheet when the budget is tight. For large AC cables, enter the reactance from the data sheet, and treat the DC resistance as a lower bound because skin and proximity effects raise the AC value.
For DC and single-phase AC, the voltage drop equation is Vd = 2 × I × L × R, where L is the one-way length and R is the resistance per unit length: the 2 covers the out and back conductors. For balanced three-phase, the 2 becomes √3, and Vd is the line-to-line drop.
Enter the one-way distance from source to load. The calculator adds the return path, so 5 m of cable means 10 m of conductor in a DC or single-phase circuit.
It depends on what the load tolerates. The NEC mentions voltage drop for most circuits only in informational notes, which are not mandatory, and the IAEI describes 5 % as the maximum generally suggested for a circuit; 3 % is a common design target. Low-voltage electronics often need less, so work back from the load’s minimum input voltage.
828 mV at 20 °C, which is 6.9 % of a 12 V supply. The cable dissipates 8.28 W.
Yes. Copper’s resistance rises about 0.393 % per °C from 20 °C, so a conductor at 75 °C has 1.216 times the 20 °C resistance. NEC table values are quoted at 75 °C for that reason.
In a balanced three-phase circuit the phase currents sum to zero, so no current returns on a neutral and each line drops I × R to its load. Line-to-line voltage is the difference of two phases 120° apart, which is √3 times the per-phase drop.
Use a thicker wire, shorten the run, raise the supply voltage so the same power needs less current, or run conductors in parallel. Doubling the voltage at constant power halves the current and cuts the percentage drop to a quarter.
Yes. Volt drop, voltage loss and voltage drop all name the voltage lost along the conductors between the source and the load. The cable’s power loss is shown next to it.
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