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Voltage Drop

Enter the current, the one-way length and the source voltage and get the voltage drop (V and %), power loss and the minimum gauge that meets your target. Supports DC, single-phase and three-phase, in copper or aluminum. Resistive model: ignores reactance and temperature.

A
m

Distance from source to load (not the round trip). The calculation already doubles the conductor for DC/1φ.

V
System

DC and single-phase share the formula (round trip). In a balanced three-phase system the drop uses the √3 factor.

Conductor material

The label on the right shows the resistance of the chosen gauge in Ω/km at 20 °C.

Typical criterion: 3 % on branch circuits, 5 % overall (feeder + branch).

redzilla.cl — vdrop
 
Voltage drop
Voltage at the load
Power loss
dissipated in the conductor

Recommended gauge

Drop
%
Min. gauge

Drop by gauge

GaugeΩ/kmDrop (V)Drop (%)

meets the target exceeds the target

How it is calculated · formulas and assumptions

1. DC and single-phase (1φ), accounting for the round trip: Vdrop = 2 · I · (L/1000) · R, with R in Ω/km and L in meters.

2. Balanced three-phase (3φ): Vdrop = √3 · I · (L/1000) · R.

3. Percentage: %drop = Vdrop / Vsource · 100. Voltage at the load = Vsource − Vdrop.

4. Power loss: P = I² · Rtotal, with Rtotal = 2·(L/1000)·R for DC/1φ and ≈ √3·(L/1000)·R for 3φ.

5. Aluminum is modeled as copper × 1.64. Resistive model: it ignores the cable's reactance and the variation of R with temperature. For long runs or reactive loads, check your code.

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How it works

The calculator estimates voltage drop in a conductor from the load current, the one-way length (source → load) and the source voltage. It uses the classic resistive model with copper resistances at 20 °C: for DC and single-phase it applies Vdrop = 2 · I · (L/1000) · R (the factor 2 covers the round trip of the conductor), and for balanced three-phase Vdrop = √3 · I · (L/1000) · R, with R in Ω/km. Aluminum is modeled as copper × 1.64.

With that it reports the drop in volts and percent (%drop = Vdrop / Vsource · 100), the voltage reaching the load, the power loss P = I² · Rtotal and, comparing against your target percentage (typical NEC criterion: 3 % on branch circuits, 5 % total), it recommends the minimum gauge in AWG or mm² that complies. The model ignores reactance and temperature, so for long runs or highly reactive loads you should verify against your local code.

Example: 16 A over 40 m at 230 V with 10 AWG copper

  1. 10 AWG copper resistance: 3.277 Ω/km.
  2. DC/1φ: Vdrop = 2 · 16 · 0.040 · 3.277 = 4.19 V.
  3. Percentage: 4.19 / 230 · 100 = 1.82 %, leaving 225.8 V at the load.
  4. Since 1.82 % ≤ the 3 % target, the gauge complies; the table also shows which smaller gauges would still comply.

Frequently asked questions

What percentage of voltage drop is acceptable?
The usual recommendation (based on the NEC, informational note to 210.19) is a maximum of 3 % on the branch circuit and 5 % total including the feeder. Sensitive loads such as 12/24 V DC powered cameras usually demand less, because at low voltage the same percentage leaves very few volts of margin.
Is the length measured one way or round trip?
You enter the one-way length, from source to load. The formula already doubles the conductor for DC and single-phase (factor 2) and uses √3 for balanced three-phase, so you must not multiply the distance yourself.
What happens if I use aluminum instead of copper?
Aluminum has a resistivity about 64 % higher than copper, and the calculator models it exactly that way (copper × 1.64). In practice, for the same drop you need roughly two AWG sizes larger in aluminum than in copper.
Why is the drop worse at 12 V than at 220 V with the same current?
The drop in volts is the same (it depends on current, length and resistance), but the percentage is computed over the source voltage: 4 lost volts are 1.8 % at 220 V and 33 % at 12 V. That is why long 12/24 V DC runs demand much thicker gauges.
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