FreeToolbox24

Voltage Drop Calculator

Enter a two-wire DC circuit and the resistance of one conductor from its datasheet. Estimate voltage lost along the outgoing and returning cable.

m
A
V
Ω/km
Use the cable datasheet at the intended operating temperature; example values are not a cable recommendation.

Includes the waste margin explained below. Check your product label for coverage and sizes.

A two-wire resistance model

A loaded cable loses some supply voltage because its conductors have resistance. This calculator follows the current through an outgoing conductor and an equal returning conductor. Enter the distance from supply to load only once: the equation doubles it to include the complete circuit. The resistance entry describes one conductor, not the entire loop. Obtain that value from the cable manufacturer at the expected operating temperature. The prefilled numbers are arithmetic examples and do not identify a particular cable size.

Equations and supported circuit

Loop resistance = 2 × length × conductor resistance ÷ 1000; voltage drop = current × loop resistance

For metric measurements the resistance is in ohms per kilometre and length is in metres. Imperial measurements pair ohms per thousand feet with length in feet. The denominator therefore remains 1,000 in either mode. Percentage loss is the drop divided by supply voltage, multiplied by 100. Cable heating in this model is current squared times loop resistance. The calculation is intended for a steady two-wire DC load with equal conductor resistances. It does not include connector resistance, startup current, earth-return paths, AC reactance or three-phase arrangements. A predicted drop exceeding the supply voltage is rejected as an inconsistent setup.

Worked example with a known resistance

A 30 metre one-way run carrying 20 amps has conductors specified at 2 ohms per kilometre. The complete loop resistance is 2 × 30 × 2 ÷ 1,000 = 0.12 ohms. The voltage drop is 20 × 0.12 = 2.4 volts. On a 230 volt supply, that leaves 227.6 volts at the load and a drop of approximately 1.04 percent. Dissipation in the two conductors is 20 squared × 0.12 = 48 watts. A separate imperial example uses 100 feet, 0.6 ohms per thousand feet and the same 20 amp load: the loop resistance and absolute voltage loss happen to be identical.

Use the result as one design check

Increasing conductor temperature increases resistance, so a cold resistance measurement can understate operating losses. Cables also have current limits set by insulation, installation conditions, grouping and protective devices. A low voltage drop does not prove that a cable can safely carry the entered current. Compare the estimated terminal voltage with the equipment requirements and have the complete installation checked against applicable rules. The resistive relationship is consistent with the conductor-resistance method in the Schneider Electric installation guide; its AC formulas additionally account for phase angle and reactance.

Frequently asked questions

Do I enter the return cable length?

Enter the one-way route only. The calculator doubles its resistance for an equal outgoing and returning conductor.

What resistance value should I use?

Use the manufacturer value for one conductor at the relevant operating temperature. Check whether it is quoted per kilometre or per thousand feet.

Can this size an AC motor circuit?

No. Motor starting current, power factor and AC reactance require a fuller circuit model. This page models steady two-wire DC resistance.

What happens at zero current?

The voltage drop and cable power loss are zero. Supply voltage still appears at the load within the ideal resistance model.

Does a small drop mean the wire is safe?

No. Ampacity, protection, temperature and installation conditions require separate checks. This calculation alone does not approve a wire or an installation.

Last reviewed: 2026-10-03 · How we calculate