DC voltage drop
ΔV = 2IR
The factor of two represents the outgoing and return conductors when the entered length is one-way.
Calculate conductor voltage drop for DC, single-phase AC, or three-phase AC circuits using copper or aluminum AWG conductors. Temperature-adjusted resistance, optional reactance, parallel conductors, load voltage, and I²R losses are included.
Access: Free to use, no installation, and No account required. Calculations run in your browser.
For DC, power factor and reactance are ignored. Length is one-way; the engine applies the correct return-path factor for DC/single-phase and √3 relationship for three-phase.
For DC, the resistive approximation is ΔV = 2IR. For single-phase AC the calculator uses ΔV = 2I(R cosφ + X sinφ). For balanced three-phase systems it uses ΔV = √3 I(R cosφ + X sinφ).
Conductor resistance is adjusted from its 20 °C value using the material temperature coefficient. Parallel conductor count divides the effective branch resistance and reactance.
The displayed current density is a screening metric, not ampacity. Actual conductor sizing must also account for insulation rating, installation method, ambient temperature, bundling, terminal ratings, fault current, protection, applicable electrical code, and equipment-manufacturer requirements.
Use the Voltage Drop Calculator to estimate conductor voltage loss, percent drop, load voltage, resistive heating, and current density for DC, single-phase AC, or balanced three-phase AC feeders.
Voltage drop is the reduction in conductor-end voltage caused by current flowing through conductor impedance. The calculation uses AWG geometry, material resistivity, conductor temperature, one-way length, parallel conductors, and the AC power-factor/reactance terms when applicable.
A low voltage-drop percentage is not by itself a complete conductor-sizing decision. Ampacity, insulation temperature rating, installation method, bundling, termination ratings, overcurrent protection, code rules, and fault-current capability still need to be checked for a real installation.
ΔV = 2IR
The factor of two represents the outgoing and return conductors when the entered length is one-way.
ΔV = √3 I(R cosφ + X sinφ)
The phase relationship replaces the two-conductor factor used for single-phase circuits.
% drop = 100 ΔV / Vs
Percent drop normalizes the calculated loss to the source or line voltage.
A 20 A DC load is 6 m one way from a 12 V source. The conductor is 10 AWG copper at 30 °C with one conductor per polarity.
Result: The load receives about 11.2 V in this example, which illustrates why low-voltage DC systems can require unexpectedly large conductors.
Each case below has a hand-checkable answer, so a wrong engine change shows up immediately.
Case: Set current to 0 A.
Expected: Voltage drop and I²R loss should both return zero.
Case: Compare one conductor with two identical parallel conductors.
Expected: Effective resistance and resistive voltage drop should be approximately halved.
There is no single universal limit for every system. Use the project specification and applicable electrical code, then check both feeder and branch-circuit performance.
No. It reports voltage drop and current density, but code ampacity depends on insulation, ambient temperature, bundling, installation method, terminations, and other rules.
The engine applies the return-path factor internally for DC and single-phase circuits and the √3 relationship for balanced three-phase circuits.
Voltage drop here is computed by the same conductor routine used in the Electrical Design Workbench, so a correction to resistivity or temperature handling reaches both surfaces at once.
Open the source workbench →Read calculation and source methodology →