What is voltage drop?
Voltage drop is the reduction in voltage between the source and the load as current flows through conductors and their electrical impedance. Longer runs, higher current and higher conductor resistance generally increase the drop. In AC systems, conductor reactance and load power factor can also matter.
A useful calculator should therefore do more than multiply distance by current. It should clearly define whether distance is one-way or round-trip, distinguish single-phase from three-phase circuits, expose the resistance assumptions and avoid confusing voltage-drop performance with conductor ampacity.
The basic DC voltage-drop formula
For a two-wire DC circuit with equal outgoing and return conductors:
Vdrop = 2 × I × R × LHere, I is current, R is conductor resistance per unit length and L is the one-way circuit length. The factor 2 accounts for the outgoing and return path.
Single-phase AC voltage drop
For a simple resistive approximation, single-phase two-wire AC uses the same round-trip factor:
Vdrop ≈ 2 × I × R × LFor more detailed AC work, resistance alone may be insufficient. Southwire's cable guidance notes that AC voltage-drop calculations can require AC resistance, reactance and the load power factor.
Three-phase voltage drop
For a balanced three-phase system, the familiar line-to-line approximation uses √3 rather than the two-wire factor:
Vdrop ≈ √3 × I × (R cosφ + X sinφ) × LWhen reactance is ignored and power factor is treated as unity, this reduces to:
Vdrop ≈ √3 × I × R × LWhy the calculator asks for one-way length
Many voltage-drop tools—including Cerrowire's—ask for the one-way circuit length. The formula itself then accounts for the return path in a two-wire circuit.
Voltage-drop percentage
The absolute voltage drop is useful, but percentage makes circuits at different voltages easier to compare:
% voltage drop = voltage drop ÷ source voltage × 100A 3 V drop on 120 V is 2.5%, while the same 3 V drop on 240 V is 1.25%.
Load-end voltage
The estimated voltage available at the load is:
Load-end voltage ≈ source voltage − voltage dropThis is a steady-state estimate. Starting currents, utility variation, transformer impedance, upstream feeder drop and changing loads can produce different measured values.
Conductor resistance and circular mil area
The conductor-size mode uses the familiar resistivity relationship:
R = K × L ÷ CMK is a resistivity constant, L is conductor length and CM is conductor area in circular mils. Larger conductor area lowers resistance and therefore lowers voltage drop for the same current and distance.
The built-in model uses approximate copper and aluminum resistivity values at 20°C or 75°C for planning. It is not a substitute for the actual AC resistance data of a specific cable.
Why conductor temperature matters
Metal resistance rises with temperature. A conductor operating warm has greater resistance than the same conductor near room temperature, so a room-temperature voltage-drop calculation can be optimistic under load.
The simple mode lets you select an approximate temperature basis. For design work, use conductor data appropriate to the actual cable and operating conditions.
Copper versus aluminum
Copper and aluminum have different resistivities. For the same conductor area, aluminum generally has higher resistance, so equivalent voltage-drop performance often requires a larger aluminum conductor.
That does not mean this calculator should automatically choose a replacement size. Ampacity, termination ratings, connector compatibility and installation rules must also be checked independently.
AWG and kcmil
American Wire Gauge sizes become physically larger as the gauge number decreases. Beyond 4/0, large conductors are commonly expressed in kcmil. Cerrowire notes that kcmil and the older term MCM both refer to thousands of circular mils.
The conductor-size mode maps common AWG and kcmil sizes to circular-mil area solely for the resistive voltage-drop model.
Custom AC impedance mode
For AC circuits, the custom mode is the more transparent option when cable data is available. Enter AC resistance R, reactance X and load power factor.
Effective impedance term = R cosφ + X sinφwhere cosφ is the power factor and:
sinφ = √(1 − power factor²)The calculator then applies the single-phase factor 2 or three-phase factor √3.
Why power factor changes AC voltage drop
With an inductive load, current is not perfectly in phase with voltage. The resistive and reactive parts of cable impedance therefore contribute differently to voltage drop.
Southwire's technical guidance specifically includes resistance, reactance and power factor in its AC approximation. This is why a calculator that assumes pure resistance for every AC circuit can hide an important assumption.
What resistance units should I enter?
In feet mode, custom resistance and reactance are entered in ohms per 1,000 ft. In metric mode they are entered in ohms per kilometre. The calculator converts the one-way circuit length to the matching basis before applying the formula.
Parallel conductors
When equal conductors are properly connected in parallel, the effective resistance and reactance of the parallel set decrease approximately in proportion to the number of identical parallel paths.
Reffective ≈ Rsingle ÷ number of parallel conductorsParallel conductors are governed by installation and code requirements. The calculator only models their electrical effect after you provide the number of parallel paths.
Worked example: 120 V, 15 A, 100 ft
Suppose a two-wire circuit carries 15 A over a 100 ft one-way run. If the effective conductor resistance is 2 Ω per 1,000 ft, the round-trip resistance is:
2 × 2 Ω/kft × 0.1 kft = 0.4 ΩThe voltage drop is:
15 A × 0.4 Ω = 6 VThat equals 5% of 120 V, leaving an estimated 114 V at the load. This example illustrates the math only; it does not determine whether the conductor is acceptable for the circuit.
Worked three-phase example
For a 400 V balanced three-phase circuit, 30 A load, 50 m one-way length, 0.8 Ω/km resistance, 0.08 Ω/km reactance and 0.9 power factor:
sinφ = √(1 − 0.9²) ≈ 0.436 Zeffective = 0.8(0.9) + 0.08(0.436) ≈ 0.755 Ω/km Vdrop ≈ √3 × 30 × 0.755 × 0.05 ≈ 1.96 VThe percentage drop is about 0.49% of 400 V.
Voltage drop is not ampacity
This distinction is critical. Ampacity addresses how much current a conductor is permitted to carry under specified conditions. Voltage drop addresses circuit performance. A conductor can produce an attractive voltage-drop result and still be unsuitable because of ampacity, temperature, termination, protection or installation requirements.
Why conductor sizing needs more checks
Real conductor selection can involve insulation rating, ambient temperature, number of current-carrying conductors, termination temperature limitations, overcurrent protection, continuous loads, motor requirements, short-circuit conditions, installation method and local rules.
Cerrowire's separate ampacity calculator illustrates how many parameters can be involved, and Southwire likewise warns that site-specific product approval remains with the responsible professional and authority having jurisdiction.
The 3% and 5% figures
In U.S. practice, 3% and 5% are commonly encountered voltage-drop design recommendations in NEC-related guidance: approximately 3% for a branch circuit and approximately 5% total for feeder plus branch circuit to the farthest outlet, depending on the applicable context.
These are not universal worldwide limits, and voltage-drop provisions can differ by circuit type and jurisdiction. The calculator therefore treats the percentage field as an optional comparison target that you control.
Why the comparison limit is user-controlled
A fixed green/red “pass” indicator can be misleading. A circuit may have a project specification tighter than 3%, or a different regulatory requirement may apply. Some sensitive equipment also needs tighter voltage performance.
Enter the target applicable to your project. The result merely reports whether the calculated percentage is above or at/below that number.
Motor circuits and starting current
Motors can draw substantially more current during starting than during normal operation. A steady-state current calculation may therefore understate the momentary voltage dip during startup.
Motor voltage-drop and starting-performance studies can require source impedance, motor characteristics and other system data. Do not treat a normal-running result as a motor-starting analysis.
LED lighting and sensitive electronics
Low-voltage lighting and electronics can be more sensitive to conductor drop because each lost volt represents a larger percentage of the source voltage. Driver behavior and allowable input range also matter.
Use the equipment manufacturer's voltage requirements rather than assuming a generic percentage is always acceptable.
Solar, battery and other DC systems
The two-wire resistive formula is useful for many simple DC runs. In low-voltage battery or solar circuits, even a small absolute drop can represent a significant percentage and waste power as heat.
But conductor ampacity, protective devices, polarity, fault current and equipment-specific rules still require separate checks.
Power loss in conductors
For a resistive path, conductor loss is related to current squared:
Ploss = I² × RpathThis explains why higher current can sharply increase losses. However, the calculator's primary purpose is voltage drop, not thermal conductor design.
Long runs
Cerrowire notes that voltage drop becomes increasingly important on long runs such as circuits to outbuildings or pumps. Longer length increases total conductor resistance directly.
For very long or complex AC circuits, distributed cable effects and system modeling may justify more advanced engineering methods than the simplified equations here.
Multiple loads along one circuit
A circuit with loads tapped at different points does not carry the same current along its entire length. A single “total current × total length” calculation can therefore overstate or misrepresent individual segment drops.
Calculate each segment with the current actually flowing through that segment, then sum the voltage drops along the path to the load being evaluated.
Unbalanced three-phase systems
The standard √3 formula assumes a reasonably balanced three-phase system. Significant phase imbalance, neutral current, harmonics or nonlinear loads may require phase-by-phase analysis.
The calculator does not model those advanced conditions.
Why measured voltage can differ
Actual load voltage can differ from the calculation because source voltage changes, conductor temperature changes, connections add resistance, loads vary and upstream feeders or transformers also have impedance.
A calculation is a design estimate; field measurement under appropriate safe procedures is a different form of verification.
Common voltage-drop mistakes
- Entering round-trip distance when the formula already uses a factor of 2.
- Using the two-wire factor for a balanced three-phase line-to-line calculation.
- Ignoring power factor and reactance when they materially affect an AC circuit.
- Using nominal voltage instead of the voltage relevant to the load calculation.
- Treating voltage-drop performance as proof of ampacity or code compliance.
- Using room-temperature resistance for a hot conductor without recognizing the assumption.
- Assuming 3% is a universal legal limit.
- Ignoring motor starting current or distributed loads.
- Mixing Ω/kft with metres or Ω/km with feet.
How to verify a voltage-drop calculation
First confirm the circuit topology: DC two-wire, single-phase or balanced three-phase. Confirm that the entered distance is one-way. Verify current and nominal voltage.
Then verify conductor data. For a basic estimate, check conductor material and area. For AC work, use cable-manufacturer resistance and reactance data when available. Finally, compare the result with the performance target and separately perform every required ampacity, protection, termination and installation check.
Frequently asked questions
What is the formula for voltage drop?
For a simple two-wire resistive circuit, Vdrop ≈ 2IRL. For balanced three-phase, Vdrop ≈ √3IRL. AC impedance calculations can add resistance, reactance and power factor.
Do I enter one-way or total wire length?
Enter one-way circuit length. The two-wire formula accounts for the return path with its factor of 2.
What is a good voltage-drop percentage?
It depends on the applicable standard, project and equipment. U.S. NEC-related design guidance commonly references 3% branch-circuit and 5% feeder-plus-branch recommendations, but those figures are not universal.
Can this calculator choose my wire size?
No. It calculates voltage-drop performance for a size you enter. Safe conductor selection also requires ampacity and other electrical-design checks.
Does it support aluminum wire?
Yes, in the simplified resistive model. Actual cable data should be used when greater precision is required.
Does it support three-phase?
Yes, for balanced three-phase voltage-drop estimates using the √3 relationship.
Why is there a power-factor input?
In AC circuits, resistance and reactance contribute according to the phase relationship between current and voltage. Power factor provides cosφ for the approximation.
Is voltage drop the same as power loss?
No. They are related but different. Resistive power loss follows I²R, while voltage drop is the reduction in voltage along the circuit.
Can I use this for motor starting?
Not as a complete motor-starting study. Starting current and source/system impedance can require more detailed analysis.
Is the result code compliant?
The calculator does not make that claim. Electrical requirements vary by jurisdiction and application, and voltage drop is only one part of circuit design.
Final note: use this calculator to make voltage-drop assumptions visible—not to replace electrical design. Confirm one-way length, load current, system type and conductor impedance, then separately verify ampacity, overcurrent protection, terminations, installation method, equipment requirements and the rules that apply to the project.