Conductor losses
What is voltage drop?
Voltage drop is the reduction in electrical potential that occurs when current flows through a conductor with resistance.
Every real wire has some resistance. Longer conductors, smaller conductor areas, higher current, and materials with greater resistivity all produce a larger voltage drop.
Main equation
Voltage drop formula
Voltage dropVdrop = 2 × I × ρ × L ÷ A
- Vdrop is the voltage lost across the complete conductor loop.
- I is current in amperes.
- ρ is conductor resistivity in ohm-metres.
- L is the one-way conductor length in metres.
- A is conductor cross-sectional area in square metres inside the equation.
Because this calculator accepts conductor area in square millimetres, it converts the entered value to square metres before applying the equation.
Complete circuit path
Why does the formula include a factor of two?
In a simple two-wire circuit, current travels from the source to the load and then returns to the source. Both conductors contribute resistance.
The calculator therefore uses twice the entered one-way length. You should not manually double the cable distance before entering it.
Solve every variable
Rearranged voltage drop equations
CurrentI = Vdrop × A ÷ (2 × ρ × L)
One-way lengthL = Vdrop × A ÷ (2 × I × ρ)
Cross-sectional areaA = 2 × I × ρ × L ÷ Vdrop
Resistivityρ = Vdrop × A ÷ (2 × I × L)
Worked calculation
Voltage drop example for a copper cable
Consider a 10-ampere load connected by a copper conductor with a one-way length of 20 metres and an area of 2.5 mm².
- Current: 10 A
- One-way length: 20 m
- Conductor area: 2.5 mm²
- Copper resistivity: 1.724 × 10⁻⁸ Ω·m
Substitute the valuesVdrop = 2 × 10 × 1.724 × 10⁻⁸ × 20 ÷ 2.5 × 10⁻⁶
ResultVdrop = 2.7584 V
The complete conductor loop has a resistance of approximately 0.27584 Ω and loses approximately 27.584 watts at 10 amperes.
Material properties
Typical conductor resistivity values
Approximate resistivity values near room temperature include:
- Copper: approximately 1.724 × 10⁻⁸ Ω·m
- Aluminium: approximately 2.82 × 10⁻⁸ Ω·m
Actual values vary with temperature, purity, alloy composition, and manufacturing conditions. Use the specification supplied for the conductor when accuracy is important.
Design relationships
What increases voltage drop?
- Increasing current increases voltage drop proportionally.
- Increasing conductor length increases both resistance and voltage drop.
- Reducing conductor area increases resistance and voltage drop.
- A material with higher resistivity produces a greater voltage drop.
- Higher conductor temperature usually increases resistance.
Model limitations
When this calculator should be used carefully
This tool uses a simplified resistive conductor model. It does not automatically include:
- AC reactance or impedance
- Power factor
- Three-phase correction factors
- Temperature-adjusted resistance
- Connector and termination losses
- Code-required conductor sizing rules
For permanent building wiring, industrial systems, high-current equipment, or safety-critical installations, follow the applicable electrical code and consult a qualified electrical professional.
Common questions
Voltage drop calculator FAQ
What formula does the voltage drop calculator use?
The calculator uses Vdrop = 2 × I × ρ × L ÷ A. The factor of two represents the outgoing and return conductors in a two-wire DC circuit.
Why is conductor length entered as one-way distance?
You enter the distance from the source to the load. The calculator automatically doubles that distance to account for the complete electrical loop.
Which resistivity value should I use for copper?
A common reference value for copper at approximately 20 degrees Celsius is 1.724 × 10⁻⁸ ohm-metres. Actual resistance changes with temperature and material composition.
Can this calculator determine cable size?
Yes. Select cross-sectional area as the unknown and enter the permitted voltage drop, current, one-way length, and conductor resistivity.
Does the calculator support AC circuits?
It models resistive round-trip conductor loss and is most suitable for simple DC or low-reactance single-phase applications. AC impedance, power factor, and three-phase systems require additional calculations.