Electrical Engineering

Voltage Drop Calculator

Estimate the voltage lost along a cable before it reaches a load. Enter the supply voltage, load current, one-way cable length and conductor size to compare wiring options for DC equipment and balanced AC circuits.

Your inputs

AWG: use 0 for 1/0, -1 for 2/0, -2 for 3/0, -3 for 4/0. Range: -3 to 40.

Calculated on your device. Raw input values are not sent to analytics.

Your estimate

Above 5%: review the design. This is an advisory threshold, not a universal legal limit.

Voltage drop
0.69 V
Voltage drop percentage
5.747 %
Estimated voltage at load
11.31 V
Conductor cross-sectional area
2.5 mm²

Engineering estimate at 20 °C, excluding cable reactance, heating and connection resistance. AC uses the resistive longitudinal component R × PF. For low PF or long/large AC cables, use an impedance-based study. Local codes vary; use manufacturer data and a qualified professional for safety-critical cable selection.

Above 5%: review the design. This is an advisory threshold, not a universal legal limit.. Voltage drop: 0.69 V. Voltage drop percentage: 5.747 %. Estimated voltage at load: 11.31 V. Conductor cross-sectional area: 2.5 mm²

How voltage drop is calculated

Voltage drop is the difference between voltage at the source and voltage available to the load while current flows. Cable resistance converts some electrical energy to heat. A longer or thinner conductor has more resistance, so the same current produces a larger drop.

This calculator uses resistance per metre R′ = ρ / A, where A is conductor area in mm². With one-way length L in metres, DC drop is 2 × I × R′ × L. The factor 2 includes the outgoing and return conductors; do not enter round-trip length.

For single-phase AC, the resistive longitudinal approximation is 2 × I × R′ × L × PF. Balanced three-phase AC uses √3 × I × R′ × L × PF and line-to-line supply voltage. At PF = 1 these reduce to the simple resistive formulas. Voltage drop percentage is 100 × drop / supply; estimated load voltage is supply minus drop.

Copper, aluminum and conductor size

The assumed resistivities at approximately 20 °C are 0.01724 Ω·mm²/m for copper and 0.0282 Ω·mm²/m for aluminum. These are reference estimates, not cable manufacturer resistance tables. Aluminum needs about 1.64 times the cross-sectional area for the same resistance under these assumptions.

AWG area is calculated from diameter d = 0.127 × 92^((36 − AWG)/39) mm and A = πd²/4. Enter -3 for 4/0, -2 for 3/0, -1 for 2/0, or 0 for 1/0. One foot is exactly 0.3048 metres.

AC assumptions and limits

AC cable impedance also includes reactance. A fuller longitudinal estimate replaces R′ × PF with R′ × cos φ + X′ × sin φ for a lagging load. Reactance is omitted here, so low power factor can understate actual drop, especially on long runs or large conductors. PF = 0 gives zero resistive longitudinal drop, not zero physical impedance.

The model assumes fixed current, equal outgoing and return conductor areas and a balanced three-phase load. It excludes temperature rise, joints, source impedance, harmonics and starting transients. A negative load-voltage result indicates that this approximation is unsuitable, not a realizable operating voltage.

Interpreting excessive voltage drop

The above-5% indication is an advisory prompt to review a design, not a legal limit or a pass/fail assessment. Allowable drop depends on equipment, application and jurisdiction. Low-voltage DC systems can lose a substantial percentage even when the drop in volts seems small.

Long cable runs, small conductors, heavy loading, hot cables and deteriorated connections can contribute to excessive drop. Larger cable reduces resistance, while shorter routes reduce the length over which voltage is lost. Have a qualified professional investigate safety-critical installations using local codes and manufacturer specifications.

Examples

12 V DC circuit

Input
12 V, 10 A, 5 m one-way, 2.5 mm² copper, PF 1
Result
0.690 V drop; 5.747%; 11.310 V at load.

24 V DC circuit

Input
24 V, 5 A, 20 m one-way, 2.5 mm² copper, PF 1
Result
1.379 V drop; 5.747%; 22.621 V at load.

230 V single-phase circuit

Input
230 V, 16 A, 30 m one-way, 4 mm² copper, PF 1
Result
4.138 V drop; 1.799%; 225.862 V at load.

400 V three-phase circuit

Input
400 V, 32 A, 50 m one-way, 10 mm² copper, PF 1
Result
4.778 V drop; 1.194%; 395.222 V at load.

Frequently asked questions

What is voltage drop?

It is the reduction in voltage between a source and a load caused by current flowing through circuit impedance. This calculator estimates the cable resistance contribution.

How much voltage drop is acceptable?

There is no universal allowable percentage. Check equipment requirements and the applicable electrical code; the calculator’s 5% advisory threshold is not a compliance limit.

Does cable length increase voltage drop?

Yes. At fixed current, conductor area and temperature, doubling the one-way length doubles the calculated resistive drop.

Does larger cable reduce voltage drop?

Yes. Doubling conductor cross-sectional area halves resistance and the calculated resistive drop. Cable sizing must also account for ampacity, installation conditions and protection.

How do copper and aluminum compare?

At the reference temperature and equal area, the aluminum constant used here gives about 64% more resistance than copper. Actual cable data may differ.

How is three-phase voltage drop calculated?

For a balanced load, use √3 times line current times one-way conductor resistance times power factor for this resistive approximation. Enter line-to-line voltage to calculate the percentage.

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