Use one-way length. For AC, you can add power factor and reactance for better estimates.
These are illustrative examples. Site conditions and codes may require different limits.
| System | Material | Area | Length | Current | Voltage | Typical note |
|---|---|---|---|---|---|---|
| Three-phase | Copper | 25 mm² | 50 m | 80 A | 400 V | Common for small motors and panels |
| Single-phase | Aluminum | 35 mm² | 60 m | 90 A | 230 V | Check drop targets for sensitive loads |
| DC (two-wire) | Copper | 50 mm² | 30 m | 120 A | 48 V | Low voltage systems need larger conductors |
Resistivity at 20°C: copper 1.724×10⁻⁸ Ω·m, aluminum 2.82×10⁻⁸ Ω·m.
R = ρ · L / A, where L is length (m), A is area (m²).Temperature correction:
RT = R20(1 + α(T−20)).
DC two-wire:
Vdrop = I · (2R · L).Single-phase:
Vdrop = 2I(R·cosφ + X·sinφ)L.Three-phase:
Vdrop = √3 I(R·cosφ + X·sinφ)L.
Two-wire loop loss:
Ploss = I² · Rloop, with Rloop = 2R·L.Three-phase copper loss:
Ploss = 3 I² R L (per-phase conductor resistance).
These formulas estimate conductor losses. Real installations may also include connection resistance, harmonics, cable grouping, soil/air cooling, and code-specific derating.
- Choose system type and conductor material.
- Enter one-way cable length, current, and system voltage.
- Select conductor size using mm² or AWG, then set temperature.
- For AC, enter power factor and optional reactance.
- Press Calculate to see voltage drop, percent, and losses.
- Use CSV or PDF downloads for reporting and records.
Voltage drop targets on job sites
Design teams commonly manage drop to protect motors, lighting levels, and control voltage. A practical field target is 3% for branch circuits and 5% for the full feeder-to-load path, then confirm against local rules and equipment tolerances. This calculator reports both volts and percent so crews can judge whether a run risks nuisance trips, dimming, or slow motor starting.
What “cable loss” means electrically
Cable loss is resistive heating created by current flowing through conductor resistance. The power lost is proportional to the square of current (I²R), so doubling current increases heating four times. For long temporary supplies, tower cranes, pumps, and welders, these losses can be large enough to justify a size change even when ampacity is adequate.
Material and size comparisons
Copper has lower resistivity than aluminum, so for the same area it produces less drop and less heat. Aluminum is lighter and often cheaper, but typically needs a larger cross-section and careful termination practices. The tool lets you compare materials and use parallel runs; two identical runs roughly halve resistance, lowering both percent drop and watts.
Temperature and AC effects
Resistance rises with conductor temperature. The calculator applies a temperature coefficient to move from a 20°C baseline to your expected operating temperature. For AC circuits, the optional power factor and reactance inputs refine voltage drop using the impedance term (R·cosφ + X·sinφ). This matters on long runs with inductive loads such as motors.
Turning watts into yearly cost
When you enter operating hours and tariff, the tool converts watts of loss into annual energy (kWh/year) and an estimated cost. This supports procurement decisions: a modest upsizing can reduce losses for equipment that runs daily, while a short-duration temporary feed may not justify extra copper. Document results via CSV or PDF for submittals. For bids, capture distance from panel to load, include return conductors where applicable, and note any parallel sets per phase clearly.
1) Does the length need to be one-way or round-trip?
Enter one-way length from source to load. The calculator applies the correct return path internally for DC and single-phase circuits, and the standard three-phase line-to-line method for three-phase.
2) What should I use for conductor temperature?
Use a realistic operating conductor temperature based on installation conditions. Higher temperatures increase resistance and voltage drop. If you do not know, 30–40°C is a reasonable planning starting point for many sites.
3) When should I enter reactance and power factor?
Use them for long AC runs with inductive loads such as motors, pumps, and compressors. If you leave reactance at zero, the tool provides a resistive-only estimate that is conservative for some layouts.
4) How do parallel runs affect the result?
Parallel runs increase effective cross-sectional area. Two identical runs per conductor roughly halve resistance, reducing voltage drop and I²R heating. Keep installation and termination practices consistent across runs.
5) Is this the same as ampacity or breaker sizing?
No. This tool estimates voltage drop and resistive losses. You must still size conductors and protection for ampacity, ambient conditions, grouping, insulation ratings, and applicable electrical codes.
6) Why can the percent drop be high at low voltage?
At low system voltage, a few volts of drop becomes a large percentage. This is common on 48 V or 24 V systems, so larger conductors or shorter runs are often needed to keep equipment within tolerance.