Cable Current Loss Calculator

Calculator Inputs

Phase type

Conductor material

Supply voltage (V)

Load power (kW)

Power factor

One-way cable length (m)

Conductor area (mm²)

Operating temperature (°C)

Parallel runs

Operating hours per day

Operating days per year

Energy cost per kWh

Custom resistivity (Ω·mm²/m)

Custom temperature coefficient (1/°C)

Custom resistivity and coefficient apply only when the material is set to Custom.

Example Data Table

Example assumptions: single phase, copper, 230 V, 5 kW, power factor 0.92, 40 m one-way, 50°C, one run.

Cable area (mm²)	Resistance (Ω)	Estimated load current (A)	Voltage drop (%)	Power loss (W)
4	0.38547	22.729	3.809	199.145
6	0.25698	23.022	2.572	136.200
10	0.15419	23.261	1.559	83.428
16	0.09637	23.398	0.980	52.758

Formula Used

1) Temperature adjusted resistivity
ρ_T = ρ₂₀ × [1 + α × (T - 20)]

2) Cable resistance
Single phase loop resistance: R = 2 × ρ_T × L / (A × n)
Three phase per-phase resistance: R = ρ_T × L / (A × n)

3) Design current
Single phase: I = P / (V × pf)
Three phase: I = P / (√3 × V × pf)

4) Estimated current at the load
This page uses a constant-impedance approximation.
Single phase: I_load = V / (Z_load + R)
Three phase uses phase voltage with per-phase resistance.

5) Voltage drop and power loss
Single phase: ΔV = I_load × R, P_loss = I_load² × R
Three phase: ΔV = √3 × I_load × R, P_loss = 3 × I_load² × R

The calculator reports current loss as the difference between the design current and the estimated load current under that approximation.

How to Use This Calculator

Choose single-phase or three-phase operation.
Select copper, aluminum, or custom conductor data.
Enter system voltage, load power, and power factor.
Enter one-way length, conductor area, and temperature.
Add parallel runs if conductors are split.
Enter annual operating time and energy cost.
Press calculate to show results above the form.
Review graph, summary table, and export files.

FAQs

1. What does cable current loss mean here?

This calculator estimates the reduction in available load current caused by cable resistance. It compares ideal design current with estimated current after resistive voltage drop under a constant-impedance assumption.

2. Why does temperature increase losses?

Higher conductor temperature increases resistivity. More resistance causes larger voltage drop and greater I²R heating. That reduces load voltage and raises energy waste.

3. Why is one-way length used?

One-way length matches common cable schedules. The calculator internally converts it to loop resistance for single-phase circuits. Three-phase calculations use per-phase conductor length.

4. Does a larger conductor reduce loss?

Yes. A larger cross-sectional area lowers resistance. Lower resistance reduces voltage drop, heating, annual energy loss, and current reduction.

5. When should I use custom resistivity?

Use custom values when working with nonstandard conductors, manufacturer-specific data, or design studies that require exact material properties instead of default copper or aluminum assumptions.

6. What is current density useful for?

Current density helps compare thermal loading against conductor area. Higher values often mean more heating and tighter design margins, especially in enclosed or high-temperature environments.

7. Are these results suitable for final code compliance?

No. Use this page for engineering estimates and screening. Final cable sizing must follow applicable electrical codes, installation methods, ambient corrections, grouping factors, and manufacturer data.

8. Why does the graph use cable length?

Length is usually the strongest geometric driver of resistive loss. The graph quickly shows how voltage drop percent and current loss percent rise as the cable run becomes longer.