Conductor Cross-Section from Allowable Voltage Drop
Conductor Cross-Section from Allowable Voltage Drop
Instead of checking voltage drop after picking a conductor size, this calculation works in reverse: given a maximum allowable voltage drop, it solves directly for the minimum conductor cross-sectional area needed to stay within that limit over a given run length and current.This is the more practical direction for many designs, since the allowable voltage drop is usually fixed by code or by equipment tolerance, and the conductor size is the one variable still open to the designer — solving for area directly avoids the trial-and-error of guessing a wire size and checking its drop.
The minimum cross-section is A = (2·ρ·L·I)/Vd, twice the resistivity times the length times the current, divided by the allowable voltage drop. where ρ is the conductor resistivity, L is the one-way run length, I is the load current, and Vd is the maximum allowable voltage drop.
Inputs
rho_cu := 0.0175 Ohm·mm^2/m=1.75e-08 kg·m^3/A^2/s^3
L_run := 40 m=40 m
I_load := 30 A=30 A
Vd_max := 6 V=6 V
Rearranging the voltage-drop formula to solve for area gives the smallest conductor cross-section that keeps the drop within the allowed limit for this current and run length.
A_min := (2·rho_cu·L_run·I_load)/Vd_max=7e-06 m^2
Results
A minimum area of about 7 mm² points toward a standard 10 mm² conductor once the next commercially available size is selected, since the calculated minimum must always be rounded up, never down. If the run were longer or the current higher, the required area would grow proportionally, which is exactly why long feeder runs to remote loads often need noticeably larger conductors than their current alone would suggest.
Assumptions & limitations
- Single-phase, two-conductor circuit (factor of 2 for out-and-back run).
- Conductor resistivity at a standard reference temperature (not derated for operating temperature).
- Result must be rounded up to the next standard commercial conductor size.