Cable Size Calculator

Find the minimum recommended wire gauge (AWG) or cross-sectional area for copper and aluminum conductors. This tool evaluates both thermal ampacity and voltage drop constraints.

Cable Size Calculator

Results

Recommended Cable Size8 AWGSized for Ampacity limit
Design Load Current:37.5 A
Cable Ampacity (75°C):50 A
Actual Voltage Drop:4.67 V (1.95%)
Conductor Resistivity:0.778 Ω/1k ft
Sizing Criteria

Conductor selected must carry 1.25 × Continuous Load + Non-Continuous Load, and must have <= 3.00% voltage drop under full load.

How Cable Sizing Works

Choosing the right electrical cable size is not just about matching the load current. A properly sized conductor must satisfy two distinct engineering criteria:

  1. Thermal Ampacity Limit: The cable must carry the design current safely without generating excessive heat that could melt its insulation.
  2. Voltage Drop Limit: The cable must be large enough so that resistance doesn't drop the voltage too much at the end of the line.

Continuous Load Code Sizing

Under codes like the National Electrical Code (NEC), circuits are designed with safety margins. A load that runs continuously for 3 hours or more must have the conductor and overcurrent protection sized at **125%** of that continuous load:

Idesign = (Icontinuous × 1.25) + Inon-continuous

Sizing Case Example

Scenario: Sizing a 120V single-phase lighting feeder running a constant 16 Amp continuous load over a distance of 120 feet, with a 3% voltage drop limit. We will use Copper THWN (75°C terminals).

Step 1: Check Thermal Ampacity

Calculate the design current: 16A × 1.25 = 20 Amperes.
Reviewing standard wire ampacities:
* 14 AWG Copper is rated for 20A.
So, the minimum size for thermal safety is 14 AWG.

Step 2: Check Voltage Drop Constraint

The allowed drop is 120V × 3% = 3.6 Volts.
If we test 14 AWG ($R \approx 3.07\ \Omega$/1k ft) at 16A load:
Vd = (2 * 120ft * 16A * 3.07) / 1000 = 11.79 Volts (9.8%). This fails the 3% limit.

If we test 10 AWG ($R \approx 1.24\ \Omega$/1k ft):
Vd = (2 * 120ft * 16A * 1.24) / 1000 = 4.76 Volts (3.96%). Still fails 3%.

If we test 8 AWG ($R \approx 0.778\ \Omega$/1k ft):
Vd = (2 * 120ft * 16A * 0.778) / 1000 = 2.99 Volts (2.49%). This succeeds!

Conclusion

The recommended wire is 8 AWG Copper. Sized up due to voltage drop over distance.

Frequently Asked Questions (FAQs)

What is the difference between AWG and kcmil wire sizes?

American Wire Gauge (AWG) measures smaller wire diameters (numbered from 18 AWG up to 4/0 AWG where smaller numbers are thicker). For larger industrial feeds, sizes are measured in kcmil (thousand circular mils) or MCM, where circular area indicates thickness.

Why does aluminum require larger wire sizes than copper?

Aluminum has roughly 61% of the conductivity of copper. Because it has higher resistivity, you must use a larger cross-sectional aluminum wire to carry the same current without overheating.

What is the standard allowed voltage drop limit?

Standard codes (like NEC) recommend keeping voltage drop to 3% or less for branch circuits and 5% total for the entire distribution feeder. Some utility specifications mandate less than 2% for critical services.

Related Calculators

Voltage Drop Calculator →Ohm's Law Solver →