How this instrument works
A 100 A circuit is usually a feeder, not a light switch: a subpanel out to a detached garage or workshop, a hot tub, an EV charger circuit, or a service upgrade. At that current, wire sizing has two separate jobs pulling in different directions — keeping the conductor cool enough that its insulation survives (ampacity), and keeping the far end of the run close enough to supply voltage to actually work (voltage drop). This instrument solves only the second job, turning current, distance, and an allowed percentage of loss into the minimum copper cross-section, measured in circular mils, needed to keep that voltage drop within budget.
The arithmetic starts from Ohm's law, V = I·R, with resistance itself equal to a material constant K times length over cross-sectional area, R = K·L ⁄ CM. Combine the two and solve for CM instead of R, and the length in that expression carries a hidden multiplier of two — not because the run is measured wrong, but because current does not stop at the load. It returns to the source along a second conductor, so 50 feet from panel to subpanel puts 100 feet of current-carrying copper into play, and the formula bakes that doubling in so the length field can just ask for the distance you can walk off with a tape measure. K, fixed at 12.9 ohm-circular-mils per foot for copper, is where the metal itself enters the sum, expressed in the one unit — circular mils — that every AWG table already uses.
For a feeder this size specifically, voltage drop is rarely what decides the wire — ampacity is. The 75°C ampacity column rates 6 AWG copper at only about 65 A, nowhere near enough to carry 100 A safely no matter how short the run or how generous the drop budget; plain 3 AWG earns the 100 A rating, and continuous loads carry a further 125% multiplier that typically pushes the practical minimum higher still. This calculator answers a narrower question than 'what wire do I buy' — only how much copper keeps the drop under budget — and on a feeder this size that answer is usually a smaller number than the ampacity table would ever let you install.
- Enter the circuit current in amps — the steady load the feeder will actually carry, not the breaker's trip rating.
- Enter the one-way wire run length in feet, measured in one direction only; the formula accounts for the return conductor internally.
- Set the system voltage to the circuit's actual supply — 120 V, 240 V, or another value entirely.
- Set the allowed voltage drop as a percent — 3% is the common electrician's target for a single branch circuit or feeder.
- Read the allowed voltage drop in volts and the minimum wire size in circular mils, then check that figure against an ampacity table too before buying cable.
Worked example — sizing a 100 A feeder on a 240 V system
Take a detached-garage subpanel drawing 100 A, fed 50 feet from the main panel on a 240 V system, sized to the common 3% drop guideline many electricians target for a feeder run. The allowed drop in volts comes first: VD = 240 × 3 ÷ 100 = 7.2 V — the most the feeder may lose to its own resistance before the subpanel bus reads meaningfully under 240 V.
Then the cross-section itself: CM = 2 × 12.9 × 100 × 50 ÷ 7.2 = 17,916.7 circular mils, a floor that rounds up to 17,917. On the voltage-drop math alone, 6 AWG copper — about 26,240 circular mils — already clears that number. But 6 AWG carries only about 65 A on a standard 75°C ampacity table, well short of the 100 A this feeder actually needs, so ampacity overrules the drop calculation entirely; a real 100 A feeder over this distance typically lands on 2/0 AWG copper, about 133,100 circular mils, once continuous-duty rules and code minimums are applied — leaving this formula's 17,917-cmil floor almost beside the point for the final choice.
Questions
Why does the formula multiply the wire length by two?
Current doesn't stop at the load and vanish — it returns to the panel along a second conductor, so a run measured as 50 feet from source to load actually puts 100 feet of copper in the current's path. The built-in factor of 2 accounts for that return trip automatically, which is exactly why the length field asks only for the one-way distance you can walk off with a tape measure.
What is a circular mil, and why does the wire industry use it?
It's the cross-sectional area implied by a conductor's diameter once that diameter is measured in mils, thousandths of an inch, with no pi required to get there. The convenience compounds: AWG size charts, resistivity constants like the 12.9 used here, and cable ampacity ratings are all published in circular mils, so the calculator's output can be checked straight against a parts catalog without converting anything first.
Does doubling the wire run double the required wire size?
Yes, exactly. Resistance is proportional to length, so for the same current and the same allowed drop, circular mils scale linearly with distance: a 100 A, 240 V circuit needing 17,917 circular mils at 50 feet needs 35,833 circular mils at 100 feet — precisely double, with everything else held fixed.
What happens if I tighten the allowed voltage drop percentage?
The required wire gets thicker. Halving the allowed drop from 3% to 1.5% on the same 100 A, 50-foot run cuts the volt budget from 7.2 V to 3.6 V and doubles the minimum circular mils, from 17,917 to 35,833 — a stricter drop budget always costs more copper for an identical circuit.
The result is 17,917 circular mils — is 6 AWG wire fine for a 100 A feeder?
No. 17,917 circular mils is only the voltage-drop floor, and 6 AWG copper, about 26,240 circular mils, would clear that particular number. But 6 AWG is rated for roughly 65 A on a standard 75°C ampacity table, far short of the 100 A this circuit carries, so ampacity — not voltage drop — sets the real minimum here. A 100 A feeder typically needs something closer to 2/0 AWG copper once ampacity and continuous-duty rules are applied; voltage drop only becomes the binding constraint on much longer runs or lower currents.
Does this formula work for aluminum wire?
Not without changing K. The value 12.9 ohm-circular-mils per foot is specific to copper at typical operating temperature; aluminum's higher resistivity puts its K closer to 21.2, so plugging aluminum into a copper-K calculation understates the wire size needed and leaves the circuit dropping more voltage than planned. Match K to the actual conductor metal before trusting the result.