
Written and maintained by the PhDino author · Last reviewed 21 September 2026 · Every calculator used here is tested against independent reference values · how PhDino checks its numbers
Work out the current an individual load draws, then size the feeder that runs from the main panel out to a garage or workshop subpanel so the far end still gets full voltage.
Adding a subpanel for a garage, a workshop or a shed is really three sizing questions asked in order: how much current do the loads draw, what conductor is rated to carry it, and is the voltage still good at the far end of what is often a much longer run than a typical branch circuit. This guide starts small, with one load, and works out to the feeder.
As with the generator guide, this is meant to help you understand the numbers behind a subpanel install, not to replace an electrical permit or a licensed electrician where your jurisdiction requires one. Subpanel work touches your main service, and getting the grounding and bonding wrong there is genuinely dangerous.
A subpanel is a second distribution point fed by a single feeder from the main panel. It moves the breakers close to the loads, so a workshop needs one heavy cable from the house instead of a dozen light ones. The feeder does not have to carry every breaker's rating added together. It has to carry the current the loads can actually draw at the same time, and the main panel has to have the capacity to give it.
That is why this project asks three questions in order. How much current do the loads draw, worked out load by load? What conductor and breaker are rated for that current, which is an ampacity question answered from code tables? And is the voltage still good at the far end of the run? The calculators here answer the first and the third. The second is a code lookup, and the answer is the larger of what ampacity and voltage drop each demand.
A modern subpanel feeder has four conductors: two hots, a neutral and a ground. In the subpanel the neutral and the ground go to separate bars. They are tied together only at the main service, because a second connection would let normal load current wander onto the ground wires and metal enclosures. A separate building also normally gets its own grounding electrode. Details vary by code edition and jurisdiction, so treat this as the outline your electrician or inspector fills in.
A workshop 240 ft from the house is to be fed by a 60 A, 240 V subpanel feeder. The loads that could run together are a 4,500 W, 240 V heater, a 240 V compressor drawing 12 A, and two 120 V tool circuits of 12 A each, one on each leg. The question is what conductor the feeder needs.
| You enter | Value |
|---|---|
| Voltage (0 = solve for it) | 240 V |
| Current (0 = solve for it) | 0 A |
| Resistance (0 = solve for it) | 12.8 Ω |
| The calculator returns | Value |
|---|---|
| Voltage | 240.00 V |
| Current | 18.75 A |
| Resistance | 12.80 Ω |
| Power | 4,500.0 W |
A heater is a resistor. Its nameplate says 4,500 W at 240 V, so its resistance is V² ÷ P = 240² ÷ 4,500 = 12.8 Ω. Leaving Current at 0 tells the calculator to solve for it: I = V ÷ R = 18.75 A, and the power check comes back at 4,500 W.
The same 4,500 W on a 120 V circuit would draw 37.5 A, twice the current in the same wire. That is why big loads run at 240 V: half the current means much less heating and voltage drop in the conductor.
| You enter | Value |
|---|---|
| Current | 42.75 A |
| Length | 240 ft |
| Resistance | 0.3951 Ω/1000ft |
| System Voltage | 240 V |
| The calculator returns | Value |
|---|---|
| Voltage Drop | 8.11 V |
| Percent Drop | 3.38 % |
| Within Limits | 0 yes/no |
Each leg carries the 240 V loads plus its own 120 V circuit: 18.75 A for the heater, 12 A for the compressor and 12 A for the tools on that leg, which is 42.75 A in total. That fits inside a 60 A feeder, and it is the current we check for voltage drop.
The usual conductor for a 60 A feeder is 6 AWG copper, whose resistance is 0.3951 Ω per 1,000 ft (the wire-table figure at 20 °C). The calculator doubles the run for the return path, so over 240 ft the feeder loses 8.1 V, which is 3.4 percent. That is over the 3 percent guide, and the far end of the run would see only about 232 V.
Voltage drop is just Ohm's law applied to the wire. The loop is 240 ft out and 240 ft back, so its resistance is 0.3951 × 2 × 240 ÷ 1,000 = 0.1896 Ω, and V = I × R = 42.75 × 0.1896 = 8.1 V. The calculator saves you the multiplication, not the idea.
| You enter | Value |
|---|---|
| Current | 42.75 A |
| Allowed V Drop | 3 % |
| Cable Length | 240 ft |
| System Voltage | 240 V |
| The calculator returns | Value |
|---|---|
| Required Area | 29.56 kcmil |
| Next Standard Size | 41.74 kcmil |
| Actual V Drop | 2.12 % |
The handoff carries the current, length and voltage across. For a 3 percent limit the copper area needed is 29.6 kcmil, and the next standard size is 41.7 kcmil, which is 4 AWG, with an actual drop of 2.12 percent. On this run voltage drop, not ampacity, sets the conductor: 4 AWG instead of the 6 AWG that the 60 A rating alone would have allowed.
Ampacity is still a separate check, and 4 AWG copper clears it easily for a 60 A breaker. Neither calculator knows about aluminum, which has roughly 1.6 times the resistance of copper for the same size, so an aluminum feeder needs a conductor a size or two larger to match these results.
The chain turned four load currents into a design current of 42.75 A per leg, showed that the obvious 6 AWG conductor would drop 3.4 percent over 240 ft, and landed on 4 AWG. The cheap part of a long feeder is the copper: trenching and conduit cost far more than the difference between two conductor sizes, so on a buried run it usually pays to go up a size now rather than dig it up later.
What the chain left out is worth stating plainly. The main panel's capacity, the demand calculation your code requires, the breaker sizes, the conduit fill, the grounding electrode and every inspection are outside these calculators. They are the reason this is a job for a permit and, in many places, a licensed electrician.
These are the errors that pass a quick look and show up months later.
Work inside the main panel involves live bus bars that stay energized even with the main breaker off, and anything on the utility side of the meter is never a do-it-yourself job. Most jurisdictions require a permit and an inspection for a subpanel, and many require a licensed electrician.
An electrician can also do the load calculation that decides whether your service can take another 60 A at all, which this project cannot. If the service is near its limit, that answer changes the whole plan before a single cable is bought.
Because the current has to get to the load and back. On a balanced 240 V feeder the two hot conductors carry it out and back, so the loop that loses voltage is twice the one-way distance. The length you enter is the one-way run.
Add up what can run at the same time now, then ask what you might add: a welder, a heat pump, an electric-vehicle charger, which can alone draw 40 A or more for hours. Conduit and trenching are the expensive part, so many people pull a larger feeder than today's loads need. Your service capacity limits how far you can go.
The common guidance is about 3 percent for the feeder or for a branch circuit, and about 5 percent from the service to the furthest outlet. It is a recommendation rather than a hard rule, but it splits the budget sensibly: the feeder here uses most of its 3 percent, leaving the branch circuits their share.
Yes, with terminals rated for aluminum and the right installation practice, and it is common for large feeders because it is cheaper. Its higher resistance means a larger conductor for the same voltage drop, and these calculators assume copper, so size up.
It compares the percentage drop with 3 percent and reports yes or no. Treat it as a screening flag for the feeder alone, not as a code compliance check, and remember that hot conductors have higher resistance than the 20 °C values used here.
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