Why is 12 AWG limited to 20 amperes when the table shows more?

NEC 240.4(D)(5) sets a hard ceiling on overcurrent protection for small conductors that overrides the ampacity in Table 310.16.

What breaker size does a 25 ampere continuous load need?

At least 125 percent of 25 amperes, or 31.25 amperes, which rounds up to the next standard size of 35 amperes per 240.6(A).

Is 10 AWG copper enough for this circuit?

No. 10 AWG has the ampacity but 240.4(D)(7) limits it to a 30 ampere device, and this circuit requires 35 amperes.

Can I use 12 AWG copper for a 30 amp, 240 volt branch circuit with a 25 amp continuous load?

Short answer: no. Two separate rules kill this one, and either one alone is enough.

The first is the small conductor rule. Section 240.4(D)(5) limits overcurrent protection for 12 AWG copper to 20 amperes, regardless of what Table 310.16 says the conductor can carry. A 30 ampere breaker on 12 AWG copper is a violation on its face, and the length of the run and the raceway type do not change that.

The continuous load calculation

The second rule is the continuous load multiplier. A load that runs three hours or more is continuous, and 210.19(A)(1) requires the branch circuit conductors to have an ampacity of at least 125 percent of the continuous load before any adjustment or correction factors are applied. Section 210.20(A) applies the same 125 percent to the overcurrent device.

25 amperes times 1.25 is 31.25 amperes. Section 240.6(A) lists the standard breaker sizes, and the first one at or above 31.25 amperes is 35 amperes. So the breaker is 35 amperes, not 30.

Now find a conductor. Section 110.14(C)(1)(a) holds terminations on equipment rated 100 amperes or less to the 75 degrees C column. In that column of Table 310.16, 10 AWG copper is 35 amperes, which clears the 31.25 ampere requirement. But 240.4(D)(7) caps overcurrent protection for 10 AWG copper at 30 amperes, and you need a 35 ampere device. That pushes you to 8 AWG copper at 50 amperes in the 75 degrees C column, protected at 35 amperes. Section 240.4 permits a conductor larger than the minimum.

Voltage drop over 75 feet

Run the numbers before you assume the distance is a problem. Chapter 9 Table 8 lists 8 AWG stranded uncoated copper at 0.778 ohms per 1000 feet. For a 240 volt single phase circuit, voltage drop equals 2 times the one way length times the current times the resistance per foot. That is 2 times 75 times 25 times 0.000778, or 2.9 volts. On 240 volts that is 1.2 percent, well inside the 3 percent branch circuit figure suggested in the Informational Note to 210.19(A).

So the 75 foot length is not what disqualifies 12 AWG here. The small conductor rule and the continuous load multiplier are.

What about derating in the EMT?

If the circuit is two ungrounded conductors plus an equipment grounding conductor, you have two current carrying conductors. The equipment grounding conductor is not counted per 310.15(E). Two current carrying conductors is below the four conductor threshold in 310.15(C)(1), so no adjustment factor applies. If you pull several 240 volt circuits through the same EMT the count climbs fast and the adjustment table will bite.

Field takeaway

12 AWG copper is a 20 ampere conductor, full stop. For 25 amperes continuous at 240 volts, use 8 AWG copper on a 35 ampere breaker. Size the equipment grounding conductor from Table 250.122, which calls for 10 AWG copper at 35 amperes.