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My question is pretty simple. I've seen several forums suggest that you should have the max power on any outlet be 80% of what the circuit is rated for (so for example, if you have a 15 A 120 V outlet, then you shouldn't use more than 80% x 15 A x 120 V = 1440 watts).

Is this really the case? I ask because in my situation, I have a 240 volt outlet running on a 30 ampere dedicated circuit where I want to run some equipment continuously. I am certain that there will be no other equipment running on that circuit. Given this, should I expect to only run 80% x 30 A x 240 V = 5,760 watts on that circuit, or may I use the full 7,200 watts continuously?

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    I don't have the latest Code but I sense a very good reading of the Code is about to happen from ThreePhaseEel. The answer is "maybe"... – Harper - Reinstate Monica Oct 20 '17 at 23:26
  • If I'm not mistaken (which I might be, since I'm not looking at my code book right now), you size the overcurrent protection 125% of continuous loads. However, the conductors can be sized to the load itself. So you might have to bump the breaker a size up, but you might be able to get away with 30 amp conductors. Ambient temperature, and circuit length also have to be taken into account. – Tester101 Oct 21 '17 at 0:11
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    The answer is undefined because there's no universal relationship between amps and watts due to the power factor. When sizing circuits, you use amps or VA, not watts. – user71659 Oct 21 '17 at 6:54
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Use the 80% rule for continuous loads (because the breaker will, even if you don't)

While 210.22 would seem to indicate that you have the full 30A available to you:

210.22 Permissible Loads, Individual Branch Circuits. An individual branch circuit shall be permitted to supply any load for which it is rated, but in no case shall the load exceed the branch-circuit ampere rating.

you have to consider that the breaker may have other ideas, as stated in 210.20(A):

(A) Continuous and Noncontinuous Loads. Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the rating of the overcurrent device shall not be less than the noncontinuous load plus 125 percent of the continuous load.

Exception: Where the assembly, including the overcurrent devices protecting the branch circuit(s), is listed for operation at 100 percent of its rating, the ampere rating of the overcurrent device shall be permitted to be not less than the sum of the continuous load plus the noncontinuous load.

The reason why is because garden-variety breakers made to UL 489 (and their counterparts in fuse-land) will eventually trip (or blow) if you run 100% of their rated current through them for hours on end. While there is such a thing as a 100%-rated breaker, they are typically only found in industrial work.

Furthermore, the branch circuit wiring needs an 80% derate for continuous loads as well, as per 210.19(A)(1):

(1) General. Branch-circuit conductors shall have an ampacity not less than the maximum load to be served. Conductors shall be sized to carry not less than the larger of 210.19(A)(1)(a) or (b).

(a) Where a branch circuit supplies continuous loads or any combination of continuous and noncontinuous loads, the minimum branch-circuit conductor size shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load.

(b) The minimum branch-circuit conductor size shall have an allowable ampacity not less than the maximum load to be served after the application of any adjustment or correction factors.

Exception: If the assembly, including the overcurrent devices protecting the branch circuit(s), is listed for operation at 100 percent of its rating, the allowable ampacity of the branch-circuit conductors shall be permitted to be not less than the sum of the continuous load plus the noncontinuous load.

So, you're limited to 5760W for an on-all-the-time (or at least longer than a few hours at a time) load. Non-continuous loads (say a large well pump motor), though, can pull the full 7200W from the circuit as the duty cycle of the load provides adequate time for things to cool off between runs.

  • Strictly speaking, isn't 5760 VA available? If you try to connect 5760 W of non-resistive loads, you'll exceed the current limit. This seems relevant because in another question, the asker mentions large computer loads. – user71659 Oct 21 '17 at 6:25
  • @user71659 -- modern computer supplies have a quite good PF; furthermore, the reactive definition of PF does not hold for nonlinear loads to begin with – ThreePhaseEel Oct 22 '17 at 14:10
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    The PF of computer power supplies under Energy Star is only .8-.9 depending on load. Under nonlinear loads, you still have a power factor, it just uses the RMS definitions, just like how you calculate voltage on a square wave inverter. However, this is all irrelevant to your answer. The asker should not size the loads on a circuit by assuming the PF is 1, because the circuit is likely to be overloaded. – user71659 Oct 22 '17 at 16:54
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The answer is your circuit can provide anywhere from 0 to 5760 W without tripping. Unless you are running only incandescent light bulbs or heaters both without any variable control, you will not get 5760 W.

As another answer gives, you are allowed to draw 80% of the amp rating continuously, which in your case is 80% of 30A = 24 A. A fundamental mistake is to multiply 24 A by 240 V and get 5760 W. This is incorrect. The circuit can provide 5760 VA.

Due to a property known as the power factor, a load can draw more energy than it consumes in part of the AC cycle, and gives it back to the line in the other half. This means current will circulate without being consumed by the device, but using ampacity by still heating up the wires. However, the power company (in a residential setting) bills you by the power actually consumed, W, which is why devices have it labeled.

For example, this LED bulb I have says "120 V 14.5 W 234 mA". If you multiply voltage by current, you get 28 VA, showing that it draws more current than is consumed, and we would say it has a power factor of 14.5/28=0.51. If I used the wattage rating where I should have used A or VA, I would put twice as many bulbs as safe on a circuit.

In conclusion, add the loads up by nameplate amps or if available by the VA rating. See this guide, for example. The wattage rating is irrelevant for sizing in most cases.

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