panhandle444
Member
- Location
- oklahoma
the maximum volt ampere loading permitted on a 20 ampere branch circuit with a continuous load would be
20x120x1.25 percent=3000va. Correct?
20x120x1.25 percent=3000va. Correct?
VA continuous load question
- Thread starter panhandle444
- Start date
- Status
- Location
- Massachusetts
JFletcher
Senior Member
- Location
- Williamsburg, VA
Alternatively, 80% of 2400VA (20x120) also equals 1,920VA. Division by 125% (5/4ths) is the same as multiplication by 80% (4/5ths).
- Location
- Massachusetts
Of course it is, but the OP and the NEC reference 125% so that is what I chose to show for clarity.
JFletcher
Senior Member
- Location
- Williamsburg, VA
Agree. It's just that some people learn differently. I can do it either way, tho I'm sure there are folks more comfortable multiplying by .8 than dividing by 1.25.
2400
------
1.25
or 2400 x .8
I can do the latter in my head easier. and when I see 2400/5/4, I convert it mentally to 2400 x .8. Actually, I've memorized the answer, and know 16A is the max continuous load on a 20A circuit. That's a third way to do it. (20 x 4/5 = 80/5 = 16).
woodduder
Senior Member
- Location
- West Central FL.
To the OP. Just remember that when you are having a continuous load, you put more heat into the conductors and over current protection so you need to have a result that is lower than 20 amps x 120 volt which = 2400 VA.
You cannot allow more VA when you know the load is continuos. You have to reduce the allowable VA to compensate for the heat generated from the 3 or more hours of load.
Sent from my SM-G920V using Tapatalk
You cannot allow more VA when you know the load is continuos. You have to reduce the allowable VA to compensate for the heat generated from the 3 or more hours of load.
Sent from my SM-G920V using Tapatalk
2Broke2Sleep
Senior Member
- Location
- Florida
Could someone provide the code reference stating you cannot have a continuous load more than 80% of a branch circuit's rating? This would imply you have the ability to reduce the draw on the appliance in use. How do you reduce the current draw on an appliance pre-assembled with plug n cord ready out of the box? I know we use 80% of circuit size as a good rule of thumb, but I haven't seen it as rule in the code book.
My understanding is: 20A x 120v = 2400VA is the load. Since it is a continuous load you would need a conductor rated at 20a x 1.25 or 25amps. "...minimum branch conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load."
My question is your terminals are rated at 75 degrees and you ran 12AWG THW (or any THHN for that matter), would that be code compliant? The OCPD wouldn't change, but the code is satisfied (even 240.4[D] ) to carry the calculated load now.
My understanding is: 20A x 120v = 2400VA is the load. Since it is a continuous load you would need a conductor rated at 20a x 1.25 or 25amps. "...minimum branch conductor size, before the application of any adjustment or correction factors, shall have an allowable ampacity not less than the noncontinuous load plus 125 percent of the continuous load."
My question is your terminals are rated at 75 degrees and you ran 12AWG THW (or any THHN for that matter), would that be code compliant? The OCPD wouldn't change, but the code is satisfied (even 240.4[D] ) to carry the calculated load now.
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Most breakers, especially of the branch circuit sizes, are not rated for 100% with a continuous load. So a 20 amp breaker can only carry 16 amps (80% of a 20 amp breaker).
You are stating an entirely different scenario than the op. He asked what is max continuous VA on 20A circuit. Your question is about a continuous 20A-rated load.
210.21(A) states "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." Minimum OCPD is 20A × 125% = 24A, rounded up to next standard size 25A.
240.4(D)(5) you cannot use #12 copper with greater than a 20A breaker.
2Broke2Sleep
Senior Member
- Location
- Florida
Got it. I overlooked 210.21 Would 10 AWG (romex/thhn/thw) with a 25 amp breaker satisfy this installation? Or as I stated previously, a 20A breaker rated at 100% with 12AWG THW?
Ummm... meant that to be 210.20(A)
Yes to first question.
To use a 20A, 100%-rated breaker requires the entire enclosure (panel board, e.g.) to be rated for 100% operation. You will also discover as part of achieving that rating, you must use all 90°C-rated conductors.
Last edited:
- Location
- Massachusetts
What is the load you are trying to supply?
2Broke2Sleep
Senior Member
- Location
- Florida
Hypothetically a cord and plug stereo system attached to a 20 duplex....(obviously a violation in the case of a 25 amp OCPD)
Hmmm... a cord-and-plug-connected sound system that draws 20A continuous. Not possible unless all you play is pink noise at max volume for more than 3 hours at a time. :happyno:
- Location
- Massachusetts
That was what I was getting at, I was not sure if you were aware a 25 OCPD can not be used with a 20 amp receptacle.
Beyond that I agree with smart$
- Location
- Massachusetts
I used to hook up concert power a lot and while killing time sometimes I would put my amp clamp on the feeders during the show.
It would be all over the place, and the (Harmonic?) currents on the sound power feeder neutral would often equal to the ungrounded conductor currents.
(208Y/120 source and distribution)
When I provided a 300 KVA trailer mounted generator for the lighting feed to an outdoor show it was fun watching it bounce around on it's springs to the beat of the music due to the lighting load changing. This was back in the 90s so the lighting rigs were all 1,000 watt PAR cans.
- Status