i agree, but we still have 40Amps going through the device and 40Amp generates a lot of heat. So i guess a duplex receptacle will be able handle the heat that is generated from 40Amps.
20 amp duplex recepticle
- Thread starter arnettda
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i agree, but we still have 40Amps going through the device and 40Amp generates a lot of heat. So i guess a duplex receptacle will be able handle the heat that is generated from 40Amps.
- Location
- Chapel Hill, NC
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- Retired Electrical Contractor
Yes if the tabs are broken. You are saying that each female part can only carry 10 amps. So if I plug in a 12 amp unit in a 20 amp receptacle it wouldn't last. Think of a duplex with the tabs removed as two single receptacles. Each one is rated 20 amps.
Allso with the op's post a toaster and waffle iron are not going to be on very long. I can see the plastic getting hot if 2-20 amp circuits were going full steam but in this situation I don't see an issue.
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Cow
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- Eastern Oregon
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- Electrician
I'd do it and wouldn't think twice about it. If they didn't want you to feed duplexes with two circuits, why do they have tabs on the side that can be removed?
Hi all, been a while, but I was researching MWBC and this thread came up.
A few posts back, someone was discussing amps (and thus 'poor' connection watts) per device per strap etc...
If one broke the brass side 'jumper', wired a 20 amp MWBC, and measured current - I believe there would still be a max of 20 amps (but at 240 volts) at any point. Think matched loads, neutral flow is '0', and the 20 goes through 1 load, the neutral 'jumper', and then through the other load. A marginal connection could 'heat' a few ohms at 20 amps for X watts, and then again at another connection, etc, maybe 6 times.
The other scenario - 2 separate circuits, both 'jumpers' broken out - could be 20 amps on each device with the 'same' 120's or 2 different 120's (Potentially 8 poor connections?) I have to think that if that was not allowed, there would be no reason to have a breakable jumper on the neutral. The breakable jumper seems to imply that the total device can handle the heat from 20 amps twice - either from a MWBC, or two circuits.
Trying to help...
Now to my interest...
I'm trying to understand practicality of the previous rule of tied breakers for MWBC if landed on same strap.
I see it explained as for worker safety.
Seems that it assumes a worker will not see the broken 'jumper' before before pulling out the strap (but after turning off only breaker #1).
Is that really any different than a double switch (single strap) with 2 different feeds? Or a Cmas light receptacle (1/2 room circuit, other half global switched circuit, but not MWBC's)
Is there a better reason?
THX
A few posts back, someone was discussing amps (and thus 'poor' connection watts) per device per strap etc...
If one broke the brass side 'jumper', wired a 20 amp MWBC, and measured current - I believe there would still be a max of 20 amps (but at 240 volts) at any point. Think matched loads, neutral flow is '0', and the 20 goes through 1 load, the neutral 'jumper', and then through the other load. A marginal connection could 'heat' a few ohms at 20 amps for X watts, and then again at another connection, etc, maybe 6 times.
The other scenario - 2 separate circuits, both 'jumpers' broken out - could be 20 amps on each device with the 'same' 120's or 2 different 120's (Potentially 8 poor connections?) I have to think that if that was not allowed, there would be no reason to have a breakable jumper on the neutral. The breakable jumper seems to imply that the total device can handle the heat from 20 amps twice - either from a MWBC, or two circuits.
Trying to help...
Now to my interest...
I'm trying to understand practicality of the previous rule of tied breakers for MWBC if landed on same strap.
I see it explained as for worker safety.
Seems that it assumes a worker will not see the broken 'jumper' before before pulling out the strap (but after turning off only breaker #1).
Is that really any different than a double switch (single strap) with 2 different feeds? Or a Cmas light receptacle (1/2 room circuit, other half global switched circuit, but not MWBC's)
Is there a better reason?
THX
mjf
Senior Member
readydave8
re member
- Location
- Clarkesville, Georgia
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- electrician
I'm not convinced that 40 amps would be going through device but I might be wrong.
If I had a fully loaded 20 amp, 240 volt receptacle, I clamp amprobe on each hot, each shows 20 amps. Total 20, not 40. Same thing at 2/pole breaker. So what's the difference here?
I'm not trying to argue, only to learn, please be polite.
if you have two separate single receptacle and each receptacle has its own 2 wire circuit with 20Amps on it, you would have 40amps correct?
hurk27
Senior Member
- Location
- Portage, Indiana NEC: 2008
No this is not correct, each circuit only has 20 amps, there is no connection between the circuits or receptacles (tabs removed) so this in the same as if you had two single receptacles in a two gang box, side by side, any point of connection on either receptacle would only be 20 amps, you don't add the current of different circuits together, just like you don't add the current of a 100 amp two pole breaker to say you have a 200 amp breaker, you will still only have 100 amps of usable current.
i agree that you don't add them up but both circuits have equivalent to 40 amps that are going through the receptacle. another words there is 40 amps being supplied through the panel.
this is what i meant.
Joethemechanic
Senior Member
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- Hazleton Pa
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- Electro-Mechanical Technician. Industrial machinery
The thing is with the tab/tabs broken at any point in the circuit there is only 20 ampa flowing even if there are two 120V 20 amp loads plugged into it.
If you don't look at it that way and look at it as 40 amps, then your I^2R heating calculations get screwed up.
20 squared is 400
40 squared is 1,600
Say each blade to slot connection is 0.25 ohms (example not real) for a total of 0.50 ohms for each plug
Total heating for each plug would be 20^2 = 400 x 0.50 for 200 watts of heating at each plug and a total heating of 400 watts for both plugs.
If you call it 40 amps you get 40^2 = 1600 x ?????????
There is no resistance in the circuit that has 40 amps flowing through it
If we double the resistance in any given circuit, we double the heating
If we double the amperage in any given circuit, we quadruple the heating.
A duplex with the tabs broken only doubles the heating
If you don't look at it that way and look at it as 40 amps, then your I^2R heating calculations get screwed up.
20 squared is 400
40 squared is 1,600
Say each blade to slot connection is 0.25 ohms (example not real) for a total of 0.50 ohms for each plug
Total heating for each plug would be 20^2 = 400 x 0.50 for 200 watts of heating at each plug and a total heating of 400 watts for both plugs.
If you call it 40 amps you get 40^2 = 1600 x ?????????
There is no resistance in the circuit that has 40 amps flowing through it
If we double the resistance in any given circuit, we double the heating
If we double the amperage in any given circuit, we quadruple the heating.
A duplex with the tabs broken only doubles the heating
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When the hot tab or both tabs are removed you no longer have a duples receptacle, you have two single receptacles side by side. If you are wondering whether or not the metal or plastic that these two single receptacles are mounted to can handle the resisual heat that may be trasfered to it, the answer is yes.
Roger
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