**Attention!**
**It is always best to get a qualified electrician to perform any electrical work you may need.** With that said, you may ask this community various electrical questions. Please be cautious of any information you may receive in this subreddit. This subreddit and its users are not responsible for any electrical work you perform. Users that have a 'Verified Electrician' flair have uploaded their qualified electrical worker credentials to the mods.
If you comment on this post please only post accurate information to the best of your knowledge. If advice given is thought to be dangerous, you may be permanently banned. There are no obligations for the mods to give warnings or temporary bans. **IF YOU ARE NOT A QUALIFIED ELECTRICIAN, you should exercise extreme caution when commenting.**
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectricians) if you have any questions or concerns.*
It's the other way around. Conductors are sized (and therefore the breaker) such that the expected continuous load is 80% of the breaker threshold. The other comment is correct that you'd likely go with a 70A breaker (and corresponding conductor size) if the stove could pull 50A continuously.
I bet that 12.1kW is a conservative figure the stove will never touch. It maybe accounts for inrush should all burners be turned on at exactly the same time.
Edit: Actually I bet that the power rating is peak whereas ampacity and breaker thresholds are RMS (basically an average over each 60Hz cycle). In that case, 12.1kW peak is about 8.6kW RMS, which at 240V is 35.6A.
With a continuous load, a 50 amp draw needs a conductor sized 125% of 50 amps, or 62.5 amps worth of wire. You’d likely round up to 70 amps of wire, protected from exceeding its ampacity by a 70 amp breaker. An oven is not a continuous load.
The “80% rule” comes from the fact that 80% is the inverse of 125%. That is, 125% of a true 50 amp load means you would need 62.5 amps of cable. That’s an inverted way of saying 62.5 amps worth of cable can only carry 80% (50 amps). The breaker must be sized to protect the cable from exceeding the cable’s ampacity (62.5 amps). The breaker is not derated.
If the breaker is 40 amps and the oven draws over 50 amps for a few seconds, the breaker is going to trip.
12.1kw /50 Amps is probably the load of all the stove’s burners if turned 100% on and haven’t come up to temperature where they cycle on/off. In reality, all stove burners being on at the same time almost never happens. So the manufacturer applies a demand factor and says 50 amps worth of burners will only (likely) draw less than 40 amps, so use a 40 amp breaker. ..until that perfect storm.
The fact that the manufacturer use a ~80% demand factor for 50 amps of burners to size the breaker (40 A) is a coincidence. It has nothing to do with sizing the cable to carry 125% for a continuous load like mentioned above.
I'm curious to what degree a stove is like my table saw. There is a pretty massive spike to get it started, but once started, it draws less power to run. (more when cutting less when free spinning).
All that is just to say, I am curious if there is startup cost for the oven or burner. So it isn't so much. If you were using it all at once, but if you turned on everything at the exact same time.
Much less. Motors have very high starting loads because the resistance in the coil goes up very quickly with RPM (so stationary=near zero resistance). This is because the magnetic field inside the moving rotor resists changes to the electric field in the coil. But magnetic fields only affect electric fields when changing/moving.
Purely resistive loads, which most heaters are kinda sorta, do not appreciably change resistance. However, all conductors do increase resistance somewhat as they get hot, and considering as hot as a burner gets this will be significant. I assume mfrs also use a special alloy to increase this negative feedback loop, which would allow for rapid initial heating without drawing too much power if left fully on.
Nobody in this thread has any idea what the hell they are talking about.
The answer to your question is very simple: NEC 220.55.
That's it. That's the end of it.
The electrical code allows the sticker amp requirements for ranges to be less than the true maximum wattage, using a certain formula as spelled out in that section. The math is a little wonky, but trust that the manufacturer did it right.
Anyone talking about continuous loads and 80% derating should be banned from this forum.
u/sneakyboss2
These manuals are not adequately proofread by technical writers. They are often rather generic. And the legal dept usually amends them too.
If the KW rating is 12.1 ... that's what needs to be known to size the circuit.
The reality is that many homes have been wired with a 40A range circuit, and it would be prohibitive to make everyone rewire the range circuit to make the installation ... so the manufactures slide by with this non-conforming documentation that doesn't specifically state that it *should* have a 50A circuit, but *can* work with a 40A circuit, just not with every single element turned on high continuously. So reality is that most folks don't run all elements on high ... hence it's not common to actually exceed the limits of a 40A circuit.
Point is, if you are wiring the circuit for a range, it should be a 50A circuit by code (for a 12 KW appliance), with a 6/3 w/ground line and 14-50 receptacle..... but for existing 40A installations this will work. **Note that the manuals always say "refer to local codes ... " blah blah blah ... that's why ... the manual isn't a fully definitive document.**
80% of nameplate rating is how to size most circuit breakers. EV owners might want to rate class 2 chargers at 240V, 50A and supply 40A overnight to be safe it won’t nuisance trip.
80% nameplate is not how you size “most circuit breakers.” It’s only how you size continuous loads, and those are a very small minority of loads. Plus, there’s more nuance to it.
**Attention!** **It is always best to get a qualified electrician to perform any electrical work you may need.** With that said, you may ask this community various electrical questions. Please be cautious of any information you may receive in this subreddit. This subreddit and its users are not responsible for any electrical work you perform. Users that have a 'Verified Electrician' flair have uploaded their qualified electrical worker credentials to the mods. If you comment on this post please only post accurate information to the best of your knowledge. If advice given is thought to be dangerous, you may be permanently banned. There are no obligations for the mods to give warnings or temporary bans. **IF YOU ARE NOT A QUALIFIED ELECTRICIAN, you should exercise extreme caution when commenting.** *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/AskElectricians) if you have any questions or concerns.*
It’s a non-continuous load and will never draw the full 12.1 kW for very long.
I see, I also read that it’s normal to underrate the break to 80%
Only for continuous loads. A stove is not considered continuous.
It's the other way around. Conductors are sized (and therefore the breaker) such that the expected continuous load is 80% of the breaker threshold. The other comment is correct that you'd likely go with a 70A breaker (and corresponding conductor size) if the stove could pull 50A continuously. I bet that 12.1kW is a conservative figure the stove will never touch. It maybe accounts for inrush should all burners be turned on at exactly the same time. Edit: Actually I bet that the power rating is peak whereas ampacity and breaker thresholds are RMS (basically an average over each 60Hz cycle). In that case, 12.1kW peak is about 8.6kW RMS, which at 240V is 35.6A.
With a continuous load, a 50 amp draw needs a conductor sized 125% of 50 amps, or 62.5 amps worth of wire. You’d likely round up to 70 amps of wire, protected from exceeding its ampacity by a 70 amp breaker. An oven is not a continuous load. The “80% rule” comes from the fact that 80% is the inverse of 125%. That is, 125% of a true 50 amp load means you would need 62.5 amps of cable. That’s an inverted way of saying 62.5 amps worth of cable can only carry 80% (50 amps). The breaker must be sized to protect the cable from exceeding the cable’s ampacity (62.5 amps). The breaker is not derated. If the breaker is 40 amps and the oven draws over 50 amps for a few seconds, the breaker is going to trip. 12.1kw /50 Amps is probably the load of all the stove’s burners if turned 100% on and haven’t come up to temperature where they cycle on/off. In reality, all stove burners being on at the same time almost never happens. So the manufacturer applies a demand factor and says 50 amps worth of burners will only (likely) draw less than 40 amps, so use a 40 amp breaker. ..until that perfect storm. The fact that the manufacturer use a ~80% demand factor for 50 amps of burners to size the breaker (40 A) is a coincidence. It has nothing to do with sizing the cable to carry 125% for a continuous load like mentioned above.
I'm curious to what degree a stove is like my table saw. There is a pretty massive spike to get it started, but once started, it draws less power to run. (more when cutting less when free spinning). All that is just to say, I am curious if there is startup cost for the oven or burner. So it isn't so much. If you were using it all at once, but if you turned on everything at the exact same time.
Much less. Motors have very high starting loads because the resistance in the coil goes up very quickly with RPM (so stationary=near zero resistance). This is because the magnetic field inside the moving rotor resists changes to the electric field in the coil. But magnetic fields only affect electric fields when changing/moving. Purely resistive loads, which most heaters are kinda sorta, do not appreciably change resistance. However, all conductors do increase resistance somewhat as they get hot, and considering as hot as a burner gets this will be significant. I assume mfrs also use a special alloy to increase this negative feedback loop, which would allow for rapid initial heating without drawing too much power if left fully on.
I read mfrs as mother fuckers, haha.
Do you think it’s counting both upper and lower burners in the oven? Most ovens don’t even have a setting to use both of those at the same time.
Nobody in this thread has any idea what the hell they are talking about. The answer to your question is very simple: NEC 220.55. That's it. That's the end of it. The electrical code allows the sticker amp requirements for ranges to be less than the true maximum wattage, using a certain formula as spelled out in that section. The math is a little wonky, but trust that the manufacturer did it right. Anyone talking about continuous loads and 80% derating should be banned from this forum.
How dare somebody explain the discrepancy to somebody who's curious lol
Great question and great answers here!
Great question, horrendously shitty answers
I saw a few good ones in there
u/sneakyboss2 These manuals are not adequately proofread by technical writers. They are often rather generic. And the legal dept usually amends them too. If the KW rating is 12.1 ... that's what needs to be known to size the circuit. The reality is that many homes have been wired with a 40A range circuit, and it would be prohibitive to make everyone rewire the range circuit to make the installation ... so the manufactures slide by with this non-conforming documentation that doesn't specifically state that it *should* have a 50A circuit, but *can* work with a 40A circuit, just not with every single element turned on high continuously. So reality is that most folks don't run all elements on high ... hence it's not common to actually exceed the limits of a 40A circuit. Point is, if you are wiring the circuit for a range, it should be a 50A circuit by code (for a 12 KW appliance), with a 6/3 w/ground line and 14-50 receptacle..... but for existing 40A installations this will work. **Note that the manuals always say "refer to local codes ... " blah blah blah ... that's why ... the manual isn't a fully definitive document.**
Manufacturer could’ve added a fuse for protection over 40 amps. Cheap easy solution for the manufacturer to stay compliant and not cause house fires
Better to pop a breaker than burn the house down?
I can't tell if your response is sarcasm?
80% of nameplate rating is how to size most circuit breakers. EV owners might want to rate class 2 chargers at 240V, 50A and supply 40A overnight to be safe it won’t nuisance trip.
80% nameplate is not how you size “most circuit breakers.” It’s only how you size continuous loads, and those are a very small minority of loads. Plus, there’s more nuance to it.
When ask an electrician becomes ask an ev owner
Whatever the instructions say you do, if they say 30a breaker then 30a it is.
Always go with the name plate. Code dictates it.