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I am not trying to distort or insult and I am sorry if you take it that way. All I am trying to do is to get you to see that a 50 amp breaker will only deliver 50 amps of current be it 240 or 120. It is not 50 plus 50 at 120 volts.Originally Posted by Homeownerinburb
Here is what you said;
If what you are saying that 50 plus 50 equals 100 amps then just how much do you think the neutral would be carrying?
Now if your answer is zero then there would be 50 amps on one side and 50 amps on the other side not a total of 100. All the current flow from one leg would be returning on the other leg if there was no current on the neutral.
If I used an ammeter and checked the amperage on the two legs of the feeder and found 50 amps on one leg and 40 amps on the other leg there would be 10 amps on the neutral and the entire 240 volt circuit would be drawing 50 amps not 90 amps.
There would be 40 amps of 240 volt current and 10 amps of 120 volt current even if everything on these feeders was 120 volt circuits.
If you are saying that on one side of the circuit there is 50 amps what you are saying is that there is 50 amps returning on the neutral. If you say that the other side is carrying another 50 amps then this current is returning on the neutral. If you are saying that 50 plus 50 equals 100 then this is saying that one side has 50 amps on the neutral and the other side has 50 amps on the neutral and this equals 100 amps on the neutral.
If you are saying that the two sides would balance the neutral and there would be no current on the neutral then just how is this current returning? It must be returning on one of the hot conductors.
No insults are intended just trying to get you to understand that the two don’t add up.
Jw, you truly are the master of this, I always read your threads because they are always insightful, I love them. Just for the heck of it, I will tell you, for a minute there I thought I was at work with something you said. lol. You said, " I am sorry you feel that way..." I just recently, said that to a very irate person who didn't understand too, I was not trying to be rude or insulting, only helpful. Sometimes, it is what it is.
For the love of gawd, stop being so damned pedantic.
If you could uncork your illusion that you are perfect for one moment, you will see that I am just telling this guy that a "50 amp service" does not mean he can only have 50 amps of demand at 120 volts.
He has two sides of the phase and each can carry 50 amps.
So if he has two loads that each demand 45 amps at 120v he does not need a 100 amp service to accommodate them.
Do you enjoy being tiresome?
Just trying to insure that the information given is true.
But my friend this is just what a 50 amp service is.
This is an untrue statement. It is one 50 amp circuit not two sides that carry 50 amps each.
this is true, he will only need a 90 amp service
I am here to insure that any and all information given is true and safe.
A 50 amp feeder is just that, 50 amps. It is not 50 amps at 120 volts plus another 50 amps at 120 volts for a total of 100 amps at 120 volts.
If there was 50 amps at 120 volts on one side and another 50 amps at 120 volts on the other side the neutral would have 100 amps on it.
If you are going to say this is untrue then pray tell me how these amps are returning to the transformer.
Edited to add;
For the sake of this discussion let’s take away the trip curve of breakers and assume that a 50 amp breaker will trip at 51 amps.
I have a panel that is supplied by 240 volts at 50 amps.
I load the panel to 50 amps of 120 volt loads and all this load is on one side of the 50 amp breaker. I now load the other side of the breaker with a wall clock and the breaker trips. The total load on that 240 volt two pole breaker will equal 50 amps plus one wall clock.
What can’t be done is to install a two pole 50 amp breaker and load one side to 50 amps at 120 volts and then add another 50 amps at 120 volts to the other side of the breaker at 120 volts to where there would be a total of 100 amps of 120 volt load. It just doesn’t work that way.
Take this same feeder at 50 amps and install two 100 watt lights one on each side of the breaker. The total load on that feeder will be 200 watts with or without the neutral. As long as each side of that breaker has the exact same amount of current the neutral is not needed. A simple experiment you can do is use two lights at 100 watts each and connect them to a 240 volt circuit in series without a neutral and check the voltages at each bulb. You will find that each bulb has 120 volts of drop which equals a total load on that 240 volt circuit of 1.66 amps although each bulb only pulls .83 amps each. Now install the neutral and the load on the feeders will still be 1.66 amps.
Using this same circuit that has a amperage draw of 50 amps on one leg and a 40 amps on the other leg and the neutral will have 10 amps of current flowing through it. You will have a total of 40 amps of 240 volt series circuit and 10 amps of 120 volt parallel circuit but the very second that either side reaches 51 amps the circuit will open (this is if we take away any trip curve of the breaker).
Taking your analogy of a 50 amp breaker allowing 50 amps of 120 volts on one side and then adding another 50 amps to the other side would be a total of 100 amps of 240 volt series circuit.
I agree with you that a 50 amp feeder will be more than he will ever use but I don’t see any reason not to install 250 kcmil copper conductors and a 250 amp disconnect out there if that is what he wants to do.
Last edited by jwelectric; 09-09-2012 at 04:53 AM.
After giving what you have said some thought I believe I am beginning to grasp what you are trying to say. I think you are saying one thing and I am interpreting it as something different.
A 2 pole 50 amp breaker will handle 12000 watts of electrical energy at 240 volts therefore either side of this breaker will handle 6000 watts of electrical energy to ground. Or as the way you are putting it 50 amps of current on one side and then 50 amps of opposing current on the other side. This would be 50 amps to ground on one side with 50 amps to ground of opposing current on the other side which would be 50 amps of current at 240 volts.
If this is true then I stand corrected. What I was thinking is 50 plus 50 or a total of 100 amps off of this one 50 amp breaker which cannot happen.
Am I now correct in what you are stating? I have concluded this from your comment earlier (post 28) about the 60-40 ratio on the same 50 amp breaker.
Last edited by jwelectric; 09-09-2012 at 08:19 AM. Reason: spelling
OK,
So over the weekend I did my install, I have one wire to each breaker on the house panel box, and one wire to each breaker in the garage panel box. From the house panel box, I have one wire from each of the buss bars, (one neutral, and one ground) to the box in the garage. I did not install the grounding bolt (or whatever the bolt is called I`m sure I`ll get beat up for not knowing its name) so in the garage I have one buss bar for neutral, and one buss bar for ground. On the ground buss bar, I have a bare wire to one grounding rod, and I ran it thru the connector, and have the second grounding rod 10 feet away. In the garage I have one 100 amp main breaker, then there is one 40 amp 220, one 15 amp 110 (Lights), one 15 amp 110 breaker (Wall outlets). I put theses all on the same side of the panel box hoping to balance the load somewhat.
Last night I had 4 fluorescent tube lights (total 8 bulbs) 1 radio, 2 outside 60 watt lights and a box fan. When I fire up my welder and struck an arc, the 40 amp breaker in the house tripped. So I did the obvious, turned off the outside lights, radio, and fan, and was able to weld.
I figured since I used 90 amp wire I would not have very much amperage loss on the 40 amp breaker. Or is it the other way, since I have wire rated for 90 amp, am I loosing amperage from the 40 amp breaker in the main panel?
I purchased 50 feet of cable, and have between 4 and 5 feet cut off, so length of run is not a factor. All breakers are brand new never used.
Is there something I missed in the connection? I have my electrical inspection tomorrow, and want to make sure I have not missed something.
FYI, the only reason i have a 40 amp breaker in the main panel is that the store was out of 50`s, and I want the trench back filled this week. I will be getting a 50 when they get more in.
Thanks
Last edited by knied1; 09-10-2012 at 07:29 AM.
Put a 90 at the house and a 50 to feed the welder.
Next spring when i upgrade the service to my house from 100 amp to 200 amp i am planning to do that......I just needed to get power into the garage for the winter, and wanted to make sure what i was running would accomidate future upgrade.
Thanks for the input.
go on and get the 90 amp breaker now.
ok, ill have to hold off until after my permit is satisfied..... He made a point of saying i can not send more then half my service to the garage / out building.
Thanks again guys for your help
Ask for a code reference as this is not true. You could install a 1000 amp feeder and overcurrent device in that 100 amp panel if you so choose
Finally! Gawd! Yes? Why did I not just speak in watts? Possibly because the OP sounded a little less up to speed on the whole concept?
My point: almost every time I have put a sub in a garage for a home handyman or hero gear head, and I tell him a 30 amp (or 50) is going to serve his demands, he wants to challenge me until I explain that he actually has two legs of power at 30 or 50 amps in a three wire circuit, not one, as might at first blush appear to be the point.
Could be a local rule? Many of my local cities would not permit NM in single family homes up till about 10 years ago.
This is very true. I remember a few years ago working in Fl. and the local would not permit ground rods due to the water table. They required a concrete encased electrode.
I find myself complicating issues sometimes simply due to the fact that I stand in the classroom.
When I teach a class on transformers, which is where we are at now, and how they relate to a dwelling unit service we discuss 120 verses 240 volts that is supplied by the transformer.
I present the question to the class which is correct for a 200 amp service; the service will be 200 amps or 400 amps if the two legs are added together. Most answer that it would be 400 amps if added together which is incorrect.
The most 120 volts that can be achieved would be 200 amps. When the second leg is added to it the addition would be +200 added to -200 or if both legs were loaded to the max there would be no 120 being used at all. When we load one side to 10 amps and then the other side to 5 amps even if what are used are 120 volt appliances then the service will only see 5 amps at 120 and 5 amps at 240.
After going back and reading your post I see that what you were saying is that if each 120 volt appliance being used at one time that the total draw on the feeders would be just that and you did not say that each leg would be carrying twice the rating of the conductors or breaker. As a matter of fact you were only addressing each branch circuit instead of the total amperage being on the feeder.
My mind went to the load on the neutral which I guess was easy to see and how that any balanced portion of the 240 volt feeder would be a series 240 volt circuit instead of the parallel 120 volt circuit as you were addressing.
I am slow but I usually get there.
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