Thread: Determining the amperage mat a main disconnect fuse block

Originally Posted by JOHN PAVAN

It really seems that people are talking past each other. I wonder if the following isn't a better way to look at it. As far as asking electricians, I don't know how many of them haven't insisted to me, when I was supplying pool pumps, that it is cheaper to run things on 240 than 120! When I did the watt calculation for them, they scratch their heads, and look dubious.

The service is 60 Amps @ 240 Volts- which would allow a maximum of 14,400 watts - A * V = W. If you pulled 60 amps at 120- you get 7,200 watts, two such circuits would yield the same maximum Watts- 14400. Watts is the total measurement of electricity used- Power- after all Watts as in Kilowatts is what you get billed for.

I think it is customary to list the rating of the service at 240 Volts, so we say 60 Amps. One could say it is 120 Amps @ 120 volts, but we all know that is not the custom. If we insist that it is 60 Amps @ 120 Volts, that can't be right, because we know that we can pull 14400 watts with two 60 Amp 120 volt circuits. The gentleman who said the wire size should be checked is of course correct, because either the wire size or the breaker could be the limiting factor, and of course the breaker should never exceed the wire capacity

Thank you John for pointing out the math aspect of this issue and showing that two legs of 120 is the same capacity as one leg at 240.

Vern, look at the following link. Insert a 60 amp 120 volt load on both legs A and B of the panel and tell me where the current is flowing to. If you said the neutral you would be correct. Now how much is flowing through the main breaker. How much is flowing through breaker A, how much through breaker B? Answer 60 and 60.

http://www.samlexamerica.com/support...chCircuits.pdf

Originally Posted by Vern Heiler

And as Roland said, " Do you want to talk about amps or power. Cause if you are talking about power you have just changed the subject."

My challenge stands; someone, anyone, draw a circuit diagram showing where the current goes that is different than the one I posted! Because with mine, there can not be 120 amps on a 60 amp service with 60 amp breakers!

So your contention is that one could only get a maximum of 7200 Watts if everything is on single pole circuits?

Originally Posted by JOHN PAVAN

So your contention is that one could only get a maximum of 7200 Watts if everything is on single pole circuits?

Just draw the circuit would you! Then you can put me out of my misery.

I was too quick to answer that! 120 volt loads on opposite legs will react as though they were 240 volt loads as far as current is concerned. So if the loads are balanced you will have the full wattage available, still only 60 amps total.

Originally Posted by Scott Bray

I inspected an apartment in a building with 200 apartment units, The main disconnect for the sub panel of the apartment is located at the electrical meter in the basement of the building. There are two 60 amp cartridge type fuses in a fuse block as the disconnect. I stated that it was a 60 amp service and the building supervisor stated no it is a 120 amp service because you combine the two legs of the service. I am confusing my self, I always thought you rated the disconnect by the weakest link, which would be the 60 amp fuse. Which is correct, is it a 60 amp service or 120 amp service ? Thanks in advance.

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y.

Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Vern, you're getting yourself stuck on trying to apply DC electrical theory and a simple circuit for same and trying to apply (your "directional arrows on your 'diagram'") which doesn't "fly" with AC. Unlike DC current on a simple circuit with a purely resistive load, AC current doesn't flow "one way" from a positive battery terminal with a negative for reference - each conductor pulses the exchange in a "pull me/push me" fashion - on your scope you'll see a flat line for each half cycle for each conductor with center tap reference. Not all loads are purely resistive, in fact in todays "usual" residence with other than coil/resistive electric range, other than electric WH and other than electric resistive heat, few loads are "purely resistive". Modern electronics, including the electronics involved in compact fluo., televisions, etc. motor loads refrigeration, etc. multitude of rectiifiers & power supplies, none of which are purely resistive loads (most inductive, and PF increasingly "of issue") for large multi-unit buildings.

Likely the main power feeder is 60 amps 208/120. IIRC the OP is in CT. Although early utility in No. NY and CT may have been DC "back in the day" the entire US has been on a 60 Hz AC standard for over a century (for other than RR, aircraft & marine) and Con Ed does not supply DC power to residential occupancies.

Be that as it may, there is no way the load side feeder from the fuse block to the residential unit can be called a "service". Presuming it is a dedicated feeder and not a polyphase riser tapped or fed-through, at best it could be called a "main power feeder" for the occupancy.

I don't recall the OP having indicated the "meters" were "utility owned and/or controlled" for that matter either, not that it matters.

60 amp fuse on each of two split half-phase legs is still a 60 amp supply (and likely each are split half phases from poly-phase supply) The "Service point" before the feeder to the unit not in the unit.

If it was drawn it would not make it true. Each breaker or fuse will add capacity. Since there are two legs able to supply 7200 watts each, the total capacity of the service is 14,400 watts. (60*120)=(7200) *2 = 14,400 watts.

Vern, pull one of the fuses out of the block and tell me how many amps can flow through the remaining leg. How many legs can support that current? How many amps can flow through both legs?

Originally Posted by Jim Port

Thank you John for pointing out the math aspect of this issue and showing that two legs of 120 is the same capacity as one leg at 240.

Vern, look at the following link. Insert a 60 amp 120 volt load on both legs A and B of the panel and tell me where the current is flowing to. If you said the neutral you would be correct. Now how much is flowing through the main breaker. How much is flowing through breaker A, how much through breaker B? Answer 60 and 60.

http://www.samlexamerica.com/support...chCircuits.pdf

Did you read this part? "The maximum current flowing in the common Neutralwill be limited to the breaker capacity (Maximum current will flow in the common Neutral when one of the split phase branch circuitsis not loaded
and the loaded split phase branch circuit is drawing itsfull rated capacity)."

Originally Posted by H.G. Watson, Sr.

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y..

Who cares what the voltage is? The question was how much power could be supplied through a 2 fuse 60 amp pullout disconnect. Stick to the subject and stop speculating.

Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Why do you think this has anything to do with the original question?

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Originally Posted by Vern Heiler

Did you read this part? "The maximum current flowing in the common Neutralwill be limited to the breaker capacity (Maximum current will flow in the common Neutral when one of the split phase branch circuitsis not loaded
and the loaded split phase branch circuit is drawing itsfull rated capacity)."

Correct, the neutral current will never be more than the ungrounded current in a MWBC. If only one leg is flowing the maximum current and the other leg is flowing 0, the current on the neutral is still limited to the breaker maximum.

Where do you think the extra current would be coming from?

Originally Posted by H.G. Watson, Sr.

First of all, as you indicated this is a multi-unit (200 - two hundred unit) apartment unit building AND that the main power feeder (not "service") is protected by a fuse block, it is highly unlikely that the power supplied is 120/240, and is most likely 208/120Y.

Furthermore since the most likely situation is that heat is supplied by other than pure electrical resistance heat, and that cooking is more likely to be supplied by other than electrical resistance coil - again multi-phase for load diversity for the BUILDING SERVICE is the most likely, even with individual metered supplied by the utility.

Vern, you're getting yourself stuck on trying to apply DC electrical theory and a simple circuit for same and trying to apply (your "directional arrows on your 'diagram'") which doesn't "fly" with AC. Unlike DC current on a simple circuit with a purely resistive load, AC current doesn't flow "one way" from a positive battery terminal with a negative for reference - each conductor pulses the exchange in a "pull me/push me" fashion - on your scope you'll see a flat line for each half cycle for each conductor with center tap reference. Not all loads are purely resistive, in fact in todays "usual" residence with other than coil/resistive electric range, other than electric WH and other than electric resistive heat, few loads are "purely resistive". Modern electronics, including the electronics involved in compact fluo., televisions, etc. motor loads refrigeration, etc. multitude of rectiifiers & power supplies, none of which are purely resistive loads (most inductive, and PF increasingly "of issue") for large multi-unit buildings.

Likely the main power feeder is 60 amps 208/120. IIRC the OP is in CT. Although early utility in No. NY and CT may have been DC "back in the day" the entire US has been on a 60 Hz AC standard for over a century (for other than RR, aircraft & marine) and Con Ed does not supply DC power to residential occupancies.

Be that as it may, there is no way the load side feeder from the fuse block to the residential unit can be called a "service". Presuming it is a dedicated feeder and not a polyphase riser tapped or fed-through, at best it could be called a "main power feeder" for the occupancy.

I don't recall the OP having indicated the "meters" were "utility owned and/or controlled" for that matter either, not that it matters.

60 amp fuse on each of two split half-phase legs is still a 60 amp supply (and likely each are split half phases from poly-phase supply) The "Service point" before the feeder to the unit not in the unit.

H.G. If you can get current to flow in two directions at the same point in time in the same conductor you will be the first to do so. The potential difference along the secondary at any point in time is +-+-+- and it could be a series of batteries for that point in time.

"Where do you think the extra current would be coming from?"

That was my question to you, I've always said 60 amps is the maximum. Where do you think the sixty first amp comes from? (And I diagrammed my current flow)

Sixty amps is able to flow through each fuse. You have two fuses. How much is 2 x 60?

Lets try this another way again. Say you have 1500w toaster oven on leg A and are warming dinner. Now you also want a cup of coffee and start your 1500w coffee maker that is on leg B. How many amps are being used, 12.5 or 25 amps?

Originally Posted by Jim Port

Sixty amps is able to flow through each fuse. You have two fuses. How much is 2 x 60?

Lets try this another way again. Say you have 1500w toaster oven on leg A and are warming dinner. Now you also want a cup of coffee and start your 1500w coffee maker that is on leg B. How many amps are being used, 12.5 or 25 amps?

The 60 amps on leg A is the same 60 amps that is on leg B in a closed circuit that is a total of 60 amps.

Originally Posted by Vern Heiler

The 60 amps on leg A is the same 60 amps that is on leg B in a closed circuit that is a total of 60 amps.

That is 60 amps producing 3000 watts of power. Good night gentlemen....I gotta get up in the morning.

You didn't answer the question. How many amps are being used by the two 1500 watt appliances?

Originally Posted by Jim Port

You didn't answer the question. How many amps are being used by the two 1500 watt appliances?

Oh sorry...12.5. And the reason this works is that the two appliances are treated as one 240 volt load with none of the current on the neutral. I'm really off to bed this time.

Originally Posted by Vern Heiler

Oh sorry...12.5

WRONG. 12.5*2 = 25, not 12.5.

I will gladly stick with my electricians.

Here's an attempt to explain why you can get 120 amps from two 60 amp fuses. To explain this I will use as an example a double pole 60 amp circuit breaker. We know this breaker is connected to both legs in the panel. We also know the voltage will be 240 volts AC across the legs. Now we put a 60 amp load on the breaker. Follow me Vern? Good! If you calculate the voltage by the amperage, (240 volts x 60 amps = 14400 watts). The panel doesn't give a hoot that I'm using a two pole breaker that's connected to both legs. The fact is I'm consuming 14400 watts. Am I still good Vern. OK. Now, we know we can only get 60 amps from one leg and we know this leg gives us 120 volts in respect to neutral. this calculated out (60 amps x 120 volts = 7200 watts). Say what? 7200 isn't very good, we need 14400 watts. So where do we get the other 7200 watts to make 14400 watts? Lets use both legs. LEG A (60 amps x 120 volts = 7200 watts + LEG B (60 amps x 120 volts = 7200 watts)). Fact, we have 7200 watts available on each leg. Fact, if we use both legs to feed the double pole circuit breaker we will have 14400 watts total. Here is the proof your looking for. The proof is we can add together the current on each leg to achieve the 14400 watts we need. there is no dispute about this. Its real, its fact, its true. But, if this is possible, that means each leg must be compatible to be use this way. It must also mean each leg is independent of each other. So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect. This works and I proved it. Now, its fact we have 7200 watts per leg. Calculating this out works like this, (7200 watts / 120 volts = 60 amps). If we can add leg A to leg B for a sum of 120 amps, there MUST be 60 independent amps per leg. They MUST be compatible to work together. And in fact, if we use both legs regardless of anything else, there is useable, consumable, and undisputable 120 amps available. A sixty amp, 240 volt panel has 120 amps to be used as we see fit which is the same as saying a sixty amp 240 volt circuit breaker can supply 14400 watts. Each leg supplies 60 amps, independent of each other, without interfering with each other, with out relying on each other. I proved it earlier in this post. A drawing or diagram isn't necessary. Its simple math. Please except this as a vial, friendly explanation to help you understand.

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Originally Posted by Mike Borchardt

Here's an attempt to explain why you can get 120 amps from two 60 amp fuses. To explain this I will use as an example a double pole 60 amp circuit breaker. We know this breaker is connected to both legs in the panel. We also know the voltage will be 240 volts AC across the legs. Now we put a 60 amp load on the breaker. Follow me Vern? Good! If you calculate the voltage by the amperage, (240 volts x 60 amps = 14400 watts). The panel doesn't give a hoot that I'm using a two pole breaker that's connected to both legs. The fact is I'm consuming 14400 watts. Am I still good Vern. OK. Now, we know we can only get 60 amps from one leg and we know this leg gives us 120 volts in respect to neutral. this calculated out (60 amps x 120 volts = 7200 watts). Say what? 7200 isn't very good, we need 14400 watts. So where do we get the other 7200 watts to make 14400 watts? Lets use both legs. LEG A (60 amps x 120 volts = 7200 watts + LEG B (60 amps x 120 volts = 7200 watts)). Fact, we have 7200 watts available on each leg. Fact, if we use both legs to feed the double pole circuit breaker we will have 14400 watts total. Here is the proof your looking for. The proof is we can add together the current on each leg to achieve the 14400 watts we need. there is no dispute about this. Its real, its fact, its true. But, if this is possible, that means each leg must be compatible to be use this way. It must also mean each leg is independent of each other. So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect. This works and I proved it. Now, its fact we have 7200 watts per leg. Calculating this out works like this, (7200 watts / 120 volts = 60 amps). If we can add leg A to leg B for a sum of 120 amps, there MUST be 60 independent amps per leg. They MUST be compatible to work together. And in fact, if we use both legs regardless of anything else, there is useable, consumable, and undisputable 120 amps available. A sixty amp, 240 volt panel has 120 amps to be used as we see fit which is the same as saying a sixty amp 240 volt circuit breaker can supply 14400 watts. Each leg supplies 60 amps, independent of each other, without interfering with each other, with out relying on each other. I proved it earlier in this post. A drawing or diagram isn't necessary. Its simple math. Please except this as a vial, friendly explanation to help you understand.

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Mike thank you for your attempt to get 120 amps through a 60 amp wire. Please accept my picture of reality.

"So how do we have independent legs coming from the same source? The answer is the legs are out of phase. As a matter of fact, there 180 degrees out of phase. Perfect."

Mike, what you are describing is the voltage in reference to neutral, not the current. If you have a 240 volt panel, no neutral is required and the voltage across the two legs is in phase, do you think you reverse half of the current just because you put a neutral in? Doesn't happen!

Took awhile to find this from a member from here. Haven't seen Roger here in a few years.

Originally Posted by Jim Port

Took awhile to find this from a member from here. Haven't seen Roger here in a few years.

Well Roger is probably collecting his Nobel Prize for getting current to flow in two directions at the same time. (and what happens to the neutral when it is carrying double the amps of either leg?)

Originally Posted by Vern Heiler

(and what happens to the neutral when it is carrying double the amps of either leg?)

If you notice the two hots are fed from opposite legs of the panels so it will not carry 2x the current.

Do you feel that a 15 amp breaker will affect a 20 amp circuit on the other leg of the panel?

It shows 10a on one leg and 5a on the other leg with 15a on the neutral. Just bump the numbers up to 60a on each leg and tell me what is on the neutral. And they can't go in both directions at the same time on the neutral.

The diagram clearly states in the box that the neutral is carrying 5 amps, the difference between the 10 on A and the 5 on B.

Simplify this, draw two independent 60 amp 120 volt circuits, each with a neutral. You now have 2 hots (ungrounded) and two grounded conductors. How much current can flow through one hot? How much can flow through the other hot? What is the combined current through both hots at 120 volts? Can you turn off either hot without affecting the operation of the other circuit?

If both legs are carrying 60 amps or any other balanced value and are on opposing legs, the neutral is carrying 0. The opposing currents offset each other. This is why the neutral does not need to be twice as large as the ungrounded conductors.

Um, no! It shows 10a going from right to left and 5a going from left to right on the same wire.

Originally Posted by Jim Port

If both legs are carrying 60 amps or any other balanced value and are on opposing legs, the neutral is carrying 0. The opposing currents offset each other. This is why the neutral does not need to be twice as large as the ungrounded conductors.

Jim, can you dispute anything on the last drawings I posted. Does the current go in the same direction on the wires at any given point in time. Does the current going in equal the current coming out.

Your biggest problem is mixing current and voltage. The 180 phase shift in your diagram is voltage. The voltage is more positive than the neutral on top and less positive on the bottom, but all one single current has developed the potentials. Its just that we are measuring from the middle to the ends.

Those are the opposing amplitudes and are subtractive, not additive.

Please answer the questions I posted above. If you do the math correctly you should come up with the same answers as myself, Robert M, John P and Mike.

I really want to hear you explain what the affect of turning off one leg has on the other.

If you had a 2" hose connected to Hoover Dam and added another 2" hose would you get the same amount as the single hose? Same concept.

Originally Posted by Jim Port

Those are the opposing amplitudes and are subtractive, not additive.

Please answer the questions I posted above. If you do the math correctly you should come up with the same answers as myself, Robert M, John P and Mike.

I really want to hear you explain what the affect of turning off one leg has on the other.

If you had a 2" hose connected to Hoover Dam and added another 2" hose would you get the same amount as the single hose? Same concept.

Jim, in my last attachment, fig 1 is one leg turned off. Don't let facts get in the way of your reality.

I am done, no hard feelings, just done!

I'm not about to choose a side, but you guys are to be commended.
Nearly 100 post on this and no name calling, slanders, or put downs.

Vern, sorry to see that you could not see the facts regardless of the examples presented or by whom. If you had looked at this as 2 parallel 120 circuits it would have been easy to see how to have 120 amps available through the disconnect. The math was shown several times.

Originally Posted by Jim Port

Vern, sorry to see that you could not see the facts regardless of the examples presented or by whom. If you had looked at this as 2 parallel 120 circuits it would have been easy to see how to have 120 amps available through the disconnect. The math was shown several times.

I understand the math, I have forgotten more formulas than most if not all electricians have ever seen. Microwave radio technician and main frame computer repair for over 20 years.

Ignore the transformer and focus on 2 60 amp single pole circuits. Each can flow 60 amps. You say you can understand the math, but cannot accept that 60 x 2 = 120?

Originally Posted by Jim Port

Ignore the transformer and focus on 2 60 amp single pole circuits. Each can flow 60 amps. You say you can understand the math, but cannot accept that 60 x 2 = 120?

First do you agree the two diagrams are 60 amp circuits?

Originally Posted by Robert Meier

Your diagrams do not actually depict what were talking about which is a 3-wire circuit.

We are going to get there! Bear with me.

This is as simple as it gets. I'm having a real hard time you don't see this. Each load can take 60 amps. I'm going to word this very carefully. In reference to the second drawing - This is what exists in a main pannel less the transformer. The secondary of the transformer with a center tap makes two 120 volt individual circuits both providing 60 amps in reference to the center tap / neutral or ground. The center tap is essential for creating two seperate 60 amp supplies. Please note the word seperate. Seperate means not of the same body, individual, not together. Each leg in the main panel is powered by is own power source in reference to neutral which means they are individual sources seperate from each other. Not of the same body, individual, not together. Each source supplies 60 amps. There are two seperate sources. 60 amps on leg A and 60 amps on leg B. I can put a load on leg A of 60 amps. I can also put a load on leg B of 60 amps. The total amount of amperage I can load from the panel is 120 amps. Please read and think about the last statement. The total amount of amperage I can load from the panel is 120 amps. Not from each leg, The panel. The total load of both legs. Your drawings are nonsense. You show series resistance and label each one 60 amps. Your diagrams relate more to DC then AC. The arrows show DC paths. Its sad that you are so obstinate in trying to prove something that is wrong instead of listening to what others are telling you, or you just don't understand whats being said. This is not criticism but If you truly believe everyone who posted there response is wrong, I suggest you consult a professor or engineer in the electrical field (not an electrican) with your drawings and get a second opinion before producing any more arguments to your beliefs.

Originally Posted by Rick Cantrell

I'm not about to choose a side, but you guys are to be commended.
Nearly 100 post on this and no name calling, slanders, or put downs.

I was thinking the same thing. Very glad for the forum that the discussion could continue without degrading due to frustration. Good demonstration that it is possible. Great example for others to follow.

Have enjoyed the ping-pong match.

Have a few questions for you guys.

1) How do you typically name the service cable AWG 4/3 coming from the triplex connection to the meter ? ____________________.

2) How would you describe a service cable that was AWG 4/2 ? ______________

3) Using the answer from question # 1 . What is the total number of Amps that the service can supply? ___________

4) Using the answer from question #2 . What is the total number of Amps that the service can supply? ___________

5) If you had two separate service panels and each was serviced by only one leg of a 4/3 service cable how many Amps would each box be able to supply ? ___________________

6) If a property needed 128 Amps to satisfy the the load.
__A) What would you expect to see as a service cable ?___________
__B) What would you expect to see as a main service panel ? ___________________
__C) How many and what size breaker would you to find ? _____________

I understand your frustration.

Originally Posted by Garry Sorrells

I was thinking the same thing. Very glad for the forum that the discussion could continue without degrading due to frustration. Good demonstration that it is possible. Great example for others to follow.

Ah but so few "Likes" from the gallery. Nothing scheduled for the afternoon, so going sailing. Race start is 6:30...

Originally Posted by Vern Heiler

What examples? Anyone can say something is so, proving it is another story. I continue to wait for your or Roberts diagram showing where the 120 amps comes from and goes to. (I would not use your electricians).

In the Mike example (if that is the diagram you are talking about), the current on leg A is the same current that came from or goes to leg B not additive. You need to read things a little closer!

"Turn either breaker off and the load will drop to 60 amps on one leg and 0 on the other leg. The load left on continues to function properly." The amps will drop to the load on the not tripped side of the neutral and is shifted to the neutral that was carrying 0 amps.

I will have to say you have a lot of staying power. No one will ever produce a diagram or the math that will show the 120 amps exists because there isn't any such animal. I wonder if they balance their bank account using the same math Probably why this countries' banking system is in such a mess..

Roland the math and how you could get 120 amps from that panel was described numerous times in this thread. That panel can supply 120 amps at 120 volts or 60 amps at 240 volts. Use Ohms Law and you will see how it is the same amount of watts.

One circuits ability to flow current does not affect another circuit. If it did you would be limited to whatever the smallest circuit breaker installed was. I have asked VH to explain how he thinks it is possible, but so far he has not answered the question. He also could not add up the amps flowing on 2 single pole circuits and get the correct answer.

Originally Posted by Robert Meier

I agree and you could say it 100 more times and still not convince those who cannot grasp the simple concept. There are numerous examples that already in this thread.

I know what you mean!! This discussion was never about power or watts or volt-amps. It was about whether you can get 120 amps from 2 60 amp breakers on opposite phases. The answer is no and will remain no even to "those who cannot grasp the simple concept". Whan the direct question was asked about whether or not everyone wanted to talk about watts, etc. There was no clear answer. Only repeating that the 120 amps was valid. No one here can draw a circuit, calculate or measure 120 amps from a 60 amp service..

Originally Posted by Robert Meier

I agree and you could say it 100 more times and still not convince those who cannot grasp the simple concept. There are numerous examples that already in this thread.

I apologize for continuing this so long. I was hoping to use this as a teaching moment and that the students would finally grasp the subject and understand what they were missing. Oh well, you can't reach everyone. Sad.

I should have realized that some were not getting it when the load of 2 12.5 amp circuits still equaled 12.5 amps.

Let me put into dollars since everyone should under stand money.. You put $60 into your account (or pocket) you immediately take $60 dollars out for something. Some of you would reach in your pocket and expect to find 120 dollars. The net result is 0 dollars. Same math with this circuit..no 120 amps anywhere...because it is additive using "absolute" values. Try putting it on a number line and move your finger back and forth, it will never spike up to 120..

And some are trying to confuse the issue by saying this is not a DC circuit and its not but it doesn't make any difference. It would work with AC or DC.

I would ask those that believe that there is only 60 amps available from that panel to post the same question asked the same way on an electrical forum and post a link to that discussion.

Originally Posted by Roland Miller

Let me put into dollars since everyone should under stand money.. You put $60 into your account (or pocket) you immediately take $60 dollars out for something. Some of you would reach in your pocket and expect to find 120 dollars. The net result is 0 dollars. Same math with this circuit..no 120 amps anywhere...because it is additive using "absolute" values. Try putting it on a number line and move your finger back and forth, it will never spike up to 120..

And some are trying to confuse the issue by saying this is not a DC circuit and its not but it doesn't make any difference. It would work with AC or DC.

Ok, lets use a pair of pants that has two pockets that represent the two legs in a panel. Put 60 dollars in each pocket. How much money do you have on your person? Oh wow, must be 120 bucks.

Originally Posted by Robert Meier

Try this: Typical 1Ø, 100 amp service, 120/240 volts, has only two fully loaded 20 amp circuits both on the same leg how many total amps do those two circuits draw?

Almost 4 hours later and still no answer. Looks like a stumper.

Originally Posted by Jim Port

Almost 4 hours later and still no answer. Looks like a stumper.

And my answer is: The problem is about apples. Why is someone introducing a question about oranges? The question Jim is referring to is about oranges. The source and the circuit was sufficiently defined early on so lets try to answer that question before moving on. The question (as a reminder) is about a single phase, 120/240 volt service with 60 main. How many amps is the maximum at any time? Not watts, not VA not all on the same phase.....or for those of you who are stuck, draw, measure or calculate it to prove it will be 120 amps. (you can't BTW)

Oh and BTW---the answer is +60 and (-60). Because when one phase is positive the other is negative. It does not add up to 120 amps. Never more than 60..

The wattage (VA) is positive in both cases because VA=E x I and VA= (-E) x (-I) both positive answers. Basic 5th grade math. but we are not talking about this..

Again the concept that each leg at 120 volts can supply 60 amps eludes.

Thankfully the electricians understand this.

Could you answer Roberts question above above the two 20 amp breakers on the same leg?

From post #6, where the debate started.

Originally Posted by Roland Miller

At 120 volts you still only have 60 amps available.

After that is was people trying to show the math errors.

Originally Posted by Roland Miller

I will have to say you have a lot of staying power. No one will ever produce a diagram or the math that will show the 120 amps exists because there isn't any such animal. I wonder if they balance their bank account using the same math Probably why this countries' banking system is in such a mess..

There diagram will work just fine! Just needs a little help.....

Congratulations, you finally see using Diagram 1 that it is possible to flow 120 amps @ 120 volts.

Originally Posted by Jim Port

Congratulations, you finally see using Diagram 1 that it is possible to flow 120 amps @ 120 volts.

I don't know if I should call the fire department or the scientific community and tell them to hold there nominations, we have a winner of the Nobel prize without a doubt.

Not exactly breaking news or a fire hazard, single pole circuits have operated safely like that for years.

Originally Posted by Jim Port

Not exactly breaking news or a fire hazard, single pole circuits have operated safely like that for years.

How exactly do you explain getting twice as much current through the single neutral/ground wire in the second diagram without it glowing red?

Because the hots are from opposing legs of the panel they would only carry the difference between the two hots, not the sum of the hots. Every single phase 120/240 panel operates the same way. The service is one large multi-wire circuit.

Originally Posted by Vern Heiler

There diagram will work just fine! Just needs a little help.....

This diagram ignores basic transformer theory and continues to compare apples and oranges..

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Originally Posted by Vern Heiler

How exactly do you explain getting twice as much current through the single neutral/ground wire in the second diagram without it glowing red?

He can't explain it because the current will never exceed 60 amps on the neutral.

One of the principles I have learned over the years in Math, Electrical Engineering and teaching is you must keep track and do the math operations on the unit labels all the way through the problem or the answer will be meaningless- as it this example. I have also learned the you can't mix apples and oranges to come up with something of meaning such as in a lot peoples thinking here. And I have also learned that a politician is always right even when they are wrong . They just have to convince a lot of people they are right...(doesn't make them correct)...

Originally Posted by Roland Miller

One of the principles I have learned over the years in Math, Electrical Engineering and teaching is you must keep track and do the math operations on the unit labels all the way through the problem or the answer will be meaningless- as it this example. I have also learned the you can't mix apples and oranges to come up with something of meaning such as in a lot peoples thinking here. And I have also learned that a politician is always right even when they are wrong . They just have to convince a lot of people they are right...(doesn't make them correct)...

I agree. The biggest problem some are having is trying to apply ohms law before understanding the circuit. You must understand the circuit before applying the math.

Originally Posted by Jim Port

Because the hots are from opposing legs of the panel they would only carry the difference between the two hots, not the sum of the hots. Every single phase 120/240 panel operates the same way. The service is one large multi-wire circuit.

I think it is interesting how you try to convince everyone there is 120 amps somewhere and then you post this in complete disagreement with yourself???? I think you have confused yourself.

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Originally Posted by Mike Borchardt

This is as simple as it gets. I'm having a real hard time you don't see this. Each load can take 60 amps. I'm going to word this very carefully. In reference to the second drawing - This is what exists in a main pannel less the transformer. The secondary of the transformer with a center tap makes two 120 volt individual circuits both providing 60 amps in reference to the center tap / neutral or ground. The center tap is essential for creating two seperate 60 amp supplies. Please note the word seperate. Seperate means not of the same body, individual, not together. Each leg in the main panel is powered by is own power source in reference to neutral which means they are individual sources seperate from each other. Not of the same body, individual, not together. Each source supplies 60 amps. There are two seperate sources. 60 amps on leg A and 60 amps on leg B. I can put a load on leg A of 60 amps. I can also put a load on leg B of 60 amps. The total amount of amperage I can load from the panel is 120 amps. Please read and think about the last statement. The total amount of amperage I can load from the panel is 120 amps. Not from each leg, The panel. The total load of both legs. Your drawings are nonsense. You show series resistance and label each one 60 amps. Your diagrams relate more to DC then AC. The arrows show DC paths. Its sad that you are so obstinate in trying to prove something that is wrong instead of listening to what others are telling you, or you just don't understand whats being said. This is not criticism but If you truly believe everyone who posted there response is wrong, I suggest you consult a professor or engineer in the electrical field (not an electrican) with your drawings and get a second opinion before producing any more arguments to your beliefs.

If the diagram with this post was mathematically correct (valid) you should be able to hook this circuit up in the lab and measure 120 amps on the neutral. And that won't happen cause it will be 0 amps when they are combined.

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Originally Posted by Vern Heiler

I agree. The biggest problem some are having is trying to apply ohms law before understanding the circuit. You must understand the circuit before applying the math.

Yes, You have it nailed. When you don't understand the whole package it is impossible to analyze it..

Originally Posted by Jim Port

Again the concept that each leg at 120 volts can supply 60 amps eludes.

Thankfully the electricians understand this.

Could you answer Roberts question above above the two 20 amp breakers on the same leg?

I know the answer to this question, but the question and the answer have nothing to do with the discussion.. yah know--the apples and oranges thing again

In the 120 volt circuits there are two ungrounded conductors, one for each circuit carrying 60 amps.

Originally Posted by Jim Port

In the 120 volt circuits there are two ungrounded conductors, one for each circuit carrying 60 amps.

​SHOW ME THE MONEY!

Originally Posted by Vern Heiler

​SHOW ME THE MONEY!

This is the same thing I have been saying since the start of this thread. Two 120 volt legs at 60 amps equals the ability of that panel to supply 120 amps worth of load. Nothing has changed, including two peoples lack of understanding.

Originally Posted by Roland Miller

I know the answer to this question, but the question and the answer have nothing to do with the discussion.. yah know--the apples and oranges thing again

The answer is at the heart of the question. Post your answer please.

I also have not seen the link posted to any electrical forum.