# Total amperage in a service panel

## Main Question or Discussion Point

Hello everyone. This is my first post, I hope I'm in the right place.
I'm a home inspector and belong to a group who posts topics about various different things including electrical questions. The question was asked about an electrical panel which housed two 60 amp main fuses. The service entering the panel is single phase 120 volt on each of two legs with a neutral. There for the panel can supply voltages of 120 volts (one leg and neutral) or 240 volts across both legs. This is a common setup in most homes. The inspector stated this is a 60 amp service, which is true. The home owner says its a 120 amp service, which is a false statement. The size of the service is always based on the size of the main circuit breakers or main fuses. This is fact. Here's where the argument started, Some say if you put a 120 volt 60 amp load on leg A and a 120 volt 60 amp load on leg B, the total consumption of current is 120 amps. Others say its not possible to draw more than 60 amps total no matter what the configuration is because the main fuses are only rated at 60 amps. I'm asking for clarification and proof of one statement over the other. Thanks.

## Answers and Replies

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Drakkith
Staff Emeritus
Science Advisor
Looking at the diagram at the following link: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/hsehld.html

It appears that since each leg is separated from the other and uses their own circuit breakers, you could indeed pull 120 amps total.

I don't know the standard convention is for naming the service, but I'd say it's a 60 amp service since no single appliance can pull more than 60 amps at a time without blowing a breaker. (Although I don't know how the 240 v receptacle works)

nsaspook
Science Advisor
The 120vac 60A circuits are in series from the transformer output so the total current flow is 60A across the 240vac hot legs with 60A loads on each 120 volt circuit. The neutral connection to ground on the center tap keeps the 120 volt legs balanced with unequal series loads.

Thank you both for your reply. As much as others and myself have argued the same facts you have pointed out, It's not enough information. I believe 60 amps can be drawn from each leg at 120 volts for a total of 120 amps. The 240 volts can only pull 60 amps which would be the equivalent to 120 amps at 120 volts. This is indeed considered a 60 amp service panel. To change the minds of others, I need mathematical proof or an argument to prove 120 amps is achievable. I totally agree with what your saying but I don't have the absolute proof to change the opinions of others. I hope i'm not asking for the impossible. If anyone can give me undisputable proof it would be appreciated. Thanks again

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Nugatory
Mentor
(Although I don't know how the 240 v receptacle works)
You can't draw more than 60A of 240v - put a 240v load on the system and the current "flows in" through one main breaker, "out" through the other.

The handles of the two 60A main breakers are tied together (at least under the US electrical code), as are the handles of the two breakers on the two hot legs going to the 240v receptacle. This ensures that in an overcurrent situation, both legs will be disconnected - opening one breaker is enough to break the 240v circuit and stop the overcurrent from setting something on fire, but leaving the other leg hot would make for a nasty surprise for someone trying to repair the system (and can damage the equipment connected at that receptacle if the breaker opened because of a ground fault instead of an over current).

The diagram also has a small mistake in the 240v receptacle - that's supposed to be a green grounding wire, not a white one, out of the ground plug. The white wire would only be present if this were a receptacle for a combined 120/240 appliance, and then would be four prongs and four wires: hot 1, hot 2, white grounded and green grounding.

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Drakkith
Staff Emeritus
Science Advisor
Thank you both for your reply. As much as others and myself have argued the same facts you have pointed out, It's not enough information. I believe 60 amps can be drawn from each leg at 120 volts for a total of 120 amps. The 240 volts can only pull 60 amps which would be the equivalent to 120 amps at 120 volts. This is indeed considered a 60 amp service panel. To change the minds of others, I need mathematical proof or an argument to prove 120 amps is achievable. I totally agree with what your saying but I don't have the absolute proof to change the opinions of others. I hope i'm not asking for the impossible. If anyone can give me undisputable proof it would be appreciated. Thanks again
Not sure I understand.

If you can pull 60 amps from each circuit, which it seems like you can since each circuit has a separate breaker, then it's as simple as adding them together.

Nugatory
Mentor
I believe 60 amps can be drawn from each leg at 120 volts for a total of 120 amps. The 240 volts can only pull 60 amps which would be the equivalent to 120 amps at 120 volts. This is indeed considered a 60 amp service panel. To change the minds of others, I need mathematical proof or an argument to prove 120 amps is achievable.
If you just look at the current flow into and out of the house, what the meter is metering, you'll see that there is no difference between the 60A at 240v and the 120A at 120v case - either way, 60 amps is flowing through both hot legs. That's why we call this a 60A service.

As for winning an argument... The electrical code (at least in the US) says that's a 60A service not a 120A one. And right or wrong, logical or not, the codebook wins the terminology argument every time.

Nugatory
Mentor
Not sure I understand.

If you can pull 60 amps from each circuit, which it seems like you can since each circuit has a separate breaker, then it's as simple as adding them together.
Look carefully at that diagram - when you're drawing 60 amps on each of two 120v devices on different legs, the breakers are connected in series not in parallel (there's no neutral current back to the transformer at all if the two loads are balanced). And it doesn't make much sense to add the flows between two breakers in series.

Drakkith
Staff Emeritus
Science Advisor
Look carefully at that diagram - when you're drawing 60 amps on each of two 120v devices on different legs, the breakers are connected in series not in parallel (there's no neutral current back to the transformer at all if the two loads are balanced). And it doesn't make much sense to add the flows between two breakers in series.
How can that be if the neutral wire is the one completing the circuit from all the outlets? (Not talking about the 240 v outlet here, just the 120 v ones)

Nugatory
Mentor
How can that be if the neutral wire is the one completing the circuit from all the outlets? (Not talking about the 240 v outlet here, just the 120 v ones)
No neutral current back to the transformer, I said. The current flows in through one of the 60A main breakers, through the receptacle, through the white wire back to the neutral tie block, from there through the white wire to the outlet on the other leg, then out the other hot leg through the breaker on the other hot leg. That makes the two 60A breakers a cleverly disguised series circuit.

Edit: of course with alternating current this talk of "in" and "out" is bogus - that's why I used scare-quotes in #5.

Drakkith
Staff Emeritus
Science Advisor
No neutral current back to the transformer, I said. The current flows in through one of the 60A main breakers, through the receptacle, through the white wire back to the neutral tie block, from there through the white wire to the outlet on the other leg, then out the other hot leg through the breaker on the other hot leg. That makes the two 60A breakers a cleverly disguised series circuit.

Edit: of course with alternating current this talk of "in" and "out" is bogus - that's why I used scare-quotes in #5.
So that means that you have 60 amps flowing through each leg of the circuit because it's one big series circuit at that point?

davenn
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So that means that you have 60 amps flowing through each leg of the circuit because it's one big series circuit at that point?
yup ... the 2 legs of the AC to the 240V outlet are in series its just a plain loop
out of one side of the transformer through the 60A CB through the 240V device plugged in out and through the second 60A CB and back to the transformer, thus completing the loop

Dave

Drakkith
Staff Emeritus
Science Advisor
yup ... the 2 legs of the AC to the 240V outlet are in series its just a plain loop
out of one side of the transformer through the 60A CB through the 240V device plugged in out and through the second 60A CB and back to the transformer, thus completing the loop

Dave
Sure, but I think we're talking about the 120 v outlets when you have equal loads on each circuit.

davenn
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ya ya I see where you are coming from
Sorry, the earlier intermixed 240V comments were a sidetrack

you're saying that you could be drawing up to 60A on the outlets on Hot#1
and also another up to 60A total on Hot#2

So as a result a total of 120A could be drawn from the transformer ( if it could supply that)
60A on each 1/2 winding

Dave

Drakkith
Staff Emeritus
Science Advisor
ya ya I see where you are coming from
Sorry, the earlier intermixed 240V comments were a sidetrack

you're saying that you could be drawing up to 60A on the outlets on Hot#1
and also another up to 60A total on Hot#2

So as a result a total of 120A could be drawn from the transformer ( if it could supply that)
60A on each 1/2 winding

Dave
I have no idea. I think Nugatory is saying that you can draw 60 amps through both circuits at the same time and still have only 60 amps through the transformer since both are in series.

russ_watters
Mentor
Not sure I understand.

If you can pull 60 amps from each circuit, which it seems like you can since each circuit has a separate breaker, then it's as simple as adding them together.
Not if the voltages are different. Mike is correct: 60@120+60@120=60@240.

Another way to think about it is this: there are two "hot" wires going into the panel, covered by a single breaker. The two 60a feeds however each cover only one wire. Essentially, that makes each one half of a 60A, 2 pole breaker. Incidentally, that's probably a code violation anyway since you can't shut off the whole incoming service with one flip.

Drakkith
Staff Emeritus
Science Advisor
Not if the voltages are different. Mike is correct: 60@120+60@120=60@240.
I am so confused...
What does 240 v have to do with this?

nsaspook
Science Advisor
Another way to think about it is this: there are two "hot" wires going into the panel, covered by a single breaker. The two 60a feeds however each cover only one wire. Essentially, that makes each one half of a 60A, 2 pole breaker. Incidentally, that's probably a code violation anyway since you can't shut off the whole incoming service with one flip.
The code requires a ganged breaker main breaker (shown on the diagram) so both disconnect at the same time. As I and others have said when the two 120ac loads are matched there is no neutral current so the effect is a series 240vac loop with a resistive voltage divider with each center tapped output providing the same amount of current (60A) in series. When the loads (resistance) are unbalanced the transformer rebalances (current*resistance) the circuit voltage by providing more current from one of the tapped sides and then we see neutral current as the difference in current between the two 120vac circuits which is never more than 60A but could be zero so no wire has more than 60A at any time.

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jim hardy
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I am so confused...
What does 240 v have to do with this?
Forget voltage for a minute.

Freeze frame your thinking. At any instant AC current is DC, that is it flows only one direction.
The mistake is in assuming that the neutral can carry the sum of the two fuse currents.

Each fuse is 60 amps.
At any instant, current entering through one fuse must exit through either the other fuse or through the neutral wire.
So in that instant you can't get more than sixty amps in through the 60 amp fuse.
At any instant , current entering through the neutral wire and whichever fuse is positive must exit through the other 60 amp fuse.
So in that instant you can't get more than sixty amps out.

So at any instant you cannot have any more than 60 amps entering and leaving, regardless of direction through neutral wire...

So it's a 60 amp service.

A 240 volt load at 60 amps would spin the electric meter the same as 2-120 volt loads at 60 amps each. At 240 volts the current flows between the two legs, no neutral involved. If a 120 volt load on leg 1 and 120 volt load on leg 2 both drawing 60 amps for a total of 120 amps, what would be the mathematical equation to show the relationship between drawing 240v 60a and 120v 120a? In other words, how can I put into a mathematical formula that would equate 240v 60a is the same as 2-120v 60a for a total of 120 amps? Thanks.

A 240 volt load at 60 amps would spin the electric meter the same as 2-120 volt loads at 60 amps each. At 240 volts the current flows between the two legs, no neutral involved. If a 120 volt load on leg 1 and 120 volt load on leg 2 both drawing 60 amps for a total of 120 amps, what would be the mathematical equation to show the relationship between drawing 240v 60a and 120v 120a? In other words, how can I put into a mathematical formula that would equate 240v 60a is the same as 2-120v 60a for a total of 120 amps? Thanks.
If your criteria for being equal is that the energy meter should spin at the same rate for the 240V@60A and 2x120V@60A loads, then you need to show that they consume the same amount of real power, since that's usually what the energy meter of a residential customer will respond to.

Assuming a constant power factor ##\phi## for both loads, the 240V@60A load will consume real power at the rate:
$$P_1 = 240 \cdot 60 \cos(\phi) \, \mathrm{W}$$
The 2x120V@60A load will consume real power at the rate:
$$P_2 = 2 \cdot 120 \cdot 60 \cos(\phi) \, \mathrm{W}$$
So you have ##P1 = P2##. Is that what you're looking for?

Hi Miles
I believe the equation you expressed is what i'm looking for if the formula proves its possible to draw 120 amps from two 60 amp circuits ( leg 1 + leg 2). Here's the problem, some say you can't add the two legs together to achieve 120 amps of consumable energy. Others say you can. My argument has been if I had three 20 amp circuit breakers on leg 1 and three 20 amp circuit breakers on leg 2, I could conceivably draw 120 amps at 120 volts even though the main fuses are only rated at 60 amps. Question, is my thinking wrong or is there a way to show this can actually be the case? The formula may have to include the phase difference between the two legs with respect to the center tap/neutral or phase difference between the two legs or both. I really appreciate the help. thanks

Averagesupernova
Science Advisor
Gold Member
This is so simple to me. You cannot ever, no matter what, draw more than 60 amps. Take two 120 volt batteries and wire them in series for 240 volts. Now take two 7200 watt light bulbs rated at 120 volts and wire them in series. Hook the two bulb network to the 240 volt battery. Do the math you will find that 60 amps are being sourced by the two batteries. I would assume there is no argument here. You will have a resistance of 2 ohms per light bulb. Equal current flowing in each battery, do the math again and you will find that there is 120 volts across each light bulb. Now connect a wire between the node where the two batteries are hooked together and where the two bulbs are connected together. There will be no change. No current will be flowing in this wire. There is not suddenly 120 amps here or anywhere in this scenario. A service with 60 amp fuses is a 60 amp service no matter what. Even if it were only a 120 volt service with a single 60 amp fuse, it is still a 60 amp service. As Jim Hardy pointed out earlier, this is a snapshot in time of AC.
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It never ceases to amaze me how residential power such related subjects causes such mystery and confusion on this forum.

Here's the problem, some say you can't add the two legs together to achieve 120 amps of consumable energy.
Current is not energy or power. Drawing 60 A at 120 V from each leg means you're consuming 2*120*60 VA worth of apparent power from the grid. It's true that you would consume the same amount of apparent power by drawing 120 A at 120 V from one leg (if the installation allowed it), 2*120*60 VA = 120*120 VA, but that doesn't mean, in the case with the 60 amp breakers, that you actually have 120 A worth of current flowing in any one leg.

My argument has been if I had three 20 amp circuit breakers on leg 1 and three 20 amp circuit breakers on leg 2, I could conceivably draw 120 amps at 120 volts even though the main fuses are only rated at 60 amps.
Your breakers won't change the fact that you're not drawing 120 amps from any one leg, you're drawing 60 amps from two legs, at most.

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Hi Miles
Let me ask the question this way cause i'm trying to understand. If there's a 60 amp 120 volt load on each leg, would this be equivalent to a 60 amp load at 240 volts? My understanding be it right or wrong is if you double the voltage the current is reduced by half. The panel is rated 60 amps at 240 volts not 120 volts. Therefor at 240 volts I have 60 amps and if I reduce the voltage by half (120 volts) I double the current - not per leg - but total amperage available. Which means if I have a 60 amp load on each leg I would consume the equivalent of 120 amps. 60 amps at 240 volts must be the same as 120 amps at 120 volts. Where am I going wrong with this because it makes sense but may not be correct. Also, let me ask this. A fuse has a voltage rating as well as an amperage rating. If I have a 240 volt 60 amp fuse, how many amps would it take to blow the fuse at 12 volts?