Total amperage in a service panel

[Mike Borchardt]

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.

 

[Drakkith]

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]

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.

 

[Mike Borchardt]

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

 

[Nugatory]

(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.

 

[Drakkith]

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]

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]

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]

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]

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]

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]

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]

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]

ya you 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]

ya you 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]

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]

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]

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.

 

[jim hardy]

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.

 

[Mike Borchardt]

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.

 

[milesyoung]

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?

 

[Mike Borchardt]

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]

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.
-
It never ceases to amaze me how residential power such related subjects causes such mystery and confusion on this forum.

 

[milesyoung]

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 anyone 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 anyone leg, you're drawing 60 amps from two legs, at most.

 

[Mike Borchardt]

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?

 

[Averagesupernova]

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?

Yes.

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.

No, the panel is rated at 60 amps PERIOD! It is also rated at a certain voltage. The voltage rating is significant to arc-over and things of this nature. No current carrying conductor in a 60 amp panel should ever be expected to carry more than 60 amps.

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?

You say it makes sense but it does not. The example I gave in a previous post should lay it out pretty solid. In the case of a 60 amp fuse with a voltage rating of 240 volts being run on a 12 volt system it should blow at the same current it would on a 240 volt system. It is not good practice to do this and I will not clutter this discussion up telling you why. Another time.
-
I think you are confusing some things. Alot of single phase induction motors can be wired for 120 VAC or 240 VAC. When you run them at the low voltage they require double the current to get the same work done as they would at the higher voltage but that does not apply here to service panels. A service panel is simply a bunch of glorified conductors with some safety built in. It is not a transducer like the motor is.

 

[jim hardy]

Time for some "reducto ad absurdum"..

So you still need a formula?

Assume Kirchoff's current law holds:
AMPS_fuse1+AMPS_neutral + AMPS_fuse2 = 0

Assume that your customer is right, he can get the current through the fuses to add to a total of 120 amps,

what must be the per fuse current?
Now what two numbers , neither of which is bigger than 60, add to 120 ?
2X = 120 has only one solution. So it has to be sixty amps per fuse.

And all 120 amps will have to leave through the neutral to satisfy Kirchoff.



That requires the currents through the two fuses to be in phase, for current is an arithmetically complex number(has both magnitude and direction) therefore requires consideration of the angles. In EE we use phasor notation which is very similar to vector notation used in basic physics . As you know, vectors add directly only when they have same direction. (I'll try to consistently call them vectors hereafter but I'm getting old...)

Now let us define voltage from FUSE 1 to NEUTRAL as V_1,n and voltage from FUSE 2 to NEUTRAL as V_2,n

V_1.n = 120∠0 because that's a logical starting point, we'll use it as zero angle for our phasors(vectors)

and V_2,n = 120∠-180
(V_n,2 would be -120∠0)

We know that each fuse must have current with amplitude 60
what about their angles?
Well, the angles must be the same else the current vectors will not add to 120 - they'll add head to tail and have sum equal to third leg of a triangle with each current vector as one side..

Okay so what two impedances would, with 120 volts of opposite phase impressed across them ,, produce currents of sixty amps in phase?
Well amplitude would be 120Volts/60Amps = 2 Ohms

So let's try just straight two ohms with no angle (resistive)
IF I_Fuse1 = (120∠0) /2 = 60∠0
and I_Fuse2 = (120∠-180) /2= 60∠-180
THEN
I_Fuse1 + I_Fuse2 = 60∠0 + 60∠-180 = 0 (try it on paper !)
Oh really? the two resistive currents add to zero?
Hmmmmm... (scratch head icon) that won't work - Their real components cancel one another.
(SO THAT's what everybody means when they say the neutral just carries the unbalance ! )


SO----
Our two impedances need to shift our two currents such that they have no real component, for the real components cancel out making the sum less than 120 amps.

Hmmmmm... Maybe,,, Shifting one current behind its driving voltage by 90 degrees and shifting the other ahead by 90 degrees would remove the real components, and if we pick the directions right will bring the two currents into phase so they'll add.

Let's try retarding fuse 1's current by making its load two ohms of pure inductance
and advancing fuse 2's current by making its load two ohms of pure capacitance

and we'll name those impedances Z₁ and Z₂ (okay names?)

Z₁ = 2∠+90
Z₂ = 2∠-90

Now
I_Fuse1 = V_1,n / Z₁
I_Fuse1 = (120∠0) / (2∠+90) = 60∠-90

I_Fuse2 = V_2,n / Z₂
I_Fuse2 = (120∠-180) / (2∠-90) = 60∠-90

I_Fuse1 + I_Fuse2 = 60∠-90 + 60∠-90 = 120∠-90

Aha ! Now neutral current is 120 amps purely inductive.

How cool is that ! And how useful ! Purely reactive current, be it inductive or capacitive, transfers no power so is abjectly useless to a homeowner.
That is why it is called "Imaginary" current, as opposed to "real" current which does transfer kilowatts

and "real" current was shown above to have phase angle of 0 or 180, and maximum sum of 60.

By algebra above only imaginary currents can sum to more than 60.
so bluntly tell your client he needs a reality check.

Perhaps a younger, sharper member can streamline the specious train of thought above

but I did try to answer the question. Probably it'll simplify to a one line inequality.

I apologize for my rusty algebra. Hope it was at least entertaining!

old jim

 

[256bits]

Time for some "reducto ad absurdum"..

Not so absurd. Each 120 volt branch is 180 degrees out of phase and each can carry a maximum current of 60 amps. If the leg load is balanced the neutral carries no apparent current - ie current from one leg is in one direction, the other leg carries current is in the opposite direction and the both currents cancel.

My panel is rated 125A 120/240v AC. It has two columns of 8 spots for breakers. Each column could conceivably carry 8 x 15A = 120 A at 120 volts. Populating all spots would conceivably mean that the panel can supply 2 columns x 120A per column = 240A at 120 volts if the circuits in the home are split evenly between the hots. But each hot circuit can only have a maximun load of 120A.

The problem I bet stems way back to the old old 30A breaker box with 2 15 amp fuses and the 2 wire knob wiring. These I think operated on only 120v and people needed electricity mainly for lights and one or 2 outlets, with only one 'hot' being supplied to the hone. Carry that forward to the nowadays panel of 100a and 120/240 v rating and you can see the confusion. Is the amperage rating displayed on the panel for only a single 120v leg with one 'hot' or for both 120v legs with 2 'hots'? It is not so readily apparant.

A 60A service would require (I think that this is correct ) a 10HP utility transformer ( 10 HP = 7450 watts, or 60A 120v ), whiich is again only per one 'hot' wire leg. If each hot carries 60A then the transformer still doesn't melt or its circuit breaker blow, since the wiring in the transformer from hot to hot is still only 60A.

Perhaps a backwards compatability rating issue more than anything else,

 

[Drakkith]

Not so absurd. Each 120 volt branch is 180 degrees out of phase and each can carry a maximum current of 60 amps. If the leg load is balanced the neutral carries no apparent current - ie current from one leg is in one direction, the other leg carries current is in the opposite direction and the both currents cancel.

Okay, now I know where I went wrong. I was thinking current would flow through the neutral line, back to the transformer, and then split and into each leg of the transformer. But that's nonsense, right?

 

[jim hardy]

Okay, now I know where I went wrong. I was thinking current would flow through the neutral line, back to the transformer, and then split and into each leg of the transformer. But that's nonsense, right?

It's neither nonsense nor a hard absolute invariant fact.
Go back to the original picture you posted in #2.

The hard invariant fact is that any current leaving that transformer winding must get back to it by hook or crook. That's Kirchoff.


Let's freeze frame our thinking again, this time at 90 degrees into that one cycle of sinewave shown, which is the first peak.
Hot #2 is positive and hot #1 is negative.
That's what a center-tapped winding does , relative to its centertap it tries to push current out one end and pull it in the other.
I'll use conventional current here, ie unlike electrons it's pushed from positive toward negative.
If you prefer [STRIKE]electron[/STRIKE] negative current flow the analysis is similar. But Sophie will give us "that look"...

(Perhaps the following would have been more intuitive had the picture shown a 120 volt load connected on hot#1 side of panel - draw one in if you like, but I can imagine it..)

Let's observe some current that has got "pushed" out hot #2 and through a load on that side of panel into the neutral tie busbar. It's that big white metal strip where all the neutral wires are landed.
Upon arrival at the neutral it sees two paths back to the transformer:
1. It can return via neutral
2. It can return through one of the loads connected to hot #1.


Which path will that current choose? Or will it split? How will it ever make up its mind??
Well, let's let Kirchoff analyze that too.

Any current that gets pulled into hot#1 through a 120 volt load on hot#1's side of the panel must enter that load via its neutral connection.

Aha !
So in our freeze-frame mental picture we have current being pushed into neutral through loads on hot #2 side,
and being pulled right back out through loads on hot #1 side.

That current from neutral into hot #1 had no choice - its fate was determined by the loads connected to the panel.

So - current arriving at neutral can and will return to transformer via a load to opposite side of transformer, but if there's not enough loads connected there it can always take the neutral .

Kirchoff says sum of all currents entering and leaving neutral is zero
so whatever current is left over after the loads are satisfied returns to the transformer via neutral,
and upon reaching its centertap they DO split exactly as you suggested.

Draw a circle around neutral - it's just another node.
Draw a circle around either half of the transformer winding - it's just another node.

Any help? I sure had fun...

thanks !

old jim

 

[milesyoung]

I went back and reread your posts and I think this is the crux of the matter:

I believe 60 amps can be drawn from each leg at 120 volts for a total of 120 amps.

I have to ask you:
What if, on the transformer secondary, you instead had 240 separate 1 V windings with a 60 A breaker on each. That could supply the same amount of apparent power as a single 240 V winding with a 60 A breaker, right?

Would you then say you have a total of 14400 amps available or call that a 14400 amp service panel?

 

[256bits]

Okay, now I know where I went wrong. I was thinking current would flow through the neutral line, back to the transformer, and then split and into each leg of the transformer. But that's nonsense, right?

From the diagram you posted, Hot1 at the top, neutral, Hot2 at the bottom
You could measure voltages as 240v - 120v - 0v,
or as 120v - 0v - (-)120v
where currents and voltages are in phase.

If you measure the bottom leg as Hot2 to neutral ie as 120v - 0v, then that means it is just 180 degress out of phase from neutral to Hot2. ie 0 to (-)120v.

Current through the transfomer secondary would always be in one direction eiither Hot1 to Hot2 or revesred Hot2 to Hot1 depending upon the state of the cycle. so if you have loads on Hot1 and Hot2, when Hot1 current flows out of neutral, the current from Hot2 is flowing into the neutral and cancel out.

You could have a transformer where both currents from both legs flows through the neutral in the same direction. If the second leg is wiried in the counter direction as to the first leg, then any current from Hot1 or Hot2 would flow into or out of neutral and both currents would add. If both loads are equal the neutral would carry double that of Hot1 or Hot2 and the neutral wire would have to be of a greater thickness to compensate for I squared R heat.

 

[Averagesupernova]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads. If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg. Also, the XC and XL load as you described form a series resonant circuit. Any more thoughts on this? I feel silly questioning a guy with your background.

 

[jim hardy]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads. If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg. Also, the XC and XL load as you described form a series resonant circuit. Any more thoughts on this? I feel silly questioning a guy with your background.

Please do question me, I tend to reverse things often. Mild case of autism...

I never even thought about their being resonant. That might be a clue...
will have to chew on that a bit.
Were it not for the neutral connection we'd surely have a series resonant circuit with its substantial voltage gain. Probably the neutral connection prevents that.

Regarding phase, which was your question -
remember that wrt neutral the two driving voltages are 180deg out of phase, and that's a half turn apart... so currents a quarter turn behind one and a quarter turn ahead of the other could be in phase.
If I've invented some untrue math to justify that I haven't yet caught my mistake. I'll not be embarrassed or offended if you find it though, I'm here to learn things.. especially about myself.
The brain can accept things that are quite untrue and then invent justifications for them. Eternal vigilance...

If you find an error please post !

old jim

 

[milesyoung]

Jim Hardy, something doesn't look right with your example of capacitive and inductive loads. Can't put my finger on your math though. My crude sketches show current canceling in the neutral with XC and XL loads.

I'm not Jim, but I hope you won't mind my comments.

If you have equal impedances on both legs, the currents in both will be 180° out of phase (since the voltages from the center-tapped transformer are 180° out of phase), so you won't have any neutral current.

If you, for some odd reason, wanted to maximize the neutral current, you could advance the current in one leg by 90° and retard it by 90° in the other so the current phasors "line up" instead of being in opposite directions, which is what Jim did by specifying the load impedances as:

Z1 = 2∠+90
Z2 = 2∠-90

If I am correct, that means a service feeding lots of 120 V inductive loads will have a lot more current on the neutral than what it was designed for. And of course there is a much greater chance of this than there is with large capacitive loads on one leg.

As long as you keep the loads on both legs balanced, be they inductive or capacitive, there can't exist any neutral current.

In practice, in a typical residential home, any unbalance is probably not going to be of any real consequence. Jim's example was pathological in nature, i.e. extremes in terms of current amplitude and load unbalance.

Also, the XC and XL load as you described form a series resonant circuit.

The two loads are on separate legs, they can't resonante. They share a common return, but they can't exchange energy.